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Old talk

I've just deleted this (hidden) comment from the article in order to discuss it here:

The problem with most of what follows here is that a photon is not a particle per se and doesn't move at the speed of light...photons are quanta, gauge bosons if you will...as they describe a single field mode they are nonlocal..

Of course, a photon travels with c. There is no contradiction here. Actually, the notion of velocity makes more sense in the particle picture than in the wave picture (where you have to sort out the amibuity of phase velocity and group velocity.

Also, a photon with energy uncertainty describes a not well-selected field mode.

The article overstates the wave-particle problem generally, as it in no way contradicts the photon picture. I wil think about a good formulation to make clear, that in quantum field theory, saying that the photon is a quantum does not imply that it is a particle in the old sense of being a localized corpuscular point thingy.

Sanders muc 21:03, 26 May 2004 (UTC)

I have added my tuppence, and tried to make the article a bit more "layman friendly". Someone with a better grasp of Foch states had better look at that part - I suspect that it could be rather interesting if explained a little further.

I have moved some text under a new "Properties" heading - lots of the text under "Symbol" was not about the symbol at all, and some text on physical propeties was buried in "Creation".

Comments welcome. -- ALoan 17:55, 27 May 2004 (UTC)

I have deleted the generally in the first sentence. Some notes about that:

• Electro-magnetic radiation can travel over unlimited distances. In gauge theories infinite range of a interaction boson is associated with the property of it having mass zero. There are actually several experiments confirming this and giving thery small upper bounds for the photon mass. (cf. Review of Particle Physics, publisherd by CERN, for an overview)

• In high-energy physics it is custumary to call mass' the rest mass of a particle and energy its total energy (rest mass (= rest energy) and kinetic energy).

• So a photon's moving mass is of course not zero because a photon carries energy, and energy is mass according to E=mc^2. Hence a photon is e.g. subject to gravitation (cf. gravitational lensing). But this is usually called the photons energy not its mass. Note especially, that this mass/energy is not always tiny, but can be quite substancial.

• Of course, a photon always travels exactly with the speed of light, per definitionem: A photon is light, hence its speed is the speed of light.

What you might have been thinking about is that a photon in matter travels at a speed smaller than the speed of light in vacuum. But that's then still the speed of light. And strictly speaking, a photon within matter is not a photon anyway, but a polariton, as correctly remarked at the end of the article.

-- User:sanders_muc (a physicist)

A Photon is a particle of very little mass that likes to travel at the speed of light. After all that it is what light is, photons. These particles appear to be everywhere. They carry energy with them when they are let loose. Photons can come from such places as combusting fuel (gas). In order to see fire, photons are given off. How exactly this mechanism takes place, we do not know. There are many theories on how such occurrences take place, but until you can prove something it is nothing more than a bedtime story.

Science is all bedtime stories, eh? In a main article like photon?

As a result of their size photons can pass through several different types of matter (with certain structures, one being glass) without even seeing it.

This isn't true. Photons, even photons of visible light, interact with glass - you can tell because they move slower through it, are refracted by it, and so forth. And size has nothing to do with it, as all subatomic particles are more than small enough to slip through glass.

Changed 'diffraction' back to 'refraction'. It really is refraction of light that causes rainbows. I'll explain the mechanism when I get around to doing rainbow.

Is it really true that photons are slowed down when moving through a medium? I thought that they are constantly absorbed and reemitted, so it's not really one and the same photon moving through and being slowed down. --AxelBoldt

Not sure about in QM, but IIRC in classical oscillator theory, photons move at c (i.e. vacuum speed of light) between atoms. They're not really absorbed and re-emitted by the atoms, though. What happens is that the electric field of the photon drives charges into oscillation, and those oscillating charges radiate a field which is slightly out-of-phase with the photon. The superposition of the photon and the radiated field is slightly retarded w.r.t. the original field, and so the photon is 'delayed' a bit at each atom. On a large enough scale, this looks like the photon is slower. -- DrBob

The photon is a 0 mass boson responsible for transmitting the electromagnetic force and producing visible light. While virtual photons, or photons that have a probability to exist due to the uncertainty principle carry the E&M forces, a changing electromagnetic field creates an electromagnetic wave, which when quantized, can be broken up into particles, which are real photons. Because of their massless nature, photons travel at a speed of C and changes in energy are manifested only as changes in frequency, the higher the frequency, the greater the energy. Frequencies falling between a certain range detectable by the human eye are called visible light.

But on to some other points brought up in previous entries.

According to GR, gravity can be described as curved space-time, and objects that "fall" in a gravitational field are simply taking streight paths through curved space, known as geodesics. Photons, although they have no mass, and therefore should not alter the geodesics themselves, still travel along a special type of geodesic, called a null geodesic, which is basically a path of zero distance through space-time.

At first glance, this may not seem to make sense, since we can clearly see that light moves from one place to another. But notice that I didn't say the distance through space is 0, but rather the distance traveled through the 4 dimensional space-time continuum. Due to the odd sort of geometry used in relativity, the total distance of any particle through space-time is equal to x^2 - y^2 - z^2 - t^2, where x,y,z, and t are all in the same units. "How can this be?" you ask, "since x,y, and z are spacial coordinates and t is a time coordinate?" Don't forget that a fundimental principle of relativity is that time is simply another dimension of space, and can therefore be expressed in spacial units, with c, (the speed of light) as the conversion factor, c units of space per one unit of time (note this means we could also write the above equation as x^2 - y^2 - z^2 - ct^2).

As for two photons having mass, I believe this is possible in a scenario where the two have momenta in opposite directions, creating a standing wave. Since E^2 = p^2c^2 + m^2c^4, or if you preffer, for each individual photon E = pc, but since p2 (momentum of the second particle) can be described as -p1, the momentum of the entire system is equal to zero. However, there is still energy present, which is equal to 2hf, (where hf is planks constant times the frequency, which equals the energy of one photon), so(2hf)^2 = 0^2 + m^2c^4, which reduced to 2hf = mc^2, and m = 2hf/c^2. -- Mark Palenik

Could you someone write this for someone who has never taken Physics in college? quatum? excitation? superposition? -- Taku 18:58, 21 Sep 2003 (UTC)

I think there was some kind of controversy about whether light was a photon or a wave. Now the currently held belief I think is that you look at it however it makes the most sense for the experiment you are doing. Like, maybe there's really no difference between a massless particle and a wave? A short wave is a high frequency, and packs more punch in a shorter amount of time so it's a high energy photon. Is there such a thing as a high-intensity low-frequency beam? Or does it have to be higher frequency to fit more energy in a narrower beam? --Luke Parrish 18:48, 21 September 2003 (-0600)

This is especially for late editors of this article, namely TakuyaMurata and Pizza Puzzle. Thanks for editing. But please make changes that are consistent. If you want to bring in front some facts you judge more important then adjust the article for them, especially remove these informations in other places so that the article has some natural flow and obeys a structure. Also try to avoid contradictions. The present state of this article is currently the worst that can come out from cooperative edition: things are obviously a heap of independent editors, repeating and tellings things and their converse. To the reader the result is complete confusion. If the previous text didn't make sense to non-physicist, now opening as 'a relatively massless particle; which generally travels at the speed of light' it doesn't make sense even to physisists. Particle is vivid picture for the layman but actually very poor concept to a physicist. Even on Misplaced Pages there is no decent article for this. Quanta of excitation (we could say energy) with a proper link for quanta, is not only exactly true, I think it is also very understandable by anybody. Generally travelling at speed of light is confusing (and actually false, if it doesn't travel at the speed of light it is not a true photon anymore). Relatively massless is also confusing. May be we should think of splitting this article in two, one vivid, particle, massless, light picture for layman, and a proper scientific article covering this issue at depth. I am not sure this can be done in a single article since this would need to assert false or at best vague all the previous material for non-scientific audience. I understand the previous form was not adequate but its present form I find worst. Thanks to discuss and I hope we improve this important article tackling an important concept.

what is a photon, really

This article and the vast majority of other such material about light, overlook completely what a photon actually is.

Firstly, a charged particle is surrounded by an electric field, which extends outwards in all directions, reaching a distance of at least 13.7 billion light years. When a charged particle is accelerated, due to conservation of energy and angular momentum, the change in movement must be conveyed to the electric field which surrounds it. This change propagates at the speed of light, and is in fact, what we call light. Any light we see is caused by a change in an EXISTING electric field. Thus, a photon probably should be called a virtual particle, since it does not really exist like a regular particle independent of its creator. It is as virtual as other such particles like phonons, solitons, or polaritons. If we observe a photon emitted long ago from a distant star, we are observing the change in the electric field of the particle which emitted the photon. There is no particle which is a photon. The quantization of light is solely an effect of the quantization of charge and the particulate nature of the electrons used to observe and emit light. There might, however, be a quantization of space-time, which would also lead to the observation of quantized light.

Furthermore, a photon is not a packet of energy with a simple frequency or polarization. A wave of light can find itself in any shape allowed by the propagation of change in an electric field. This means that a photon isn't necessarily a sine wave, or even periodic. It just so happens that we are designed to detect periodically changing electric field in a certain range of frequencies called visible light. And we've created devices which are designed to detect light of many other frequencies. But we shouldn't limit our thoughts to the special (albeit useful) case of the periodically changing electric field.

I use the term electric, rather than electromagnetic field, since it has been shown that the observation of a magnetic field is a special relativistic effect upon charged particles in motion with respect to an electric field.

In view of quantum theory, the electromagnetic field is made of truly virtual photons, and when a change is propagated through the field we observe "real" photons.

Would anyone else like to comment? Like why aren't these ideas common knowledge? And why does the idea of a photon as corpuscle still exist? (please spare commentary on the photoelectric effect, unless it can be shown that this is not the result of interaction with particles of definite charge and cross-section) --D. Estenson II 08:20, Mar 22, 2005 (UTC) (BTW,IANAP,BWBSS)

Yes, "photon" carries misconceptions, and even physicists aren't immune. Photons are quanta, not particles, since "particle" carries conceptual baggage (the word "particle" unfortunately implies something like a very small grain of sand.)

A good collection of recent articles by physicists is from the Optical Society of America : THE NATURE OF LIGHT: WHAT IS A PHOTON?

The articles have some some suprising stuff: the Photoelectric Effect does not prove the existence of photons, Nobelist W. Lamb takes a very dim view on the whole photon concept, and even as late as 1966 some physicists had a running bet whether quantum phenomena (Lamb Shift) could be calculated without recourse to quantized EM fields. Other articles in the collection show where the "photon" concept is crucial in physics. --Wjbeaty 04:16, Mar 23, 2005 (UTC)

FYI, here is a link to the abstract for the original article from the Optical Society of America. The Nature of Light: What is a Photon?. There you can find a link to the article in pdf format. Thanks for pointing out these articles. --D. Estenson II 10:08, Mar 23, 2005 (UTC)

I support most of the said. Even if we don't know, what a electromagnetic field oder matter is, and even, if we don't know, what momentum and energy is, we believe: the amount of energy and momentum doesn't change, but can be moved from field to matter. This exchange of energy and momentum we call a photon. This will solve some problems with imagination. Field no longer consist of photons, but are excitated by "photonic events". Such an event is a change in action of size h. Action is, according to my knowledge, the only appearance, that is realy quantisized. Energy quantization depends on the properties of a "measuring instrument". Quantisation of action NOT. This is the key! ErNa 08:48, 14 November 2005 (UTC)

Truth is, photons are virtual particles, which means they really aren't anything at all. The concept of a photon is just a conceptual placeholder to describe what happens when matter interacts with an EM field and how that field interacts with matter. Dan 18:08, 9 August 2006 (UTC)

Who named the photon?

Wave-Particle duality is not unique to the photon and the EM field in QM. Every particle, for example the electron, behaves like a wawe and under the right conditions this can be (and was) detected experimentally.

Elad Tsur 13:30, 1 June 2004 (UTC)

Photon and Action

Is it true, that, whenever a photon is emitted or absorbed, the emitter/absorber state changes by h (Planck's quantum of action)?217.229.215.59 14:35, 17 May 2005 (UTC)

Photons carry mass?

Photons do not have mass. Photons have energy. Some claim that E=mc means that they are the same thing, but this is not a widely accepted theory. Gravity simply couples energy, of which mass is a type. Photons do not have that type of energy. It is interesting to note that two photons can have mass, even though one does not.--BlackGriffen

Photons don't even "have" energy. They have the property that they can transfer a certain amount of angular momentum from one charged particle to another in a certain amount of time. True, this is a fine distinction in semantics, but one I think should be discussed. --D. Estenson II July 8, 2005 09:08 (UTC)

Does it matter if one or both of them is Catholic?

--- BlackGriffen: you write "It is interesting to note that two photons can have mass, even though one does not." I'd be very interested to know how this works. -- SJK

In relativity mass and energy are the same thing. It is customary to separate out kinetic energy, which stems from the motion of the system in question as a whole, and invariant mass, which stems from its simple existance. In composite systems, though, there can be binding energy which raises the mass relative to the sum of the components. Actually, though, I'm not sure how photons can get stuck together like that... --- My bad, guys, I think. But it's nice to see that Misplaced Pages is working in removing the (accidental!) piece of bullshit that I put in. Any chance of expanding upon why photons don't have mass, and why they're affected by gravity, which only affects mass-carrying particles if they don't?

And is invariant mass just a better, non-flawed description for rest mass, or are they something subtly different? -- Same thing, but invariant mass is the preferred term, as rest mass carries the strong connotation of the mass at rest and that doesn't make any sense for a photon.

With thanks, Dave McKee.

I'm not sure how two photons having mass arises. I recall two things a theoretical physicist said at a colloquium: 1, two photons can have mass; 2, it has something to do with the physical arrangement. My guess would be that when two photons are moving in opposite directions through the same space, they can create a standing wave. At least, two hertzian waves can. In order to be standing and still have energy, it must have mass. Don't quote me yet, though.

"Any chance of expanding upon why photons don't have mass, and why they're affected by gravity, which only affects mass-carrying particles if they don't?"

Certainly. I don't recall the experimental evidence, but here is the theoretical background:

p = γm_ov
E = γm_oc
γ = 1/√(1-(v/c))

Photons travel at c, right? Well, sticking that in the above equations, which all particles (including photons) must obey, yields an infinite γ. If the photons have any finite non-zero mass, they would have an infinte energy and momentum! Last time I was knocked over by a photon... ;) In all seriousness, the reason they are affected by gravity is because gravity doesn't couple with mass, it couples with energy (of which mass is a type).

The idea that energy and mass are the same comes from one of these two fallacies:

mv = γm_ov
mc = γm_oc

leading to:

m = γm_o

Essentially, an attempt to redefine mass so that the old equation for momentum works, the idea that gravity couples to mass still holds, and that it was consistent with the new rest mass energy formula (sticking in v = 0). The only thing Einstein said was that the total energy of the system is:

E = γmc

So you might say that in order to measure mass you need to bring the system being measured to rest. Already the idea of a photon having mass runs in to trouble since a photon can't be brought to rest (the easiest way to imagine trying to stop a photon is by trying to move your reference frame at c in the same direction as the photon. The dopler effect and other relativistic spatial transformations will reduce the frequency to zero).--BlackGriffen

Hopefully recent edits and links clarify the situtation. -- Beland 06:27, 14 August 2005 (UTC)

I threw the picture in

Hope you like it; the article needed some sprucing up.

I think a better picture given the caption would be two next to each other: a photograph of diffraction lines from the double-slit experiment with laser light, next to picture of a (maybe x-ray) micrograph showing single points of light (resulting from the photon collision with the detector). I'm no wikipedia expert, but such pictures may already be somewhere on here. The current picture, on the other hand, does nothing to enhance the understanding of photons, or how both their wave-like and particle-like features can be demonstrated. --D. Estenson II 11:09, Mar 23, 2005 (UTC)

Good point. I'll look around and see what I can find (and legally upload to Wiki). -- Zalasur 07:53, Mar 24, 2005 (UTC)

Here is a good illustration of diffraction of laser light, and it's already on wiki: (single-slit diffraction of laser light, with intensity graph) --D. Estenson II 12:32, Mar 24, 2005 (UTC)

• • Sorry, I think the picture should be removed. It makes people think that the wave behaviour of photons is spatial, and not electromagnetic. It's a common belief among students (and perhaps people in general), mainly because of pictures showing photons moving up and down as it propagates.

Yes, the photon wave is in a sence spatial, but the picture is still misleading. //Johan Falk, physics education researcher, Sweden

Mr Falk, please explain how the photon's wave is spatial in a sense. AFAIK, a photon follows a straight line through empty space (ignoring gravity and other complications). While it is in transit, the electric field oscillates (grows more and less intense) where the photon is along the path. This can be visualized using a sine wave with the x-axis representing space (or time, with appropriate conversion) and the y-axis representing electric field strength. But many laymen misinterpret such a graph as saying the photon wiggles up-and-down as it travels through space. This is the interpretation that should be avoided at all costs, and which the picture only reinforces. Am I mistaken? --D. Estenson II 19:20, Jun 20, 2005 (UTC)

To partially answer my own question, since individual photons are polarized, they have a spatial orientation to their electric oscillations. --D. Estenson II 20:11, Jun 20, 2005 (UTC)

Coherence edit and adding simpler examples

This is probably not the right form of address:

In layman's terms, photons are the building blocks of electromagnetic radiation: that is, a photon is a "particle" of light, although, according to quantum mechanics, all particles, including the photon, also have some of the properties of a wave.

Nowadays we simply say "mass", meaning either of the "invariant mass" or "rest mass", since the old-style "relativistic mass" is deprecated because it is likely to cause confusion.

For overall consistency, bolds are for vectors, and not required to define symbol usage.

-Guest April 23, 2005

Polarization

This talk page or section is in a state of significant expansion or restructuring. You are welcome to assist in its construction by editing it as well. If this talk page has not been edited in several days, please remove this template. If you are the editor who added this template and you are actively editing, please be sure to replace this template with {{in use}} during the active editing session. Click on the link for template parameters to use. This article was last edited by 74.134.239.77 (talk | contribs) 18 years ago. (Update timer)

This article needs a section on polarization. I may add one within a few weeks if no one else does. --D. Estenson II 20:16, Jun 20, 2005 (UTC)

• Yes, and there's nothing about a photon wave as E × B; having electrical and magnetic components. — RJH 5 July 2005 22:45 (UTC)

Always an electron?

A photon is always created by an electron? Never by proton-proton bremsstrahlung, or by a positron? I'm just curious about that statement. Thanks. — RJH 5 July 2005 22:42 (UTC)

A photon can be created by the acceleration of any charged particle, not just an electron. --D. Estenson II July 8, 2005 08:48 (UTC)

History

This talk page or section is in a state of significant expansion or restructuring. You are welcome to assist in its construction by editing it as well. If this talk page has not been edited in several days, please remove this template. If you are the editor who added this template and you are actively editing, please be sure to replace this template with {{in use}} during the active editing session. Click on the link for template parameters to use. This article was last edited by 74.134.239.77 (talk | contribs) 18 years ago. (Update timer)

This article could use some content on the history of the photon, or at least some links. See e.g. photoelectric effect. -- Beland 06:30, 14 August 2005 (UTC)

what photon is?

"Eenergy of a photon can equaly be shown in another way:

where T=1/f is a period of a photon. In this form it seems that Planck`s constant h might be interpreted as the energy of a ( "standard" ) photon with T=1s ( photon lasting 1 second ); and that ( single ) photon as ( single ) EM vawe, with period inversly proportional to its energy."

I am not physicist, it`s only my imagination, scientific method is somthing else. I expect comments.Sslavko 09:16, 12 September 2005 (UTC)

I'm not sure I'm following your objections(?). Please reiterate informally. Thanks. El_C 02:55, 15 September 2005 (UTC)

Understanding of the nature of photon and Planck`s constant is closely related with much more important question of understanding quantum processes and quantum world in the whole. Mathematical equations give correct results and predictions, but nobody knows what some of them describe in real, everyday world at microscopic level. Some most obvious examples are double-slit experiment, quantum entanglement, wave/particle duality and, quantum loops, strings and so on. It appears that quantum world is governed by rules quite different from our everyday expirience.

Expresion for the energy of the photon E=hf describes its energy, and is experimentaly confirmed, but tells nothing about what photon is. But, because f=1/T, the same can be writen as E=h/T, where T is the period of the vawe (in seconds). From the equation E=h/T it is seen that the single oscillation of electromagnetic wave lastig one second, contains energy of h , that lasting 0.1 sec contains h / 0,1=10 h etc. The greater available energy by craetion of photon, the shorter period of oscillation. On the other hand, Planck`s constant, in this case at least, is action (energy) of elecromagnetic wave (oscillation, photon) lasting one second ( „standard“ photon).

I wish to induce philosophical consideration with the intention to show that quantum and macroscopic events, although different in some aspects, are governed by the same laws and bz the same logic as human mind. I expect in the first place participation of laymen, because experts have their own view, but any contribution of experts is wellcome, of course.

Playing with numbers and equations is a pleasent activity. At least I think so... And sometimes it even produces interesting and ground-breakig results. De Broglie hypothesis is something of the kind. But notice that Misplaced Pages's purpose is not to do any kind of first hand investigation and I've never eard of "standard photon" nor I could find any reference in a fast web search (other than "standard photon-something").
«that quantum and macroscopic events are governed by the same laws» is beyond the scope of this article, and is probaly covered somewhere else, namely at Hawking radiation. Nabla 21:24, 19 September 2005 (UTC)

This argument is well known. To me it seems to be just a "killer". The biggest nonsense, documented in an paper, that can be referenced, will have a mirror in Misplaced Pages, but, the most simple conclusion from very basis knowledge will be "first hand investigation".

It is very clear: in relation to a photon, we can not talk of "frequency". For: it is not possible to determine by measurement the slope of the electromagnetic field of an single photon event (Heisenberg). To talk of frequency needs at list two zero crossings. But who ever calculated the fourier transform of a wavelet know, that the result is another wavelet. So we always have a spread spectrum, we can not talk of a frequency.

Everything becomes very clear, if we understand, that any electromagnetic field can be quantisised. But, imagine:

I have a cubic box with black body radiation inside of a certain temperature. The dimension of the box defines the lowest possible energy & frequency of a photon. The temperature defines the highest possible energy of a photon: temperatur and volume of the box determine the energy content of the field, and all the energy can be focused (for an very short moment) into a single "Photon"

Now, in any moment, I freeze the field to determine the number of photons of different possible energies. I receive a sprectrum. But, whenever I determine the dimension of the box, there is a uncertainity and so all the energies will be lightly shifted. This will result in a different spektrum. This shows very clear: the electromagnetic field is not consisting of photons, but is created and by photonic processes. A photon is just the process of exchanging energy and puls between electric charge and field. And this process is quantisised in the manner, that the change in action is always equal h. Therefore, it takes more time to exchange less energy. I never found a reason, to see photons as particles. Who else did? ErNa 06:13, 17 October 2005 (UTC)

What am I missing?

If I'm reading this article right, light is made out of photons, and photons cannot decay on their own. So why is it that when I turn a flashlight off, the light goes away? What happens to all the photons that had been produced by the flashlight? -- Mwalcoff 00:00, 20 October 2005 (UTC)

Photons can be absorbed (and turned into kinetic energy) when they collide with normal matter. (Sometimes an electron is knocked free by this process, hence the photoelectric effect.) --Jasonuhl 07:39, 14 November 2005 (UTC)

Breaking some people's bubbles?

I have issues with a specific statement...(and most of this article...)

"In some respects a photon acts as a particle, for instance when registered by the light sensitive device in a camera."

A massless "particle" could not possibly contain enough "information" to render color or even shading onto film. Only a wave function could effect that.

Photons cannot be particles, simply because if they are without mass they would be nothing. If they have mass, they violate the General Theory of Relativity. If energy directly equals mass, as E=MC2 represents, then a photon must have mass as it is capable of doing work (read as having energy). An item of mass that would be traveling at the speed of light would become infinitely massive, said Einstein, and we all know that our universe is not one infinitely massive photon. The only solution for this quandry is to realize that photons do not exist. Light is only a wave function.

I am going to take this one step further and invalidate almost all of the calculations made within this discussion thread. As light is not a particle, and only a wave function carried through mass, light cannot travel through a true vacuum of relatively large volume. C cannot be a constant, because it specifies that light has a set speed within a vacuum, but if light can't travel through a vacuum, then how can the constant be valid? It can't be. There are no human beings that can prove me wrong on this either, as there exists nowhere in the universe a true vacuum of relatively large volume. (yes, there is true vacuum everywhere, as the space in between all singularities, but the distance between massive objects is still relatively small. Massive objects meaning all items of mass, such as single atoms.) All of modern physics is built upon a theory that is not proven, or provable, but can be proven wrong. Theories are not fact. Facts are facts. You cannot build fact using parts that are still only theories.

Kevin E. Carman 11-18-05

Sorry, but you just don't know enough physics to make a cogent argument on this issue. Pretty much everything you are saying is incorrect, but I'm not going to spend the next two hours writing a detailed rebuttal. Go take a course in modern physics.

Clue: in modern notation E≠mc, and mass does not increase with velocity (that would not work with General Relativity). Massless particles carry momentum and energy just fine.--Srleffler 08:06, 23 November 2005 (UTC)

Inertia increases with velocity, and at C mass would have infinite inertia. Massless particles could not carry energy at all, as E=MC^2 means that energy equals mass. The simplest logic would also dictate that something without mass, would actually be nothing. If energy is mass, and vice versa, something without mass could have no energy. If you would like to discuss relativistic mass versus mass at rest, you would have to assume that at some point the photon would have mass in either state of motion. By description, photons cannot have mass, and therefore they cannot actually be able to do any work (aka have energy). I can provide upwards of 2 dozen very strong logical arguments against the existence of the photon, but I suspect your logic would not be strong enough to make the connections I would provide, based on your reply to my comment. If you choose to correct me on using E=MC^2, I only used it because everyone is familiar with it. The real equation is E=M, because energy would be a direct conversion. The speed of light has no bearing on this conversion factor, and thereby not be necessary. For example, 1amu would be capable of X amount of work, if it were to be a complete conversion. (IMHO, I completely disagree with the premise of the formula itself, as it implies that mass can actually "change" into something other than simple kinetic energy, and it can't. Mass does not simply go away, as many assume.) Have scientists actually "captured" a photon? Nope. They can't because they do not exist. If you doubt me, provide us proof of humans capturing a photon in it's energetic state. I feel pretty safe that you cannot.

Kevin E Carman

You argue that photons can't be particles because they have no mass. Note that photons have relativistic mass but no rest mass. At the speeds that photons travel, one could conceivably think of them as a tiny particle (like a neutrino which was previously thought to be massless, but have been since found to have a rest mass). As long as the particle is moving, it exerts a gravitational field and has momentum - just like any mass would. The quote "light is massless" is thrown around too freely, and people who hear it most likely don't understand the difference between rest mass, and relativistic mass. Fresheneesz 22:40, 7 April 2006 (UTC)

Kevin, here is where your error arises from, in fact this is a common error a lot of people make in relation to the whole E = mc^2 thing, the complete equation for this is E^2 = (pc)^2 + (mc^2)^2, where E,m and c have the usual meaning and p is the momentum of the particle, which in the case of light is expressed by the de Broglie expression i.e wavelength = h/p. Now as you see a particle need not have a rest mass (m) to be able to exert a force (so to speak). I could have just said "two words, photoelectric effect" but that I thought might be inappropriate, also this would clear it up for you and help your understanding of the matter (no pun intended :-P ). Hope this helped. Pubuman 16:24, 10 August 2006 (UTC)

Photon mass

I hesitate to write this after the section above, which frankly looks crankish to me, but there doesn't seem to be any other place to put it.

What is the status of the possibility that photons have nonzero rest mass? I remember seeing an old Scientific American article on it, showing how Maxwell's equations would have to change, among other things. As I understand it, current observations have put an upper bound on photon rest mass, but have not ruled out its existence. However, supposedly (I don't understand why), the discovery of a magnetic monopole would refute a nonzero rest mass for photons.

Anyone know anything about this? Have there been any theoretical or experimental developments that bear on the question since the article I remember (which probably would have been from the '70s or so)? --Trovatore 00:23, 23 November 2005 (UTC)

If a photon were to have mass while it was at rest, it would defy the theory of general relativity. If a photon were to have mass while it was in motion, it would also defy GR. Mass would require infinite energy to accelerate it to C, so if photons existed as mass before they were emitted from electrons and such, then it would have required infinite energy to "propel" that photon to C. We know that electrons do not lose energy, and certainly not infinite energy, when they emit photons. Therefore, to stay within the parameters GR sets out for mass and velocity, photons can never within their lifetimes have mass. This also means that because they can never have mass, they can never represent energy.

The only possible exception to this would be the ability of a body of mass actually going at C and emitting an object of mass. The energy to accelerate that body of mass to C was already invested in the mass itself, so it would take no further energy to emit something at C. The problem is that no mass can go C, within the parameters of GR. Photons cannot accelerate themselves, and since no objects of mass can go C to emit something at C, there is nothing that can propel them to C.

Kevin E Carman

I'm afraid I don't believe you know enough physics to give a useful answer to my question. For those who are interested, the question is discussed at Misplaced Pages:Reference desk/Science#photon mass. (I don't know how long the discussion will stay there or whether it will eventually be archived.) --Trovatore 19:17, 6 December 2005 (UTC)

You haven't provided any substance to back up your claim that I do not know enough physics to back up my commentary. You assume that GR is valid, and what I am saying is that it is atleast partially invalid. Light only travels in wave form. The only reason why you cling to the idea of the photon is because most of GR "needs" it to exist to propogate energy without a carrier of some source. If you are willing to not insult me with snide remarks that imply I do not know about what I speak of, I can enlighten you to a complete unified field theory that is far more functional, and logical, than what GR offers.

Please enlighten us with your complete unified field theory. Even better, you could submit it to a peer-reviewed scientific journal for publication. Best of all would be to provide us with a citation of an existing article in such a journal where we can all read about it. -- ALoan (Talk) 20:07, 6 December 2005 (UTC)

I plan on publishing it. I am currently writing it for exactly that. Kevin Carman

The discussion at the reference desk has been archived at Misplaced Pages:Reference_desk_archive/Science/December_2005#photon_mass. Someone might like to incorporate the information from there into the article. --Trovatore 19:19, 2 January 2006 (UTC)

Energy vs frequency

The energy of a photon is directly proportional to its frequency, correct? But this only applies to quanta, correct? A continuous sine wave of amplitude A at 100 MHz doesn't have "more energy" than a similar wave at 10 MHz, does it? The power level and amplitude would be the same. Perhaps the higher frequency wave has more energy per quanta, but fewer quanta? (And what would a radio quanta look like? Or is this one of those "don't go there" wave-particle duality kind of questions?)

I've heard this used in reference to sound (higher frequencies have "more energy"), and I don't believe it. Seems like a misunderstanding to me (though my own understanding is limited). — Omegatron 20:30, 22 December 2005 (UTC)

The power of a classical wave is proportional to its amplitude. This does mean that a higher frequency wave of equal power is made up of fewer photons. -- Xerxes 21:37, 22 December 2005 (UTC)

Right, energy E=hv applies to photons only. For example, a microwatt of gamma rays would have a few high-energy photons, while a microwatt of 60Hz would have way way more photons. On the other hand, with acoustic waves of the same energy but different frequencies, the energy is proportional to pressure, but I think the air moves far less for higher frequency waves. With a few watts at 30Hz you can see the loudspeaker cone vibrating perhaps 1mm back and forth, and the pressure variations in the waves are tiny, but if a 30KHz tweeter moved that much, the ultrasound wattage might heat your skin. Maybe they're referring to transducer motion, where a vibrating object with constant amplitude motion would put out higher pressure excursions and higher wattage at higher frequencies. --Wjbeaty 21:46, 22 December 2005 (UTC)

but I think the air moves far less for higher frequency waves

Hmmm... isn't sound pressure directly related to particle displacement, regardless of frequency?

Nope.

${\displaystyle \xi ={\frac {p}{Z\cdot \omega }}}$

Where ${\displaystyle \xi }$ is displacement, p is pressure, ω is angular frequency, and Z is acoustic impedance. You learn something new every day. — Omegatron 22:45, 22 December 2005 (UTC)

Omegatron and Xerxes: the power of a classical wave is proportional to the square of its amplitude, and you are correct that if you compare two waves of different frequencies but the same power, the one with higher frequency is made up of fewer photons. A side point: the exact number of photons in the wave is not necessarily well-defined. Under certain pretty common circumstances, the number of photons present becomes uncertain.

Photons don't really "look like" anything. If you want to think about it quantum mechanically, there are neither waves nor particles, just states. Waves and particles are convenient abstractions that make things easier to understand, but don't always work well. In particular, quanta need not be small. It's perfectly possible to construct an experiment in which a "photon" is spread over a region a meter or more across.

Mechanical vibrations, including sound, can be treated quantum mechanically, and it's not uncommon in that context to talk about "phonons"—quanta of sound. This doesn't alter the fact that sound is a mechanical wave. Quantum mechanics does always give the correct answers for "classical mechanics" problems. We say that it "reduces" to classical mechanics in the "classical limit". Phonons are very commonly used in describing vibrations ("sound") in crystals.--Srleffler 21:55, 22 December 2005 (UTC)

Yeah, I've read our article on phonons many times and still don't understand it. I'll keep trying... — Omegatron 22:35, 22 December 2005 (UTC)

I would like to see some information about photon angular momentum, perhaps contrasted to spin. Spiral interfference pics and an explanation would be nice too. Also, I do not know if anyone is familiar with the red shifting out of existence 'paradox' (1. where does its spin go? 2. relativity says that a photons energy is not fixed, that it is relative to the observer, so can a photon be non existent to one observer and real to another? How is this different from the Unruh effect?)

Wow! I'm 100% surprised that Omegatron doesn't understand something. I'm so used to it being me.

"A at 100 MHz doesn't have "more energy" than a similar wave at 10 MHz, does it?"

What exactly do you mean by "simliar wave"? If their number of photons was "similar" then 100MHz would have more energy, if their power was similar, then 100MHz would have less photons. The same is true with sound, for the same power, the air would be pushed and pulled farther for lower frequencies. Given the same physical amplitude of air "push and pull", a higher fruency would have more power. I'm guessing i'm missing the entire point of your question. Fresheneesz 23:26, 7 April 2006 (UTC)

Perceived? Detected?

The caption says that photons can be perceived as particles or waves. "Perceived" implies with our senses. Should this be changed to detected? Experimentally observed? Bubba73 (talk), 05:43, 12 January 2006 (UTC)

No, perception reflects how the experimenter interprets what is detected. The photons are not "detected" as either particles or waves, but one may interpret or perceive the detected result to imply a wave-like or particle-like nature.--Srleffler 06:43, 12 January 2006 (UTC)

How about "observed" then? Or how about rephrasing it to say that they can behave as particles or as waves, depending on the experiment/situation? Bubba73 (talk), 16:03, 12 January 2006 (UTC)

The thing is that photons are neither particles nor waves, but those are abstractions that allow us to think about it more easily. In some situations the wave model is simpler, and in others, a particle model is. Fresheneesz 23:28, 7 April 2006 (UTC)

rephrase?

I think the sentence "Photons are always moving, and photons in a vacuum always move at the vacuum speed of light with respect to all observers, even when those observers are themselves moving." needs to be rephrased. There is too much information there for one sentence. I propose replacing it with something like:

• Photons area always in motion, and move at a constant speed in any medium (optics). The speed of a photon in a vacuum is known as the speed of light. This speed is the same with respect to all observers, even when those observers are themselves moving. Bubba73 (talk), 16:56, 12 January 2006 (UTC)

I'm fine with breaking up the sentence, but it would be better to avoid talking about what light does in a medium. Whatever you say, in this context, isn't going to be quite true. Light doesn't always move with a constant speed in any given medium. It's easy to come up with counterexamples, like birefringent crystals. There is also an issue about what exactly photons are doing in a medium. It's not clear that the photon that comes out is the same one that went in. Just restrict the statement only to vacuum, and it will be fine.--Srleffler 17:32, 12 January 2006 (UTC)

You can ommit some problems of understanding, if you just take for true, what you know: energy and momentum is transfered between matter and elmag field in certain quantities: this is what we call "photon". Whatever the field is, whatever matter is: it doesn't matter! If the field is exitated in place A at time T, there is a probability to exite a atom at place B after time (B-A)/c. That's all. It needs no "moving photons". ErNa 08:54, 13 January 2006 (UTC)

In case you missed it, the title of this article is "Photon". --Srleffler 13:44, 13 January 2006 (UTC)

What did I miss? In physics, the photon (from Greek φως "phos", meaning light) is the quantum of the electromagnetic field, and further quantum: a "light quantum", being a unit of light (that is, a photon). Can you tell me, what a "unit of light" is? Do you know more properties of a photon? ErNa 21:36, 13 January 2006 (UTC)

Actually, yes I do. It carries both momentum and angular momentum, for two examples. That's beside the point, though. I agree with you that one can describe light quantum mechanically without recourse to "moving photons". Moving photons are, however, a good simple conceptual model for explaining the behavior of light to people who aren't ready for the full quantum mechanical treatment yet, or who don't need it for what they are doing. Lots of physics is still done by thinking about moving photons. Since this page is Photon, it is an appropriate place to deal with light in terms of the "moving photons" model, rather than a more detailed quantum mechanical treatment. --Srleffler 02:26, 20 January 2006 (UTC)

Thank You, I intended to provoke an answer and it worked! Now, we have three properties. Any more? Do you think, it is possible do open an article on photon, that make the quantum mechanical treatment plausible to normal people, which will not be immediately deleted?

Is it known how photons interact with particles? Since their postulated to have no charge (which seems odd to me), then they must interact in some other way. Fresheneesz 23:55, 7 April 2006 (UTC)

Quantum entanglement

isnt there somthing to do with seperatin photons, and then when one gets annilated, so does the other one, no matter how far apart they are, this recently came up while taking to a physicit i know.—The preceding unsigned comment was added by 172.202.236.88 (talk • contribs) .

Not exactly, but it sounds like you're talking about quantum entanglement. --Jasonuhl 22:37, 19 January 2006 (UTC)

The theory that two distant or adjacent photonic particles could share a bond would be possible and highly likely. If there was more than theoreticle proof of the existence of a photonic bond the subject could yeild results but since there is no proof it is simply discredited. —The preceding unsigned comment was added by 142.165.110.167 (talk • contribs) .

Not sure what you're talking about. Quantum entanglement is well established and has been demonstrated in many labs worldwide.--Srleffler 04:15, 14 February 2006 (UTC)

Mass

Mass. Photons haven't got any. As this is an article about photons, the article should (and does) say so.

There are other interesting things to say about mass. Some of these are in mass in special relativity. As this article is not about mass, these things are not relevant. -- Xerxes 01:13, 12 February 2006 (UTC)

I think the text you reverted was there to take care of a possible terminological confusion. Contemporary physicists usually use "mass" to mean "rest mass", but this has not always been so. To someone expecting "mass" to mean "relativistic mass", the current text is possibly misleading.

Moreover, as I understand it, even a nonzero rest mass for the photon has not really been ruled out on theoretical grounds; all that has been determined is an upper bound, which is explained in the paragraph immediately following the one you edited. The two paras are currently a bit at cross purposes; perhaps the first one should be edited to explain that physicists generally assume photons have zero rest mass, but etc etc etc. --Trovatore 01:48, 12 February 2006 (UTC)

Yep. Worse, current popular science books sometimes use "relativistic mass", despite the fact that this is no longer current scientific usage. I didn't revert his change, though, because perhaps this unfortunate usage should not be encouraged. The article on mass which is linked there does explain this issue.--Srleffler 02:03, 12 February 2006 (UTC)

The obsolete term "relativistic mass" is synonymous with the non-confusing term "energy". Use of "relativistic mass" should never be encouraged. Zero photon mass is theoretically necessary due to gauge invariance; however, one can never be totally certain that it's exactly zero. Current experimental bounds are really small: less than 10 eV. It might be nice to work that into the article somewhere, but it's really only of interest to the small group of physicists working to make the bound ever-smaller. Feel free to bet the farm that it's exactly zero. -- Xerxes 03:06, 12 February 2006 (UTC)

You don't have to encourage the use of "relativistic mass", but you ought to have language that explains the situation to readers who might be confused. There already is text about an upper bound for photon rest mass; what's needed is a better transition from the previous para, which claims photons have "no" mass. Quibble on that: Of course they have a mass, even if it's zero. So "zero (rest) mass" would be a better way of putting it. --Trovatore 03:10, 12 February 2006 (UTC)

I think the term is dead enough now that merely mentioning it confuses the non-confused and does little to alleviate confusion amongst the confused. I do agree on "zero" versus "no"; duly changed. -- Xerxes 03:27, 12 February 2006 (UTC)

We don't need to mention relativistic mass. We can simply say that light is affected by gravity, has momentum, and has inertia. Quite simply, if we weighed a photon the same way we weigh ourselves, we would find that it has mass.

Another thing to note is that it is still in question whether photons actually have zero rest mass, or whether they are simply very very very light. This is noted on the page that the consensus upper limit on photon mass is 6*10^-17 or something. Fresheneesz 23:40, 7 April 2006 (UTC)

Article unnecessarily long and potentially confusing

Most of what's written in this article about "photons" is equally true about light in general and classical electrodynamics specifically. Radiation pressure is a classical effect. Calling an antenna or my microwave oven an emitter of photons seems to be an abuse of the language that brings to mind billiard balls. Similarly, talking about photons in the context of General Relativity, a nonquantum theory, is odd.

Why not merely devote this article to treating the photon as a fundamental particle (gauge boson/force carrier) and as the quantum of light?

Does Willis Lamb read Misplaced Pages? 09:30, 20 February 2006 (UTC)

Photons and color

I am unfortunately not a particle physicist, but I am trying to learn. What exactly is the relationship between an individual photon and its wavelength, e.g. color? I know wavelength times frequency equals the speed of light and all that. I also know that color perception (e.g. eyesight) is variable. But are there, for example, red photons, blue photons, or infrared photons? When light is separated by a prism, does that mean that "white" photons went in one side and individual photons for each color of the rainbow came out the other side? Someone please answer! oneismany 22:52, 23 February 2006 (UTC)

For the most part, you can think of each photon as having a specific frequency. The reality is a little more complicated than that, but that will do for most purposes. The relationship between the wavelength and the frequency depends on what medium the light is travelling through. When light enters a medium like water or glass, it slows down and its wavelength increases. The frequency stays the same. Don't try to imagine how a discrete "particle" of light can have a "wavelength". It won't help you understand this. What is helpful, though, is to understand that a photon's frequency is proportional to its energy. A photon with higher frequency has more energy than a lower-frequency photon.

Color is a perception created in the human brain by photons being detected by the retina in your eye. It is somewhat correlated with wavelength and frequency, but not entirely. For example, the well-known phenomenon that you can mix colors to obtain other colors (e.g. red and blue give purple) is entirely an artifact of the human visual system. Physically, if you mix red light with blue light, you have a mixture of red and blue light with no other colors present. The eye and brain perceive this mixture as purple.

So, there are sort of red photons and green photons and blue photons, but when you see green light (for example) you can't be sure there are any "green" photons there—you might be seeing a mixture of two other colors. There are no brown photons or pink photons, as those colors do not exist in the optical spectrum. Yes, there are infrared and ultraviolet photons. When light is separated by a prism, photons with a whole range of "colors" from red to violet (and beyond) go in, and they are separated in space by the prism. If photons with some range of frequencies are missing from the light that goes into the prism, the corresponding colors will be missing from the spectrum that comes out of the prism.

If you see green light it's because there are "green photons" because green is a primary color (same as red and blue). If, on the other hand, you see yellow light, you can't be sure whether it's due to "yellow photons" or a mix of red and green. There is an interesting experiment that uses this to highlight the variation in color perception between people. You have a source of "pure" yellow light, and a source of mixed red/green light. You then adjust the red/green ratio (by turning a knob or such) until you think that the two yellow lights look identical. You then show the result to someone else, who will almost certainly tell you that the two yellows are not identical! Similarly, when other people adjust the colors until they think they are identical, they will probably not look identical to you. Itub 16:22, 24 February 2006 (UTC)

Thank you, that is very helpful. Now, I wonder what is the mechanism for the color of shadows. Shadows are colored only among multiple sources of light with distinct colors. If you block red light, you get green shadows. If you block blue light, you get orange shadows. If you block purple light, you get yellow shadows. (And vice-versa.) What do you get when you block ultraviolet or infrared light? oneismany 16:32, 26 February 2006 (UTC)

Cancer-free and cold shadows? We humans can't perceive that stuff anyway... Melchoir 21:10, 26 February 2006 (UTC)

moving

The article says ...always move at a constant speed with respect to all observers, even when those observers are themselves moving. With respect to what are these observers moving? Should it say "with respect to each other"? Bubba73 (talk), 17:01, 24 February 2006 (UTC)

With respect to whatever you please. That's part of the point: there is no absolute reference for motion. You always have to specify with respect to what you are moving. Light is special, in that no matter what you pick as your reference, and no matter which direction you move in or how fast, the speed of light that you observe is always the same. You can be moving toward the source at 90% of the speed of light, and the light will be coming towards you at the speed of light relative to you. A friend can be moving away from the same source at the same time, and she will percieve the light overtaking her from behind, moving at the speed of light relative to her. This is extremely odd and unexpected behavior, but there is solid evidence that it's true, and relativity theory explains how it all works in a consistent way.--Srleffler 03:58, 25 February 2006 (UTC)

Yes, but the old version that you reverted to is still wrong. I'll take a crack at it. Melchoir 04:13, 25 February 2006 (UTC)

I like your version.--Srleffler 06:05, 25 February 2006 (UTC)

But my point is that if you have an observer and a photon and nothing else in the universe, you can't tell if the observer is moving. Bubba73 (talk), 04:25, 25 February 2006 (UTC)

Yes. I reverted your change, though, because it's irrelevant whether the observers are moving with respect to one another. You can pick any reference you please. Melchoir's new wording is better: instead of referring to the observers as moving, he merely says that the speed of light is constant regardless of their velocity. That covers all cases.--Srleffler 06:05, 25 February 2006 (UTC)

I like the new version better too. The old version said that they were moving, but not with respect to what. It could have specified "the light source', "other inertial reference frames", or something - but it idn't specify anything. It mentioned only a photon and observer. You and I known what the article intended to say (but didn't say), but many people viewing the article wouldn't know that. (i.e. "You can pick any reference you please." - the article didn't say that, but needed to say something to clarify that.) Bubba73 (talk), 01:00, 26 February 2006 (UTC)

Photons exerting gravity

• "Photons have zero mass and zero electric charge, but they do carry energy, momentum and angular momentum."

I wanted to change this sentence a bit, not only to reflect that photons are not for sure 100% massless, and also to note that they do everything mass should do - exert gravity, are affected by gravity, and exert a force on particles they hit. I wanted to make sure that I was correct in those assumptions.

• "Photons are deflected by a gravitational field twice as much as Newtonian mechanics predicts for a mass traveling at the speed of light."

? How would newtonian mechanics predict anything for a supposedly massless object? Fresheneesz 23:46, 7 April 2006 (UTC)

Absolute certainty is not part of science. Photons are massless to within very tight experimental constraints, and there are solid theoretical reasons to think they must be exactly massless. There is no point in obfuscating the situation by adding weasel-words. Newtonian mechanics predicts that everything falls at a constant rate, independent of mass; one may assume that this also holds in the massless limit. -- Xerxes 17:15, 8 April 2006 (UTC)

Thats not my point. This article already addresses the uncertainty of a photons mass. However, I was still making sure that the reset of my assumptions are correct. It is a very common mistake with people to be confused by the descrepensies of saying light is "massless" when it interacts with things like a mass should. People think "why does light bend if it has no mass?" or "how can light push a solar sail if it has no mass?" people don't understand that light has properties that act exactly as if it had mass. I think this very common misconception should be emphasized in this article.

I was particularly confused with my second bullet pointed quote. Fresheneesz 03:55, 10 April 2006 (UTC)

People's misunderstanding of general relativity is not really a topic for the photon article. In any case, it has little bearing on whether or not photons have mass. In particular, a photon does not act "exactly as if it had mass"; it acts exactly as if it had zero mass, because it has zero mass. Momentum and energy are properties wholly independent from mass. -- Xerxes 04:38, 10 April 2006 (UTC)

No, people's misunderstanding of LIGHT *is* really a topic for the photon article. People associate mass with gravity, bounching, momentum, etc. Light HAS momentum, and can bounce, and is affected by gravity. This should be noted in order for people to more fully understand what it means for light to be massless. This isn't about making things too specific, or expanding an unrelated issue into an article where it doesn't belong. Light is misunderstood, and this article should address that. Fresheneesz 19:33, 10 April 2006 (UTC)

"...deflected by ..gravit .. twice as much as Newtonian mechanics predicts..."

• "Photons are deflected by a gravitational field twice as much as Newtonian mechanics predicts for a mass traveling at the speed of light."

Since this isn't getting answered in the above post, I'll reitterate this. This sentence doesn't make sense to me. How does Newtonian mechanics predict the deflection of light by gravity? And what "mass" is it talking about (since the article says light is massless)? Fresheneesz 19:37, 10 April 2006 (UTC)

To reiterate the answer (posted above), Newtonian mechanics predicts a deflection that is independent of mass. (For a detailed derivation, see Newtonian gravitational deflection of light revisited.) In that sentence "mass" is being used figuratively to mean "test particle" (often called a "test mass" in gravitational contexts). -- Xerxes 20:04, 10 April 2006 (UTC)

Yea that link doesn't help. How can deflection of gravity be independent of mass - especially in newtonian physics? What *was* it dependent on? Fresheneesz 10:19, 20 April 2006 (UTC)

Could you be more specific? What about the detailed derivation of the appropriate equations is not helpful? Note that the deflection does depend on the mass of the deflector, it just doesn't depend on the mass of the deflected object (the light, in this case). To be really perfectly fair, there ought to be a teeny-tiny correction due to the deflector being nudged by the gravity of the deflected object. Since the deflector is usually a star, it doesn't move very much. However, such a term would depend on the mass of the deflected object. -- Xerxes 14:24, 20 April 2006 (UTC)

Ahh, I see, maybe you're right. I don't have time to read it again (I only skimmed it the first time), but i'll see if I can make sense from it. In any case, that part in this article isn't very clear since most people aren't familiar with newtonian bending of light. Is there a page on that on wikipedia? I'm going to red link "Newtonian mechanics predicts" for now. Fresheneesz 21:57, 20 April 2006 (UTC)

To a first approximation (in general relativity), the gravitational force exerted on an object moving transversely (perpendicular to the radial direction) by the Sun is proportional to E+P·V where E is the energy of the object (including both the energy of its rest mass (if any) and its kinetic energy), P is its linear momentum, and V is its velocity (transverse). If the object is moving at the speed of light, then P·V is equal to E which is why you get a doubling. A more complex expression is needed, if the motion is not transverse. JRSpriggs 03:52, 10 August 2006 (UTC)

Imaging in a camera does not require a wave description of photons

I don't think this a good example of wave behaviour. Imaging can be described with ray optics and thus is amenable to a particle description. This only breaks down when the diffraction-limit is reached but that is a secondary effect to the actual image formation. Why don't we use the standard example of interference (e.g. Young's double slit or diffraction at the edge of a shadow)?--J S Lundeen 22:15, 13 April 2006 (UTC)

A single photon

Can anybody confirm/refute this (somewhat paradoxical) interpretation of the photon? Consider a single photon emitted from some source. Is it not true that light is by definition EM radiation that ALWAYS propagates in every direction? (Every direction in a vacuum assuming no matter blocks it.) This means that the "particle" interpretation of a photon is quite misleading as the single photon is ALWAYS as sphere of EM disturbance whose radius increases with time (at the speed of light), which means the photon "exists" everywhere on the sphere at once. This is quite difficult to imagine if thinking of a photon as what most people consider a particle (a very small thing moving about in space with a definate position at any given time). Is this sort of like what is meant when, in explaining the wave/particle duality of matter/light, that the particle is "smeared" out over the wave front?

It is also interesting to consider the photon after a long time, when the radius of the EM sphere is very very large, meaning the photon "particle" exists simultaneously over an ever increasing spatial area. Because of this, I would almost expect that the energy of the photon decrease with time, as it has to "cover" more area. Is it not true though that in any given direction the energy of a photon does not decrease with time (or stated another way the magnitude of the EM disturbance does not decrease?) as the wave continues to propagate forever? It seems almost like getting something for nothing: while the energy at every point on the ever increasing sphere remains the same, the amount of space affected by this energy is ever increasing implying that the total energy of the photon in all of space is ever increasing with time. Perhaps this is just a property of PURE energy (which if I understand correct is what light is) as it relates to space and time that just does not sit well if one tries to give matter-like properties (a particle) to something that is just not matter. Is this a seriously flawed interpretation of quantum theory or an illumination of the paradoxical nature of it all? -Kyp4 19:51, 28 April 2006 (UTC)

Yes it is : ) . The photon model says that light does *not* "always propoage in every direction". However, for a point source, the probability that the photon will travel in any given direction is the same - thus for a large number of particles, statistics dictates that the light will form a virtual sphere of photons. Fresheneesz 06:15, 8 May 2006 (UTC)

Cleaning up the photon article

It seems to me that this article is not in very good shape. One of the first things that needs to be done to clean it up is to decide whether this is an article about photons or an article about light. (Obviously, it should be the former.) As an article about photons, we really ought to throw out all the bits dispersion and whatnot that are related to classical optics (except as it pertains specifically to the quantization of light). Also, the parts about the speed of light (in vacuum or not) are pertinent to light rays, but irrelevant to the photon. As a massless particle, the photon must always travel at c. The superposition of photons with some excited states of condensed matter are interesting, but need to be better phrased in the language of quantum mechanics, not of optics, to be appropriate for this article. Anybody agree or disagree? -- Xerxes 18:34, 1 May 2006 (UTC)

I completely agree. Melchoir 19:46, 1 May 2006 (UTC)

I agree too. However, I think we should have some explanation as to how the photon model agrees with the wave model. Fresheneesz 06:30, 8 May 2006 (UTC)

Photons have relativistic mass - ie gravitational attraction and momentum

People are *always* confused about the "masslessness" of light. Light does have the property of the outdated concept of relativistic mass. I want to write a little blurb - not on relativisitic mass (which everyone says is a bad concept) - but on the gravitational attraction and momentum of light. I would also like to mention the "effective mass" of light, which is derived from its gravitational attraction and momentum - like any other mass is.

I don't think anyone can really understand light without knowing that *while moving* it acts like it has mass. I have a couple sources I could site to support this view:

Anyone like to give me the thumbs up? - Fresheneesz 06:28, 8 May 2006 (UTC)

Absolutely not. "Acts like it has mass" is meaningless nonsense. Photons act exactly as if they have zero mass; they do have zero mass. -- Xerxes 15:17, 8 May 2006 (UTC)

Did you not read what I wrote? I was not talking about rest mass. The *only* way humans can detect somethings mass is through detction of momentum and gravity. Photons have both of these properties while in motion. Please read what I write if you're going to call it nonsense. Fresheneesz 01:36, 9 May 2006 (UTC)

This is also nonsense. General relativity indicates that objects are deflected according to the geodesic equation for a metric determined by the stress-energy tensor. Mass is only a small part of the stress-energy tensor; gravity is due to many things other than mass. The way you actually determine mass is by the root squared difference between energy and momentum. -- Xerxes 16:18, 9 May 2006 (UTC)

What is a "root squared difference"? The bottom line is that photons have momentum and are effected by gravity, and that isn't addressed very well in this article. Like I have said before, there are millions of people that simply *do not* understand that photons can be "massless" *and* be effected by gravity, or push things. Fresheneesz 05:28, 10 May 2006 (UTC)

Root squared difference: ${\displaystyle m={\sqrt {E^{2}-p^{2}}}}$. I think the article is very clear that photons are affected by gravity. Associating that effect with mass only served to propagate a falsehood. -- Xerxes 16:50, 10 May 2006 (UTC)

Do you deny that photons bouncing around in a box will make the box heavier than if it had no photons inside? Fresheneesz 05:28, 10 May 2006 (UTC)

While a box full of photons may have mass, a single photon does not. Since boxes of photons are not everyday sorts of objects, they are not mentioned in the article. See, for more detail, photon gravity on the topic of boxes of photons. -- Xerxes 16:50, 10 May 2006 (UTC)

Of course you don't really need the box; two photons, not going in exactly the same direction, constitute a system having nonzero invariant mass. That seems like a fairly "everyday object" to me. I would think it might rate a mention here. --Trovatore 17:05, 10 May 2006 (UTC)

It's interesting, but is it about photons? I'd say it's something that belongs in an article about mass or binding energy or some other relativistic phenomenon. I mean, you could say the same thing about a box of neutrinos (modulo neutrinos actually having a tiny mass). -- Xerxes 19:40, 10 May 2006 (UTC)

Neutrinos do have mass though, which makes anyone considering it be less than confused about why it interacts with gravitational fields etc the way it does. Photons are the *only* thing in science that most people have heard of that has no invarient mass. Given that it involves photons, it deserves at least a mention on this page, and like you said, probably on pages involving relativistic phenomenons, and articles about mass. People aren't going to read the entire wikipedia, so multiple mentions of the same thing should appear whereever relevant. And I think it is relevant in this case. Fresheneesz 10:34, 11 May 2006 (UTC)

BTW, I like the link you gave - it says very nicely exactly what I was trying to say. Fresheneesz 10:35, 11 May 2006 (UTC)

Last batch of edits

This certainly did not meet any set of standards we discussed here on the topic of light and gravity. The most glaring error was the incorrect use of the term "effective mass". Please refrain from adding bad physics to the article. -- Xerxes 21:10, 14 May 2006 (UTC)

Yes, I did some research to try and justify the much less used (or incorrectly used) definition of "effective mass". I found a couple cases, but over all, I realized I shouldn't have used that particular phrase. Relativistic mass is looked down upon, and so I tried a different term - which I hadn't realized had a precise scientific meaning.

However, I had many other edits that I think are relevant and helpful. Please discuss other "glaring error"s that you have issue with. Fresheneesz 18:43, 15 May 2006 (UTC)

I also noticed that you reverted my edits en-masse. I would very much appreciate if you discriminate between my good edits and my bad edits, and take out the crap - but not the good stuff. I hardly can think that all of my edits were bad. Reverting en-masse is an action taken far too often. I would ask that you please revise your edit, and discuss the issues you have with what you actually want reverted. Fresheneesz 18:48, 15 May 2006 (UTC)

Origin of the term photon

The original Misplaced Pages article was correct. The term "photon" was coined by Gilbert Lewis in a 1926 letter to the editors of the journal Nature. The 1920 lecture by Planck, as posted by nobelprize.org, is a translation from the original German. In German, Planck uses the term "lichtquantum", meaning literally "light quantum". So this usage does not pre-date Lewis's 1926 coining. -- Xerxes 14:32, 18 May 2006 (UTC)

Yes. I've actually looked at the 1926 article. I don't have it right now, but I can vouch that it does say something like "I hereby propose the term 'photon'" (paraphrased from memory). Itub 18:31, 18 May 2006 (UTC)

mass of photon and "gas"

Xerxes, you put the phrase "gas of photons" under that header, and I think thats just a bit misleading. Some people might misinterpret that to mean that photons constitute a gas. In the most common sense of the word, photons don't make up a gas. Comments? Fresheneesz 02:49, 24 May 2006 (UTC)

Photon gas is a perfectly cromulent physics concept. See, for example, these hundreds of papers in the arXiv. -- Xerxes 03:30, 24 May 2006 (UTC)

Ok, I created a page explaining somthing about a photon gas, and linked to it. Fresheneesz 17:34, 24 May 2006 (UTC)

I guess our theme-song for the box of photon gas wouldn't be by Mason Williams....

Seriously this is over-complicated. One of the reasons a single photon has no invariant mass is that you can change its energy to anything you like, just by jumping to a different inertial frame. So what invariant mass would it have, if it had one? You can make its energy go as close to zero as you like.

The nifty and tricky thing about PAIRS of moving objects, so long as they aren't co-moving (they have some relative velocity) is that there's no reference frame you can pick, to make all their energy go away. You're always STUCK with some, no matter which single frame you pick. All you can do is find the frame where the total energy of the two particles is MINIMIZED. In that frame, you'll find each particle headed away from you (or toward you, one or the other) with exactly the same MOMENTUM. So the total system momentum is zero. In that frame, energy is the least. And in that frame, the total mass of the system is the invariant mass. Multiply by c^2 and you have the "rest energy" of the system, which is also the total energy of the system (since momentum is zero). It's the minimal energy you can't avoid, because it's a system. Notice that it includes the sum of the rest energies of the particles, plus some kinetic energy you can't get rid of by changing inertial frames.

Now for the fun part. This is also true for two photons. For every pair of photons (let's pretend they start from the same point), so long as they aren't headed the same exact way, there's some inertial frame where both photons are headed "exactly" away from (or toward) the observer, and both have the same frequency, and the same momentum. *That* is the frame in which the energy of the PAIR is the least. And in that frame, the pair of photons have a mass, and that mass is the "invariant mass" of the pair. Their momenta cancel, and are zero. Which means their mass is their /c^2. Just as advertized. You don't need a box, but if you had one, that box would force both photons into bouncing around until the zero-total momentum frame was very close to the initial rest frame of the box. In that frame, that's how much extra the box would weigh, by comparison with empty (on average, after many photon bounces transfer momentum to the walls). See below for detail.

Actually, if you have box, you can weigh just ONE photon. Does that mean the photon has a mass? No. It means you still are stuck with a SYSTEM of two objects (box + photon), which you can look at from many inertial frames, until you find one in which the total momentum of the two is zero. What would that look like? In that frame, if the photon is moving the right with momentum p = E/c, then the box is moving to the left with the same momentum m*v. Obviously v will be small. When the photon bangs off the other side, they trade directions. If you weigh this collection you see it jumps from side to side (jiggles) as the photon hits, but the net downward force can be calculated, because the one photon suffers aberration in a g field, and hits each wall at a slightly downward angle. Downward momentum transferred gives you extra weight, and over and above the m*g of the box, you find that this extra weight is *g where E is the energy of the photon in your zero-momentum frame. So the single photon seems to act like it has mass of E/c^2, but only because of your prejudice that the weight of a box has to equal the sum of the rest masses of the things in it. It doesn't. In a sense, you're not weighing the photon, but weighing the photon's kinetic energy. If the box was full of gas molecules you'd weigh their kinetic energies in the same way, along with their rest masses. But kinetic energy is all the photon has. No rest mass. Still, you weigh just the energy, and you can do it for ONE photon, so long as it's part of a zero-momentum system. Sbharris 04:25, 9 June 2006 (UTC)

The idea of a 'gas' in physics is widely used and in fact there is a special mathematical set of tools used to treat such gases. This can include a photon gas as well as an electron gas, and others, and is not related to a state of matter - but rather a tenuous collection of particles. This is not only useful, but it is also illuminating when exploring the fundamental physical interactions of elementary particles, including photons. It is therefore important to mention this in the article and even more so to mention that a particle 'gas' is not to be confused with a state of matter. For peer-reveiwed scientific literature on: an electron gas click here, a photon gas click here, mathematical tools for general particle gases click here. Since there are peer-reviewed scientific journals illustrating how widely used this concept is, and since Wiki is an encyclopedia that reports as exhaustively as possible on the state of different subjects, then we should include photon gases in this article, electron gases in the electron article, and perhaps even "particle gases" in the particles article. It does require a bit of care on the reader's part but it is real information that exists in science and so people should have access to it for their education and edification. Cheers, Astrobayes 22:24, 26 June 2006 (UTC)

Note about pair production

I had writen that a single photon can't turn into a particle or particles with rest mass, for the reason that the energy of a single photon is not defined, it's rest mass is zero, and rest mass (invariant mass) is conserved in reactions. However, I neglected to mention that of course single photons can turn into particles with rest mass by interaction with another particle. In that case, it is because the photon provides an extra invariant mass to the system which IS defined.

This is very much the same case as a single photon in the box, above. The photon and the box now defines a 2-object SYSTEM, in such a system has increased invariant mass due to the photon, and so the energy available from the photon to produce rest mass (in the form of other particles) is now defined. In the case of a photon interacting with a nucleus or other charged object in pair production, much the same thing happens. Since we have a system, the single photon's energy can be defined: if it's 1022 kev in the center of momentum frame of the 2-particle system, that's enough to make an electron-positron pair. If it's less than that, it can't happen. That second particle is necessary not just to absorb the extra momentum when the photon offloads its energy, but also acts as an objective guage to how much "rest-mass convertable" energy the photon "has." Otherwise it's just a matter of observer.

Single particles with rest mass can decay all by themselves, because they have some store of minimal energy which isn't kinetic. For a photon there's no such thing, so in a sense they all need "help" from an external reference to tell how much "kinetic" energy they have. Sbharris 22:41, 9 June 2006 (UTC)

For a single photon... what you're saying here seems right. For two photons, you *can* create particles with rest mass... it's done all the time at the B-factory here (this is my third edit to this section - I am finally satisfied with this link I provide here :D ). I know it may seem like fantasy that from two massless Bosons you can create positive-mass Leptons and Hadrons but it's done all the time. At the PEPII B-factory (link above) the Upsilon "4S" resonance is exploited to do just this very thing. I've added this photon article to my watchlist. I have experience in this and it has always been an interest of mine. I look forward to contributing to this article thus. Astrobayes 21:57, 26 June 2006 (UTC)

Luxons??

Good for you, Xerxes. What a needless word. Sounds like the particles that come out of the top of the big black pyramid casino in Las Vegas. Luxons. Yeah. And I may add to critics that while luxon may be a perfectly cromulent word, its use does not embiggen this article. Steve 00:28, 26 June 2006 (UTC)

I actually cannot see why luxon is not in this article...it gets its definition by photons and gluons alone. I also like how photons being gauge bosons gets like no play. --HantaVirus 20:33, 9 August 2006 (UTC)

In 100,000 physics papers I searched, there are 2 mentions of this word. So it's a word for which there is no pressing need (physicists say "massless" when they need to lump gluons and photons together), and which isn't much used. Let's not use Misplaced Pages to give it any airtime, or soon it will creap into the language like some horrid particulate neologism the writers of ST:TNG made up, especially for use by Ensign Crusher to save the day with. One of the papers which used it had a scheme for making tachyons too, which lets you know how odiferous the associations are. And let us not forget tardyons, which sound like something from Dr. Who. Or maybe they are the guage particle which explain school lateness, something like Higgs do for inertia. Surely a grand unified theory of this is needed . And bradyons as the field particle of bad 70's TV sitcoms.

But opinions differ. If you think the word is cool, start your own website for it. As for myself, you remind me to tag the luxon wiki article for deletion, unless somebody can come up with a physics literature reference for it. I'm sure one exists (so it will be kept, probably), but having to FIND it may teach luxon-boosters out there a lesson. SBHarris 21:45, 9 August 2006 (UTC)

Photon Creation Section Missing Critical Information

X-ray photons are created whenever a charged particle accelerates, a well-known process called synchrotron radiation. When we even have an article at Wiki devoted to this very subject, how could x-ray photon production by this well-known process get overlooked? I've added a comment in the "creation" section about the production of x-ray photons by synchrotron radiation. Also, photons are also created in particle collisions, also called particle annihilation. This isn't mentioned either but I hesitate to add a comment for it because we have a section in this article specifically labeled "annihilation" and I feel it may be confusing to some readers without knowledge of particle physics. What are your thoughts on how to include photon-production from different kinds of particle annihilation (there are many)? Thanks. Cheers, Astrobayes 22:33, 26 June 2006 (UTC)

One final comment on my addition: while I know the article *did* mention the acceleration of charged particles, my desire and intention is to illuminate that this is an x-ray photon production process that is so widely familiar and used that it has a name - and this information is important for anyone who has either not taken a modern physics course or who is not familiar with advanced electromagnetism or particle physics. My addition was not to suggest that the article did not contain a cursory reference to acceleration of charged particles but rather that the link between this general process, and the popular method of producing x-rays was highlighted. Cheers, Astrobayes 22:38, 26 June 2006 (UTC)

Go for it. Be bold in edits-- they can only get reverted by others if wildly unpopular. And if good additions, they stay. I recall once being given a tour of the LLNL national synchrotron light source, which is basically this big electron synchrotron in UC Berkley, run to do nothing but produce X-rays, which are tapped off in beamlines for various materials science projects which need them. Whole thing was run by one master 486 computer at the time. :) These days I'm sure it takes a supercomputer, at least. But of course you're right that synchrotrons are very important sources of high intensity coherant X-rays. The mechanism is acceleration of charge, but the frequencies obtaineds are due to odd and specific relativistic effects, as you know. So they are worth mentioning for that alone.Steve 23:25, 26 June 2006 (UTC)

Polaritons

Are photons in matter really polaritons? A polariton is the composite particle that results from strong coupling between a dipole carrying particle (usually a quasi-particle) and a photon. The characteristics of the polariton are a combination of the composite particles. So in the case of matter, this other particle would in general be an atom (or molecule). But the dispersion of a photon in a matter is not a combination of the atomic dispersion and the vacuum photon dispersion. Moreover, I've never heard anyone refer to a photon travelling through matter as a polariton (other than very particular situations such as in EIT, or in a cavity). Could someone provide a reference? --J S Lundeen 17:48, 22 July 2006 (UTC)

After asking around and discussing with my colleagues, I see how a photon could be referred to as a polariton in a material. For photons in glass, it is definitely not usually discussed in this manner, nor is the optical dispersion typically calculated in this way. But it doesn't look like it is incorrect (in fact it looks like a very general explanation, applicable to a wide variety of materials).--J S Lundeen 15:46, 25 July 2006 (UTC)

Photons at rest mass

According to the article, photons have a rest mass of zero but have a small relativistic mass while they are moving. There's just one thing I don't understand about that. Hypothetically, if I were to somehow travel at the speed of light and then fly next to a photon, then relative to me, the photon would be at rest. However, as stated earlier, photons have a rest mass of zero. So, if I were to fly next to a photon, wouldn't it then blink out of existence?

In a sense. If you fly in the same direction as a photon sufficiently fast, then you can redshift its energy down to the point where no reasonable detector in your frame can detect it. Melchoir 02:59, 29 July 2006 (UTC)

It might be more correct to say that single photons in free flight have an undefined energy, since you can make their energy (or relativistic mass) anything you like, by choosing your inertial frame (you do that for particles, too-- for single particle the kinetic energy is anything you like, but zero only in ONE frame). But such a photon has no rest mass because it's never at rest. Only when you get two or more photons going in two or more different directions, do you get "stuck" with a minimal energy for the system, which will occur in just one of all possible frames, as with the single massive particle. This frame happens to be the frame in which the photons have no net momentum (it all cancels out). In that frame, they all have minimal energy, when you add it up. And THAT energy is their invariant mass, as a system. It's the energy in the system which is available to make massive particles with rest mass. But note that unless you have two or more photons, that energy doesn't appear. There is no way for nature to "decide" how much energy a single photon has available to make rest mass, unless it has a reference point frame (a second particle) to do it. And as soon as you choose that frame by introducing a second particle with rest mass at rest in that frame, you define the photon's energy FOR that frame, and thus define how much energy you have to work with, if it interacts with the second particle. For a system of two photons (where neither photon can be at rest, but both can be flying off in exactly opposite directions), it's the zero momentum frame (the center of momentum frame) which defines the minimal (rest) energy. I'm not sure that's clearSBHarris 03:25, 29 July 2006 (UTC)

Dubious "in vacuo" statement

Strait removed some of this paragraph, saying it was dubious and uncited:

Since photons move at the speed of light, by relativist time dilation they do not take any time to get from their source to where they are finally absorbed; that is, they have zero lifetime but can travel arbitrarily far. The emission and absorption events are at zero space-time interval. From this point of view, first articulated by Gilbert N. Lewis in 1926, the photon's energy never exists in the vacuum, but transfers from the source to the absorber without delay.

And shortened it to:

Since photons move at the speed of light, no proper time passes for them. The emission and absorption events form an interval of length zero in space-time.

It's not clear to me from the short version just what it that would be considered dubious. As to citation, I've placed copies of these two papers temporarily on my personal web site . Gilbert N. Lewis, "The Nature of Light," Proc. National Academy of Science, Vol. 12, pp.22–29, 1926; and Gilbert N. Lewis, "Light Waves and Light Corpuscles," Nature, Vol 117, pp.236–238, Feb. 13, 1926. I haven't searched to see exactly what statement I was paraphrasing above, and perhaps someone else would like to have a go at it to make it less dubious. Dicklyon 20:22, 3 August 2006 (UTC)

The dubious things were:

• "they have zero lifetime". First, the word "lifetime" is ambiguous. If it means log_e(2)*half-life, then it's definitely wrong. If it means "time that it exists" (which I think is what was meant), then it is only true from the photon's perspective, which was already stated.

Good point. I perhaps over-reacted to the definition I saw somewhere that said photons have infinite lifetime (since they don't decay). I figured that by analogy with how partical lifetimes are measured, which is in their own frame of reference, if a photon is to be assigned a half-life or a lifetime, then the value must be zero, or at least shorter than the lifetime of any particle traveling at a finite speed. In some sense, though, I guess I put it there to provoke an explanation. The "lifetime" in this sense is less than a cycle of the frequency.

Well, I think it is meaningless to assign the photon a lifetime (meaning log(2)*half-life). But FWIW, the Particle Data Group says that it is "stable" . I agree with them that this makes more sense than saying it is zero if you are forced to pick one or the other. --Strait 02:46, 4 August 2006 (UTC)

Stable? If that means it doesn't decay, OK. But on the other hand, there is no time so short that that you can't find an inertial frame in which the photon lasts longer than that. That is, it lasts shorter than any particle does, if you pick a reference frame moving from emitter to absorber at close enough to the speed of light. It's not at all like a particle that is long-lived in the sense of continuing to exist in its own frame. Dicklyon 04:09, 4 August 2006 (UTC)

• "the photon's energy never exists in the vacuum". I don't know what that could mean. I did a slow skim of the papers that you linked to and couldn't find that statement.

It means that there is no finite region of space of and time that excludes the emitter and absorber than can be said to have the energy of the photon in it. I guess I read that into what Lewis said, too. But if you look at the uncertainty principle, with photons have a definite frequency, the positional uncertainty is infinite, so includes the emitter and absorber; the energy is in their states, even during their quantum transition when they are in a mixture of states beating at the frequency of the photon.

Errrmmm... while perhaps there is some sense to this viewpoint, it is certainly not how most physicists think about it. They would say that the energy of the photon certainly exists between the emitter and the absorber. Perhaps a discussion of this in the article would be ok, but it needs to present the more usual viewpoint first and clearly contrast the two. --Strait 02:46, 4 August 2006 (UTC)

Good idea. I think I'll work on a section such as "Photons as transactions" to contrast this view with the more "Copenhagan" view. Dicklyon 04:09, 4 August 2006 (UTC)

• "but transfers from the source to the absorber without delay." See the first point.

In any case, I think the interesting point in Lewis's papers is not this, but the issue of whether or not a photon can be emmitted without having a predestined absorber. --Strait 23:06, 3 August 2006 (UTC)

Indeed, that is the interesting point, except that the concept of "predestined" is not time-symmetric. Following his logic, the reversibility of all known physical laws would make your question equivalent to the issue of whether or not a photon can be absorbed without having a predetermined emitter.

I may not be expressing it quite right, but this idea of Lewis is what later became Feynman-Wheeler theory and Cramer's Transactional interpretation and Carver Mead's "Collective Electrodynamics", none of which disagrees with standard QED in any way that I'm aware of, but is a difference of interpretation that can provide some useful insights.

Dicklyon 01:28, 4 August 2006 (UTC)

In a box? With a fox?

What's up with the mass of a photon in a box? It says "Although a single photon has zero mass, multi-particle objects including photons may collectively have mass. For example, a mirrored box containing a gas of photons, or even a single photon, with total energy E will have greater mass (by Δm = E/c²) than if the box did not contain photon(s)." Now if you wait a second until the photons are absorbed, is the mass the same, or different? What does it mean to have a photon in a box anyway? Photons are emitted and absorbed. So who put them there? If they were inserted from an external source, then the added mass is the same as the mass of the photons themselves, is it not? Reflection absorbs photons and radiates new ones. Do some of these have zero mass? Can someone explain how to make sense of this section? Dicklyon 04:44, 6 August 2006 (UTC)

We have a mirrored box so we don't have to talk about absorption, because it just makes things more complicated. But no, absorption doesn't change the mass of the system.

. In a box with a single photon bouncing back and forth, the COM frame is the center-of-mass frame, the same as if the box had a superball or gas molecule bouncing from side to side. The box in free space will wiggle back and forth about a point, and that point is the center of mass. An inertial frame with observer stationary with respect to that point, is the COM frame. Okay?

. Now the total box mass in the COM frame is the mass of the empty box plus E/c^2 where E is the photon energy in this frame. So in that sense the trapped photon has mass which is E/c^2.<g>But what if the photon isn't trapped, you ask? Suppose somebody puts a hole in the box ahead of it, or it's just flying through? Answer is that the answer does not change. There is a still a COM frame, and in that frame the system of photon-plus-box still has mass equal to mass of empty box plus E/c^2 for the photon. The difference is this mass isn't easily accessible by putting the thing on a scale, which is why we trapped the photon. However, if you calculate the invariant mass of photon+box system, you get what the mass WOULD BE, if you weighed it "at rest" (ie weighed it in the COM frame). So doing that calc can give you the rest mass of a particle before it decays-- you just figure out the energy of the decay products in the COM frame and the mass of the whole shebang, including rest mass of the particle which was the origin of it all, is E/c^2, where E is the COM total energy. SBHarris 05:08, 6 August 2006 (UTC)

So the explanation for the photon without a box not have a mass is that you can't be in its center-of-mass system, and that if you approach it the photon energy goes to zero? Dicklyon 05:12, 6 August 2006 (UTC)

Exactly right. It's the presence of the box (which we have to assume has SOME energy or mass itself) that gives you a 2-particle system, so that there's a certain SYSTEM energy you can't get rid of by choice of frame (you can chase the photon, but you'll find your box gaining energy when you do it-- a losing game). The (minimal) energy for the whole system is your (invariant mass) * c^2.SBHarris 05:33, 6 August 2006 (UTC)

Photon in the box is very poorly written

Please :

a. delete

b. correct in order to contain a correct explanation of what is going on:

The correct explanation of the "photon in a box" issue is that the trajectory of the photon inside the box gets slightly curved in the gravitational field such that there is a vertical component of the photon momentum p_tangential=p*sin(theta) where p=E/c is the momentum and theta is a very small angle. At each collision with the vertical walls of the box the photon transmits a downward force dp=dp_tangential/dt to the box that can be (mis)construed as an increase of mass equal to dp/g where g is the gravitational constant. Please remove any misleading references to Δm = E/c². Such ridiculous stuff can only further the misconception that the photon has non-zero mass that can be measured.

Thank youAti3414 05:03, 6 August 2006 (UTC)

COMMENT: I see you don't read your talk page. The reason somebody reverted your edit, is that it indeed misses the point. You have the MECHANISM for how the photon energy in a box shows up as an extra weight g* (E/c^2). But this extra weight is not at all "pseudo" weight. It's as real as any weight and it's due to the mass of the trapped photon(s) bouncing back and forth. Single free photons have no mass, but single trapped ones do, since they are forced into a 2-object system, which has a COM frame, which gives the photon energy an invariant mass.

You'll notice, BTW, that your explanation applies to any particle in a box, not just a photon. This is the mechanism by which all trapped particles show their mass or weight in a system. Note that for a relativistic massive particle, this mass is proportional to relativistic momentum γmv, which means that the "mass" or "weight" measured for it is not just the rest energy, but the total energy. So the kinetic energy of the particle gets "weighed" too, and shows up as mass of the box. Interesting stuff, yes? SBHarris 05:10, 6 August 2006 (UTC)

I think the original explanation was terrible, where is this gedankexperiment coming from anyway? What is its relevance? Is there any such experiment that has ben reported in a peer refereed journal? If not, why is it included? I also think that the explanation based on g* (E/c^2) makes no sense whatsoever. In effect, it encourages the people that claim that the photons have an equivalent mass m=E/c^2 which is totally wrong. Come to think of it, the whole page misses the fact that current experiments put a limit of 6*10^-17 eV on the photon mass (with newer experiments set to lower this limit). This limit is in total contradiction with any type of E/c^2 derivation. If you don't like my explanation of "photon in the box" could you please remove the entry altogether? It is very misleading and badly written as originally set. Thank you Ati3414 05:21, 6 August 2006 (UTC)

I was looking for an explanation in the absense of gravity, hopefully relating to when the energy mass is considered to be a mass and when it is not. I think the gravity effects you are describing are beside the point. Dicklyon 05:09, 6 August 2006 (UTC)

Right, they are beside the point, and so is the box, for that matter. Two photons, not going in exactly the same direction, constitute a system having nonzero invariant mass. --Trovatore 05:14, 6 August 2006 (UTC)

Yep. The box is only used to illustrate this fact, since in a g-field, the invariant mass of a closed system can be simply determined by weighing the thing in the COM frame.SBHarris 05:20, 6 August 2006 (UTC)

Then why aren't you writing an entry on systems of photons instead of the very badly written "Photon in the box". Two of us already complained about the shoddy story, almost simultaneously. You cleaned it up a little by removing some errors and now the entry is almost incomprehensible. Someone reading the entry might conclude that the mass of the photon is E/c^2 a classical rookie mistake by antirelativists. Please write a scholarly piece or take the "photon in the box" out. altogether.Ati3414 05:34, 6 August 2006 (UTC)

The entry uses one photon to illustrate a point, because the box then becomes the other mass in the system. One photon has no mass. One photon which is part of ANY system in which there is ANY other mass, contributes a photonic mass to the system, E/c^2. Just as a particle with rest mass M and kinetic energy T contributes (M+T)/c^2 mass to such systems (in the COM frame). The point is that ANYTHING in there gets its energy counted as mass. Okay? Part of the reason I'm doing this, is because it's poorly understood out there. I'm having problems with you, looks like. So you should be grateful. SBHarris 05:43, 6 August 2006 (UTC)

No need to resort to personal attacks. By insisting on the posting, you simply reinforce the belief of many crackpots who think that m_photon=E/c^2 (or, expressed in eV, m=hf), notwithstanding that there are experiments that clearly refuted that idea. You can have your entry, try to make things a little clearer and don't take it personally. As it is currently written, is very bad , though I must say that after my interventions you cleaned it up quite a bit. So, I will put in an entry, that is sorely missing on the mass of the photon, such that there is no possibility of confusion.Ati3414 06:38, 6 August 2006 (UTC)

Weight/mass of single photon in a box bouncing up and down

ONE photon has a mass (and a weight) in relativity, if you trap it in a box so that it bounces around, and put the box on a scale. The weight and mass turn out to be due to the gravitational Doppler shift. Here, I'll do the calc (excuse my editing):

Put one (well localized) gamma photon of ground energy E_o = hf_o in a perfectly gamma mirrored box (okay, it's Gedanken) of height x, in a gravitational field g. Let it bounce up and down, hitting the top and bottom of the box, each bounce after time t=2x/c. Each bounce transfers 2 times the photon momentum E/c. Now the force that the photon exerts on box top (or bottom) is dP/dt = Force = F = (2E/c)/(2x/c) = E/x.

First error: I took another look, this is obviuosly wrong, the "dt" in the calculation of force has nothing to do with the time taken by the photon to travel up and down the box , t=2x/c. The "dt" is actually the very short time during which the photon bounces of the wall and reverses momentum. F=dP/dt where dt is the short (and unknown) interval of momentum reversal. You are confusing a finite time interval , t with an infinitesimal one , dt. So, your derivation seems totally bogus, this is why it comes up with E/c^2.Ati3414 04:11, 8 August 2006 (UTC)

Sign you comments, please. If you don't like dP/dt note that this is average force (pressure time area) averaged over the time of many photon impacts, so you can just as well use ΔP/Δt because we're averaging the momentum transfered over long periods of time and many photon impacts (that is what the ratio I use does-- do I have to spell it out?). Your sniping is ridiculous here, as you have to do exactly the same thing to get an average downward momentum transfer with time in the side-to-side force = weight, in the example you wanted to put in the article. SBHarris 17:19, 6 August 2006 (UTC)

No, I didn't , I used the proper dt. I did not mix in an abitrary time interval as you do in your derivation. I think that your entry has no business in the page since it is clearly based on personal research and it is wrong to boot. Ati3414 18:50, 6 August 2006 (UTC)

It's not an arbitrary time. It's the average time over which the momentum transfer takes place. You get the same momentum transfer rate per second as you do per hour. Which means the same force. Only if you go to very small times < x/c does it get lumpy. SBHarris 20:22, 8 August 2006 (UTC)

The difference in photon force (ΔF) between the top and bottom of the box is the "weight" of the photon, as measured by the scale the box is on, and THAT is due to the Doppler change in energy, (ΔE)/x.

Second Error: Nothing gives you the right to attribute "wight" to the photon. A resulting downward force, yes, but "weight"? Nothing gives you the right to assign it weight. You are preparing the ground to assign mass to the photon and I will expose this trick here. We all know that there is such a thing as a pressure radiation (see Einstein 1905).Ati3414 04:11, 8 August 2006 (UTC)

Hello? Yes, the weight is here the difference between radiation pressures. And bouyancy force in water is due to the difference in water pressures with height. Here's a shock: Did you know you can calculate the weight of a cube of gas (never mind a photon gas) by integrating the difference in gas pressure on a bottom vs. the top of a vessel? Same with a container of water, for that matter. SBHarris 20:22, 8 August 2006 (UTC)

Nothing permits you to mix in the Newtonian notion Weight=m*g , you are mixing relativistic calculations with Newtonian ones. You are attributing a the force due to the variation of radiation pressure to weight, with no justification whatsoever. Ati3414 02:14, 9 August 2006 (UTC)

Any permanent increase in force on a scale in a g field, we call "weight." The justification is the definition of the word. The scale says the box weighs more due to its content of photons, and so it does. What, do you think this is some kind of magical rocket motor? SBHarris 02:42, 9 August 2006 (UTC)

Now ΔE is just h*(Δf) where f is the photon frequency. From Einstein's formula, the Δf from a gravitational shift through a constant field g and height x, is given by Δf = f_o. Here f_o is the frequency at ground level. Substituting gives: Weight = ΔE/x = h(Δf)/x = h(f_o)/x. Cancel x's and note that hf_o = E_o which is the energy of the the photon at the ground, and you have:

Weight = (E_o) g/c

If you equate the weight as mg, then the "mass" you weigh for the photon is m = E_o/c.

Third and most fatal error: You are now mixing relativistic calculations with the Newtonian ones. You are saying : Weight=g*"something" so "something" must be the "mass" of the photon. Sloppines and sleigh of hand can produce any result (especially if you already know the result you are after). Ati3414 04:11, 8 August 2006 (UTC)

Some things are the same for both Newton and Einstein. The fact that mass is proportional to weight is one of them. Also the fact that mass is proportional to inertia. There's no way to increase the passive weight of a box, with nothing but gravity acting on it, except by increasing its mass and inertia. That's what weight is: m*g. By suggesting not, you're embarrassing yourself SBHarris 20:22, 8 August 2006 (UTC)

For a photon going from side to side you have to figure the g field as an acceleration producing a relative box-side photon lateral velocity, then do the calc from relativistic aberration. But the answer comes out the same. SBHarris 02:42, 9 August 2006 (UTC)

BS, I challenge you to do it and you tried to weasel out by asking me to do it. I am not the one supporting the crackpot idea that m=E/c^2. So, for the third time, show the proof or cease and desist on your claims Ati3414 02:14, 9 August 2006 (UTC)

Okay, a calculation for mass contribution of a single lateral photon in a rocket and a box, showing increase in ineria, therefore weight, of E/c^2. See last section of this page. All for you. Afterall, I used your suggested method, even though you don't believe its result. SBHarris 02:42, 9 August 2006 (UTC)

This may not be true. If you really wanted to prove that the photon trapped in a box imparts "mass" to the system, you would need to demonstrate mathematically that this happens INDEPENDENT of the direction of photon motion. Since you botched the simpler case (up/down) motion, the chances are slim that you will be able to make your point for the side to side case. But , please try (and try to do it without sneaking a peek at my previous solution to this situation)67.170.224.36 04:00, 8 August 2006 (UTC)

You gave no previous formal solution-- just a few handwaves. If you'd actually worked the problem, you'd have had to deal with the things you just criticized me over, namely having to integrate over many impacts. So let's see you do it. Post it at the end of the TALK here, as a section. I'll give you a week, and if you can't, I will.SBHarris 20:38, 8 August 2006 (UTC)

THe point is that you botched your solution and you twisted the math such that it gives an answer (E/c^2) that is 17-19 orders of magnitude larger than the experimental data. You refuted your own calculations, remember? Try taking care of your errors, then we;ll talk. Especially the mish-mash of relativistic and newtonian mechanics. Also the fact that your solution is for one direction only, it should be direction invariant. Be a good sport, take out the embarassing E/c^2 "answer" , it makes the post ridiculous. Ati3414 20:50, 8 August 2006 (UTC)

A photon may not have mass in and of itself, but photon energy contributes mass to all systems it's part of. SBHarris 05:39, 6 August 2006 (UTC)

A photon contributes energy and momentum to the system into which is being incorporated. The extra energy manifests itself as radiation pressure which in turn may manifest itself as a force. No need to go past this point and venture into the murky waters of "relativistic mass", rest mass", etc. Change the post to drop any mention of "mass contribution" and I will stop objecting to your posts.

You got this idea from an ill-inspired entry in the John Baez FAQ. The FAQ is known to have errors, I have personally convinced him in the past to change something that was wrong in the test theories of relativity paragraph, I contacted him again asking to correct the paragraph about the mass of the photon. I see no reason why the FAQ will not be corrected. Ati3414 04:11, 8 August 2006 (UTC)

Looks very similar to my original post. Would you care to put it on the front page such that people don't get the mistaken idea that photons have mass? We are only talking about photons getting WEIGHTED, correct? Which was my point. Also, I think that the case with the photon moving side to side was treated correctly in my post that you removed.Come to think of it, what is the point of the whole "Photon in a box"? It is a gedankexperiment, no real application (we already know about radiation pressure). Why confuse people with it? Come to think of it, after looking at your explanation, I believe that it is wrong. You calculate the radiation pressure on the bottom of the box (2E/c)/(2x/c). But there is EQUAL pressure on the TOP of the box, so in reality the box is "jumping" up and down on the balance platter, so on average, you should see ZERO extra weight. This is a very bad entry for an encyclopedia, would you please remove it? Thank you Ati3414 05:46, 6 August 2006 (UTC)

There is no equal pressure on the top of the box for Doppler shift reasons explained.

OK, I see that.Ati3414 06:17, 6 August 2006 (UTC)

I didn't remove your example. Somebody else who agreed with me did (yep, another crank here). And since we're not only talking about photons being weighed, we don't put the mechanism for this in the article (though it's fun for the talk page, so I did it). Photon energy DOES contribute E/c^2 mass to systems, in the COM frame. Not just weight. Actual, really real mass.SBHarris 06:25, 6 August 2006 (UTC)

How could it? You are only "weighting" your system in the gedankexperiment. You cannot infer that mass has been added based on the fact that in reality you have a differential in radiation pressure between the top and the bottom of the box.To say that the photons are contributing "Actual,reallly real mass" is philosophically abhorrent. They are contributing energy and momentum to the open system into which they have been injected. The resultant system has a higher energy and a higher momentum. We can WEIGH the new system and we find the weigh to be higher than the weigh prior to photon injection. This does not mean that mass has been created, that the photons have contributed "Actual, really real mass". I believe that the entry is very bad and that it should be removed, notwithstanding your very interesting explanation. Ati3414 06:18, 6 August 2006 (UTC)

I was under the impression that in standard physics, the three kinds of mass, total energy mass, inertial mass, and gravitational mass, are regarded as being always equal for all closed systems. Are you saying that's not so? Dicklyon 06:27, 6 August 2006 (UTC)

Look, the same explanations inevitably arise with gluons, which have no more mass than photons (ie, they are massless). But most of the mass of nucleons is kinetic energy of quarks and energy of massless gluons. IOW, a hadron is just a box, being weighed. So shake any ordinary object. 1% is rest mass of quarks and electrons. The rest of the mass (that other 99%) is the sort of stuff we're discussing here. So it's important as a topic. SBHarris 06:25, 6 August 2006 (UTC)

I agree it is important, this is why it needs to be written properly or not included at all.Ati3414 06:47, 6 August 2006 (UTC)

We have written it properly. Massless gluons in a box (nucleons) contribute mass to the system. Massless photons when trapped in the same way, contribute mass to THEIR system. How much shorter, cleaner, and clearer could be it be? Massless particles can, and do, contribute mass to systems. Period. SBHarris 07:47, 6 August 2006 (UTC)

You mean that you cleaned it up after yesterday's criticisms? This is a more correct statement. Even so, read by a neophyte is still bad: it implies a photon mass of 3eV , in clear disagreement with the current experiments (and with the official .gov site) that place a limit of 6*10^-17 eV on it. There must be an error in the calculations of your gedankexperiment. 67.170.224.36 14:10, 6 August 2006 (UTC)

It implies nothing of the sort, and there is no error. The mass measured in systems due to trapped photons is not the mass of the photon but the mass of their energy contribution. Same as for gluons. SBHarris 17:21, 6 August 2006 (UTC)

I took another look, this is obviuosly wrong, the "dt" in the calculation of force has nothing to do with the time taken by the photon to travel up and down the box , t=2x/c. The "dt" is actually the very short time during which the photon bounces of the wall and reverses momentum. F=dP/dt where dt is the short (and unknown) interval of momentum reversal. You are confusing a finite time interval , t with an infinitesimal one , dt. So, your derivation seems totally bogus, this is why it comes up with E/c^2. You should do the right thing and eliminate the entry. It has no place in an encyclopedia since it apperas to be the outgrowth of your personal interpretation of a college exercise. Ati3414 18:52, 6 August 2006 (UTC)

If not so, the local gravity field would disappear POOF every time an electron met a positron. Do you really think it does? That was a serious question. SBHarris 06:25, 6 August 2006 (UTC)

mention relativistic mass

I think the basic underlying problem is this: What the section really is, is an attempt to explain that photons have nonzero relativistic mass, without referring to that unfashionable concept by name. I understand that contemporary physicists don't find the concept particularly useful, and I understand (at least reasonably well) why they don't, and that's fine. Nevertheless, from a historical point of view, the concept was used rather recently, and it's not as though it's been refuted; it's just been found not to be the most convenient way of organizing information.

Since many readers probably have heard explanations that implicitly used relativistic mass, and want to know how those relate to the descriptions they're reading in this article, I think at least a brief summary (giving a link to the excellent discussion in the mass in special relativity article) would be appropriate. The summary should also mention that the concept of relativistic mass is not much used these days, but should not imply that it's a bad, evil concept. --Trovatore 18:19, 6 August 2006 (UTC)

It is a bad, evil concept. Probably it's behind the rising tide of global terrorism. I hate relativistic mass because it's frame-dependent. We have relativistic energy to do its job. But it has been noted that the special thing about the COM frame, is that it's where the relativistic mass of the system, and the ordinary "rest mass" (or really invariant mass), are the same. So E=mc^2 holds no matter which kind of mass you're talking about. Which means the "relativistic mass" of photons counts HERE in this frame as rest mass (of the system = box), but not so in other frames. Perhaps I need to just SAY something like that, here. SBHarris 18:25, 6 August 2006 (UTC)

Just say they have nonzero relativistic mass and be done with it; detailed discussion isn't really about photons per se and can go in other articles. It doesn't matter whether you or I think relativistic mass is the right way of thinking about things. It's the way workers in the field did think of it, until, I don't know, 1950 or so? and it's still in plenty of popular literature. It shouldn't be implied to readers that those writers were spouting gibberish, except in the cases that they were (which I'm sure are abundant, but this isn't one of them). It's just a different way of describing the same facts. --Trovatore 18:33, 6 August 2006 (UTC)

Ati3414, please desist from removing references to relativistic mass. Your hostility to the usage notwithstanding, the point needs to be addressed, descriptively rather than as an advocate for contemporary usage. It would certainly be correct to note that the usage is not very current, but it isn't wrong, and readers who are used to it need to know what has happened to it. --Trovatore 04:53, 7 August 2006 (UTC)

The usage of "rest" and "relativistic" mass has been abandoned in all modern texts. It has been replaced with "proper" mass, the only mass that current relativistic textbooks and papers deal with. Relativistic mass has been expunged. "Photon relativistic mass" is even worse, there is no such thing.

In addition to the embarassing terminology, the section is plain wrong, looks like someone's favorite college book exercise. Wiki does not allow for personal exercises to be turned into postings. To make matters even worse, there is no credible reference as to where the posting came from (textbook , paper published in a peer refereed journal, etc) To make matters even worse, SBharris' attempt to prove the statement can be easily refuted , his proof is not correct. Either way, why have a college book EXERCISE as a post in an encyclopedia?

So, why use wiki to perpetrate incorrect notions, take out the posting and we are done.Thank you for your cooperation. I think that the two of us have worked together in the past removing some crackpot statements from other areas of relativity (Luminiferous Aether). Why does it take so much effort in removing wrong posts? Ati3414 06:22, 7 August 2006 (UTC)

It's not my exercise. It's part of the USENET physics FAQ and has been for many years. This is written by professional physicists. Do read and learn, please. SBHarris 17:07, 7 August 2006 (UTC)

By the same author: "You should not use this to justify the statement that light has mass in general." I agree with you, read and learn.

Gedankexperiments with botched explanations do not constitute science. And they have no room in a work that aspires to be an encyclopedia. Ati3414 17:20, 7 August 2006 (UTC)

Can you guys tell what is the "modern" alternative to this concept? I don't understand what is motivating this dispute, since I haven't see Ati3414 provide an alternative accounting except in terms of "weight" which seems to only confuse the question of mass. Dicklyon 05:13, 7 August 2006 (UTC)

I suppose the contemporary approach is just to describe the momentum, energy, etc, using the formulas that work when you plug in rest mass instead of relativistic mass. I am not arguing against that approach; I kind of see why contemporary physicists prefer it.

However Ati3414 is simply wrong when he refers to relativistic mass as an "incorrect notion". It's not even a different notion from the contemporary one—it not only predicts exactly the same outcome to every conceivable experiment (which should satisfy instrumentalists), it even has exactly the same underlying noumena, for the hardcore realists. It's purely a difference in language and bookkeeping.

Indeed, but as pointed out, it's a concept which is redundant. We have total energy/c^2 to do the job of "relavistic mass". Meanwhile, what we measure as mass in our daily lives and work with as "mass" is actually invariant mass. It only HAPPENS to be the same as relativistic mass in the COM frame, but otherwise it's not the same thing at all. Since this quantity is separately conserved and invariant for all observers, and is usable for all the old concepts of mass, let's use it for "mass" (relativistic mass is not usable for mass in many cases where you have motion == for example you can't make a black hole from relativistic mass unless you're in the COM frame, and then it's also invariant mass as well).SBHarris 17:07, 7 August 2006 (UTC)

Look, I'm not arguing with you on this point. Physicists prefer to use invariant mass and that's fine with me. But that does not erase the historical notion of relativistic mass, and it doesn't make it wrong. I don't know that we need as elaborate an explanation as you've provided; I'd be happy with a brief footnote, after the text saying that the photon mass is zero, pointing out that there is a historical notion of relativistic mass, which is not zero for the photon, and pointing the reader to mass in special relativity. --Trovatore 17:21, 7 August 2006 (UTC)

The statement that the photon has nonzero relativistic mass is a correct statement. Ati3414 may wish that the notion of relativistic mass had never been formulated, but it was formulated, and it is not WP's role to rewrite history to conform to someone's prescriptive notions. Some neutral statement about the photon's relativistic mass must be made; I don't insist on SBHarris' version, but something needs to be there. --Trovatore 15:33, 7 August 2006 (UTC)

What has to be is something that is supported by experiment and references, not a botched grad school exercise as the "Photon in a box". Wiki tries to be an encyclopedia, not a collection of exercises and answers sheet. As to your statement that "the photon has nonzero relativistic mass" , you will need to back it up with published experimental references or retract it. 12.36.122.2 17:13, 7 August 2006 (UTC)

Obviously you don't understand the definition of relativistic mass, which is relativistic energy/c^2. And READ the *&%%$ing reference, please. SBHarris 17:23, 7 August 2006 (UTC)

You are resorting to personal attacks again, which means that you are loosing the scientific argument. You mean your own personal interpretation of the term relativistic mass based on a FAQ? Since when are YOUR PERSONAL interpretations of FAQ entries scientific references? Besides, your definition comes at odds with wiki's. See here: http://en.wikipedia.org/Relativistic_mass Same reference you gave above, have you read it? Here is what wiki says:

The relativistic mass M is then formulated as:

${\displaystyle M=\gamma m\!}$

where

m is the rest mass, and

${\displaystyle \gamma ={1 \over {\sqrt {1-u^{2}/c^{2}}}}\!}$ is the Lorentz factor,

u is the relative velocity between the observer and the object, and

c is the speed of light.

Nothing to do with YOUR INTERPRETATION, actulally at ODDS with YOUR INTERPRETATION No wonder that wiki is such a mess. Ati3414 17:40, 7 August 2006 (UTC)

Your math is a mess. The equations you're using do not apply to photons, which have no rest mass.SBHarris 17:53, 7 August 2006 (UTC)

It is not MY math, it is lifted off the wiki page that YOU referenced as the definition of the antiquated term called relativistic mass (or mess?). You never anwered my challenge: "You mean your own personal interpretation of the term relativistic mass based on a FAQ? Since when are YOUR PERSONAL interpretations of FAQ entries scientific references?" Ati3414 18:11, 7 August 2006 (UTC)

Your notation is a mess, and you didn't lift it off the Wiki, or the equations would come out properly. See those symbols ${\displaystyle math}$? They exist for a reason. I'm not going to fix it for you. It's educational, so you do it yourself.

You are using equations which do not apply. The wiki on relavivistic mass specifically states: Another downside of this approach is that since γ depends on velocity, observers in different inertial reference frames will measure different values, which can be complicated. It should also be noted that these equations apply to matter and they are unsuited for photons: with v=c and m=0, γm of a photon is undefined. Got that? No disagreement, no contradiction. Now, as for the other, what are you asking for? The invariant mass of systems of photons is a topic of several published literature cites. Is that what you want? The relativistic mass of a photon is a concept not used much these days, but for a very long time was a common term in the physics literature. If I find you a cite, will you zip it? There's a reason you were blocked for a month. Your behavior is obnoxious. SBHarris 18:27, 7 August 2006 (UTC)

I was blocked because I challenged cranks that pass themselves as "editors" by editing out their nonsense. The point is that you send the wiki page referring to "relativistic mass". I pointed out that the wiki term m_relativistic=m_rest*gamma comes at ODDS with your personal interpretation inferred from the John Baez FAQ. So your post on the "photon in the box" has only the scientific support that is inferred from a FAQ for beginners, right? No experiment, no peer reviewed paper? As such, it has no place in wiki. Ati3414 19:00, 7 August 2006 (UTC)

I see you've learned something about the math editor. Good. Now continue in student mode, please. The invariant mass of pairs of photons is such a common concept in physics that it's textbook material. Therefore it's rather perverse to demand "experimental" or "peer reviewed" discussion of a student level concept. Peer reviewed papers are long past this concept, and take it for granted. I gave you a reference on another page, but here's another example from dozens of abstrants and papers you'll find in the physics arXive: . Are you really suggesting that pairs of photons have an invariant mass but a system of a photon and a box does NOT? A pair of photons from a positron-electron anihilation obviously has the same invariant mass as the electron and positron, since invariant mass is conserved. Or are you arguing against that, also? Frankly, it's difficult for me to tell what your problem is. You don't like relativistic mass in connection with photons, and you don't like invariant mass in connection with photons, and one gets the impression you don't think photons have mass in any circumstances at all. But since mass turns into photons, that would mean that mass would have to disappear. Since the equations of general relativity do not allow for that (no discontinuities in the G field), it would be an obvious problem. Fortunately, it's one we don't have. SBHarris 01:26, 8 August 2006 (UTC)

I restored the section again, then made some minor changes. In particular, I wrote it to make sense for the case of what happens after the photons are absorbed. The energy is still in the box, so the mass doesn't change. Dicklyon 01:37, 8 August 2006 (UTC)

Thank you. After Ati3414 deleted the section yet again, I restored it yet again. That's revert #2 for me. One more delete from him and he runs up against WP:3R. I leave it to the community overlooking this article to deal with him on this matter from here. SBHarris 01:54, 8 August 2006 (UTC)

compromise

p.31 of Concepts of Modern Physics, by Arthur Beiser, has a pretty good explanation that doesn't invoke relativistic mass, but rather assigns the energy (of the photon in our case) to the rest energy of the system, which therefore becomes part of the rest mass of the system. Maybe such wording would go down better with him? The result is the same, if I understand correctly. You can "search inside the book" on Amazon for energy mass. Dicklyon 02:03, 8 August 2006 (UTC)

It explains the observations fine, but doesn't solve the main problem, which is that it doesn't point out to readers confused by the claim that photons have zero mass, that this doesn't mean they have zero relativistic mass. That needs to be pointed out. Once it is, I have no great attachment to the whole "photons in a box" section, and I don't really care if it goes away. --Trovatore 02:14, 8 August 2006 (UTC)

I didn't find the term "relativistic mass" in the books I looked in, which is why I looked for "energy mass" and settled for what I found. Can you recommend a book that explains it more like you prefer? I'm not sure what it is that you think is the main problem that is not explained well now. Dicklyon 03:49, 8 August 2006 (UTC)

You won't find "relativistic mass" and you won't find "rest mass" anymore.They are outdated, incorrect, misleading. This was one of my points.

My second point is that it is plain wrong to talk about "photon mass". QED mandates that the photon is a massless particle and there are many experiments that confirm it. So, talking (or even worse, writing) about "photon mass" encourages all the crackpots that think that the "mass" of the photon is hf (i.e. about 3eV) when in reality the current experiments set an upper limit at 6*10^-17. I edited the post in such a way that it avoids any of the pitfalls and any of the tell-tale signs of old terminology. Hopefully this will be an acceptable compromise to all parties. Ati3414 04:28, 8 August 2006 (UTC)

I made another edit, to try to make it more clear, and put the term "relativist mass" in quotes for what it is called. And I took out the complicated bits about the inertial and weight, as they really are beside the point; I just summarized the point that energy mass, inertial mass, and gravitational mass are the same (so far as standard physics knows, I should have added). Dicklyon 03:58, 8 August 2006 (UTC)

And then I added the example of the system of two photons that come from electron–positron annihilation. If anyone disagrees, please speak up; I might have it not quite right. Dicklyon 04:18, 8 August 2006 (UTC)

He has perverted the section again to a very different view, while ignoring my pleas for discussing it here. I reverted it, since it seem unreasonable for a person to cite something that he is disparaging. It appears that his problem is with the word "mass" for all the usual manifestations that everyone else calls "mass". Or is there some other explanation that anyone knows, since he's not saying? Dicklyon 04:29, 8 August 2006 (UTC)

"He" has not perverted anything. "He" has been trying to get you to understand that physics has changed since you learned it and since someone made an ill fated post in 1997 on the John Baez FAQ. "He" has explained this to you numerous times and has shown the errors in the derivations of such effects. Read above my answer to SBharris derivation. "He" has tried to compromise and phrase things in such a way such that the entry doesn't fly in the face of current understanding and such that it doesn't get laughed by current grad students. Ati3414 04:53, 8 August 2006 (UTC)

Oops, I see I did miss a talk entry above. Sorry. Still, there's nothing dangerous about the old terminology that associated energy and mass via E=mc^2. Dicklyon 04:31, 8 August 2006 (UTC)

E=mc^2 is not even correct. Why don't you let it be? You can't have decisions over posts that you only have partial and obsolete understanding. Leave things alone, let others decide.Ati3414 04:53, 8 August 2006 (UTC)

E=mc^2, using m for invariant mass or rest mass, still continues to be perfectly correct in the CM or COM frame, which is the frame we usually work in, and is certainly the frame we always are in when we're using scales to measure masses. So we note that and pass on. Heat an object, and E goes up and m goes up and E=mc^2 remains true. Very handy. E(total) is directly weighable in the COM frame. That's sort of the point of this whole discussion. And why it needs to go in the Wiki. SBHarris 22:23, 8 August 2006 (UTC)

So, you are saying that E=mc^2 is correct for photons. this is how you got your naive "solution" m=E/c^2, a classical rookie mistake. Any self-respecting textbook will tell you that E=pc for photons (http://www.lassp.cornell.edu/~cew2/P209/part11.pdf). Good diversion, though, got an insight how you "manufactured" the result for your "proof". Ati3414 22:41, 8 August 2006 (UTC)

E=mc^2 is correct for systems of photons in the COM frame, yes. It's correct for single photons and one other particle, so long as the COM frame is used. It's not correct for single free photons, but I never said it was. One rarely meets single free photons anyway.SBHarris 00:52, 9 August 2006 (UTC)

There is nothing like persisting in a rookie mistake. E=mc^2 does not apply to the photon. E=pc is what applies. Crack open a more modern book, you are showing your ageAti3414 02:14, 9 August 2006 (UTC)

OK, I'll back off for now and see who else cares about this. Can you recommend a good book to explain why (and when) E=mc^2 got to be wrong, or incomplete, as an expression of the equivalent of mass and energy? Maybe I'm still stuck in teh 20th century. I need a source besides you to explain how the old view is no longer acceptable. Dicklyon 05:02, 8 August 2006 (UTC)

Try this , it is free and comes from Harvard, it is pretty much the norm for modern teaching of relativity . Thank you for being so nice to me.

http://www.courses.fas.harvard.edu/~phys16/Textbook/ch11.pdf Have a look at chapter 11.8..... http://www.courses.fas.harvard.edu/~phys16/Textbook/ch12.pdf

Ati3414 05:24, 8 August 2006 (UTC)

Thanks for the references. But I don't buy your point. That book clearly states that "Some treatments of relativity refer to the mass of a motionless particle as the “rest-mass” m0, and the mass of moving particle as the “relativistic mass” mrel = γ m0." So it admits what is "conventional", which is what the rest of us have been saying. Then it goes on to argue for a different way that it claims is better, and supports that argument by a ridiculous strawman about Newtonian stuff, without even recognizing that Newton's law is usually stated as the more sensible F=dp/dt. And it never really addresses the question of whether the total rest mass of a system that has internal energy needs to account for that energy. The guy who wrote it is "Lecturer on Physics & Assistant Head Tutor, Assistant Director of Undergraduate Studies," with a Ph.D. just 10 years ago; not exactly the kind of authority who is likely to change the conventions in the field of physics by sheer force of personality. I have no problem with the concept of redefining mass to only have rest mass, and phase out relativistic total mass, and to redefine the meaning of E=mc^2 accordingly. More power to him; and to you; but not here, not now, not by using Misplaced Pages as the podium. Write a paragraph that describes this as a new alternative view, reference the book, but leave the other part in the conventional form. Dicklyon 05:55, 8 August 2006 (UTC)

This is just Harvard, do you want to see the MIT or Stanford website on the subject? I understand how difficult it is to admit that your views are outdated. The newer generation has a new view, wiki needs to reflect the most up to date view. (A small correction F=dp/dt is not Newtonian, it is SR. F=ma is the Newtonian view). Ati3414 06:05, 8 August 2006 (UTC)

As any student of Newton will tell you, F = dp/dt is how Newton defined it also.SBHarris 20:53, 8 August 2006 (UTC)

328 new books, many with "outdated" view

I'm OK admitting my views are outdated if it can be shown to me. But when I search recent books I find many that don't seem to have gone this way. Maybe your view is just too "with it" to be encyclpedic? Why not write it up as a paragraph about the new view, instead of denying the conventional view? Books 2003-2006 with "relativistic mass" mentioned

Pleading ignorance is not a valid argument. Books change much later,they take a lot longer to edit, reissue, etc. I think that your argument is specious and you know it. Look at how things are being taught today, books will follow suit:

http://www.lassp.cornell.edu/~cew2/P209/part11.pdf

and here:

http://www.courses.fas.harvard.edu/~phys16/Textbook/ch11.pdf Have a look at chapter 11.8....Ati3414 06:20, 8 August 2006 (UTC)

I don't dispute that some people are teaching it your way. That's no reason to deny the conventional way, which is still being put into current books, when writing a pedia article on it. So give it up; your radical approach is inappropriate here. It's fine to talk about the new way, but not to exclude the conventional way. If you are really "though with this BS" as you said on my talk page, then so long. Dicklyon 06:38, 8 August 2006 (UTC)

Because an encyclopedia is supposed to convey the most up to date view of science. It is supposed to correct older, outdated views. If one wilfully conveys the outdated view, in the context of such an encyclopedia being easy to update instantaneously, it turns the encyclopedia into a joke. If a joke is what you want, then continue to wallow in the blissful ignorance. After all, all that matters is that you have your way, scientific truth is secondary. Ati3414 06:47, 8 August 2006 (UTC)

There isn't any difference on "scientific truth"; this is language and bookkeeping. It's like the way people are taught to subtract by borrowing. My generation was told that, when you have to borrow, you take one away from the digit on top; an earlier generation was taught to add one to the digit on the bottom. (Tom Lehrer has a song about this, among other things.) That's what relativistic mass/invariant mass is like: Do you multiply by γ before or after you do something else?

Teachers of elementary courses (including first-year university courses) often simplify the story to save time. Algebra teachers teach their classes that "you can't divide by zero"; there's no need for them to bother with the Riemann sphere, which would just confuse their students anyway. That may be what's going on with these texts. Or maybe the professors really believe that one bookkeeping method is right and another is wrong; as silly as that would seem to me, we can certainly report that view. But we should not expunge the historically well-attested account in terms of relativistic mass, simply because it's inconvenient for a few instructors' lesson plans. --Trovatore 07:01, 8 August 2006 (UTC)

Sorry, it goes much deeper than that. From a professional physicist poit of view, the "photon in the box" is a joke, trying to convey what's not there, a mass increase of the system. As scientists , or wanna-be scientists, the people that put it together should know. But heck, you are all happy with the outcome, your collective POV prevailed, this is all that counts. I no longer give a s####Ati3414 07:09, 8 August 2006 (UTC)

I believe you have it wrong. What about the case of the massless gluons in nuclei? When you fission a U235 nucleus, and get a set of other particles where a lot of gluon energy has gone into kinetic energy, if you catch those particles and weigh them they add up to less than the U235 you started with. Isn't that because the U235 mass includes the energy of the massless gluons that are part of it? Isn't that just like the photons in the box? One of us is still confused. Dicklyon 15:11, 8 August 2006 (UTC)

If Ati seriously believes that the box plus photon doesn't have a higher invariant mass than the box without the photon, then he's simply in error. I had thought he had given up on that point and was now arguing about what should be called "mass"; looks like I was wrong about that. --Trovatore 15:30, 8 August 2006 (UTC)

And if you're waiting for Ati to explain how massless gluons contribute to the mass of the gluon-container that is a nucleon (or any hadron) then you're going to be waiting awhile. This contradicts his worldview, therefore he's going to ignore it. Modern physics or not. That is how you tell reasonable from unreasonable people in a discussion. When the man with the fixed delusion is brought up against a perfect counterexample, he just shuts down like a locked-up computer running Windows. But his mind is not changed. The reasonable person, on the other hand says: "Wups, looks like I was wrong." SBHarris 15:39, 8 August 2006 (UTC)

Hmm, you may be crediting me with more reasonableness than I really intended. I was saying I was wrong about what Ati3414 was arguing, not about what should be called mass. My position on that remains that, yes, as I knew, physicists no longer like to describe things in terms of relativistic mass, and they have good reasons and that's fine, but that doesn't make the descriptions in terms of relativistic mass wrong, and we should mention those descriptions, neutrally and briefly, for the benefit of readers who would otherwise be confused. --Trovatore 21:17, 8 August 2006 (UTC)

On the idea of mentioning relativistic mass as a previous notation, sure. That's why there's a section about it in mass in special relativity. Relativistic mass as a notion, should be mentioned, then the advantages and disadvantages of the notation noted, and finally the fact that the physics field is generally moving away from it, tending to use energy instead. Although not without carnage. :) You know, one of the worst problems with "relativistic mass" is that when people decided to discard it, some people seem to have been left with the impression that heat and radiation, which had relativistic mass but no rest mass, somehow now are left massless in all ordinary senses of the word mass. But that's not true. All energy continues to have rest mass in the system COM frame, and that includes all energies of the things which were composed wholey or in part of "relativistic mass". Heat, light, and kinetic energy in systems continues to exhibit mass when the system as a whole is at rest, and it remains just plain old ordinary mass of the kind you weigh. According to E = mc^2, end of story. Doing away with "relativistic mass" hasn't changed this remarkable fact. So thanks, Dr. Einstein. It's been more than a century now since you clued us, and some people, by god, still don't understand it. And apparently this Ati3414 is one of them. SBHarris 01:10, 9 August 2006 (UTC)

Please carry on with the personal attacks, they form a very good scientific argument. In between personal attacks maybe you find a few minutes to have a critical look at your mathematical "proof". I added some notes yesterday, you may have missed them.Ati3414 18:08, 8 August 2006 (UTC)

OK, I'll be the Mac, like in those fun commercials, since you brought it up. But seriously, I added one more example, of the mass of the "virtual photons" holding an atom together. Again, I seek feedback if I got this wrong, but more feedback from Ati will only serve to make me want to further belabor the point that he disagrees with. Is there any reasonable physicist out there who can tell me if I've captured the essense of truth, at least as accepted in modern physics, with these little examples? Dicklyon 17:12, 8 August 2006 (UTC)

Well, alas, Mac, I had to revert you, before you remembered (in embarrassment) that the atom has less mass (not more) as a bound system. Just as the nucleus does. But you're not wrong about the virtual photons (or for the nucleus, virtual pions too). Static electromagnetic field is DESTROYED when opposite charges are brought together, as in an atom. So the missing mass is missing virtual photons. In a bound nucleus you gain virtual photons, but you lose even more virtual pions, so the net comes out lost mass. Which of course must appear as mass someplace else (the mass of kinetic energy, radiation, etc) when the bound object is bound. Missing mass is only "missing" if the system wasn't closed and it was allowed to leak out (which is generally the case in nuclear physics, of course). That's one more oddity of how physics is taught, which is historical. Mr. PC. SBHarris 18:08, 8 August 2006 (UTC)

Yes, even Macs have their flaky moments. I stand corrected and humbled. Dicklyon 19:43, 8 August 2006 (UTC)

Ati3414's reversions

By my count, the following edits of Ati3414 are reversions within the meaning of WP:3RR :

• (restoration of slightly reworded note about section being "embarrassing" or "a disaster")
• (removal of the section altogether)
• (removal of the section altogether)
• (removal of the section altogether)
• (removal of the section altogether)
• (partial revert; removal of reference to relativistic mass)
• (partial revert; removal of reference to relativistic mass)

These are all within 24 hours (from 05:11, 07 August 2006 UTC to 04:40, 08 August 2006). The following edit misses being within the 24 hours by less than half an hour:

• (full revert to a previous version of Ati3414)

I think someone (preferably someone not a party to the debate, with admin powers) should warn him and block him on the next revert, or even without warning given the magnitude of the violation. If this behavior continues an RFC should be opened. --Trovatore 06:09, 8 August 2006 (UTC)

• Disclosure: On reviewing the record, it looks like I did it too — four reverts in the same time period. So, person I've invited above, you may wish to block me as well; that would be fair. --Trovatore 06:52, 8 August 2006 (UTC)

Certainly not, you did everything within your powers to maintain the status quo of "design by committee" so pervasive in wiki. No point in penalizing you for maintaining backward views. After all, this is what wiki is all about. Ati3414 07:00, 8 August 2006 (UTC)

Dear Il Trovatore

I understand your frustration, you don't like to be corrected. The truth is that physics has moved on since you last looked at it and the concept of mass has suffered some serious changes. So, instead of admitting that you have outdated knowledge why not ask for the banning or even better, execution of the person that does not conform to your views. See here:

http://www.lassp.cornell.edu/~cew2/P209/part11.pdf

and here:

http://www.courses.fas.harvard.edu/~phys16/Textbook/ch11.pdf Have a look at chapter 11.8.....

Heck, have it your way, it isn't worth it. Let's keep wiki a joke, the most important thing is that "editors" are never proven wrong, scientific truth is secondary. Ati3414 06:16, 8 August 2006 (UTC)

For the record there's nothing in this chapter which contradicts our basic point that the mass of a system in the COM or CM frame, is system total energy/c^2. And that includes a system of photons, which add energy and therefore mass to the system. Or a box that contains photons. Photon E adds to system mass in the ordinary E/c^2 way. Any kind of energy added to the system does, whether it's rest mass of system particles, the kinetic energy of system particles, photons injected, whatever. Anything in a box contributes to its mass. Also, any part of a system contributes to its mass, which in the COM frame is the system mass or system rest mass or system invariant mass, or whatever you want to call it. But it's what you weigh, and it's the inertia the system has in the COM frame (equivalence principle). Things get heavier when heated, and they get heavier when they absorb photons. And they get heavier even if the photons go into them and just bounce around. Which part of this don't you understand? SBHarris 16:34, 8 August 2006 (UTC)

I took a second pass at your "proof" a few days ago. have a look at it, a few sections above. as to the personal attacks that I shut down and all this stuff, you should have had a look at the answers before that. Anyways, I decided to add a non-crackpot section on the experimental mass of the photon, you must have "missed" that too.Ati3414 18:02, 8 August 2006 (UTC)

1. The experimental mass of the photon is talking about single photons, not photons which are part of systems, and so is not relevant. Kinetic energy also has mass in systems, but not in single particles, because it can be made to go to zero (or be any value) by choice of frame. This is exactly the situation with photons, except they have no residual rest mass. However, their energy contributes mass to systems.

2. As for my derivation, you still have not come to grips with the fact that you can replace dp/dt with p/t if you like (and as I did), since this momentum transfer is constant over time (just as is mean time-averaged pressure exerted by an ordinary molecular gas). So take out the calculus, use p/t, where you pick a p and a t that goes with it. And get over it. SBHarris 18:37, 8 August 2006 (UTC)

1. But your "photon in a box" talks exacly about adding E/c^2 by virtue of adding one photon to the box. You can't have the cake and eat it too.In other words, you cannot claim that adding one photon adds E/c^2 to the "mass" of the box. Looks like you provided the best argument against your own post. Finally.

Whaterver you do, never admit to error, doc. So, after you botched the calculations by mixing Newtonian concepts (no, dt/dt is not Newtonian, is SR, and you are "calculating" and oddball dp/t) with SR concepts in order to get your desired answer E/c^2 which is many orders of magnitude off you repeatedly "forget" that you contradicted your toy experiment by the posting comparing it with the recognised experiments in the field. You are 17-19 orders of magnitude off, doc!Ati3414 21:28, 8 August 2006 (UTC)

The photon does not fall as m*g in the gravitational field, so you cannot infer the "mass" of the photon from its downwards force, and your derivation needs to be direction invariance. Try the experiment with the photon going horizontally and let's see how you "manufacture" the E/c^2 "solution" again. Don't ask me, I am just a spectator for your magic acts 21:26,Ati3414 21:28, 8 August 2006 (UTC)

The photon does fall as m*g in a gravitational field. For a uniform gravity field (note the caveat) a photon falls as you'd expect for any object traveling at its speed (deflections are larger when traveling into a gravity field and then back out, but those don't concern us here-- we're weighing starlight in a box, not deflecting starlight into and out of a gravity well). And yes, while bouncing around inside a box, a photon will add an effective mass m to whatever it is that contains it, whether traveling "up and down," or "transversely" in the field. I asked you to do the calculation so you could demonstrate your superior knowledge of physics, not stand on the side and snipe at others. Come on, Mr. Critic.SBHarris 01:19, 9 August 2006 (UTC)

2.To the dp/dt replacement with p/t, I don't think so. You have several other errors as well that you are conveniently not addressing. Have a second look. Ati3414 18:45, 8 August 2006 (UTC)

Whaterver you do, never admit to error, doc. So, after you botched the calculations by mixing Newtonian concepts (no, dt/dt is not Newtonian, is SR, and you are "calculating" and oddball dp/t) with SR concepts in order to get your desired answer E/c^2 which is many orders of magnitude off you repeatedly "forget" that you contradicted your toy experiment by the posting comparing it with the recognised experiments in the field. You are 17-19 orders of magnitude off, doc!Ati3414 21:28, 8 August 2006 (UTC)

The photon does not fall as m*g in the gravitational field, so you cannot infer the "mass" of the photon from its downwards force, and your derivation needs to be direction invariance. Try the experiment with the photon going horizontally and let's see how you "manufacture" the E/c^2 "solution" again. Don't ask me, I am just a spectator for your magic acts Ati3414 21:28, 8 August 2006 (UTC)

How many different ways do I have to say it? See those dt's and dp's? Integrate over time and get rid of them. IOW, integrate over MANY bounces. Just as you would for a box containing ONE gas molecule. Which would exert an average pressure, and by virtue of this, an average mass/weight over time. Of course, over shorter times, a box with one molecule or one photon jumps like a jumping bean. So does an elevator with a man jumping up and down in it, for that matter. But in the end, over time, exactly the right extra weight shows up on the cable. Or the scale. SBHarris 20:01, 8 August 2006 (UTC)

Selective answering, congratulations.What you are doing in your hack is calculating dp/t. No matter: there are three other errors in your "proof", like your continous mixing of relativistic mechanics with newtonian mechanics. You are also conveniently missing the fact (point 1) that you have refuted your own calculations. Just take the E/c^2 of the post, would you? It looks ridiculous, you are 17-19 orders of magnitude off.Ati3414 20:06, 8 August 2006 (UTC)

No, for force I calculated p/t (or Δp/Δt if you like). There are no differentials, because with a differential left in you cannot get an answer. The mean force is a number. It has a value. Had you bothered to get a value from your dP/dt you'd have had to do the same average that I did. So go for it. SBHarris 20:59, 8 August 2006 (UTC)

Uncertainly relations

I'm a bit unclear on the uncertainty relations section. It says: "...quantum mechanics forbids the simultaneous measurement of the position and momentum of a particle in the same direction. Similarly, it forbids simultaneous measurement of the number of photons in an electromagnetic wave and the phase of that wave." I'm familiar with the first part, but how are phase and number of photons conjugate variables? It also constrains the uncertainty of the product of energy and time. Is it fair to say that since the photon's momentum and energy are definite and known values (in some situations, at least), that the position and time are completely unknown? Is there a sensible way to temper this somewhat absurd-sounding statement, or does it have some deep meaning? Dicklyon 01:26, 9 August 2006 (UTC)

See Coherent state#Coherent states in quantum optics. "Specifically, coherent light is thought of as light that is emitted by many such sources that are in phase." "Contrary to the coherent state, which is the most wave-like quantum state, the Fock state (e.g. a single photon) is the most particle-like state." "Mathematically, the coherent state ${\displaystyle |\alpha \rangle }$ is defined to be the eigenstate of the annihilation operator ${\displaystyle a\!}$." In other words, measuring the phase assumes a coherent state which requires changing the number of photons so that that number cannot be measured simultaneously. JRSpriggs 05:36, 10 August 2006 (UTC)

OK, I see the point now. But the way it is written, I don't think "the number of photons in an electromagnetic wave" makes much sense as a way to describe those incoherent states. At the least, come clarification is needed. Dicklyon 06:49, 10 August 2006 (UTC)

Making quantum mechanics clear is beyond my abilities. However, let me add that instead of "electromagnetic wave", one could refer to "a vibrational mode of the electromagnetic field". Instead of "photon", one could say "excitation of the mode" or "energy level" (actually ${\displaystyle {\frac {E}{h\nu }}-{\frac {1}{2}}}$). Instead of "phase", one could speak of "time since the beginning of a cycle divided by the period of the cycle" or "time since the beginning of a cycle times the frequency". So you see that this is equivalent to the uncertainty relationship between energy and time in the context of quantum field theory. JRSpriggs 04:07, 11 August 2006 (UTC)

Single photon trapped transversely in a rocket; inertia and weight change

A rocket of mass M flies free in space, firing its engine to give it exactly an acceleration of 1 g (as seen by the passengers). Now a photon of light with energy E (as seen by the passengers of the rocket) flies in through a porthole and begins to bounce around inside the rocket transversely, in a direction orthogonal to the acceleration of the rocket. Does the engine suddenly now need to fire a bit harder, to maintain g? Does the photon act as though it has MASS, requiring extra rocket power to now accelerate a rocket with mass M + m, where m is the effective mass of the photon? And how much would that mass be? Small (to be sure), but since photons have no mass, would it be zero? Einstein told us in 1905 that if the photon where absorbed, the rocket would gain mass equal to E/c^2. But what if it is not absorbed, but merely bounces around inside, before exiting through another porthole? Does it take extra push to accelerate it? Does the photon have inertia? And if so, how much?

Answer: It does. And it's the same as if the photon was bouncing in any direction. In this case, we call the direction toward the engine Y, and track the photon starting from the centerline axis, to a bounce at the far wall of the rocket, then return to the centerline. This is half a bounce cycle. The mean force which these cycles exert on the rocket in the Y direction (the photon weight in this system) Fy, is therefore F= P/t, where P is Py, or the momentum transferred in the Y direction by ONE photon bounce off the far wall, and t is the time for the photon to go from centerline, bounce, and return to the centerline. This answer does not change if the photon completes a hundred bounces, because P and t increase by the same amount in fixed ratio. The component Py won't be large, because the photon starts out headed exactly at the far wall, and not at all in the direction of the engine. But between the centerline and the wall, the rocket moves, so that by the time the photon hits, it's no longer moving exactly straight into the wall, but rather at an aberration angle sin(θ)= v/c (exercise left for the student, particularly why this is sin(θ) and not tan (θ), but remember the wall is moving and there is some Lorentz contraction to consider). Here v is component of velocity in the Y direction the photon picks up while going from centerline to wall, as a result of the acceleration. This process continues, with every bounce off a wall by the photon transfering force downward in the direction opposite rocket acceleration.

Now if P is the photon's total momentum to begin with, then Py = P sin(θ). We already know that sin(θ) = v/c. And v will be g * (t/2) where t is the time for the photon to go from centerline axis to wall, and back to centerline (it's t/2 because the rocket only gets half this time to accelerate while the photon goes from axis to wall). So:

Py = P(v/c) = P*g*(t/2)/c = (P*g*t) / (2c)

The average force downward toward Y due to the photon bouncing through a half cycle is P/t and in this case P is 2Py because a total reflection causes twice the momentum transfer. Putting in our expression for Py:

Thus F = 2Py/t = Pg/c

Remembering that in relativity photon P = E/c:

F = (E/c^2) * g

This is the force the photon exerts on the rocket, which is the same as the rocket exerts on the photon. The extra force the rocket motor needs to use due to the photon the rocket picked up, is E/c^2 *g, which is the same as if something with mass E/c^2 had come in through the porthole. So that is our answer. This is the effective mass of the trapped photon in this situation, until it gets out through another porthole.

What if the rocket is sitting on the pad on Earth, being weighed by a scale at 1 g. The rocket's mass is M, and its weight is M*g. If a single photon now comes in through a window and is trapped transversly in the same manner, the equivalence principle tells us that the gravitational field of the Earth will act exactly like the 1 g acceleration of the motor, and cause the photon to exert the same extra force on the rocket wall by bouncing back and forth. Thus, the rocket now weighs more due to the photon of energy E, by an amount E/c^2. The single photon not only has inertia in a moving rocket, but weight in a landed rocket. And inertia and weight is mass. If the rocket is shaped like a box, the answer does not change :) . SBHarris 02:21, 9 August 2006 (UTC)

Very clearly stated. That is in agreement with special relativity and other principals of physics as I learned it, and so far as I can tell it hasn't changed since then. Dicklyon 03:35, 9 August 2006 (UTC)

Very clever, you finished what I started and got erased. Now, the issues:

same as in the previous solution, a gratuitous F=P/t which sets things in the desired direction but not is not justifiable

This does not mean that the rocket picked up an extra mass E/c^2 , it simply means that there is a debatable (see ) resistence type force F = (E/c^2) * g exerted in the direction of rocket travel and with a sense opposite to rocket travel. Nothing in QED allows you to attribute mass to a photon, even worse, to attribute it weight.

Even if objections , were removed by some miracle , you are now stuck with your own refutation of the derivation. The well respected experiments put a constraint on the mass of ONE photon (as you well observed somewhere in a previous post) of 6*10^-17 eV . But your formula produces an equivalent em mass equal to E , which is about 3eV. So, your personal research is 17 orders of magnitude off the accepted experimental data (actually 19 orders of magnitude considering the latest experimental results). Interestingly enough, you are off by a factor of about (v/c)^2.

I liked your transposition of the problem using the EP. Ati3414 03:49, 9 August 2006 (UTC)

Answer

1) If you don't like my method, then YOU finish what you started. Do not criticize somebody else's physics unless you're prepared to demonstrate better technique. If not, you're just blowing smoke.

Well, you get it wrong repeatedly. And you keep mixing in Newtonian with SR mechanics <shrug> Ati3414 17:08, 9 August 2006 (UTC)

2) Inertia is mass. If you believe in the equivalence principle, the rocket also picks up weight. Are you really arguing that it picks up weight and inertia equal to E/c^2, but not mass? How else would you measure its mass?

Sure, I believe in EEP. This doesn't give you the right to infer (again , repeatedly) that the term in front of "g" is the photon "mass". The Proca equations on which the current experiments are based express the violation in "photon mass" , no mention of "rest mass". Have you read the papers? Ati3414 17:08, 9 August 2006 (UTC)

3) As has been carefully explained to you many times now, you are arguing apples and oranges when you argue that photons have no rest mass. We're not measuring their rest mass. A single gas molecule of rest mass m trapped in a rocket would add the inertia and weight m/c^2, but if it was moving with kinetic energy T, it would add (m+T)/c^2. A photon has no rest mass, so it only adds its kinetic energy T. Which is E/c^2. SBHarris 15:23, 9 August 2006 (UTC)

"...so it only adds its kinetic energy T. Which is E/c^2" I learn new physics from you every post. Now, seriously speaking there is no such notion of rest/relativistic mass for the photon. It was explained to you repeatedly. As such, modern teaching of relativity did away with the notion of rest/relativistic mass altogether. The paper I pointed you to talks about constraining the "photon mass", no reference to the "rest" mass. Ati3414 17:22, 9 August 2006 (UTC)

Photons do not have rest mass, but they contribute to inertia and gravitate

Hi all,

Someone asked at the Physics Wikiproject to clarify the relationship of photons and mass. I'll do my best! :) I hope everyone will be patient with my explanations.

First, photons do contribute to inertia. That's stated explicitly by Einstein in his 1909 article, The Development of Our Views on the Composition and Essence of Radiation, which I translated recently. You can find the original German text on the web, but I'm not allowed to post it at the German Wikisource because of copyright issues.

Second, photons do gravitate. Since they have energy, they contribute to the stress-energy tensor and, thus, to the curvature of space time. Hence, they act gravitationally on other objects. Conversely, they respond to a gravitational field of other objects; light's path is curved in a gravitational field, as shown (for example) by gravitational lensing. That's not surprising, since inertial mass and gravitational mass are equivalent in this sense; if we agree that light has inertial mass, it must also respond to a gravitational field.

The "weighing" of a box of photons was used by Einstein in a famous thought-experiment to refute the Heisenberg uncertainty principle, ${\displaystyle \Delta E\Delta t\geq \hbar /2}$. His idea was that a hole in a box of photons could be opened for an arbitrarily short time, allowing one photon to escape. The energy of that photon could be measured to arbitrary accuracy by weighing the box before and after the hole opened. Thus, it seemed that the uncertainty in time and energy could be made as small as possible. Bohr's refutation (delivered the next morning) was a watershed moment for quantum mechanics.

All that being said, light does not have rest mass (to the best of our present knowledge). That's not a problem, though; the rest mass of an object is its mass in a coordinate frame where the object is at rest. There is no such frame for light moving through the vacuum; it moves at c in all reference frames.

Hope this helps, Willow 12:06, 9 August 2006 (UTC)

COMMENT

It would help if you'd clarify. A photon's inertia is undefined (because it's energy is undefined) and cannot be measured quantitatively, until it's part of a system. Same for its weight. These two constitute a mass of E/c^2, which a photon adds to a system in which it is trapped (and in reference to which its energy is E, which now CAN be defined because we have defined a reference frame). And similarly a photon subtracts E/c^2 invariant mass from a system (seen in the COM frame) which emits it. SBHarris 15:28, 9 August 2006 (UTC)

Another comment written concurrently

As you can see by Ati's reply below, it doesn't help much. What you've stated is pretty much what we all know and agree on. What's left is the question of whether adding inertia and gravitation to an object is the same as adding to its mass. Most of us think it is. We all agree that light does not have rest mass. Can you address his response below in terms of how it relates to standard modern physics as you know it? Dicklyon 15:32, 9 August 2006 (UTC)

His response has nothing to do with physics as we know it, so there's not much place to compromise on the issue. If you increase the energy, inertia, and weight of a resting compound object, whether by heating it or adding one or more photons to it (which of course can be the same process-- try an infrared broiler), you have increased its proper mass, invariant mass, rest mass, and just plain old everyday mass as we and modern physicists use the word. The end. It's not complicated. Only cranks believe otherwise. SBHarris 15:50, 9 August 2006 (UTC)

Photons have energy, momentum and spin but no mass

The views that the photon contributes E/c^2 to the inertia of a system are not the view of modern physics. While it is true that Einstein demonstrated in september 1905 that a body radiating em would loose a mass ${\displaystyle \Delta m}$ where that math is related to the energy variation ${\displaystyle \Delta E=\Delta m*c^{2}}$ , the converse is paradoxically not true, the naive view that adding a photon to a system would increase its mass by E/c^2 is refuted by modern experiments. One can only tell that the total energy of the system has increased by E and its momentum has increased by p (proper mass is not additive in relativity, the so called "relativistic" mass is, but that is an obsolete term ). Turns out that the photon has zero mass as proven by modern experiments. Modern experiments look for a violation of the Maxwell's laws predicted by A. Proca in his 1937 seminal paper. The violation is of the order of ${\displaystyle mu_{g}=m_{g}*c/\hbar }$ where ${\displaystyle m_{g}}$ is the "photon mass". Any ${\displaystyle mu_{g}>0}$ would result into an observable phenomenon , confirming indirectly that ${\displaystyle m_{g}}$ is greater than zero. This would bring about a violation of gauge invariance. As it can be seen from the many other experiments -, this is not the case.Ati3414 14:05, 9 August 2006 (UTC)

References:

http://www.fourmilab.ch/etexts/einstein/E_mc2/e_mc2.pdf

http://silver.neep.wisc.edu/~lakes/mu.pdf

Eric Adelberger, Gia Dvali, Andrei Gruzinov, "Photon Mass Bound Destroyed by Vortices", arXig.org

Goldhaber, Alfred S., and Nieto, Michael Martin, "Terrestrial and Extraterrestrial Limits on The Photon Mass", Rev. Mod. Phys. vol.43 #3 pp.277–296, 1971

E. Fischbach et al., Physical Review Letters, 73,514-517 25 July 1994.

Official particle table http://pdg.lbl.gov/2005/tables/gxxx.pdf

L. Davis, A. S. Goldhaber, and M. M. Nieto, Phys. Rev. Lett. 35, 1402 (1975)

Roderic Lakes, "Experimental Limits on the Photon Mass and Cosmic Magnetic Vector Potential", Phys. Rev. Lett. 80, 1826 (1998)

J. Luo et al., Phys. Rev. D 59, 042001 (1999)

B. E. Schaefer, Phys. Rev. Lett. 82, 4964 (1999)

J.Luo et al., Physical Review Letters, (28 February 2003)

COMMENT

You actually write:

The views that the photon contributes E/c^2 to the inertia of a system are not the view of modern physics. While it is true that Einstein demonstrated in september 1905 that a body radiating em would loose a mass Δm where that math is related to the energy variation ΔE = Δm * c2 , the converse is paradoxically not true...

I wonder if you know how completely crazy that idea is? You're suggesting conservation of energy and mass in one direction for a process, but not the other direction. The same process. Give it up. You've lost this one.

And by the way, if your (true) arguments that photons have no rest mass were relevant here (which they aren't), a system wouldn't lose mass when emiting a photon, either. So you shot yourself in the foot there by even admiting that. But I understand why you don't want to call Einstein wrong, since they you'd be claiming he said something incorrect a century ago, and nobody caught him for a century until YOU came along. LOL. SBHarris 15:34, 9 August 2006 (UTC)

What else is new, you obviously can't read (or you twist whatever you are reading). Keep on reading : "One can only tell that the total energy of the system has increased by E and its momentum has increased by p (proper mass is NOT additive in relativity, the so called "relativistic" mass IS, but that is an OBSOLETE term)." This was the argument from the beginning but it doen't register with you. Try reading the modern papers and the modern way of teaching the subject, I put them on the website for you. Maybe you learn and you stop "manufacturing" mass from gedank experiments.Ati3414 15:43, 9 August 2006 (UTC)

Actually, proper mass (aka invariant mass) is additive in special circumstances, so long as you measure mass in the COM inertial frame for each object or submass, and measure the total mass in a common COM frame for the whole. Put two boxes on a scale in relativity and you get the summed mass of two boxes. Put one box on a scale and keep it closed, and its mass will stay the same no matter what goes on inside it. You could generate heat, have a nuclear reaction, blow up a bomb-- even an atomic bomb if the box was superstrong. So long as you let nothing out of the box, all the heat, light, and mass of whatever was inside, would continue to have the same mass and weight, before, during and after the reaction. SBHarris 16:08, 9 August 2006 (UTC)

A simple exercise

Let's try something really simple. Imagine your empty box, with no momentum. It has the mass:

${\displaystyle m={\frac {E}{c^{2}}}}$

Now, we add a single photon (not a pair), exactly as in the "photon in a box" wiki entry having energy e and momentum p=0 (how could that be, the single photon has nonzero momentum) to the box as in your example. The mass of the system becomes:

${\displaystyle m'={\frac {E+e}{c^{2}}}=m+{\frac {e}{c^{2}}}}$

It is correct to write the formula

${\displaystyle m'=m+{\frac {e}{c^{2}}}}$

but one SHOULD NOT call the term ${\displaystyle {\frac {e}{c^{2}}}}$ "mass", "mass increase", "mass increase due to photon". All you can say is that the mass of the system has increased by ${\displaystyle {\frac {e}{c^{2}}}}$ . You cannot attribute mass to the term ${\displaystyle {\frac {e}{c^{2}}}}$ because if you do, the gauge invariance in QED is violated. This is purely a pedagogical issue but it is very important.

Now let's try the real case, when the photon has a nonzero momentum. You must now modify to accomodate the photon's momentum. You get something quite different:

${\displaystyle m'={\frac {\sqrt {(E+e)^{2}-(pc)^{2}}}{c^{2}}}}$

Not easy to relate m' to m anymore.....What is the "photon's contribution"? Does it make sense to even mention it anymore? What about the most general case:

${\displaystyle m'={\frac {\sqrt {(\Sigma E)^{2}-(\Sigma pc)^{2}}}{c^{2}}}}$

Ati3414 22:33, 9 August 2006 (UTC)

A lovely string of muddled assertions and nonsequiturs. I was particularly impressed by

You cannot attribute mass to the term ${\displaystyle {\frac {e}{c^{2}}}}$ because if you do, the gauge invariance in QED is violated. This is purely a pdagogical issue but it is very important.

which is what happens when you fail to distinguish between rest mass and relativistic mass. Gauge invariance of QED is violated if the photon has a nonzero rest mass, and your first reference to mass is really relativistic mass. So your sentence reduces to:

You cannot attribute relativistic mass to the term ${\displaystyle {\frac {e}{c^{2}}}}$ because if you do, rest mass of the photon becomes non-zero. This is purely a pdagogical issue but it is very important.

and we see what a load of garbage it is. --Michael C. Price 23:03, 9 August 2006 (UTC)

PS it's quite easy to add zero-momentum photons to a box. Pair them up so their momenta cancel. --Michael C. Price 23:06, 9 August 2006 (UTC)

There is no such thing as rest/relativistic mass for the photon. So your counter is a practical zero. Try reading the Proca paper and try retaking the relativity classes, you are stuck in the past. Ati3414 23:09, 9 August 2006 (UTC)

"...one SHOULD NOT call the term ${\displaystyle {\frac {e}{c^{2}}}}$ "mass", "mass increase", "mass increase due to photon". All you can say is that the mass of the system has increased by ${\displaystyle {\frac {e}{c^{2}}}}$." ...Okay: I add a momentum-less photon (yes, I know; a pair of photons with equal but opposite momentums would be the better system, but I want to maintain the original situation) that has an energy ${\displaystyle {e}}$. The mass of the system increases by ${\displaystyle {\frac {e}{c^{2}}}}$. However, I cannot attribute that increase (${\displaystyle m'-m>0}$) to the introduction of the photon into the system. Um, lol? If only mass and energy were equivalent... Regardless of whether you (or a Harvard textbook) approve of the term 'relativistic mass' for all cases or not doesn’t change the fact that we can consider the system as a whole to have gained mass by the addition of a photon. If the photon adds mass to the system, then it must have had a mass prior (in the form of energy). // Sidebar: I do not see how you intend to have this community accept this so-called modern take on physics, if it truly is the modern view, by insulting people. The negative responses that you are getting are not because people think that your information is flawed. (And before responding, I would suggest reviewing the wiki-formatting of your posts (Show Preview) as well as running them through an MSWord spelling/grammar check, as your credibility tends to decrease with the increase of such errors) --HantaVirus 17:37, 10 August 2006 (UTC)

Try the above exercise with one photon. Then try it with 3 photons, one going in one direction, the other two going at 180 degrees. Come back when you have the answers.Ati3414 20:04, 10 August 2006 (UTC)

Actually, when trying to prove your point, it works best if you prove it. As you've yet to provide persuading evidence, I believe that the burden of proof still lies with you. Moreover, if, in performing the suggested calculations using your equations, I was to get a nonsensical answer (which is what I assume you are trying to get at), then that means that those equations do not accurately represent the system as stated. --HantaVirus 20:51, 10 August 2006 (UTC)

Actually, I did (see above calculations). How about you calculated something? Let's see the 3 photon arrangement, couldn't be too complicated. Ati3414 21:04, 10 August 2006 (UTC)

Don't you mean 3 equal energy photons going at 120 degrees, so their momenta add to zero? What's the point of having one go at 180 degrees from TWO others?? SBHarris 22:29, 10 August 2006 (UTC)

No , precisely 180 degrees, such that you cannot get the "COM" frame triviallyAti3414 00:03, 11 August 2006 (UTC)

On second thought, I'm pulling out. I lack both the interest and subject matter expertise to continue in this debate. If it matters to anyone, I, as a layman when it comes to relativistic physics, I understood the article as is. Good luck resolving this, guys. --HantaVirus 22:43, 10 August 2006 (UTC)

At least you admit it. Ati3414 00:03, 11 August 2006 (UTC)

Separation of two distinct issues

Hi all,

There seem to be two issues; perhaps disentangling them and solving them separately will help?

First issue: does the photon have rest mass? The answer at present (to the best of our experimental knowledge) is "no". The references you cite above (e.g., ) discuss the issue in detail. I don't have my books here, but I recall that Feynman also recounts several arguments for limits on the photon rest mass in his first autobiography Surely you're joking, Mr. Feynman!. Moreover, if we believe the standard electroweak unification theory, the photon is exactly massless, since its associated U(1) symmetry is not broken; given that the other vector bosons (W and Z) were discovered with the correct properties, there's good reason to believe in the predictions of that theory for photons as well.

Second issue: does the absorption or emission of a photon change the inertia or gravitational interaction of the absorbing/emitting body? The answer is "yes", if we believe the theory of special relativity; Einstein himself says so in the last sentence of reference that you cite: "The photon conveys inertia between the emitting and absorbing bodies." If inertial mass were lost upon emission but not gained upon absorption, that would violate the principle of time reversibility; moreover, any material system would be continually losing mass with every emission, which is not observed.

Hopefully, this separation of topics clarifies the difficulties. Willow 15:48, 9 August 2006 (UTC)

PS. I wrote this reply before all the other responses; I'll try to read those now and see what needs to be added.

Definitions of "mass"; Helpful thought experiment?

Hi, it's Willow again.

From what I can tell, the remaining difficulties seems to lie in the definition/usage of the word "mass" — they're semantic difficulties.

One definition of mass (for a massive particle) is its rest-mass, i.e., the coefficient ${\displaystyle m_{0}}$ in front of the energy ${\displaystyle E=\gamma m_{0}c^{2}}$ and momentum ${\displaystyle \mathbf {p} =\gamma m_{0}\mathbf {v} }$. By contrast for a massless particle like the photon, the energy-momentum 4-vector is given by ${\displaystyle (E,\mathbf {p} )=\hbar (\omega ,\mathbf {k} )}$. If we all agree that the photon is massless, there's no need to argue over this definition.

Another (more experimental/practical) definition of mass is by inertia or gravitation: how difficult is it to move? How strongly does it attract other bodies? I hope we now agree that the photon conveys inertia between the emitting and absorbing bodies; and that the photon is capable of gravitational interactions, since it contributes to the stress-energy tensor.

Perhaps the following thought experiment will clarify matters. Consider a very large, hollow cavity bounded by perfectly reflecting, perfectly rigid but extremely thin walls; inside the cavity are free electrons, atoms and also charges bound by springs of every resonant frequency. At thermal equilibrium, this cavity will be filled with the normal Planck distribution of blackbody radiation, no? If the cavity volume is large enough, the energy in the blackbody radiation will be much larger than that of the walls.

Now assume that the cavity is prepared in a non-equilibrium state such that it has no photons within at time ${\displaystyle t=0}$, i.e., all the energy is in the matter. As the cavity equilibrates, the energy in the photons will increase over time and that of the matter will decrease over time, until the thermal equilibrium distribution of blackbody radiation (given by Planck's formula) is achieved. Still with me?

The pivotal question is: does the inertia of the box (as seen from outside the box) change as the energy distribution within the cavity gradually equilibrates? For example, does the box gradually become easier to accelerate? The answer is "no", according to modern physics. The photons within the cavity contribute to the inertia and gravitation of the box. Expressing the pivotal question in another way: does the effective ${\displaystyle m_{0}}$ of the box vary with time as the energy in its interior equilibrates? The answer again is "no".

Hoping again that this helps, Willow 16:32, 9 August 2006 (UTC)

Thank you, this is much better. Now, how do we reflect this in the much beleagured entry "Photon in the box" that claims that: "For example, a mirrored box containing a gas of photons, or even a single photon, with total energy E will have greater rest mass (by Δm = E/c²)" ? can you clean it up ? This misbegotten sentence conjures the idea that one could see a measurable photon mass of about 3eV when, in reality all modern experiments put a limit at about 6*10^-17 or smaller. I tried to explain (in vain) that for the photon the notions of "rest" and "relativistic" mass do not exist in modern physics. Because of that the two notions have been eliminated altogether from the way relativity is being taught (see links). The current experiments, based on the Proca violation extensions to the Maxwell's equations talk about "mass", period. The difference is subtle but very important: when someone comes in with a badly written gedankexperiment and tells us that "For example, a mirrored box containing a gas of photons, or even a single photon, with total energy E will have greater rest mass (by Δm = E/c²)" and people see that the whole rest of the wiki page says clearly "the photon is a massless particle" is it to wonder that people reading the "photon in the box" get confused? To counter all the confusion I wrote the whole entry on "Photon mass, experimental limits" but I still hope that someone (maybe Willow?) can clean up the "Photon in the box". I fully agree with what you wrote above, this is the view of modern (I should say contemporary ) physics, now can you convey it by cleaning up all the refences to "relativistic/rest mass" and the ill begotten "For example, a mirrored box containing a gas of photons, or even a single photon, with total energy E will have greater rest mass (by Δm = E/c²)"?

http://www.lassp.cornell.edu/~cew2/P209/part11.pdf and here:

http://www.courses.fas.harvard.edu/~phys16/Textbook/ch11.pdf Have a look at chapter 11.8.

Thank you Ati3414 16:54, 9 August 2006 (UTC)

Hope so, too. I thought we had pretty much agreement on these points, but now Ati is saying that the energy of the photon does NOT add to the inertia of the system (in the view of modern physics). I'm perplexed as to why he would make such a claim, as nothing in his references that I've been able to find address that. Dicklyon 16:39, 9 August 2006 (UTC)

Yes, that's a good example. If the photon gas the cavity fills with, doesn't have mass, then the whole thing will get lighter as the photons leave the walls and lower the mass by E/c^2 (per Einstein) but then don't contribute EQUALLY to the externally measured mass of the object, while they bounce around inside.

And it will be medically interesting to see the response from those who wish to argue otherwise. SBHarris 16:42, 9 August 2006 (UTC)

Please, please refrain from that sort of talk; it makes me sick at heart. I'm sure that we're all here to improve Misplaced Pages and our own understanding by conversing. No one, living or dead, has a perfect understanding of nature. Let Physics speak for Herself. Willow 16:56, 9 August 2006 (UTC)

I see you're new here. In this discussion we've long gone past the point where physics is allowed to speak for itself. As you will see from the replies you are about to get to your example. SBHarris 17:29, 9 August 2006 (UTC)

No need to sling mud and get personal, I am just trying to explain thigns to you that come at odds with what you learned many years ago.There is no more such notion of "rest/relativistic" mass. Neither of them makes any sense in the case of the photon and both have been banished from modern teaching of physics. Here it is, the same explanation again:

http://www.lassp.cornell.edu/~cew2/P209/part11.pdf and here:

http://www.courses.fas.harvard.edu/~phys16/Textbook/ch11.pdf Have a look at chapter 11.8.Ati3414 17:36, 9 August 2006 (UTC)

Please stop quoting these cites. They don't discuss mass of systems except in the examples (which aren't fully referenced), and thus they really are irrelevant.SBHarris 19:36, 9 August 2006 (UTC)

Mass of systems of photons

If we could keep things off the repeated personal attack, the only thing I have been saying all along is that you cannot infer that "For example, a mirrored box containing a gas of photons, or even a single photon, with total energy E will have greater rest mass (by Δm = E/c²)" because it conveys the notion of photon having mass. And this comes to odds with the whole rest of the wiki page (and with mainstream physics). Try to accept this: since there is no "rest" vs. "relativistic" mass for the photon, it makes no sense to talk about the "photon mass contribution". It also makes no sense to talk about "rest/relativistic mass" anymore. The modern teachers have moved on. That's allAti3414 17:36, 9 August 2006 (UTC)

No, try and accept this (it's quite simple):there is no photon "rest mass", there IS a photon "relativistic mass" . Ok? --Michael C. Price 17:50, 9 August 2006 (UTC)

Just a quibble: There is a photon rest mass; it just happens to be zero. (Dilbert: "I've got a personality." Dogbert: "Let's not have that 'is zero a number' argument again.") --Trovatore 17:56, 9 August 2006 (UTC)

Just a quibble-quibble: Debatable, since the photon can never be at rest... :-) --Michael C. Price 17:59, 9 August 2006 (UTC)

OK, but the serious part of my point is that we should always phrase things in terms of whether things "have nonzero (foo) mass" rather than whether they "have (foo) mass". --Trovatore 18:01, 9 August 2006 (UTC)

There is no more such notion of "rest/relativistic" mass. Neither of them makes any sense in the case of the photon and both have been banished from modern teaching of physics. Here it is, the same explanation again:

http://www.lassp.cornell.edu/~cew2/P209/part11.pdf and here:

http://www.courses.fas.harvard.edu/~phys16/Textbook/ch11.pdf Have a look at chapter 11.8. Ati3414 18:15, 9 August 2006 (UTC)

That POV is already explained at relativistic mass. Go there. --Michael C. Price 18:33, 9 August 2006 (UTC)

It is very poorly explained there. The photon is an exception, you cannot apply to it the notions of "relativistic/rest mass" . I have been trying to get the new way of teaching physics in, would you please have a look at the referenced curses from Cornell and Harvard. I know the subject very well, I do not need more refences to the way wiki botches it. This is not only my POV , it is the POV of leading universities in the US:

http://www.lassp.cornell.edu/~cew2/P209/part11.pdf and here:

http://www.courses.fas.harvard.edu/~phys16/Textbook/ch11.pdf Have a look at chapter 11.8. Ati3414 18:52, 9 August 2006 (UTC)

What part of "relativistic mass" ${\displaystyle =E/c^{2}}$ don't you understand? --Michael C. Price 19:04, 9 August 2006 (UTC)

The same one that you seem unwilling to read on, the one that is no longer used in teaching physics because of the confusion it generates, the one that does not apply to photon:

http://www.lassp.cornell.edu/~cew2/P209/part11.pdf and here:

http://www.courses.fas.harvard.edu/~phys16/Textbook/ch11.pdf Have a look at chapter 11.8. Ati3414 19:09, 9 August 2006 (UTC)

Can't you even figure out the indent thing? BTW I have read those links, as I've already told you on my talk page. Their POV is explained at relativistic mass. --Michael C. Price 19:14, 9 August 2006 (UTC)

Sorry, this is a nonsense argument. There is no "rest mass" for the kinetic part of the energy of a particle (I'm talking about JUST the kinetic energy). However, despite that, the kinetic energy of particles in a system (2 particles or more) DOES HAVE MASS. And it can be rest mass, if the system is at rest. This kinetic energy ADDS mass to all systems which contain the moving particles. Not only their rest mass, but their kinetic energy shows up as mass in the COM frame. Kinetic energy is part of the mass of "heat." It's half of heat in solids, and it's ALL of heat in ideal gases. And yet this kinetic energy has no rest mass when you look at the particles one at a time. So where is the mass it represents? Well, the mass of the kinetic energy of particles is part of the system of particles, even though you cannot connect it with any particular individual particle (indeed it hops around from place to place as you change your reference frame). That confuses some people when they study relativity, but our job in these articles is dig out this confusion and confront it, not bury it and pretend it doesn't exist. Kinetic energy has no rest mass for single particles but it DOES have rest mass for SYSTEMS of particles when THEY are at "rest" (rest in systems means we are looking at the system from the COM frame). Get used to it. Get over it. It's modern physics.

Photons are merely a special case of the same problem, since in a sense they're ALL kinetic energy and nothing else. None has ANY definable rest mass when viewed one at a time (you can make their energy anything you like by choice of reference frame), any more than massive particles have any particular mass associated with THEIR kinetic energy, when viewed one at at time (again, you can make it anything you like). BUT the kinetic energy of particles AND photons (which for photons we conventionally call merely their energy) adds mass to any system they're part of.

All of this is ordinary mass. It has inertia, gravity, weight. It's not only relavistic mass, but also proper and invariant mass in the COM frame. We count it as heat whether it's kinetic energy or photons. And as heat, it HAS mass and ADDS mass, no matter which kind of energy it is. SBHarris 19:55, 9 August 2006 (UTC)

Perhaps the problem is merely pedagogy

It does seem like more than one person here could use a nap. ;) Let's all remain true to the physics!

The last remaining difficulty — which seems perfectly valid — is pedagogical, i.e., not with the physics per se, but with how to explain it to lay people (i.e., non-physicists), which will include most people reading Misplaced Pages. There will naturally be at least a few readers who will be confused by the apparent paradox that the photon is massless and yet is able to gravitate and to increase the inertial mass of an object that absorbs a photon. As Ati seems to be pointing out, we should try to explain to these readers as carefully as possible why this is not contradictory.

The two web-pages cited by Ati do not seem to contradict what I said earlier. The second one (in its section 11.8) argues that physics teachers should replace the term "rest mass" with "mass" (and the symbol ${\displaystyle m_{0}}$ with ${\displaystyle m}$) for pedagogical reasons, since inexpert students might try to take over other elements of Newtonian physics that are not valid relativistically. That's a valid viewpoint, but it is not universal in the physics community, since there are other confusions that can arise from using just the term "mass". Working physicists, however, do generally stick with the briefer term "mass" when speaking among themselves. Regardless of the teaching method, the essential physics has not changed, merely its nomenclature and symbols.

I'm rather pre-occupied now with family matters, but I can try to find a compromise wording, if everyone is patient enough with me. I was thinking of working more on photons, anyway; I'm in the middle of translating Planck's 1900 paper and Einstein's 1905 "photoelectric" effect paper for Wikisource. I'm kind of busy for the rest of today, though. Willow 18:42, 9 August 2006 (UTC)

I'm against "compromising" with Ati3414; that will not improve the article. We can't allow articles to be skewed just because editors with, shall we say, unusual enthusiasms show up and make a lot of noise. --Trovatore 18:47, 9 August 2006 (UTC)

I absolutely agree. You can't comprise with a crank either on the talk pages or in the article. His beef is with conventional terminology -- it has no physical insight or merit. He simply can't accept that "relativistic mass" ${\displaystyle =E/c^{2}}$. That's his problem, not ours, and articles should not suffer because of it. --Michael C. Price 18:53, 9 August 2006 (UTC)

Sure, let's leave things as they are. Ignorance is bliss. Ati3414 18:54, 9 August 2006 (UTC)

As pointed out above, the kinetic energy of particles has no "rest mass" and yet it contributes to the mass of resting systems full of such particles (like a can of heated gas). If you can deal with that paradox, you can deal with photons. Perhaps our problem is we didn't make people deal with it before they got to photons. Here, in an article on photons, it returns to bite us. Had it been thoroughly dealt with by the people arguing here in relativistic kinematics before they ever heard of photons, they'd have an easier time seeing photons as just "bare" kinetic energy. SBHarris 20:03, 9 August 2006 (UTC)

Mass of an empty box

This article fails to mention that the zero photon state also has energy and that this gives rise to the well known Casimir effect. Another consequence is that the mass of an empty box with reflecting walls is not zero (plus the mass of the walls). Count Iblis 21:08, 9 August 2006 (UTC)

E = m c²

A while back, I finished up writing at the end of the photon in a box section "consistent with the system's total energy and mass at rest satisfying E = m c²." I later took it out as redundant, but MichaelCPrice put it back modified as "each component of the composite system, including the total composite system itself, obeys E=mc², where m is the relativistic mass." Now, I've been trying to be good and avoid the concept of relativistic mass. Is it really correct here? Is it to emphasize that in the system of the box, the photon can indeed be attributed a relativistic mass, as an independent component? What are the components of the system? Is kinetic energy a component? Or it kinetic energy to associated with some unit of mass? Does the notion of components really make sense? Are we sure we want to go this far? Obviously Ati will object; anyone else? Dicklyon 04:31, 10 August 2006 (UTC)

Look up "A simple exercise" entry above. Try getting the "photon mass contribution" from formula Would be good to read the older controversy "Photons carry mass?" from further above as well. Looks like this group has been gone thru this before Ati3414 04:59, 10 August 2006 (UTC)

I don't see why you'd want to muddle the issue by adding to a system in a way that changes its momentum. It's unclear to me whether your equations demonstrate that it is not in general easy to assign mass to components such as photons, but suppose for the sake of argument that I accept that. You might be right. What I'm trying to ask is whether anybody else sees such a problem. If so, let's leave the more general statement that all of the energy in the system counts in its mass, and not try to split it up over "components". Dicklyon 05:07, 10 August 2006 (UTC)

It's not "muddling", it is the (most) general case you asked for, remember? You asked about the "components". I gave you the "components". Now that you got an "inconvenient answer" you are trying to weasel out of the situation. How does the most general formula applies to the "photon(s)in a box" now? What is the "mass contribution" in the general case? Have you read the section "Photons carry mass?". Yes or no? Don't let all this "inconvenient questions" rattle your (mis)conceptions about what is really mainstream Ati3414 13:53, 10 August 2006 (UTC)

I said I can accept it. I'm wondering if anyone else does. Dicklyon 15:55, 10 August 2006 (UTC)

Well, you can add relativistic masses and energies, so it's correct that the total relativistic mass and relativistic energy of the box is the sum of these things for the components. But that's one reason we don't use relativistic mass, because we do this job using energy, then just dividing by c^2 where we like (and after we're done). In the COM frame (as for a box) the rest mass and invariant mass happen to equal the relativistic energy and relativistic mass, SO that fact (in this special frame) allows you to get invariant mass of the box by adding up all the RELATIVISTIC energies. Which seems a bit weird to many people because you certainly can't get invariant mass by adding the invariant masses of components. SBHarris 20:38, 10 August 2006 (UTC)

Resolving physics disputes; and a consensus nomenclature?

Hi all,

My impression is that the confusion here has persisted because people argue past one another using different definitions of "mass", without proposing a concrete physical experiment that discerns the two points of view. I recommend that we all do this when we have a substantial disagreement about the physics.

Even where there's agreement about the physics per se, there also seem to be disagreements about the best method for presenting it to lay readers. A first simple step towards solving several of our difficulties may be to agree on a common nomenclature. Here's a proposed nomenclature that may help us clarify the debates and make articles more intelligible to Misplaced Pages's readers:

(1) I propose that we not use the unqualified term "mass", since it does not have a single definition and can be confusing.

(2) I propose that we use only the term rest mass (not an unqualified "mass") to represent the coefficient ${\displaystyle m_{0}}$ in the energy momentum equations ${\displaystyle E=\gamma m_{0}c^{2}}$ and ${\displaystyle \mathbf {p} =\gamma m_{0}\mathbf {v} }$. The term "rest mass" is still in common use among physicists and is found in modern physics textbooks (see below). "Rest mass" has exactly one physics meaning and, therefore, seems preferable to "mass", which has several. Moreover, "rest mass" is an intrinsic property of a particle, the one being referred to when we say that a photon is massless, or that the electron's mass is 9.1 x 10^-31 kg (0.511 MeV). Being shorter, "rest mass" seems preferable to the equivalent term "invariant mass", also in common use.

(3) I propose that we not use the term relativistic mass, since that term is not used by practicing physicists, is not taught to physics students at major universities, is generally not used in solving relativistic mechanics problems, and may confuse the readership of Misplaced Pages. To support this third proposal, I went to a friend's house and surveyed commonly used physics textbooks that cover relativistic mechanics. The following 10 textbooks made no mention of "relativistic mass" at all

1. Halzen F and Martin AD (1984) Quarks and Leptons, Wiley.
2. Wald RM (1984) General Relativity, University of Chicago Press.
3. Goldstein H (1980) Classical Mechanics, 2nd ed., Addison-Wesley.
4. Barut AO (1980) Electrodynamics and Classical Theory of Fields and Particles, Dover.
5. Rindler W (1977) Essential Relativity, Springer Verlag.
6. Misner CW, Thorne KS and Wheeler JA (1973) Gravitation, W. H. Freeman.
7. Weinberg S (1972) Gravitation and Cosmology, Wiley.
8. Landau LD and Lifshitz EM (1975) The Classical Theory of Fields, 4th ed., Pergamon
9. Symon KR (1971) Mechanics, 3rd ed., Addison-Wesley.
10. Weyl H (1952) Space-Time-Matter, Dover.

The following two textbooks mention "relativistic mass", but treat it as an intermediate quantity used merely to arrive at the 4-vector formulation of energy and momentum:

1. Feynman RP, Leighton RB and Sands ML (1989) The Feynman Lectures on Physics, vol. I, Addison-Wesley.
2. Eyges L (1972) The Clasical Electromagnetic Field, Dover.

For example, the latter reference says of "relativistic mass" that it is not fundamental and "merely introduces new terminology". Two other textbooks follow the same path for deriving the 4-vector, but do not use the term "relativistic mass"

1. Jackson JD (1975) Classical Electrodynamics, 2nd. ed., Wiley.
2. Schwartz M (1972) Principles of Electrodynamics, Dover.

Finally, two older textbooks introduce "relativistic mass" for completeness (along with longitudinal and transverse mass) but state that the energy-momentum 4-vector approach is better

1. Sommerfeld A. (1942) Mechanics, Academic Press.
2. Pauli W (1921) Theory of Relativity, Dover.

Summarizing, I have not identified a commonly used textbook that treats "relativistic mass" as an important concept in its own right. In 2/16 instances, it is used as an intermediate concept to derive the energy-momentum 4-vector, which is subsequently treated as fundamental. In two other cases, it is presented as a historical relic on an equal footing with transverse/longitudinal mass (which I hope we agree that we don't want to cover here).

On the basis of these data, it seems fair to say that "relativistic mass" is not in current use among practicing physicists. One might argue that it has pedagogical value but, speaking for myself, it seems more likely to be confusing than helpful to the majority of lay readers.

I have restricted myself to published, paper-copy textbooks on relativity, because those seem to be the most reliable sources. Lecture notes posted on the web and similar sources are often made by people who lack expertise or who may be a little careless in their wording, such as graduate students or harried professors whose research is in a different field.

Please forgive the long post, but I think many of our disagreements will be resolved if we agree on a common nomenclature and insist upon describing concrete experiments that discern different models of the physics. Willow 19:40, 10 August 2006 (UTC)

Thank you, Willow , a very sensible post.Devoid of personal attacks and pettiness.There are multiple references to "relativistic mass" throughout wiki. Now, how do we get from what you wrote to removing those references and the m_photon=E/c^2 from the same several wiki posts? Ati3414 20:09, 10 August 2006 (UTC)

Sorry, relativistic mass has to be discussed because that is how it appears in the equation ${\displaystyle E=mc^{2}}$.

Answer, is that this is not true for resting particles, and also for systems in the COM frame. There, you can still use E = mc^2 and the energy is the system rest energy (which also is the total relativistic energy) and m is the system rest mass. The main problem for this approach is that people like Ati3414 don't really believe what I just said, because it implies that relativistic energy (and mass) adds system rest mass. Including when the energy comes from a photon (or from system heat or kinetic energy). So the issue is not pedagogical, it's an issue of reality. Basically Ati3414 will reject the physicists like Baez or whoever he needs to, because he doesn't believe it. In spite of what the equations say. Look, you just SAW him say that Einstein is right that a body which emits a photon of energy E has less rest mass, by E/c^2. But he doesn't believe that a body absorbing a photon E grows in rest mass by E/c^2: That would violate his fixed belief, so he denies it. I'm afraid that at some point you have stop looking for reasons why unreasonable people have unreasonable beliefs. SBHarris 20:24, 10 August 2006 (UTC)

Sbharris wrote: "....you just SAW him say that Einstein is right that a body which emits a photon of energy E has less rest mass, by E/c^2. But he doesn't believe that a body absorbing a photon E grows in rest mass by E/c^2"

Ati3414 answers: No, this is not what I am saying. You have been told multiple times that, while the above is true and I do not dispute it, you cannot and should not infer from it that the equivalent mass of the photon is E/c^2. This has always been the issue. I have said this to you and to the others maybe 20 times by now. There is a subtle difference but an important one. Go over all the previous exchanges, it was always there Ati3414 21:01, 10 August 2006 (UTC)

No, ${\displaystyle E=mc^{2}}$ is still true for COM/resting particles, it's just that in those circumstances the relativistic mass happens to equal rest mass. Of course we both know and understand what we both mean and that we only have a minor linguistic quibble here -- it's just Ati who has a screw loose. --Michael C. Price 20:39, 10 August 2006 (UTC)

Repeated personal attacks only show that you can't argue something scientifically and only reflect on you. Since you are attacking with such violence a couple of questions:

COM frame calculation

1. what is the "COM" of a single photon?

That's easy; you can say either that it hasn't got one, and hence has no uniquely defined energy, mass, or momentum, or you can say that its COM frame moves at the speed of light, so the mass, energy, and momentum are zero. I believe the latter answer is too far stretched to be meaningful, and we've all agreed already on the former, so let's just go with that. Dicklyon 21:28, 10 August 2006 (UTC)

2. what is the "COM" of a system of 3 photons , two flying at 180 degress wrt the third one?

That's a more interesting case. If you specify the conditions more exactly, it should be easy to find the frame with respect to which the total momentum is zero. That's the COM frame. Dicklyon 21:28, 10 August 2006 (UTC)

You have enough info, two photons fly to the right, the third one flies to the left, along the same lineAti3414 23:24, 10 August 2006 (UTC)

3. For both you and/or Sbharris, go back to "A simple exercise" and calculate the "mass of the photon" for one of the more complicated cases (like the 3 photon system above). Ati3414 21:01, 10 August 2006 (UTC)

The contribution of mass due to the photon will be E/c2, for the E evaluated in the COM frame. No magic there. The (proper) mass of the photon is still zero. Dicklyon 21:28, 10 August 2006 (UTC)

Can you do some calculations? English composition doesn't count in physics, math does. Ati3414 23:22, 10 August 2006 (UTC)

COMMENT

The only people doing calculation here so far is us. Haven't see you do one yet.

Presuming all 3 of your photons to each have energy E in the lab frame, then 2 going one direction and 1 going in the other, is equivalent to one photon of energy 2E going one way, and another of energy E going oppositely (reduction to a problem of 2 photons, one twice the energy and frequency of the other, which has the same answer). The COM frame is the one in which these 2 photons have momenta (and therefore energies, since they ARE photons) summing to zero. In one direction momentum is 2E/c, and in the other it's E/c. So our COM frame is the one which chases the 2E photon so that by Doppler shift it's energy drops to equal that of E photon. Of course, the Doppler shift will increase E for one photon while decreasing it for the other.

It's easiest for me here to simply pretend we're IN the COM frame where both photons have the same energy E and frequency hv. We're looking for a velocity v to produce a Doppler shift from this frame which is large enough that one photon now has twice the frequency of the other. Let us define β = b = v/c, which I'll use just so I don't have to use the Greek letter all the time, then the finally Doppler shifted frequency UP for one photon (1+b)/(1-b) is going to equal 2 times the shifted frequence DOWN for the other: (1-b)/(1+b). Writing that out and getting rid of fractions gives

(1+b)^2 = 2 (1-b)^2

Solving for b gives b = v/c = /. This is not an easily resolvable fraction. It's also 1/2 = about 0.2071. Thus, the COM frame is the frame which is moving at 0.2071 c in the direction of the 2 photons. In that frame, the two photons have the same energy and momentum as the other photon headed in the opposite direction. I will leave what the total energy and invariant mass for this system, to the student. Ati, here's your chance to show you can do algebra.

Now, Ati, a much easier problem for you, which doesn't involve photons so you might be able to get around the mental block. Two objects of mass M move away from each other with relative velocity V. What is the kinetic energy of one mass, as seen by the other? What is the total kinetic energy as seen in their COM frame? What is the system invariant mass? If we put these objects in a can, how much increase in mass could we expect from the can if we put it on a scale in the COM frame? SBHarris 00:15, 11 August 2006 (UTC)

No calculation of the "photons equivalent mass"? Ati3414 00:20, 11 August 2006 (UTC)

It's easily done, now that we have the COM frame. I'm leaving it up to you. Show us your stuff SBHarris 00:43, 11 August 2006 (UTC)

Finish what you started please, you are not a professor though you put on the airs . Let's see your final result. Let us know when you stop changing your answers. Ati3414 01:12, 11 August 2006 (UTC)

The final answer is total energy is 2*sqrt2*E in the COM frame, for 3 photons of E in the lab frame, with invariant mass of the system therefore = 2sqrt2*E/c^2. Now your turn with the simple kinetic energy problem.SBHarris 01:35, 11 August 2006 (UTC)

SB, thanks for bailing me out here, but are you sure that the ratio (1-b)/(1+b) is in general correct, for system velocities that are getting into a relativistic range? Is it really that simple? And is it really in general OK to treat the two photons as one of double the energy? Does the doppler shift on that do the same thing to energy and momentum as it would to two photons of the original energy? Looks like it does, but I never thought of that trick. If that's all true, you just need to solve for the frequencies of the momentum-matched photons in terms of the original photons, which are just 2*(1-b)/(1*b) and 1*(1+b)/(1-b) and should be equal. But, with b = 1/3 (your original answer) or b = 0.2071 (your new answer) these are not equal. Either your original ratio needs a sqrt in it, or your solution is in error, since the correct solution of your equation is b = 3-sqrt(8) = 0.1715 according to my scratch paper. Using that, the two photon energies are sqrt(2) times the original one photon's frequency or energy. There are two of those, so the total energy in the COM system if this is right is 2sqrt(2)E, and the rest mass of the system is that over c^2, as always. It's somewhat less than 3E, as expected. I don't have confidence that the Doppler formula is exactly correct, so if this result is wrong, patch it accordingly. Dicklyon 00:57, 11 August 2006 (UTC)

ANSWER Arggh! Yes, you've caught me squaring both sides and not showing it, and then forgetting to do it to the factor of 2. The relativistic Dopper factors have square roots. Let's see if I can get it right.

Sqrt = 2 Sqrt

(1+b)^2 = 4 (1-b)^2

1+b = 2 (1-b)

So it looks like v = c/3 was right the first time.

So re-inserting you get sqrt ( / ) = 2 * Sqrt( /)

Which is Sqrt(2) = 2 Sqrt(1/2) so this works.

Because these are square roots, the E photon gets bumped up to (Sqrt2)E and the other big 2E one (or if you like the 2 E ones going the same direction) gets taken DOWN to 2E/(Sqrt2), but because Sqrt2 = 2/Sqrt2, the energies of both sides are now equal in the COM frame and so are the momenta, as advertized. And the total COM energy and therefore rest energy of the system is 2*(sqrt2)E. Divide by c^2 for invariant mass. So Ati got us to do it anyway, blast. SBHarris 01:29, 11 August 2006 (UTC)

Yes, getting the correct relativistic Doppler formula can always get you, professor. My turn:

${\displaystyle m'={\frac {\sqrt {(\Sigma E)^{2}-(\Sigma pc)^{2}}}{c^{2}}}}$

is an invariant, i.e. one should get the same result in any frame, no need for the pesky COM frames that can trip the best.

${\displaystyle \Sigma E=3E}$

${\displaystyle \Sigma p=p+p-p=p}$

Substitute , in remembering E=pc and you get

${\displaystyle m'={\frac {{\sqrt {8}}E}{c^{2}}}}$

At this point the humble student gives the self-appointed professor a D :-) Ati3414 02:01, 11 August 2006 (UTC)

I don't see why I should get a D for getting the correct answer, since of course your sqrt8 is my 2sqrt2. And IF you didn't get your problem and your final answer out of a textbook. Yes, doing invariant mass in the lab frame CAN be a shortcut if something cancels, and so it is here. But remember, you haven't done the whole problem. You didn't originally ask for the invariant mass or energy of the system--- you asked to calculate the COM frame velocity. Which I did. It might be faster to do the invariant mass first, but when you back-calculate the velocity, you're still stuck with the Doppler formula (or the equivalent) for photons, so there are no royal roads. Anyway, now your turn, on the kinetic energy problem. Show us where that kinetic energy is, in the system. How much mass does it contribute? How can it contribute any mass, if it has no rest mass? SBHarris 02:38, 11 August 2006 (UTC)

Too bad, SB, you'll have to live with getting a D from him who claimed that photons don't add to the mass of a system, even if he just showed you exactly that they do. That's a grade you can frame (in a COM frame?). Dicklyon 02:45, 11 August 2006 (UTC)

Naturally a COM frame. This is the invariant mass of a system of nothing but photons, as described, good in any frame--- and we're being told it's not a REAL mass. Proof positive you can calculate something you don't understand. SBHarris 02:58, 11 August 2006 (UTC)

Yep, too bad you can't read: "invariant" . Not mass. But you did well, you told Sbharris that relativistic Doppler formulas are different from the Newtonian ones so he stopped doing his standard mish-mash of SR and Newtonian mechanics. Carry on guys, maybe you can collaborate on some scientific project. One more thing, pervect cleaned up a little mass in special relativity. Maybe you two guys can clean up "photon in a box". You look good together. Ati3414 03:16, 11 August 2006 (UTC)

Invariant means invariant mass. See the "m'" in your formula? The m stands for "mass" SBHarris 04:47, 11 August 2006 (UTC)

Yes, it is an invariant and it does have dimensions of mass. It should not be misconstrued that the 3 photons have the mass shown by the formula.One of these days it will sink in: the photon is a singular particle , you cannot apply your ideas of massive particles because you will get weird results. Same way you cannot mishmash SR formulas with Newtonian ones. Ati3414 05:58, 11 August 2006 (UTC)

So this system has an invariance with units of mass, and it has a center of mass that we all agree on. And the invariance that has units of mass was computed from the energies of the photons that make up the system. But the system doesn't have rest mass? What am I missing here? Dicklyon 06:11, 11 August 2006 (UTC)

Correct, it doesn't. If it had, it would violate the Proca gauge. Try reading the Roderic Lakes paper I sent you, maybe it will sink in. Look at the changes that Pervect made to mass in special relativity (not profound enough, but still) Ati3414 06:27, 11 August 2006 (UTC)

Answer to WillowW

Hi WillowW,

I appreciate your well-intended and thoughtful work, but I can't fully agree. The term "relativistic mass" is historically well-attested, and the mere fact that it is not currently fashionable to use it (even if for good reason) does not mean it should be wiped from the record. As I mentioned earlier (before you came in, probably) I would be happy with a brief footnote pointing out that while the "mass" in the contemporary sense -- that is, the invariant mass -- of the photon is zero, there is an older notion of relativistic mass, which is not zero for the photon, and pointing the reader to the discussion in the relevant article. With such a footnote, neutrally worded, I would be happy to see the "photons in a box" section just go away, since (as an earlier contributor pointed out) it isn't really about photons specifically. --Trovatore 21:17, 10 August 2006 (UTC)

How do you know : "which is not zero for the photon" ? Have you run any experiment that proves that? Can you name some experiments that prove that? I concur with you on one thing: the "photon in a box" gotta go. Delete it, please. Ati3414 21:38, 10 August 2006 (UTC)

It's definitional. You might as well ask for an experiment that proves that left is the opposite direction from right. --Trovatore 21:48, 10 August 2006 (UTC)

Please refrain from using names, this goes counter wiki privacy policy. So, your answer is that you have no answer, right? Ati3414 23:21, 10 August 2006 (UTC)

My answer is, it's true by definition. Arguing with well-attested historical definitions doesn't make much sense; you can use them or not, but they are what they are. Asking for an experiment to "prove" them is a complete non-sequitur.

As for this "privacy policy"—I don't exclude out of hand that it might exist; I haven't made a systematic study of all the policies. But you're the first person I've heard mention it. Perhaps you can point me to it. --Trovatore 23:33, 10 August 2006 (UTC)

"True by definition" 1. There is only one type of mass, proper mass. 2. In QED the mass of the photon is defined to be zero. From 1 and 2 you can easily conclude that the mass of the photon is NOT E/c^2. On the other subject look here, under harassment : http://en.wikipedia.org/Wikipedia:Harassment , you have done quite a few things that qualify as harassment, you can read for yourselfAti3414 23:37, 10 August 2006 (UTC)

Pervect has put references in the mass in special relativity article attesting the term "relativistic mass". You're entitled not to use the term; you're not entitled to insist that the rest of the world forget it. The claim is in fact true by definition. --Trovatore 23:55, 10 August 2006 (UTC)

I know, I am having a sidebar discussion with him and he's mulling over what I told him. Are you bowing out? No calculations from you? I set a few problemms in front of the opposing side, would you care to tackle one? You know, physics is about math, not about English compositionAti3414 00:07, 11 August 2006 (UTC)

The calculations are trivial and beside the point. Physics is not about English usage, but there is no physics issue here; everyone with a basic, minimal understanding will get the same final answers. The only issue is linguistic usage, what you call the various quantities involved. And on that point you are not entitled to dictate that the historical usage be ignored. --Trovatore 00:28, 11 August 2006 (UTC)

That is the key point: this is a linguistic debate only -- there is no physics here. The terms are defined by common and historical usage. End of story. --Michael C. Price 01:02, 11 August 2006 (UTC)

Would you care to help Sbharris with his calculations? He seems to need help.Ati3414 01:20, 11 August 2006 (UTC)

Worth repeating. Energy with no rest mass in single particles, DOES have rest mass in systems

The "odd" fact that relativistic energy can be added to get rest mass of systems, is what allows energies which have NO rest mass (like photons and the kinetic energies of moving particles) to be counted (indeed summed up to be) the regular mass (invariant mass) of bound systems. Energy with no rest mass DOES have rest mass in a system! We've made this point as many places as we could, in part because it seems to be widely misunderstood. It's a special property of systems which does not appear for individual particles. The kinetic energy of a (single) free gas molecule doesn't WEIGH anything, because as soon as you go to a frame where it can be weighed, it goes away. And if you trap the gas molecule, you're no longer talking about a single object, but a system of molecule plus can. The kinetic energy of a particle in a can of gas, DOES weigh something. Kinetic energy is a SYSTEM property. It doesn't reside in single objects, but in systems of two or more objects. The MASS associated with kinetic energy IS THE SAME KIND OF THING. It's not really "in" a single moving object, but rather is a property of a system of 2 or more objects in motion relative to each other.

And much the same is true of photons. Single photons have no definable energy, anymore than massive objects have a definable kinetic energy--- it's all frame dependent. The kinetic energy of a single particle or a single photon can be anything you like, depending on your reference frame chosen. But get a system of two or more objects, or photons, and now you have some residual minimal energy which you cannot get rid of by choice of frame. THAT energy has mass. THAT energy contributes invariant mass to systems.

When you get this, you'll be fine. SBHarris 20:36, 10 August 2006 (UTC)

Photon geometry question

(What the -- a post not to do with a photon's supposed mass!?) I am trying to understand how to visualize a traveling photon. Is it more accurate to view, from a very casual standpoint, a photon as (1) a point-like "cart" on a sinusoidal imaginary "track", that progresses along the track, that does not leave a trail (save an EM disturbance wake ?). (2) A ribbon whose length is finite and shaped sinusoidally. (3) A ribbon whose length is infinite and shaped sinusoidal. (4) Something else? ...Does the same apply to the virtual photon that mediates the electromagnetic forces? (see fundamental forces) I'm putting my money on concept (1). Thanks to anyone who can help out. --HantaVirus 21:17, 10 August 2006 (UTC)

You're going down a slippery slope toward nonsense here. Photons don't propagate. Propagation of electromagnetic effects is exclusively a wave phenomenon. Photons are the quantized phenomena that are apparent only where photons are emitted or absorbed. That's my position, and I'm sticking to it ... for now. Dicklyon 21:30, 10 August 2006 (UTC)

The uncertainty principle comes from the Fourier analysis of waves (any waves, for EM waves are included). If you know the photon's wavelength perfectly, it's a very LONG (infinitely long) sine wave. As your knowledge of its wavelength (momentum) gets worse and worse, you get a spread in possible wavelength values, and your summed sin wave becomes a superposition of various sin waves of various wavelengths (momenta), and now you get a wavepacket which is shorter, and better localized in space. So momenta (wavelength) and localization trade off. The better you know the one, the worse off you are with the other. So when you ask: "How big is a photon?" the answer is partly "How well can you characterize it?" SBHarris 21:41, 10 August 2006 (UTC)

(Please note removal of wikimarkup from header; made edit summaries un-clickable.)

I think you're maybe not fully following the OP's question. I believe he's visualizing the photon actually moving transversally along his posited "sinusoidal path" as it travels. No, that's not the right picture. The sinus curve corresponds to electric and magnetic field strengths, not to the path of the photon. Think of the photon as going in a straight line. It's a slightly fuzzy straight line because of the uncertainty principle, but it's not a sine wave. --Trovatore 21:46, 10 August 2006 (UTC)

My point though is that the photon does not have a path. Certainly not a straight line. Consider the two-slit setup. Can you show a photon's path through it? Certainly not. But you're right in that the sinusoidal path is not sensible. Dicklyon 21:57, 10 August 2006 (UTC)

Well, this is general quantum weirdness, not about the photon path per se. The photon in the two-slit experiment is a quantum superposition of many photons (or "possible photons", if you like the many-worlds interpretation) that do have well-defined paths, and those paths are straight lines. --Trovatore 22:00, 10 August 2006 (UTC)

There's no point trying to reify the paths used for QED calculations into photons. These straight lines are not photon paths, just paths over which you can calculator phases of wave functions. If you want to view it as a superposition of photons, you can, but it's too weird for my taste. That's why I prefer approaches such as Wheeler-Feynman electrodynamics; less weirdness, more waves. —The preceding unsigned comment was added by Dicklyon (talk • contribs) 05:10, 11 August 2006 (UTC)

Don't actually know Wheeler-Feynmen ED, but there's no way around the weirdness in general, or at least no easy way. Look up the Bell inequality and the Aspect experiment. I think you might as well grasp the nettle on the photon two-slit experiment, because nothing's going to save you from non-local-realism in the long run. --Trovatore 05:29, 11 August 2006 (UTC)

The version I'm more familiar with is Carver Mead's "Collective Electrodynamics". Like Wheeler-Feynman, it uses advanced and retarded wave symmetrically. Weirdness of the Copenhagen entanglement sort is replaced by things whose arrow of time goes backwards, like the absorption of a photon influencing the emission, possibly light-years away and years earlier. It's less weird, but of course still can't be local and causal. Cramer's transactional interpretation is a similar kind of thing. Dicklyon 05:57, 11 August 2006 (UTC)

This to you is less weird than Copenhagen? Matter of taste, I guess. --Trovatore 06:07, 11 August 2006 (UTC)

All good points. That sin wave thing is really a GRAPH of how E and B vary along a 1-D path for a 1-D wave. The illustrators steal the other two spacial dimensions for this purposes, and in the process cause no end of confusion (as in the illustration that tops THIS article). The localization along that 1-D line is constrained by the uncertainty principle, though, in both transverse and longitudinal directions. As for other than 1-D paths, EM fields propagate in many forms and although some of the classical fields are vaguely visualizable as 2-D snapshots of fields moving in 3-D (like the plane waves around an antenna), none of this really works very well for photons. A photons takes all possible 1-D paths straight from HERE to THERE, and you can always only pick only ONE of them (or at least just a few) to illustrate. And then you're stuck with using your extra dimensions in the usual way, unless you just want to show a ray-trace. SBHarris 22:09, 10 August 2006 (UTC)

User:Ati3414

User:Ati3414 (User:12.36.122.2), it is clear that your modus operandi is to spread wierdness about all the ways in which everything that we've all learned about relativity is wrong (Mass in special relativity, Time dilation, Transverse Doppler effect, Ives-Stilwell experiment, etc.). I understand the frustration you must feel, when just about everything you post gets immediately reverted, but maybe that's telling you something about how disconnected your views are from the accepted mainstream. That doesn't mean you're wrong, but that might be a hypothesis you should consider more seriously. It's getting tiresome. Give it a break. Dicklyon 22:35, 8 August 2006 (UTC)

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