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== A trivial refutation of one of Dingle's Fumbles (Ref: ) == |
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== A trivial refutation of one of Dingle's Fumbles (Ref: ) == |
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On page 230 in to , Dingle writes: |
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On page 230 in to , Dingle writes: |
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: (start quote) |
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: (start quote) |
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<div style="padding: 1em; border: solid 3px {{{bordercolor|#a08040}}}; background-color: {{{color|#fff7f7}}};"> |
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<div style="margin-left: 4em; padding: 1em 0; border: solid 3px {{{bordercolor|#a08040}}}; border-radius: 10px; background-color: {{{color|#fff7f7}}};"> |
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:: {{!xt|Thus, between events E0 and E1, A advances by <math>\color{ForestGreen}{t_1}</math> and B by <math>\color{Blue}{t'_1 = a t_1}</math> by (1). Therefore}} |
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:: {{!xt|Thus, between events E0 and E1, A advances by <math>\color{ForestGreen}{t_1}</math> and B by <math>\color{Blue}{t'_1 = a t_1}</math> by (1). Therefore}} |
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::: <math>\frac{\color{ForestGreen}{\text{rate of A}}}{\color{Blue}{\text{rate of B}}} = \frac{\color{ForestGreen}{t_1}}{\color{Blue}{a t_1}} = \frac{1}{a} > 1 \qquad \text{(3)}</math> |
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::: <math>\frac{\color{ForestGreen}{\text{rate of A}}}{\color{Blue}{\text{rate of B}}} = \frac{\color{ForestGreen}{t_1}}{\color{Blue}{a t_1}} = \frac{1}{a} > 1 \qquad \text{(3)}</math> |
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Dingle should have written as follows: |
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Dingle should have written as follows: |
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: (start correction) |
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: (start correction) |
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<div style="padding: 1em; border: solid 3px {{{bordercolor|#a08040}}}; background-color: {{{color|#f7fff7}}};"> |
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<div style="margin-left: 4em; padding: 1em 0; border: solid 3px {{{bordercolor|#a08040}}}; border-radius: 10px; background-color: {{{color|#f7fff7}}};"> |
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:: {{xt|Thus, between events E0 and E1, A, which is '''not present''' at both events, advances by <math>\color{ForestGreen}{t_1}</math> and B, which is '''present''' at both events, by <math>\color{Blue}{t'_1 = a t_1}</math> by (1). Therefore}} |
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:: {{xt|Thus, between events E0 and E1, A, which is '''not present''' at both events, advances by <math>\color{ForestGreen}{t_1}</math> and B, which is '''present''' at both events, by <math>\color{Blue}{t'_1 = a t_1}</math> by (1). Therefore}} |
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::: <math>\frac{\color{ForestGreen}{\text{rate of clock not present at both events E0 and E1}}}{\color{Blue}{\text{rate of clock present at both events E0 and E1}}} = \frac{\color{ForestGreen}{\text{coordinate time of E1}}}{\color{Blue}{\text{proper time of E1}}} = \frac{\color{ForestGreen}{\text{rate of A}}}{\color{Blue}{\text{rate of B}}} = \frac{\color{ForestGreen}{t_1}}{\color{Blue}{t'_1}} = \frac{\color{ForestGreen}{t_1}}{\color{Blue}{a t_1}} = \frac{1}{a} > 1 \qquad \text{(3)}</math> |
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::: <math>\frac{\color{ForestGreen}{\text{rate of clock not present at both events E0 and E1}}}{\color{Blue}{\text{rate of clock present at both events E0 and E1}}} = \frac{\color{ForestGreen}{\text{coordinate time of E1}}}{\color{Blue}{\text{proper time of E1}}} = \frac{\color{ForestGreen}{\text{rate of A}}}{\color{Blue}{\text{rate of B}}} = \frac{\color{ForestGreen}{t_1}}{\color{Blue}{t'_1}} = \frac{\color{ForestGreen}{t_1}}{\color{Blue}{a t_1}} = \frac{1}{a} > 1 \qquad \text{(3)}</math> |
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:: {{xt|Thus, between events E0 and E2, B, which is '''not present''' at both events, advances by <math>\color{Brown}{t'_2}</math> and A, which is '''present''' at both events, by <math>\color{Red}{t_2 = a t'_2}</math> by (2). Therefore}} |
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:: {{xt|Thus, between events E0 and E2, B, which is '''not present''' at both events, advances by <math>\color{Brown}{t'_2}</math> and A, which is '''present''' at both events, by <math>\color{Red}{t_2 = a t'_2}</math> by (2). Therefore}} |
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::: <math>\frac{\color{Brown}{\text{rate of clock not present at both events E0 and E2}}}{\color{Red}{\text{rate of clock present at both events E0 and E2}}} = \frac{\color{Brown}{\text{coordinate time of E2}}}{\color{Red}{\text{proper time of E2}}} = \frac{\color{Brown}{\text{rate of B}}}{\color{Red}{\text{rate of A}}} = \frac{\color{Brown}{t'_2}}{\color{Red}{t_2}} = \frac{\color{Brown}{t'_2}}{\color{Red}{a t'_2}} = \frac{1}{a} > 1 \qquad \text{(4)}</math> |
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::: <math>\frac{\color{Brown}{\text{rate of clock not present at both events E0 and E2}}}{\color{Red}{\text{rate of clock present at both events E0 and E2}}} = \frac{\color{Brown}{\text{coordinate time of E2}}}{\color{Red}{\text{proper time of E2}}} = \frac{\color{Brown}{\text{rate of B}}}{\color{Red}{\text{rate of A}}} = \frac{\color{Brown}{t'_2}}{\color{Red}{t_2}} = \frac{\color{Brown}{t'_2}}{\color{Red}{a t'_2}} = \frac{1}{a} > 1 \qquad \text{(4)}</math> |
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:: {{xt|'''Equations (3) and (4) are consistent and say that any event's coordinate time is always larger than its proper time:<p>hence there is no reason to say that the theory requiring them must be false.'''}} |
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:: {{xt|'''Equations (3) and (4) are consistent and say that any event's coordinate time is always larger than its proper time:'''<p>'''hence there is no reason to say that the theory requiring them must be false.'''</p>}} |
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</div> |
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</div> |
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: (end correction) |
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: (end correction) |