In reality considered from one reference frame only one clock will run slower. The reason is the construction of the clock
B1------B-------B2------B3 /|\ / \ / | \ / t1/ | \ / / | \ / D L D / / |t0 \ / v--> / t2 | \ / -.A-.-.-.-E-.-.-.-C.-.-.-.-.- Figure 3 |
The time for a light signal to travel (v=0) from E to B and back is 2*t0 In this case t0 = BE / c = L /c The time for a light signal to travel from A to B and back to C is 2 * t1 The time for a "train" to travel with speed v, from A to E to C is 2 * t2 If both arrive at the same time than: t1 = t2 = t AB = D = c*t1 = c*t and AE = EC = v*t2 = v*t Using some arithmatic we get: AB^2= AE^2 + BE^2 = c^2*t^2 = v^2*t^2+L^2 c^2*t^2 - v^2*t^2 = L^2 : t^2 = L^2/(c^2-v^2) : t^2 = L^2/c^2/(c^2-v^2)/c^2 Now we get t = L/c / SQR(1-v^2/c^2) or t1 = t0/ SQR(1-v^2/c^2)) = gamma * t0 The factor SQR(1-v^2/c^2)) with v > 0 is smaller than 1. |
B3 B1------B2------.-------B2 | / \ . . / \ | / \ . . / \ | t1/ .\t1 t1/. \ t0| / . \ / . \ L D . v D / . \ | / . --> \ / . \ |/. t2 t2 \ / .\ -.A-.-.-.-.-.-.-.-C.-.-.-.-.-.-.-.C1 Figure 4 |
B2 | B1 B1 | / \ / \ |t1 t0/ \t0 t0/ \t0 | / \ / \ D L L L L | / \ / <-- v1 \ | / t3 t3 \ / t3 t3 \| .-.-.-.-.-.-.-.-.-.-.-.-.-.AC Figure 5 |
B3 /\ B2 / \ | / \ B1 | / \ / \ |t1 / \ / \t0 | / \ \ D / L | / v2--> \ | / t3 \|/ -.-.-.-.AC-.-.-.-.-.-.-.- Figure 6 |
B3 B1------B2------.-------B2 | / \ . . / \ | / \ . . / | t1/ .\t1 t1/. t0| / . \ / . L D . v D / | / . --> \ / |/. t2 t2 \ / -.A-.-.-.-.-.-.-.-A2.-.-.-.- Figure 4 |
A2 . | . . | . . | . .B3 t| . . | . . | . /. A1 | .B2 | . | ./ | ./ | . / | / .|. / | / .|B1/ | / . | / |/. |/ A------B---------------- Distance Figure 7 |
| | A1 t| | X2 / | |. . / | / . . / | X1 . | B2 | ./ , . | ,/ | . / ., | , / | . / . , | , / | . / . , | , / A1 / . ,B1 / | . /. , | / |A2/. , | / | / . , | / |/ . , |/ A---------X---------B-------- Distance Figure 8 |
| | | .t3 | |. | | . | | . | | . | | . | | | . | | | .| | t1. .t2 | . | .| | . | . | | . | . | .t0 | . | | . | . | | . | . | | . |. | L A L Figure 9 Y |
| / | ,t5 | ,/ t4. , / | . , / | . . / | . / | . . / | . xt3 t1. / | / | / | / | / | / | / |/ A---------C Figure 9 X |
Figure 9 y shows the moving light flash in the y direction.
t4 = AC/c + t3 : v*t3/c + t3 = (v+c) * t3 /c AC = v*t3 = (t4-t3)*c : t4*c = (v+c) * t3 t5 = L1/(c-v) + t1 : L1 = v* t1: t5 = v * t1/(c-v) + t1 = (v*t1 + (c-v)*t1)/(c-v) = t1 * c / (c-v) Using this information the following two fractions can be calculated:
|
v/c | t1 | t2 | t3 | t4 | t5 | f1 | f2 | f1/f2 | f2/f1 |
0.666 | 2 | 0 | 2.683 | 4.472 | 6 | 2,236 | 2,236 | 1 | 1 |
0.5 | 2 | 0 | 2.309 | 3.464 | 4 | 1,732 | 1.732 | 1 | 1 |
0.333 | 2 | 0 | 2.121 | 2.828 | 3 | 1.414 | 1.414 | 1 | 1 |
0.166 | 2 | 0 | 2.028 | 2.366 | 2.4 | 1.183 | 1.1183 | 1 | 1 |
B1------B-------B2 /|\ / | \ t1/ | \ / | \ D L D / |t0 \ v--> / t2 | \ / -.A-.-.-.-E-.-.-.-C.-. R Figure 3 |
The time for a light signal to travel (v=0) from E to B and back is 2*t0 In this case t0 = BE / c = L /c The time for a light signal to travel from A to B and back to C is 2 * t1 The time for a "train" to travel with speed v, from A to E to C is 2 * t2 If both arrive at the same time than: t1 = t2 = t AB = D = c*t1 = c*t and AE = EC = v*t2 = v*t Using some arithmatic we get: AB^2= AE^2 + BE^2 = c^2*t^2 = v^2*t^2+L^2 c^2*t^2 - v^2*t^2 = L^2 : t^2 = L^2/(c^2-v^2) : t^2 = t0^2*c^2/(c^2-v^2) Now we get t^2 = t0^2/(c^2-v^2)/c^2 or t1 = t0/ SQR(1-v^2/c^2)). The factor SQR(1-v^2/c^2)) with v > 0 is smaller than 1. You can also rewrite the last two lines and then you get: c^2*t^2 - v^2*t^2 = L^2 or c^2*t^2 - R^2 = L^2 or L = sqrt (c^2*t^2 - R^2) with R = AE |
Z | t+x .Z |\ | \ | \ | \ | \ t |-----.B | /. | / . | / . | / . |/ . t-x .P . | . | . A----------> x Figure 5A (p21) |
^ t | T B |\ | \ | \ | \ | \ 0.5T|-t2--.P | /. | / . | /t1. | / . |/ . 0 A--t2------->Z 0.5R Figure 5B (p315) |
Figure 5A page 21 shows the worldline of a particle at rest. B is an event outside this worldline. This event creates a light ray which strike the worldline at Z. P is an earlier event that creates a lightray which coincides with B. The angle PBZ is 90 degrees. Figure 5B page 315 demonstrates a non straight worldline. In this case a particle moves with uniform velocity v from A to P and back to B. In reality the particle moves along along the Z axis The angle APB is larger than 90 degrees because v |
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