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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