- #1
StarThrower
- 220
- 1
1. Let clock A and clock B be of identical construction, and let them both not be subjected to any force. Let them be at rest with respect to each other. Thus, the relative velocity v is equal to zero.
Therefore, using Newtonian mechanics or SR, the clocks tick at the same rate. For the sake of simplicity, let the two clocks be synchroninzed. Thus, if X is the reading on one clock, and Y is the simultaneous reading on the other clock, then x-y=0. As long as the clocks remain at rest with respect to each other, the clocks will remain in-sync.
Suppose that clock B is inside a uniformly accelerating ship. Thus, clock B is being subjected to an OUTSIDE force of a constant value. As this happens, the relative speed v between the two clocks will continuously increase. While clock B is accelerating relative to clock A, clock B is no longer in an inertial reference frame. Let us stipulate that clock A is always in an inertial reference frame. Therefore, whilst clock B is accelerating, an observer stationed at clock A can use the time dilation formula of SR, to draw conclusions about whether or not clock B remains in-sync with clock A.
The time dilation formula is:
[tex] \Delta t = \frac{\Delta t^\prime}{\sqrt{1-v^2/c^2}} [/tex]
[tex] \Delta t [/tex] is an amount of time measured by clock A, and [tex] \Delta t^\prime [/tex] is the corresponding amount of time measured by clock B. In Newtonian mechanics, these two quantities remain equal as clock B accelerates, in SR the quantities are not equal, and instead related by the time dilation formula given above.
As you can see, the conclusion arrived at using the time dilation formula, is that the clocks are no longer in-sync once clock B begins to accelerate. In fact, clock B begins to tick SLOWER. That means that if I am located at clock A, and my current reading is 5525, then I can be certain that clock B simultaneously reads something less than 5525 (even if I don't know exactly what it reads). In other words, I can be certain that X>Y, where X is the current reading on clock A, and Y is the simultaneous reading on clock B.
Now, let some amount of time have passed, and let it be the case that we have:
X = 1000 and Y = 400
Thus, the difference in the readings of the two clocks is now
X-Y= 1000-400 = 600
At the moment that clock B reads 400, let it be the case that the engines shut off, so that clock B is now in an inertial reference frame. Let V denote the final relative speed of these two clocks.
Is it the case that the clocks now tick at the same rate? If they now tick at the same rate then the difference in the readings "X-Y" is again constant, and its value is 600 ticks. But what does SR say about this?
We are still assuming that the time dilation formula is a true statement. Thus, an observer at rest with respect to clock B, can formulate statements about the rate of clock A using the time dilation formula. He will use the following formula:
[tex] \Delta t^\prime = \frac{\Delta t}{\sqrt{1-V^2/c^2}} [/tex]
where delta t` is an amount of time measured by clock B, and delta t is the corresponding amount of time measured by clock A. Only if SR is wrong, are these two quantities equal. Since we are assuming SR is correct, delta t` is not equal to delta t, and instead the two quantities are related through the above formula.
Hence, by SR, clock A is now ticking slower than clock B by a factor of gamma. Thus, if an observer at rest with respect to clock B waits a sufficient length of time in his frame, the two clocks should again eventually fall into sync. In other words, an observer stationed at clock B should KNOW that X>Y but that clock A which reads X is now ticking slower than clock B which reads Y, so that at some point in the future, we will have X=Y.
The problem is now mentally visible. An observer stationed at clock A, will always use the following formula to understand how the rate of clock B relates to the rate of clock A:
[tex] \Delta t = \frac{\Delta t^\prime}{\sqrt{1-v^2/c^2}} [/tex]
Where delta t is an amount of time measured by clock A, and delta t` is an amount of time measured by clock B. Thus, by SR, clock B always ticks slower than clock A. Hence, once the reading of clock B is less than the reading of clock A, it will never again be equal to the reading of clock A. Hence, there is no moment in the future, at which X=Y. Thus, we have reached the following contradiction:
After the moment at which clock A reads 1000, and clock B reads 400, there will come a moment in time at which X=Y, and after the moment in time at which clock A reads 1000, and clock B reads 400, there won't come a moment in time at which X=Y.
Logical analysis ends here.
SR is overthrown
I welcome any challengers to this perfect line of reasoning.
Therefore, using Newtonian mechanics or SR, the clocks tick at the same rate. For the sake of simplicity, let the two clocks be synchroninzed. Thus, if X is the reading on one clock, and Y is the simultaneous reading on the other clock, then x-y=0. As long as the clocks remain at rest with respect to each other, the clocks will remain in-sync.
Suppose that clock B is inside a uniformly accelerating ship. Thus, clock B is being subjected to an OUTSIDE force of a constant value. As this happens, the relative speed v between the two clocks will continuously increase. While clock B is accelerating relative to clock A, clock B is no longer in an inertial reference frame. Let us stipulate that clock A is always in an inertial reference frame. Therefore, whilst clock B is accelerating, an observer stationed at clock A can use the time dilation formula of SR, to draw conclusions about whether or not clock B remains in-sync with clock A.
The time dilation formula is:
[tex] \Delta t = \frac{\Delta t^\prime}{\sqrt{1-v^2/c^2}} [/tex]
[tex] \Delta t [/tex] is an amount of time measured by clock A, and [tex] \Delta t^\prime [/tex] is the corresponding amount of time measured by clock B. In Newtonian mechanics, these two quantities remain equal as clock B accelerates, in SR the quantities are not equal, and instead related by the time dilation formula given above.
As you can see, the conclusion arrived at using the time dilation formula, is that the clocks are no longer in-sync once clock B begins to accelerate. In fact, clock B begins to tick SLOWER. That means that if I am located at clock A, and my current reading is 5525, then I can be certain that clock B simultaneously reads something less than 5525 (even if I don't know exactly what it reads). In other words, I can be certain that X>Y, where X is the current reading on clock A, and Y is the simultaneous reading on clock B.
Now, let some amount of time have passed, and let it be the case that we have:
X = 1000 and Y = 400
Thus, the difference in the readings of the two clocks is now
X-Y= 1000-400 = 600
At the moment that clock B reads 400, let it be the case that the engines shut off, so that clock B is now in an inertial reference frame. Let V denote the final relative speed of these two clocks.
Is it the case that the clocks now tick at the same rate? If they now tick at the same rate then the difference in the readings "X-Y" is again constant, and its value is 600 ticks. But what does SR say about this?
We are still assuming that the time dilation formula is a true statement. Thus, an observer at rest with respect to clock B, can formulate statements about the rate of clock A using the time dilation formula. He will use the following formula:
[tex] \Delta t^\prime = \frac{\Delta t}{\sqrt{1-V^2/c^2}} [/tex]
where delta t` is an amount of time measured by clock B, and delta t is the corresponding amount of time measured by clock A. Only if SR is wrong, are these two quantities equal. Since we are assuming SR is correct, delta t` is not equal to delta t, and instead the two quantities are related through the above formula.
Hence, by SR, clock A is now ticking slower than clock B by a factor of gamma. Thus, if an observer at rest with respect to clock B waits a sufficient length of time in his frame, the two clocks should again eventually fall into sync. In other words, an observer stationed at clock B should KNOW that X>Y but that clock A which reads X is now ticking slower than clock B which reads Y, so that at some point in the future, we will have X=Y.
The problem is now mentally visible. An observer stationed at clock A, will always use the following formula to understand how the rate of clock B relates to the rate of clock A:
[tex] \Delta t = \frac{\Delta t^\prime}{\sqrt{1-v^2/c^2}} [/tex]
Where delta t is an amount of time measured by clock A, and delta t` is an amount of time measured by clock B. Thus, by SR, clock B always ticks slower than clock A. Hence, once the reading of clock B is less than the reading of clock A, it will never again be equal to the reading of clock A. Hence, there is no moment in the future, at which X=Y. Thus, we have reached the following contradiction:
After the moment at which clock A reads 1000, and clock B reads 400, there will come a moment in time at which X=Y, and after the moment in time at which clock A reads 1000, and clock B reads 400, there won't come a moment in time at which X=Y.
Logical analysis ends here.
SR is overthrown
I welcome any challengers to this perfect line of reasoning.