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I think the solution to the following problem proves the theory of special relativity contains an error. I have solved the problem, and can't find any error in my reasoning. Can anyone show me where my error is, if there is one?

Let us presume that we have two identical digital clocks, clock A, and clock B, at rest with respect to each other, and in an inertial reference frame. Let R1 denote the current reading on clock A, and let R2 denote the current reading on clock B. Let us define D as follows:

D = difference in clock readings = R1 - R2

If R1=R2 then the clocks are synchronous.

Let the clocks currently be synchronous.

Suppose that when the clocks both read zero, that there is a constant force F applied to clock B, for [tex] \Delta t^\prime [/tex] seconds according to clock B. Let [tex] \Delta t [/tex] denote this amount of time according to clock A. Let V denote the final relative speed of the two clocks. Assuming the theory of special relativity is correct, the times are related by the following formula:

[tex] \Delta t = \frac{\Delta t^\prime}{\sqrt{1-V^2/c^2}} [/tex]

At this point, the clocks are now in relative motion, at a constant speed V.

As can be seen from the time dilation formula, the clocks are no longer synchronous. In fact, during the time the force was applied, clock B ticked slower than clock A, so that R1-R2 is now positive rather than negative. If the clocks were still synchronous then R1-R2 would equal zero, however we are operating under the assumption that the time dilation formula is correct.

Consider two ways to bring the clocks to rest with respect to each other again.

Let T denote the amount of time for which the force was applied to clock B as measured by clock B.

Method 1: We can apply force F to clock B for time T as measured by clock B, in the direction of clock A.

Method 2: We can apply force F to clock A for time T as measured by clock A, in the direction of clock B.

Using either method, the clocks will again be at rest with respect to each other.

Here is the problem I have detected.

If we apply the force F to clock A for time T as measured by clock A, then clock A will tick slower than clock B, and after time T has passed according to clock A, the clocks will again be synchronous.

On the other hand, if we apply the force to clock B for time T as measured by clock B, then clock B will tick slower than clock A, and after time T has passed according to clock B, R1 will be greater than R2.

Thus, the final readings on the clocks, depends upon whether or not we apply the force to clock A, or clock B.

Thus, time dilation involves vector dependence on velocity. However, the time dilation formula doesn't involve vector dependence on velocity, v is just relative speed. Thus, there is an error in the theory of special relativity.

If I have made an error, then where is my error located?

I don't see any vector term in the time dilation formula, so my analysis ends in the conclusion that the time dilation formula is erroneous.

Respectfully,

The Star

**Experimental Setup**Let us presume that we have two identical digital clocks, clock A, and clock B, at rest with respect to each other, and in an inertial reference frame. Let R1 denote the current reading on clock A, and let R2 denote the current reading on clock B. Let us define D as follows:

D = difference in clock readings = R1 - R2

If R1=R2 then the clocks are synchronous.

Let the clocks currently be synchronous.

Suppose that when the clocks both read zero, that there is a constant force F applied to clock B, for [tex] \Delta t^\prime [/tex] seconds according to clock B. Let [tex] \Delta t [/tex] denote this amount of time according to clock A. Let V denote the final relative speed of the two clocks. Assuming the theory of special relativity is correct, the times are related by the following formula:

[tex] \Delta t = \frac{\Delta t^\prime}{\sqrt{1-V^2/c^2}} [/tex]

At this point, the clocks are now in relative motion, at a constant speed V.

As can be seen from the time dilation formula, the clocks are no longer synchronous. In fact, during the time the force was applied, clock B ticked slower than clock A, so that R1-R2 is now positive rather than negative. If the clocks were still synchronous then R1-R2 would equal zero, however we are operating under the assumption that the time dilation formula is correct.

Consider two ways to bring the clocks to rest with respect to each other again.

Let T denote the amount of time for which the force was applied to clock B as measured by clock B.

Method 1: We can apply force F to clock B for time T as measured by clock B, in the direction of clock A.

Method 2: We can apply force F to clock A for time T as measured by clock A, in the direction of clock B.

Using either method, the clocks will again be at rest with respect to each other.

Here is the problem I have detected.

If we apply the force F to clock A for time T as measured by clock A, then clock A will tick slower than clock B, and after time T has passed according to clock A, the clocks will again be synchronous.

On the other hand, if we apply the force to clock B for time T as measured by clock B, then clock B will tick slower than clock A, and after time T has passed according to clock B, R1 will be greater than R2.

Thus, the final readings on the clocks, depends upon whether or not we apply the force to clock A, or clock B.

Thus, time dilation involves vector dependence on velocity. However, the time dilation formula doesn't involve vector dependence on velocity, v is just relative speed. Thus, there is an error in the theory of special relativity.

If I have made an error, then where is my error located?

I don't see any vector term in the time dilation formula, so my analysis ends in the conclusion that the time dilation formula is erroneous.

Respectfully,

The Star

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