Understanding Time Dilation in Frames of Reference: A Basic Question

[abcd2357]

Suppose we have two frames of reference, frame A and frame B, which move past each other with a velocity such that $$\gamma \equiv 2$$. In frame A is clock A and in frame B is clock B.

In frame A, clock A is at rest and clock B is speeding past. As a result of time dilation, when an observer in frame A measures clock A as having an interval of $$5 s$$, this observer measures clock B as having an interval of $$5 / 2 = 2.5 s$$. Furthermore, an observer in frame B measures clock B as having an interval of $$2.5 s$$.

In frame B, clock B is at rest and clock A is speeding past. As a result of time dilation, when an observer in frame B measures clock B as having an interval of $$2.5 s$$, this observer measures clock A as having an interval of $$2.5 / 2 = 1.25 s$$. Furthermore, an observer in frame A measures clock A as having an interval of $$1.25 s$$.

This seems to imply that when an observer in frame B measures clock B as having an interval of $$2.5 s$$, an observer in frame A measures clock A as having an interval of both $$5 s$$ and $$1.25 s$$.

Where did I go wrong?

 

[Passionflower]

Suppose we have two frames of reference, frame A and frame B, which move past each other with a velocity such that $$\gamma \equiv 2$$. In frame A is clock A and in frame B is clock B.

In frame A, clock A is at rest and clock B is speeding past. As a result of time dilation, when an observer in frame A measures clock A as having an interval of $$5 s$$, this observer measures clock B as having an interval of $$5 / 2 = 2.5 s$$. Furthermore, an observer in frame B measures clock B as having an interval of $$2.5 s$$.

In frame B, clock B is at rest and clock A is speeding past. As a result of time dilation, when an observer in frame B measures clock B as having an interval of $$2.5 s$$, this observer measures clock A as having an interval of $$2.5 / 2 = 1.25 s$$. Furthermore, an observer in frame A measures clock A as having an interval of $$1.25 s$$.

This seems to imply that when an observer in frame B measures clock B as having an interval of $$2.5 s$$, an observer in frame A measures clock A as having an interval of both $$5 s$$ and $$1.25 s$$.

Where did I go wrong?

Is this homework, it seems like a 'trick' setup? The chosen intervals are completely arbitrary and not relevant to any physics, but they were chosen in such a way that they could confuse. Thing to remember is that each frame observes the other frame's clock going slow by the same factor.

 

[yuiop]

Suppose we have two frames of reference, frame A and frame B, which move past each other with a velocity such that $$\gamma \equiv 2$$. In frame A is clock A and in frame B is clock B.

In frame A, clock A is at rest and clock B is speeding past. As a result of time dilation, when an observer in frame A measures clock A as having an interval of $$5 s$$, this observer measures clock B as having an interval of $$5 / 2 = 2.5 s$$. Furthermore, an observer in frame B measures clock B as having an interval of $$2.5 s$$.

In frame B, clock B is at rest and clock A is speeding past. As a result of time dilation, when an observer in frame B measures clock B as having an interval of $$2.5 s$$, this observer measures clock A as having an interval of $$2.5 / 2 = 1.25 s$$. Furthermore, an observer in frame A measures clock A as having an interval of $$1.25 s$$.

This seems to imply that when an observer in frame B measures clock B as having an interval of $$2.5 s$$, an observer in frame A measures clock A as having an interval of both $$5 s$$ and $$1.25 s$$.

Where did I go wrong?

Try it this way. Say both clocks pass each other at x = x' = 0 and t = t' = 0. The primed coordinates refer to the frame that clock B is at rest in. Work out what the required velocity v should be, to get the required gamma factor and then work out how far B travels in 5 seconds to obtain x when t=5. You will now have values for v, x and t to plug into the Lorentz transformation equations to obtain what the time t' is according to b. You can then use the reverse transformation to see how things look from B's frame. If you have not used the Lorentz transformations before, this is a good place to start. Drawing two space-time diagrams from the point of view of each observer, might help you see what is happening too.

Briefly, the clocks are not at the same place when they are compared and when B says his clock reading of 2.5s is simultaneous with A's clock reading of 1.25s, the observer in A will not agree that those events are simultaneous in his frame. When the observer in frame A says his clock reads 5 seconds simultaneously with clock B reading 2.5 seconds, the observer in frame B will not agree that those events are simultaneous in his frame.

 

[abcd2357]

So $$\gamma \equiv 2 \implies v = \frac{\sqrt{3}}{2} c$$.

We have: $$t' = \gamma (t - \frac{x v}{c^2})$$ and $$t = \gamma (t' + \frac{x' v}{c^2})$$.

It follows that $$t' = 2 (5 - \frac{\frac{\sqrt{3}}{2} c 5 \frac{\sqrt{3}}{2} c}{c^2}) = 2.5$$. But then going in reverse, $$t = 2 (2.5 + \frac{- \frac{\sqrt{3}}{2} c 2.5 \frac{\sqrt{3}}{2} c}{c^2}) = 1.25$$.

What's wrong with that?

 

[Mentz114]

I think v=1/sqrt(3).

You've got v/c > 1, haven't you ?

 

[abcd2357]

Sorry, there were some serious typos/latex issues in that last post. Gamma should have been 2, not 3. Now v=(sqrt(3)/2)*c, which gives v/c < 1. I edited the post, so the issues should now be corrected.