I've always believed the classic derivation you see for time dilation in a intro physics course, however I also have always had the uncomfortable feeling that some things were getting brushed under the rug.

To recap, the classic derivation is as follows:

Suppose you have a light clock which measures time by reflecting light back and forth on a mirror. In the frame of the clock, the timing proceeds as you would expect, such that one time tick $\Delta t = \frac{2L}{c}$ for L the height of the clock, and c the speed of light. In the frame of some outside observer such that the clock is moving at some speed $\beta$ (in natural units) perpendicular to its height, you can use pythagoras' Theorem to find that $\Delta t' = \frac{1}{\sqrt{1-\beta^2}} \times \frac{2L}{c}$. Frame of the clock and frame of separate observer

Okay, so in some ways this is a nice first derivation because it doesn't necessarily assume a bunch of different invariances. For example, Newton would like to say that the timestep size is invariant, but that's not assumed. Newton would also like to say that all lengths are invariant, but again, that's not exactly assumed. However, L is assumed to not change between frames. Why? Is there a general justification for why L must be the same between frames?

I suppose my issue can be summed up as follows: in the clock's frame we can say $\Delta t = \frac{2L}{c}$. In the prime frame lets consider what could conceivably change between frames: $\Delta t$ and $L$. We know that every other variable ($c$ is the only other one) is invariant. Okay, so then $\Delta t' = \frac{1}{\sqrt{1-\beta^2}} \times \frac{2L'}{c}$... but what is $L'$?

  • $\begingroup$ The feeling is natural. Things aren't actually being ignored, but you aren't been told all the details yet. Like many other things in physics you have to learn to use a framework before you can study how it all fits together and investigate it's self-consistency. I write a little bit about one aspect of that in a previous answer covering longitudinal light clocks. The key point is that the structure that develops from relativity matches observation. $\endgroup$ – dmckee --- ex-moderator kitten Aug 13 '19 at 23:58
  • $\begingroup$ That said, there are ways to develop SR that don't rely on saying "don't look behind the curtain yet", but they take more powerful math and a deeper understanding of some basics structures than most students have when they are first "ready" to deal with relativity. $\endgroup$ – dmckee --- ex-moderator kitten Aug 14 '19 at 0:01
  • $\begingroup$ You’re asking about the invariance of transverse lengths. This can be discussed several ways, for example. physics.stackexchange.com/questions/404976/… $\endgroup$ – Bob Jacobsen Aug 14 '19 at 0:51

Let's assume we have another pair of mirrors $\{A',B'\}$, at rest in the primed frame, separated by the same distance as $\{A,B\}$ in the unprimed frame (the primed and unprimed observers can communicate with each other to create the same setup in their respective frame). Thus, $$ L'_{A'B'} = L_{AB}, $$ where $L'_{A'B'}$ is the distance between $A'$ and $B'$, at rest in the primed frame, and $L_{AB}$ is the distance between $A$ and $B$, at rest in the unprimed frame.

Now suppose that motion changes the observed distance between mirrors perpendicular to that motion. Let's assume for example that the observed distance between a pair of moving mirrors is smaller than the distance between stationary mirrors. In the primed frame the pair $\{A,B\}$ is moving, so the primed observer would measure $$ L'_{AB} < L'_{A'B'}.\tag{1} $$ However, in the unprimed frame the situation is symmetrical. In that frame, $\{A',B'\}$ is moving, so the unprimed observer would measure $$ L_{A'B'} < L_{AB}.\tag{2} $$ That leads to a contradiction. We could set these two pairs of mirrors on a collision course. In the primed frame, according to $(1)$ $A$ and $B$ should be able to pass in between $A'$ and $B'$. But in the unprimed frame, according to $(2)$ $A'$ and $B'$ should be able to pass in between $A$ and $B$. Both can't be right at the same time. Therefore, $$ L'_{AB} = L'_{A'B'}\quad\text{and}\quad L_{A'B'} = L_{AB}. $$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.