Special relativity clarifiyng time dilation experiment

Question

There is a well known 2 mirror clock time dilation thought experiment in STR. However it is a bit unclear to me due to its unrealistic nature. By unrealistic I mean that experiment implies that stationary observer can immediately "see" the photon trace from side without photon actually hitting the detector (eye). As far as I know you cant see photon from side without it being absorbed by detector.

So I made a variation of this experiment, your answers to which, I hope would clarify some things about STR (for me at least).

The rocket with two photon emitters S1, S2 is moving at constant speed V (close to light speed) relative to R1 photon receiver which we will consider "stationary". Outside the rocket the only source of EMR are S1, S2 photon emitters.

The questions are:

• Will photons emitted by S1 and S2 (we consider distance between them is 0) be received at R1 simultaneously from R1's reference frame.

• Does it matter what kind of clocks I use for S2 emitter?

• Suppose I use same type of photon clock but horizontal (relative to movement direction), will S1 and S2 photons arrive simultaneously from R1 frame?

• Does STR imply that all kind of clocks work the same, so if I'm viewing all possible clock types from reference frame A and they agree between themself, they all must agree when I will be viewing them from frame B, or any other frame.

Answer 1

The difficult thing to overcome when first approaching special relativity is the idea of a "True" time.

The clock in S2 will be effected just as much as the results of the light bouncing between two mirrors

You won't have a difference between when the S1 and S2 emit light because, to an observer on the ship moving at a constant velocity, they are both clocks measuring the same thing. To an observer at R1 the signals from S1 and S2 will also be identical, neither will agree with a clock that was in R1's frame of reference.

Answer 2

It gets confusing when comparing light clocks to mechanical clocks, etc. I hope this helps.

As you see above, Jack is at rest, meaning he has neither accelerated no decelerated, and thus has not changed his frame of reference. Jack uses 2 synchronized clocks to measure how much slower Jill's clock is ticking. The results are also noted above.

Now imagine that Jill's rocket is very long and that there is also a clock located at the rear end of the rocket. The distance between these clocks is the same as the distance between Jack's clocks.

Jill takes off to the left, stops, and then accelerates to the same speed previously noted, and heads back to pass by Jack. From Jill's perspective, both her clocks still seem to be synchronized. However, from Jack's perspective, he sees that Jill's rear clock is ahead of the one at the front.

Now if Jill uses her two clocks to measure how fast one of Jack's clocks is ticking, the clock time offset between her two clocks will in turn have her measurements indicate that Jack's clock is ticking slower than her clocks, and do so even though Jack has not altered his frame of reference.

With this being the case, dozens of rockets that are identical to Jill's rocket, could have left jack and then returned and passed by him, but all do so at different speeds. They all, when measuring one of Jack's clocks, would in turn get different results. But there is only one Jack, not multiple Jack's.

Answer 3

There is a well known 2 mirror clock time dilation thought experiment in STR. However it is a bit unclear to me due to its unrealistic nature. By unrealistic I mean that experiment implies that stationary observer can immediately "see" the photon trace from side without photon actually hitting the detector (eye). As far as I know you cant see photon from side without it being absorbed by detector.

I am not sure I have understood you well, but in SR an observer does not look at the moving clock (light clock) from side. Since you call it "unrealistic" I'd try to shed some light:

In special relativity an observer is not a single person or human being, who sits on a bench and observes moving clocks. Stationary observer is the whole reference frame, or the team of observers. These observers stay apart from each other at certain distance and each has a clock. These clocks have been previously synchronized by Einstein signalling method (there is also Reichenbah's, which is self - consistent). They record space and time coordinates of events in immediate vicinity, or straight in front of each of them.

https://en.wikipedia.org/wiki/Observer_(special_relativity)

A stationary observer of reference frame $S$ conducts measurements this way: He places light clock $C_1$ at coordinate $x_1$ of his frame and light clock $C_2$ at coordinate $x_2$ of his frame.

Let these light clocks have digital oscillation counters.

Then this observer sends a beam of light from clock $C_1$ towards clock $C_2$ when clock $C_1$ shows 0. He assumes, that one - way speed of light is c (Einstein synchrony convention). Since he knows distance and speed of light, he synchronizes these clocks, so as they show the same time in reference frame $S$.

https://en.wikipedia.org/wiki/Einstein_synchronisation

Imagine light clock $C'$ passes by clock $C_1$ first at moment of time $t_1$ and clock $C_2$ at moment of time $t_2$ then. At these moments, readings of the moving clock and the corresponding fixed clock of reference frame $S$ next to it are compared.

Let the counters of moving clock measure the time interval $\tau _ {0}$ during the movement from the point $x_ {1}$ to the point $x_ {2}$ and the counters of clocks $C_1$ and $C_2$ of the fixed or “rest” frame $S$, will measure the time interval $\tau$. This way,

$$\tau '=\tau _{0} =t'_{2} -t'_{1},$$

$$\tau =t_{2} -t_{1} \quad (1)$$

But according to the inverse Lorentz transformations we have

$$t_{2} -t_{1} ={(t'_{2} -t'_{1} )+{v\over c^{2} } (x'_{2} -x'_{1} )\over \sqrt{1-v^{2} /c^{2} } } \quad (2)$$

Substituting (1) into (2) and noting that the moving clock is always at the same point in the moving reference frame $S'$, that is,

$$x'_{1} =x'_{2} \quad (3)$$

We obtain

$$\tau ={\tau _{0} \over \sqrt{1-v^{2} /c^{2} } } ,\qquad (t_{0} =\tau ') \quad (4) $$

This formula means that the time interval measured by the fixed clocks is greater than the time interval measured by the single moving clock. This means that the moving clock lags behind the fixed ones, that is, it slows down.

The animation below depicts light clocks.

Sure, from the point of view of "stationary" observer $S$ beam of light moves by hypotenuse, though in "moving" frame $S'$ beam of light jumps up and down.

Good link: https://arxiv.org/abs/physics/0512013