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I've just started special relativity and my textbook says $\Delta t_{stationary} \gt \Delta t_{moving}$. It proves this with mirrors placed perpendicular to the motion as shown: enter image description here

In the above figure, two mirrors $A, B$ are inside a bus moving with velocity $v$ relative to the stationary frame.
In the moving frame(inside bus), $\Delta x = 2L$.
In the stationary frame, $\Delta x = 2D$.
Clearly the light travels more distance in the stationary frame and thus takes more time. Good so far.

Now, what happens if I put the mirror horizontally and shoot a light ray from the right side mirror as shown: enter image description here

Clearly the light travels less distance in the stationary frame and thus takes less time. This means $\Delta t_{stationary} \lt \Delta t_{moving}$. Contradiction. What am I missing?

AgentS
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  • Based on your first image, doesn't the light travels a greater distance in the moving frame? – aRockStr Dec 25 '19 at 15:03
  • How @aRockStr ? the bus is moving to the right. In moving frame the light just travels the distance $2L$. But in stationary frame it has to travel more distance.. – AgentS Dec 25 '19 at 15:05
  • The distance in the stationary frame is simply $\Delta x_{stationary}=2L,$ while the distance in the moving frame is $\Delta x_{moving}=2D.$ Since $D$ is the hypotenuse of a right-angle triangle, one side of which has length $L$, we see that $D>L,$ so $\Delta x_{moving}>\Delta x_{stationary}.$ – aRockStr Dec 25 '19 at 15:13
  • @aRockStr No. Let me give the context. The bus is moving with velocity $v$ relative to the stationary frame. In the moving frame(inside the bus), the distance between mirrors doesn't change. So $\Delta x_{moving}=2L$. But in stationary frame, the distance between mirrors changes. – AgentS Dec 25 '19 at 15:17
  • Have you taken length contraction into account? – BioPhysicist Dec 25 '19 at 15:19
  • @AgentS Ah excuse me - I think I've misunderstood which frame you've been calling the moving and which the stationary. – aRockStr Dec 25 '19 at 15:19
  • @AaronStevens haven't covered lorentz transformations yet.. but I know somewhat... please do use them to explain :) – AgentS Dec 25 '19 at 15:21
  • @aRockStr No worries. I haven't given the full context in my question as I thought it makes question look too lengthy. My mistake really. Thank you so much:) – AgentS Dec 25 '19 at 15:22
  • @aRockStr I've updated the question... hope it is clear now:) – AgentS Dec 25 '19 at 15:26
  • It might be clearer to label the frames as "ground" and "train". Motion is relative, so labeling frames as "stationary" and "moving" can get confusing. – BioPhysicist Dec 25 '19 at 16:28
  • @BenCrowell thanks that says time dilation occurs but it doesn't address the contradiction I gave: when light goes from bottom mirror to top mirror it is $\Delta t_{stationray} \gt \Delta t_{moving}$, and when light goes from top mirror to the bottom it looks $\Delta t_{stationray} \lt \Delta t_{moving}$, – AgentS Dec 25 '19 at 20:55
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    The relativity of simultaneity really ruins this thought experiment: if the 2 frame agree on the emission time, their view of the time of the reflection is completely different, even at $v<<c$ as long as the mirror is far enough away. – JEB Dec 26 '19 at 05:00
  • @JEB I see that now.. but I don't get how the relativity of simultaneity doesn't ruin the original thought experiment of full photon clock cycle...There also both the frames agree on spacetime only at the initial emission right? – AgentS Dec 26 '19 at 05:03

2 Answers2

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Your missing the second part of the the cycle.

Yes, the first half is shorter. But the light then bounces and returns back to the initial end, which by that time has moved further away. And since it’s had longer to move, the slow “tock” is a bigger effect than the short “tick” time: the net cycle time of the clock has lengthened and it runs slower.

So how can an observer moving with the clock see equal tick/tock periods? This gets to the difference in simultaneity between the two frames.

In the rest frame, you measure the time the light hits each end with two spatially separated clocks, then compare those to compute that the two directions take the same amount of time. The moving observer does the same thing and also sees equal times! How can that be? This happens because rest observer sees that the moving observers two separated clocks are not synchronized, with one reading ahead of the other.

Bob Jacobsen
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  • Hey I get the overall clock runs slower. But Im still a bit confused why we can't consider only the first half as a full event. All events must run slow right? – AgentS Dec 26 '19 at 04:39
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    This gets to the difference in simultaneity between the two frames. The moving observer and the stationary observer can’t agree on the time at two different points. – Bob Jacobsen Dec 26 '19 at 04:42
  • Ohk so for now, the time dilation proof only proves that the photon clock in the moving frame runs slower relative to the stationary frame. Not necessarily the events that are smaller than the period of the photon clock. Am I getting this correctly? – AgentS Dec 26 '19 at 04:44
  • Because the event of first half "tick" takes more time in the moving frame compared to the stationary frame.. – AgentS Dec 26 '19 at 04:48
  • Oh the period of the photon clock can be made arbitrarily small XD Does this actually prove all events run slower? – AgentS Dec 26 '19 at 04:50
  • Btw viewing the solar eclipse here in India is awesome! – AgentS Dec 26 '19 at 04:51
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    In the rest frame, you measure the time the light hits each end with two spatially _separated clocks, then compare those to see the two directions take the same time. The moving observer does the same thing and also sees equal times! This rest observer sees the moving observers clocks as not synchronized, one being ahead of the other. – Bob Jacobsen Dec 26 '19 at 04:53
  • I now totally get how this is related to simultaneity. I haven't gotten till that in my textbook but it seems the right side clocks run slower than the left side clocks! Thank you so much really appreciate it:) – AgentS Dec 26 '19 at 04:57
  • Added a bit to the answer to cover that. – Bob Jacobsen Dec 26 '19 at 06:12
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Your confusion comes from the fact that moving and stationary are relative terms. And that you are referring to distances inside the bus (moving as you say), as seen from inside the bus (from the rest frame of the bus).

Now let's say that the bus is the moving frame (relative to Earth ground) and there is a stationary frame (relative to Earth ground).

What you are trying here is a mirror clock. One tick is when a photon bounces between the two mirrors. Clearly, if the photon has to travel a longer distance between the mirrors, one tick will take more time. This means the clock ticks slower, time passes slower, but relative to what?

It is very important to understand that you can only talk about the relative passage of time, between two clocks. Let's say there is a photon clock on the moving bus (relative to Earth ground), and there is a stationary clock (relative to Earth ground).

In the moving bus, as seen from the stationary frame (this is what you confuse), a photon has to travel longer distance, 2D, thus a tick will take more time, thus time will pass slower in the moving bus as seen from the stationary frame. What we say is that in the bus, the photon clock ticks slower relative to the stationary clock.

But people (in the bus, comoving with the bus) looking at the clock in the bus, will see it tick normally.

It is only when they compare the clock to another clock (the stationary one) that they will see it tick differently. What will they see, which one ticks slower? People in the bus, comoving with the clock in the bus, will see the other clock (stationary clock) tick slower. Why?

Because speed is symmetrically relative. There is no universal reference frame. You can only talk about time dilation relative to another clock. This is SR.

Time dilation is a difference in the elapsed time measured by two clocks, either due to them having a velocity relative to each other, or by there being a gravitational potential difference between their locations. After compensating for varying signal delays due to the changing distance between an observer and a moving clock (i.e. Doppler effect), the observer will measure the moving clock as ticking slower than a clock that is at rest in the observer's own reference frame.

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

Now you are asking about when the mirrors are moving parallel to the photon's direction of propagation and the answer is the same.

The right light clock ticks slower because half of the time, its right mirror is moving away from the photon, which takes longer to catch up with it.

What if a light clock travels perpendicular to mirrors that make up the clock?

Árpád Szendrei
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