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If I stand at the middle of a bus traveling at 0.5c, will the front half of the bus appear to me shorter than the back? and, does it make any difference if there's air inside the cabin versus a vacuum?

Please note: simply attaching a reference frame to the bus and call it a day does not cut it! There is a certain distance to the front and back walls, and what the observer sees is a "delayed" version of those objects, based on the time it takes light to travel that distance. During that time, the observer changes position. So the "delayed" version of the front wall should appear closer to the "current" position of the observer. Similarly, the back wall should appear recessed.

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    $\begingroup$ How much SR have you studied? Your added note is saying the exact opposite of what happens, because by standing in the middle of the bus you've chosen that to be your rest frame. $\endgroup$
    – Eletie
    Commented Jul 22, 2022 at 10:34
  • $\begingroup$ @Eletie how does light know what my choice was? What I see is the position where the front wall was some time ago. How can I see its instant position? $\endgroup$
    – Jony
    Commented Jul 22, 2022 at 10:40
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    $\begingroup$ As per SR, the speed of light is invariant for all observers, regardless of whether the light source is moving or not. Have you had a look through the Wiki on SR, it might be useful? en.wikipedia.org/wiki/Special_relativity. If you're familiar with length contraction in SR, then you may wonder why it isn't occurring in the case you described: that's because the observer is at rest with respect to the system (the bus). An observer on the street would see distortions in the length of the bus. $\endgroup$
    – Eletie
    Commented Jul 22, 2022 at 10:44
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    $\begingroup$ "attaching a reference frame to the bus and call it a day does not cut it! There is a certain distance to the front and back walls" --- this should say "In any given reference frame there is a certain distance to the front and back walls". There is one reference frame where those distances are equal. That's the most convenient frame to work in. $\endgroup$
    – WillO
    Commented Jul 22, 2022 at 15:37

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If you are moving with the bus, the bus will appear to you to be at rest- you will not see any relativistic effects whatsoever inside the bus.

It is easy to go wrong when approaching relativity for the first time. Once you do understand it, it all makes sense, but it conflicts with out common sense notions of space and time, and you have be prepared to abandon those if you are to make headway with the subject.

Your question implicitly assumes that there is such a thing as an absolute point in space, or, equivalently, that there is an absolute frame of reference, neither of which are true.

Specifically, let's suppose that as the front of your bus passes me as I stand on the road, the driver flashes a light towards you at the centre of the bus. You seem to think that in that case, the source of the light was the point next to me on the road where the driver was when they flashed the light, and that the driver has since moved on beyond that point and so is no longer at the origin of the light flash. That's not quite true. In SR every reference frame is equally valid for the purpose of calculating the speed of light. In my reference frame, the spatial origin of the flash was indeed the point on the road where the driver was at the time of the flash, but that's not true in your frame of reference. In yours, the spatial origin of the flash was the driver, who remains a fixed distance from you. So, while you, in the moving bus, are moving towards what I consider the origin of the flash, you are stationary relative to the origin of the flash in your frame.

I will give you another example to drive home the point. Suppose you and I are standing together next to a bulb that flashes light which heads off in all directions. If you walk after the light to the right, while I walk after the light to the left, we are each entitled to consider ourselves to be still at the centre of the expanding sphere of light. That might sound entirely contrary to the common sense view that the centre of the sphere must be the place where we were standing when the bulb flashed, but that common sense idea turns out to be just an approximation that is almost exact where low relative speeds are involved but breaks down completely when very high speeds are involved.

I hope you continue to work on SR, as it is conceptually very compelling once you get the hang of it, but if you are like most people (myself included) you will find it very hard to grasp if you keep trying to reconcile it with common sense presuppositions about space and time.

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    $\begingroup$ But the point about SR is that the speed of light is the same to everyone. If you are at the middle of the bus, (ie each end is 600,000km away), the light from each end will reach you at the same time. $\endgroup$ Commented Jul 22, 2022 at 10:45
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    $\begingroup$ Remember that there is no such thing as absolute speed. The bus does not have an absolute speed of 0.5c. Its speed is zero relative to you. Its speed might be 0.99c relative to a bus heading in the other direction, or 10 meters per second relative to another bus slowly overtaking it. You need to spend more time studying the basics of SR. $\endgroup$ Commented Jul 22, 2022 at 10:47
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    $\begingroup$ No, you misunderstand. You do see the end of the bus as it was a second ago, but to you the end of the bus is always 300,000km away. In the intervening second you have not travelled closer to where the end of the bus 'used to be'. To you, the bus is stationary and you are stationary. $\endgroup$ Commented Jul 22, 2022 at 10:56
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    $\begingroup$ @jony, trust me- you need to study SR by reading books, watching videos on youtube etc. An answer to a question of stack exchange can't possibly be a substitute for that. It is rather like asking 'how do you speak French?' and expecting the answer will make you fluent. You have basic misconceptions about spacetime and until you understand the basics you won't get much satisfaction from answers on here. If I get time I will try to extend my answer to cover more of the basics. $\endgroup$ Commented Jul 22, 2022 at 11:18
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    $\begingroup$ @Jony: Also, acceleration complicates things a lot, you'd definitely not want to get in there at the level of an elementary treatment of SR. $\endgroup$ Commented Jul 22, 2022 at 21:08
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The very basic fact is that motion is relative. It is meaningless to say that the bus "travels at 0.5c"; you have to specify – with respect to what? (The Earth? The Sun? The centre of the galaxy? Something else?)

The problem, as currently stated, consists of only two objects – the bus, and the person in the middle of the bus. They are motionless with respect to each other. It is also not true to say that "the observer changes position" – the observer does not change position with respect to the only other object present in the set-up.

Whether or not the whole combination (bus+observer) happens to move with some constant velocity with respect to some third (unspecified) body, does not play a role at all. If the bus happens to move with 0.5c with respect to the Earth, say, we may as well say that the bus is stationary, and the Earth moves with 0.5c (in the opposite direction).

To conclude, the observer will not see a difference looking forward or back, as the observer is not in motion with respect to the bus.

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  • $\begingroup$ In fact, this also holds true in Galilean relativity. If you are in a bus, and the bus has no windows, there is no physical way to determine whether the bus is "moving" (with constant velocity) or "stationary". If there are two people at either end of the bus, throwing balls towards the middle (with equal force), the observer in the middle will always see the balls coming towards him with the same speed. $\endgroup$
    – printf
    Commented Jul 23, 2022 at 15:28
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Attaching a reference from to the bus and calling it a day does indeed cut it.

The bus is at rest, you're in the middle at the origin. If the bus is length $L$, you see the front and back as they were $L/(2c)$ in the past. If the bus has air in it, then the delay is $L/(2c) \times (1 + \eta)$ where:

$$ \eta = 273 \times 10^{-6} $$

Note that the claimed speed of the bus, $c/2$, is not part of the problem (because there is no absolute reference frame).

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