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Suppose Bob is travelling in a spaceship at uniform velocity of 150000000 m/s in Michael's frame of reference.

Michael sees the spaceship length contracted and it appears to be only 86.6 meters long. Michael also notices that the spaceship clock is running slow. For every 1.2 seconds on Michael's clock, the spaceship clock only registers 1 second.

If the observers are moving relative to each other, their "meters" and "seconds" are different. So, if Bob is told by Michael that the spaceship is moving at 150000000 m/s, Bob would not agree on this speed because Bob sees Michael's 'meters' length contracted and 'seconds' dilated. I hope I have it correct. Bob would convert the speed as follows: (150000000*1.15473(/(1*1.2) = 144341250 m/s

Yes, each observer would agree on the speed of light and perhaps that's why in almost every such problem velocities are stated as fractions of 'c' like 0.5c to avoid confusion. But the problem is that I don't understand how Bob and Michael could find such a 'light-operated speedometer' and how it would work, or any other practical method which would let them agree on the speed.

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  • $\begingroup$ Also,there is no way for Michael to "see" Bob's clock. $\endgroup$
    – elias2010
    Commented Apr 24, 2020 at 9:39
  • $\begingroup$ This is a follow-on question to physics.stackexchange.com/q/545489/123208 (and possibly a duplicate). $\endgroup$
    – PM 2Ring
    Commented Apr 25, 2020 at 7:58
  • $\begingroup$ @elias2010 That's a minor technical difficulty. The clocks could broadcast their time signals over radio. $\endgroup$
    – PM 2Ring
    Commented Apr 25, 2020 at 8:00
  • $\begingroup$ The signals will appear an apparent delay due to the redshift effect. $\endgroup$
    – elias2010
    Commented Apr 25, 2020 at 9:56
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    $\begingroup$ @elias2010 That isn't redshift, and the observers can easily account for time delays between signals being sent and received. That's why in SR we talk about the spacetime coordinates of events, and we assume that actual observers are smart enough to correct for time delays due to the finite speed of light. $\endgroup$ Commented Apr 25, 2020 at 15:05

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Bob would not agree that he is moving at all. As long as Bob is moving uniformly(only under which cases special relativity holds), he can say that he is at rest, and it is Michael who is moving at a speed of 150000000 m/s. Assuming Michael is also travelling at a 100m long spaceship, Bob would observe Michael's ship having a length of just 86.6 meters, and also Bob's clock to be ticking at slower(1.2 second in Bob's clock is Michael's 1 second). The situation is symmetric and thus there is no paradox! The laws of physics holds perfectly even if Bob assumes he is rest. That is the crux of Special Relativity. Also velocities are expressed in the form 0.5c, because it makes the calculation of the Lorentz factor easier.

Response to comment: The clock 'ticking' means that if there was any way for Bob to observe Michael's clock, he would see it running slower compared to Bob's own clock. One way to do this would be that Michael sets up a light clock described here and Bob has a camera(or some other apparatus like that) that clicks a photo of Michael's ship at a certain frame-rate. This way while later analysing the photos, Bob would observe the time dilation effect and is more convenient when considering speeds close to that of light.

NOTE ON THE PRACTICAL SIDE OF THINGS: I think this answer would soon get out of context, but I just want to clear up a few things. The situation above is a thought experiment. In no way can humans right now travel at 150000000 m/s. But assuming they could, one way to think about how to observe time dilation would go like this: say there are two muons or cosmic rays, that travel from the top of any frame to bottom of some platform. Let's assume these particles are visible to both Bob and Michael. Now, let's imagine the two particles to be simultaneous for Michael(i.e. happens at the same time). Now since Bob is moving towards one particle and away from another, he sees one of them first and the second a little later. If Bob and Michael both record the time they see the particles hit. Now if they meet again and compare the times they recorded the two particle hitting the platform, the values will be different. This could help them infer the time dilation caused by motion.

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  • $\begingroup$ Neither Bob can be "hearing" Michael's clock.Who could observe the situation?This is not make sense. $\endgroup$
    – elias2010
    Commented Apr 24, 2020 at 18:16
  • $\begingroup$ @elias2010 please see the edited answer. $\endgroup$
    – PNS
    Commented Apr 25, 2020 at 6:36
  • $\begingroup$ The camera use light waves.Any kind of waves could be used,they will appear an apparent delay due to the redshift effect. $\endgroup$
    – elias2010
    Commented Apr 25, 2020 at 10:03
  • $\begingroup$ The point being that it is not important how the clock is observed, but if it can be observed in any possible way, the clock would appear to be slowed down. $\endgroup$
    – PNS
    Commented Apr 25, 2020 at 12:49
  • $\begingroup$ @elias2010 I think you answered your own question. The apparent delay of light is the cause for time dilation, if you consider a constant speed of light, one of the main points in SR. Also, see my edited answer. I hope that clears it up. $\endgroup$
    – PNS
    Commented Apr 25, 2020 at 15:00

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