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Let's take an example, where A is a photon moving with the speed of light, and B is a car, that moves with half of the speed of light

At t=0, a kid in B(thinking that the car is not moving) observes that B is 4.5 million kms away from A(initial position of the photon=4.5 million kms), and both start to move towards each other. After 1 second on the kid's watch, he observes that he and the light are in contact, so that would mean the distance between them is 0 kilometres, so he would consider the final position of the photon=0km

So if he calculates the speed of the photon, by using: (initial position-final position)/time taken, i.e, 1 second on his watch

Won't he get (4.5*10^6)/1 =4.5*10^6km/second?

What else needs to be considered in this calculation?

Edit: Most of you must be downvoting because I haven't considered the time dilation. But the time dilation again is built on the fact that the speed of light is constant, isn't it?

Swaroop Joshi
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  • I'm sorry the kid should be in B – Swaroop Joshi Feb 18 '20 at 13:01
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    The axiom that the speed of light is the same in all frames is more-than-well established by comparison with experiment. From a theoretical point of view, it is an axiom, taken as given. From an experimental point of view, it is, by now, a fact. – garyp Feb 18 '20 at 13:04
  • @garyp I'm asking about what else it needs to be considered. I think I should consider the fact that it takes some time for the light wave to reach the kid's eye, hence in his view, the photon would be far behind, from where it actually is, at that point, in time. – Swaroop Joshi Feb 18 '20 at 13:10
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    Ok, but you still haven't explained how the kid in the car at B can see the photon at A, 4.5 million km away. Maybe make A a light source. You also need a way to synchronize the light flash at A happening simultaneously with the car at B starting its journey. BTW, the speed of light is (almost) 300,000 km/s. – PM 2Ring Feb 18 '20 at 13:34
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    This question is weak. It seems as if you might be asking about the transformation between two different coordinate frames (e.g., the frame of the "kid in the car," vs. the frame of an un-named observer who sees the car moving at high speed.) If this is the case, then it would be helpful if, in each case where you describe an event (e.g, the emission of the photon, the detection of the photon) you would explicitly say, in which frame you are describing it and, if you could then explicitly ask how the spacetime coordinates of some event in one frame are transformed to the other frame. – Solomon Slow Feb 18 '20 at 14:22
  • You are thinking classically. That is your problem – BioPhysicist Feb 18 '20 at 18:30

3 Answers3

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This is how the situations looks for us, static external observers – we see a car and a photon separated by $1.5c\cdot 1$s and travelling with $v_{ph}=-c$ and $v_{car}=0.5c$ velocities. After a second they meet.

However, for an observer in the car the situation is completely different: except for $v_{ph}$, every quantity mentioned above is different! We intuitively know that in this frame $v_{car}$ differs (as you observed it's 0), but we are not used to the fact that time and distance also differ between observers.

You can calculate them using appropriate Lorentz boost.

Paweł Czyż
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  • You mean the time dilation? But that again is defined, using the fact that the speed of light in all frames, is the same, isn't it? – Swaroop Joshi Feb 18 '20 at 12:47
  • Time dilation in the math of special relativity can be shown as a consequence of the constancy of the speed of light to all inertial observers yes. – Charlie Feb 18 '20 at 14:15
  • @Charlie so I should even consider length contraction? But again, don't you think this case is unique? Because we showed that time dilation occurs by assuming that the speed of light would be constant, which, to prove, again we need to assume that time dilation exists. – Swaroop Joshi Feb 26 '20 at 14:15
  • I'd recommend reading a book or lecture notes on special relativity and revisiting your question once you have. – Charlie Feb 26 '20 at 14:29
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After 1 second on the kid's watch, he observes that he and the light are in contact, so that would mean the distance between them is 0 kilometres, so he would consider the final position of the photon=0km

Well, that cannot be true. You are arbitrarily taking a value as its speed. There have been so many experiments and measurements proving that the speed of light has an upper bound and it is nearly $3 \times 10^8$ meters per second in vacuum and is slower in other media by their respective refractive indices.

You can try some of them at home to convince yourself.

There are so many successful results derived from taking speed of light as a constant as can be seen from experimental basis of special relativity and tests of special relativity. We are well justified in considering it as an axiom.

exp ikx
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and both start to move towards each other.

If a kid started moving after determining the distance from the photon, then he changed his reference frame. He should measure the distance from he photon again, because it changes due to Lorentz contraction. In your case, the newly measured distance would be exactly one light second, which would lead to no observable change in the light speed.

What is important for measuring the speed is that the time difference and distance need to be measured in the same reference frame.

Adam Latosiński
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