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I don't really understand what the primed and unprimed coordinates imply in Lorentz transformations. In all of the books, they simply derive the coordinates, assuming that an inertial frame $ S$ is at rest and $S'$ is moving at a velocity of $v$ w.r.t $S$.

Could someone please explain if the following statements are correct:

  1. $t'$ is the time that $S$ will observe in $S'$, i.e., if an observer in $S$ could see the clocks of $S'$, she will see them tell a time $t'$ when her clocks show $t$.
  2. $t'$ is the time $S'$ will see in his own clocks, and $t$ is the time he will see in clocks of $S$ when his own clock reads $t'$.
  3. If we view the situation from $S'$'s point of view, he will still observe a time $t'$ in his own clocks, and $t$ in S's clocks, only the relation between $t$ and $t'$ will change.
  4. The relation between $t$ and $t'$ will remain the same even if we look at the situation from the POV of $S'$ frame, only the variables $t$ and $t'$ will be interchanged.

I believe that (1) and (2) should both be correct, for $S$ and $S'$ should agree on the time in each other's frames, and also because we have only two variables in the Lorentz transformation, so if we can't really find the time that $S$ sees in $S'$. All in all, it should imply that what $S'$ sees in $S$ is the same as what $S'$ sees in itself, which is confusing me, since then what is loss of simultaneity even about?

I believe then it should be that whenever we talk of $S$ seeing/observing something in $S'$, we are actually talking about when she sees the events happen in her clocks, but that should mean she sees no difference between $S$ and $S'$! Help!

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2 Answers 2

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First of all, you maybe know this, but it is not actually about seeing, since light takes time to travel. More precise would be to say that if you have two events A and B, then the time passed between these two events will be measured by S to be t and by S' to be t'. If you take these events to be

A = observer in S' started his watches

B = observer in S' stopped his watches

then t will be the time that S measured between these two events and t' will be the time that S' measured between these two events (which is the time that S' watches show at B).

This situation is equivalent to your first point if clocks at both S and S' show zero at the event A and if both of them are at the same place at A. However, at the event B, they are no longer at the same place and relativity of simultaneity kicks in. Since S does not go through event A (i.e. its worldline does not intersect A), two more events come to the picture:

C = event on the worldline of S, that is simultaneous with B according to S

D = event on the worldline of S, that is simultaneous with B according to S'

Thanks to relativity of simultaneity C is not the same event as D. And here is the answer to your problem I believe:

What time S sees on his own clocks when S' clocks show time t' is time that S clocks show at the event C

What time S' sees on clocks of S, when S' clocks show time t' is time that S clocks show at the event D.

If observer at S stopped his clocks as soon as he has seen observer S' stopped his, observer S will not agree with S' that these two events happened at the same time.

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  • $\begingroup$ Thanks for your answer. David Morin in his book on STR writes that Lorentz transformations are related to two events, but I think they are used to describe the coordinates of the same event in different inertial frames. Can you tell what is it exactly, and how do the transformations modify if the origins coincide at some $t_0$ instead of 0? $\endgroup$
    – Math boi
    Aug 12, 2022 at 18:25
  • $\begingroup$ @Mathboi Assuming the origin of both frames is the same (called O), the coordinates of the event A are distances and time measured between events A and O according to given frame. Lorentz transformations just relate one set of measurements to another, i.e. if I know measurement of one observer and I know how another observer moves with respect to the first one, I can compute what the second observer measured between these two events A and O. I don't what David Morin meant by "related to two events", maybe both of you are right. $\endgroup$
    – Umaxo
    Aug 12, 2022 at 18:34
  • $\begingroup$ @Mathboi just change $t$ to $(t-t_0)$ in the formula. As I said, the coordinates are measurements between A and O. If S clocks were showing $t_O$ at O and $t_A$ at A, then the time between A and O was measured by S to be $(t_A-t_O)$ $\endgroup$
    – Umaxo
    Aug 12, 2022 at 18:38
  • $\begingroup$ Just $t$ or $t'$ also? $\endgroup$
    – Math boi
    Aug 12, 2022 at 18:45
  • $\begingroup$ @Mathboi What do you think? It should be possible to piece this together from what I wrote $\endgroup$
    – Umaxo
    Aug 12, 2022 at 18:46
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The meaning of t is the elapsed time between two events in one frame, while t' is the elapsed time between the same two events in another frame moving relative to the first.

For example, there might be two events, A and B, that are five seconds apart in the frame of S, who is in a rocket, and three seconds apart in the frame of S' who is on the Earth.

Your statements 1) and 2) are correct, in the sense that if S and S' each make a note of the times of events A and B using their own clocks, and compare notes afterwards, S will say the elapsed time was five seconds and S' will say the elapsed time was three seconds.

Now, if the two observers switched labels, so that t becomes t' and t' becomes t, then it will not change anything, other than the labelling. The elapsed times between the two events A and B will still be 5 seconds on the rocket and 3 seconds on the Earth.

What you need to remember about the relativity of simultaneity is that two frames moving relative to each other do not share a common time axis. Their time axes are tilted relative to each other. That means that a flat plane of constant time in one frame will be a sloping slice through time in another. Two events that occur at the same time in one frame will, generally speaking, occur at two different times in the other frame. To put it another way, if you synchronise all the clocks in one frame, they will all seem out of synch in the other frame, and vice versa.

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