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In my lecture notes, there are two frames S and S'. The prime frame moves with uniform velocity with respect to the unprimed frame. In this frame, she derives the time dilation equation in the following way:

She assigns $t$ as time in S and distance as $x$. Now from inverse transformations:

$$t_1= \gamma (t_1' + \frac{vx_1'}{c^2})$$

$$t_2 = \gamma (t_2' + \frac{vx_1'}{c^2})$$

Now, if the both the positions are the same, how is S' moving?

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  • $\begingroup$ What are $t_1$ and $t_2$ exactly? You never said what is being measured and what is being compared between the 2 frames $\endgroup$
    – Hugo V
    Commented Dec 23, 2018 at 4:14
  • $\begingroup$ @HugoV $t_1$ and $t_2$ are two different times of s frame. $\endgroup$
    – Nobody recognizeable
    Commented Dec 23, 2018 at 4:15
  • $\begingroup$ Ok, but your question is confusing. What do you mean by “Now, if the both the positions are the same, how is S' moving“? $\endgroup$
    – Hugo V
    Commented Dec 23, 2018 at 4:18
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    $\begingroup$ But the position is changing, what isn’t changing is $x’_1$, which is a point in S’. It’s like you are in a car that is moving relative to ground, an observer on the ground is S and you inside the car is S’. Any point inside the car ($x’_1$) isn’t changing position relative to the car, even if the car is moving. So in the car frame $x’_1$ remains the same, while for the observer on the ground, using the S frame, this point ($x_1$) wil be changing position with time. $\endgroup$
    – Hugo V
    Commented Dec 23, 2018 at 5:02
  • $\begingroup$ @hugo you should probably write that in answer. $\endgroup$
    – Nobody recognizeable
    Commented Dec 23, 2018 at 5:04

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The coordinates in the primed frame is what $S'$ measures in his or her coordinate system. So what does $S'$ see if they are holding a clock?

Well, as far as $S'$ knows it is not moving. It just sees a clock a clock go from, say, 1 second to 2 seconds, while it has remained in the same place. So while the time coordinates have changed by one second in the primed frame (i.e. $\Delta t' = 1$ s) the spatial coordinates of the clock have not changed (i.e. $\Delta x' = 0; x_2' = x_1'$).

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