0
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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 Dec 23 '18 at 4:14
  • $\begingroup$ @HugoV $t_1$ and $t_2$ are two different times of s frame. $\endgroup$ – Nobody recognizeable Dec 23 '18 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 Dec 23 '18 at 4:18
  • 1
    $\begingroup$ 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 Dec 23 '18 at 5:02
  • $\begingroup$ @hugo you should probably write that in answer. $\endgroup$ – Nobody recognizeable Dec 23 '18 at 5:04
2
$\begingroup$

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. $\Delta x' = 0; x_2' = x_1'$).

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.