1
$\begingroup$

enter image description here

in the time dilation formula, I know that $t$ is the time measured by the stationary observer between 2 events. However, is $t'$:

  1. the time measured by the moving observer on his own clock between the 2 events,

or

  1. the time measured, in the frame of the observer at rest, between the moving observer starting and stopping his clock.

I feel like although the moving observer sees the light travel a greater distance at the same speed, the time interval between each tick of his clock is also increased by the same factor so the time intervals measured by each observer must be same and $t'$ should be $2$.

However, wouldn't that allow the moving observer to define the time interval, in his own frame, as 2 different things: number of ticks on his clock (same as the observer at rest) and $\text{distance}/c$ (greater than the observer at rest.

Improve this question
$\endgroup$
4
  • $\begingroup$ sorry, typo. The 2 events are the photon striking the 2 plates $\endgroup$
    – Vulgar Mechanick
    Nov 30, 2020 at 6:41
  • $\begingroup$ I am extremely sorry for that. I have been awake for the last 24hrs trying to make sense of relativity; my mind has stopped working $\endgroup$
    – Vulgar Mechanick
    Nov 30, 2020 at 6:47
  • $\begingroup$ when did i say the observer was between them? i mean the time measured by the observer between the photon striking plate one and the photon striking plate 2 $\endgroup$
    – Vulgar Mechanick
    Nov 30, 2020 at 7:04
  • $\begingroup$ Yes, I figured eventually. I deleted the comments. $\endgroup$
    – Umaxo
    Nov 30, 2020 at 7:13

2 Answers 2

Reset to default
1
$\begingroup$

In time dilation formula, $t$ is time between two events measured in frame S and $t'$ is time between the same two events measured in frame $S'.$

In the derivation of dilation formula, one usually takes time $t$ to be the time as measured by the observer B in your picture. This is the time of one tick of B's clocks in the rest frame of the clock.

Then you look at the same two events that demarcate the tick on B's clocks from the A's perspective. The A sees that between the two events $t'$ of time elapsed. So $t'$ is the time of one tick of B's clocks in the A's frame.

So if you know that one tick of B's clock took $t$ in B's frame, the dilation formula tells you for A the elapsed time of this tick is $t'=\gamma t.$

Improve this answer
$\endgroup$
2
  • $\begingroup$ from what I've read, t is the proper time interval ( time interval measured from a frame where the events occur at the same place). Do you mean" one usually takes time t'"? $\endgroup$
    – Vulgar Mechanick
    Nov 30, 2020 at 7:05
  • $\begingroup$ @OVERWOOTCH No, I do mean what I have written. You are analyzing B's clocks as seen in frame B and frame A. $\endgroup$
    – Umaxo
    Nov 30, 2020 at 7:12
1
$\begingroup$

One is the time as measured by the stationary observer, $t$. The other is the time as measured by the moving observer, $t'$.

For the stationary observer the travelled distance is longer, $d$, than for the moving observer, $d'$. But the speed of light is same, $c$, for both. With a different observed distance but same speed the observed time must be different as well to justify the formula of $\text{speed}=\frac{\text{distance}}{\text{time}}$:

$$\text{Stationary: }\quad c=\frac dt\qquad,\qquad\text{ Moving:}\quad c=\frac{d'}{t'}$$

where $d>d'$ and $t>t'$.

So,

  1. the time measured by the moving observer on his own clock between the 2 events

is correct, yes, whereas

  1. the time measured, in the frame of the observer at rest, between the moving observer starting and stopping his clock

would be no different than $t$ because this is measured by the stationary observer.

Improve this answer
$\endgroup$
3
  • $\begingroup$ Why aren't we taking into account the effect of time dilation on the moving observer's clock? if the time between each tick of the clock increases too, wouldn't that cancel the effect of the greater distance/time? Moreover, I thought that time dilation was relative, and you could't feel your own time get dilated/ slow down. that time only APPEARED to move slowly for u TO another observer $\endgroup$
    – Vulgar Mechanick
    Nov 30, 2020 at 7:54
  • $\begingroup$ @OVERWOOTCH Ahh, I think I get your point. If I measure the scenario with my clock while stationary, but then I sit inside the cart and measure again then you argument is that I will measure the same time since the clock that I am measuring with will also slow down. This is true. So, their clocks will show the same. But imagine comparing clocks during the measurement - imagine if both closk are live-streamed to one screen - then one will have gone slower than the other. It is indeed an issue that we will not ourselves now within our local environment ... $\endgroup$
    – Steeven
    Nov 30, 2020 at 8:06
  • $\begingroup$ ... whether time has slowed or not. But the point of this theoretical derivation is not to prove what the clocks will show, but to prove that time slows down. You could redo the experiment without a clock and simply think about how much time it necessarily must take in each of the two frames. That's the point of the derivation. Good observation, by the way. $\endgroup$
    – Steeven
    Nov 30, 2020 at 8:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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