0
$\begingroup$

Before diving into my question, we first must acknowledge Einsteins two assumptions:

  1. The laws of physics are the same for every observer and reference frame.

  2. The speed of light is constant for different observers.

Background

I was trying to learn about the Lorentz Factor ( γ), and happened to come across Fermilabs video on it. Here is one question which arose to me after brainstorming for several hours on it and looking at different sources, such as CrashCourse; But I still do not get it when I compare with Fermilabs video.

Question

The distance from the flashlight to the mirror will be denoted as W and c as the speed of light. The time it will take from the flashlight back and forth will be 2W/c relative to the person inside the train.


enter image description here

Now, let us ask the observer outside the train how much time it will take for the light to travel to the mirror and bounce back to the person inside the train.

If we assume the speed of the light ray that the observer outside the train sees is c , and the train is moving with a velocity v; then using the Pythagorean Theorem, the speed from the flashlight to the mirror is $\sqrt {c^2 - v^2} $ relative to the person outside.

Therefore, the time it takes from the flashlight to the mirror and back to the person inside the train, relative to the observer outside will be 2W/$\sqrt{c^2 - v^2} $.

Now, here is my question. If both observers experience the same speed of light from the flashlight to the mirror and back, then it means that 2W/$\sqrt{c^2 - v^2} $ = 2W/c; but that is only true when v = 0! This simply does not make sense, it should also apply for objects in motion. That is the whole point of time dilation!

And that too, how is it even possible that both of the observers obtain different equations for the speed from the flashlight to the mirror if all observers experience the same speed of light, no matter their reference frame (according to Einsteins second postulate)? What does $\sqrt{c^2 - v^2} $ actually mean?

$\endgroup$

2 Answers 2

1
$\begingroup$

You're assuming the times are the same for both observers, but the whole point is that the times are not the same. The times measured by the train ($\Delta t$) and ground ($\Delta t'$) observers are different by a factor of \begin{align} \frac{\Delta t'}{\Delta t} = \frac{2W/\sqrt{c^2 - v^2}}{2W/c} = \frac{c}{\sqrt{c^2 - v^2}} = \frac{1}{\sqrt{1 - v^2/c^2}} = \gamma \end{align} This is the Lorentz factor, and it's always greater than 1 as long as $v$ is not zero. So the time dilation formula \begin{align} \Delta t' = \gamma \Delta t \end{align} tells us that for the ground observer, for whom the train is in relative motion, each "tick" of a clock on the train takes a longer time than it does for the train observer.

$\endgroup$
9
  • $\begingroup$ Hello @d_b! I just do not get one thing, why did you divide the delta t' over delta t? $\endgroup$
    – SMK
    Aug 13, 2023 at 8:33
  • $\begingroup$ And the sources which I searched up mentioned the fact that it happened in the same time frame, so how come the time are different? $\endgroup$
    – SMK
    Aug 13, 2023 at 8:34
  • $\begingroup$ @ShaunKant Because I know that $\Delta t' / \Delta t = \gamma$. $\endgroup$
    – d_b
    Aug 13, 2023 at 8:35
  • $\begingroup$ @ShaunKant I think you are either misinterpreting your sources or your sources are wrong. Could you give an example? $\endgroup$
    – d_b
    Aug 13, 2023 at 8:36
  • $\begingroup$ Please see the section "Time dilation" on Crash Course Physics. I put up a link in my question, it is in the section "Background" $\endgroup$
    – SMK
    Aug 13, 2023 at 8:38
0
$\begingroup$

From the perspective of the outside observer, the light takes the longer, diagonal path shown in your second image, at the same speed $c$. It's the component of the speed in the direction of the mirror that is smaller and equal to $\sqrt{c^2-v^2}$.

Think of it like walking diagonally across a road. It will obviously take you longer to get to the other side than if you walked straight across at the same walking speed.

$\endgroup$
2
  • $\begingroup$ Hi @Kris Walker, I was just wondering if you could specify what "the component of the speed in the direction of the mirror" is... I do not get what "component" means in this case. Thank you! $\endgroup$
    – SMK
    Aug 13, 2023 at 11:19
  • $\begingroup$ @ShaunKant the components are just the arrows labelled by $v$ and $\sqrt{c^2-v^2}$. The light's actual physical motion is in the diagonal direction at a speed of $c$, so it advances in the direction of the tracks at a speed of $v$ and in the direction of the mirror at a speed of $\sqrt{c^2-v^2}$. $\endgroup$ Aug 13, 2023 at 11:31

Your Answer

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

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