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Sorry if the title is misleading, I wasn't sure how to briefly ask the question in the title.

I have a problem where a stick of length L is moving past me at speed v, there is a time interval between the front end coinciding with you and the back end coinciding with you.

So there are four different things I have to do, two of them are:

What is the time interval your frame? (Work in the stick’s frame.) I found this to be $\Delta t={1\over \gamma v}L$. I know this solution is right, and I was going to try to use the same method and apply it to this part of the question:

Time interval in the stick’s frame? (Work in your frame.)

I'm having great difficulty trying to orient myself, and the solution is not the same as the previous example I used (which my intuition thought it would be), and I was told that a rear clock on the stick would show a time ${Lv\over c^2}$ more than the front clock. I understand that from our frame we see the time interval as ${1\over v\gamma}L$, but I really don't know where the other value for time number came from. I have been searching through my physics textbook and I can't piece this together.

Where did they get that value for time?

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  • $\begingroup$ A stick? What's wrong with a train car? $\endgroup$ – JEB Mar 13 '18 at 16:37
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Let us assume the following configuration:
Your frame $S (t, x)$
Stick's frame $S' (t', x')$
The spatial axes overlap and $S'$ is moving with velocity $v$ in the $+x$ direction.
$L$ stick's proper length

Lorentz transformation, $S \to S'$
$t' = \gamma t -\gamma (v/c^2) x$
$x' = -\gamma v t +\gamma x$

Lorentz transformation, $S' \to S$
$t = \gamma t' +\gamma (v/c^2) x'$
$x = \gamma v t' +\gamma x'$

Work in stick's frame:
$\Delta x' = -L$ (in the stick's frame you are moving to the left)
$\Delta t' = L/v$
As per Lorentz transformation $S' \to S$, we have
$\Delta t = \gamma \Delta t' +\gamma (v/c^2) \Delta x' = \gamma L/v -\gamma (v/c^2) L$

Work in your frame:
$\Delta x = 0$ (in your frame you are at rest)
As per Lorentz transformation $S \to S'$, we have
$\Delta t' = \gamma \Delta t -\gamma (v/c^2) \Delta x = \gamma \Delta t$ (1)
$\Delta x' = -\gamma v \Delta t +\gamma \Delta x = -\gamma v \Delta t$ (2)
However, as $\Delta x' = -L$, from (2) we have $\Delta t = L /(\gamma v)$, which plugged in (1) gives
$\Delta t' = L/v$

Note: be careful that the configuration does not allow for a simple application of the time dilation, as in the moving frame the time difference is measured by two different clocks.

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