Alice is moving towards a mirror with a velocity of $0.8c$. She sends a pulse of light toward the mirror when she is at a distance $L$ from the mirror and she times how long the pulse of light takes to come back to her. Bob is sitting at rest in relation to the mirror. What is the elapsed time from when Alice emits the pulse of light to when she receives it, as measured by Alice? Finally, what is the elapsed time as measured by Bob?
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$\begingroup$ I know based on the problem I know that it involves length contraction, but after more consideration, I think it may involve time dilation as well. Any help would be appreciated. $\endgroup$– Jonnam5Commented May 20, 2018 at 18:25
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You don’t need special relativity's time dilation to calculate a time between events as seen by Alice, nor between events as seen by Bob. Alice and Bob are each in their own rest frames. They see what they see.
So calculate where the mirror will be when it and the light pulse converge to a point, then calculate where Alice will be when the light from there reaches her as she moves toward it.
Where special relativity comes in is questions like “what does Alice’s clock show to Bob when the light reaches Alice”.
answered May 20, 2018 at 18:31
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$\begingroup$ In my interpretation, this is a special relativity problem, since Alice and Bob will not agree on the elapsed time this problem is asking for. For example, the distance that Alice measures from her to the mirror is not $L$, but is it less than $L$, since she is moving at $0.8c$. On the other hand, Bob measures this distance to be $L$. I know that they are both in inertial reference frames, but from what I know this still holds true. $\endgroup$– Jonnam5Commented May 20, 2018 at 18:47
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$\begingroup$ "She sends a pulse of light toward the mirror when she is at a distance L from the mirror" - no contraction there, as it's in her rest frame. She can calculate three local events: Her sending the light, the reflection, her reception of the light. If you want to transform that time to what Bob sees, you can, but you can also calculate all those points in Bob's frame directly. $\endgroup$ Commented May 20, 2018 at 18:53