Special relativity question?
Could someone give me the answer to this question?
A traveler in a very fast train holds a mirror at arm's length and looks at his reflection. Both he and an observer outside the train see the same velocity of light for the traveler's image.
Since velocity = distance/time, what has happened to the length of the traveler's arm, and the time taken for the reflection to return, as seen by the outside observer?
Question from jacaranda physics 2
Update:by time dilation, doesn't more time elapse? "as seen by the outside observer"
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"by time dilation, doesn't more time elapse? "as seen by the outside observer""
More distance traveled, as seen by the outside observer, and more time for that light to travel.
Here are the most simple forms of the equations:
duration-at-rest = duration-when-moving * gamma
distance-at-rest = distance-when-moving * gamma
.. same gamma, a number >= 1, from gamma = 1 / sqrt( 1 - (v/c)^2 )
So when you form a speed...
duration-at-rest / distance-at-rest = duration-when-moving / distance-when-moving
(the gammas cancel)
"dilation" is a really poor word choice, do not let it confuse you.
His arm contracts, and less time elapses
He drops the mirror, and you forget about it.
No, all of your answers are wrong.