Sunday, April 23, 2017

special relativity - Time dilation formula


In the time dilation equation $$t'=t/\sqrt{1-(v/c)^2}$$ where $ t'$ is the time measured by an observer in motion for the same event, where $t$ is time measured by the observer at rest.


Imagine a situation where both of the observes measure a time of 5 sec on their watches. These events (an observer seeing the fifth tick on his watch) are not simultaneous.


Now, my question: is the time taken by the observer in motion given by the equation $$t'=t/\sqrt{1-(v/c)^2}.$$ Does this signify that as time is travelling slower for the observer in motion it takes more time than the observer at rest to measure the time interval of 5 sec?




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