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Q & A: Time dilation with two ships

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Most recent answer: 08/10/2016
My question regards time dilation. Wikipedia says "time dilation is a difference of elapsed time between two events as measured by observers moving relative to each other . . ." Assume a space platform that is 'stationary' in space. Two observers in two separate space ships leave the platform, both travelling at 60% of the speed of light, in exactly opposite directions. In my opinion, even though the distance between them increases at 120% of the speed of light, the time dilation for each is the same, namely 80%. Am I correct?
- Sydney Self (age 91)
Hendersonville, nc

Hi Sydney,

It seems as though you are already familiar with the gamma factor used to calculate pure time dilation, but here it is: gamma = 1/sqrt(1-(v/c)^2) 

The gamme factor is equal to 1 divided by the square root of one minus the velocity divided by the speed of light squared.

One note is to keep in mind who is moving relative to who, and when you calculate time dilation you again must say who the time is dilated relative to. 

In this example, our "rest frame" is the space platform. We prefer to call it a rest frame because it is important to remind ourselves that there is no special frame, and calling something "stationary in space" might lead one to believe that there is a special stationary frame. So, relative to the rest frame, the clocks on the ships will be ticking 80% the rate of the rest frame clocks, and relative to the ships, the platform clocks will be ticking 80% the rate of the ship clocks. 

As you noted, according to an observer on the platform the ships are speeding off at .6c in each direction, so the distance between them is increasing at a rate of 1.2c. However, in the frame of one of the ships, the other ship is NOT moving at 1.2 c, and it cannot be because nothing can travel faster than c. So according to one of the space ships, how fast is the other space ship moving?

Fortunately we have another formula for that, namely w = (u + v)/(1+u*v/c^2)

That is, the velocity of an object moving in one frame relative to another frame is the velocity of the frame plus the velocity of the object relative to the moving frame all divided by one plus the velocity of the frame times the velocity of the object in the frame divided by the speed of light squared.

So if we are in the frame of on of the space ships, we know the platform is moving at .6c relative to us, and the other spaceship is moving at .6c relative to the platform. When we plug all these numbers in we see that the velocity of one spaceship relative to the other is .88c! And according to us, an observer on one of the ships, the clocks on the other ship are ticking at 47% the rate of our own clocks!

Sheldon S.

(published on 08/10/2016)

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