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How do you calculate the speed of an object according to the theory of relativity?
Question Date: 2010-02-10
Answer 1:

In the theory of relativity, speed is calculated in the usual way: distance covered over time elapsed. However, the tricky part is to get both, distance and time right. Here is an example:

Say yo want to travel to Proxima-Centaury, a star about 4.2 light years away, at a speed of 80% that of light.

From the definition of speed,
v = d / t,
you will take a time
t = d / v = 4.2 ly / (0.80c) = 5.25 y = 5 years and 3 months,
where c = 1 ly / y is the speed of light using ly (light year) for the unit of length and y (short for year) the unit of time. This will be the time as measured on Earth.

One of the features of relativity is time dilation, where the elapsed time according to clocks on board your spaceship, is given by

t' = t (1-(v/c)2)(1/2)
or t' = Sqrt(1-(v/c)2)
where Sqrt is short for 'Square Root of'. Using our numbers we get
t' = 5.25 y Sqrt(1-(0.8)2) = 5.25 y (0.6) = 3.15 y
so that you (along with everything on board your spaceship) will only age 3.15 years instead of the 5.25 years your trip will take according to Earth based clocks.

Since 3.15 years is less than the 4.2 years it takes light to travel from earth to Proxima Centauri, you might be tempted to conclude that you will be traveling faster than light. How do we reconcile this numbers with the fact that you are really traveling at only 80% the speed of light? The answer lies in another feature of relativity where distances contract along the line of motion. To better understand this, imagine that as you begin your journey you accelerate to 0.8c in the neighborhood of Earth (maybe just going round in circles) so that you start your trip already moving at 0.8c. Even though you are still near the Earth, when you measure the distance to Proxima-Centauri you will find that it is only (using the length contraction formula)

d' = d Sqrt(1-(v/c)2) = 4.2 ly Sqrt(1-(0.8)2) = 4.2 ly (0.6) = 2.52 ly

away from you, as if Proxima-Centauri had been magically been brought closer to you as you start the journey at a place nearby the Earth. So, what you see now is a star that is 2.52 light years away from you and approaching (remember that motion is relative) with a speed of 0.8c. To find the time of arrival of the star at your spaceship we use the same formula we used above to find:

t = d / v = 2.52 ly / (0.8 c) = 3.15 y.

So everything adds up; no strange things happening here other than, of course, relativity's funny features of time dilation and length contraction.

Answer 2:

The speed of an object is not affected by relativity. Velocity is still velocity. Now, as velocity approaches the speed of light, all sorts of interesting things happen: the object's mass increases, time goes slower, it observes distances being smaller, and there are side effects of the mass increase as well (e.g. scaling of momentum and kinetic energy with velocity becomes nonlinear). However, velocity remains as is.

Now, you might want to calculate what happens if you apply a force to an object already moving at relativistic speeds. I forget the formula for this, and can never remember it, but here is the thing to remember: momentum is conserved, and momentum always simply adds. So if you have the change in momentum, you can use the relativistic mass formula I gave you a couple of emails ago to calculate the new velocity (and direction, if the object is moving in more than one dimension).

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