Speed of light (layman)

Question

If two bodies are moving towards each other along a straight line at a speed of $0.6c$ ($c=$ speed of light) w.r.t an inertial observer, then w.r.t each body, the other body is moving faster than light ($1.2c$). If each frame of reference is equally valid, how can the two bodies explain the velocity of approach of the other? (since velocity is always relative and hence the body cannot assume that it has absolute velocity in space)

Answer 1

Because the formula for adding velocities is not simply v1+v2 as you are supposing. For everyday speeds it’s so close to a simple sum that a calculator doesn’t have enough digits to show the descrepancy. But in your case

$$s = {v+u \over 1+(vu/c^2)} $$

where u and v are both 0.6 and c is 1 in the same units, comes to …?

Someone on either object will see the other approaching at the rate of 0.882 c, not 1.2 c.

Answer 2

Using appropriate formula for adding velocities, resulting from the lorentz transformation, we arrive at the result, that in the reference frame of the train, the other train moves slower than light. Interestingly, in the reference frame of an observer standing near the tracks, both trains have speed equal to 0.6c, and the distance between them decreases with a speed equal to 1.2c. It may seem strange, but perfectly agrees with special relativity, because there is no material object moving faster than light, only the time derivative of distance between trains is greater than c. We must also remember that due to lorentz contraction, in the reference frame of a train the distance between trains is smaller than in the reference frame of the observer near the tracks. So although the distance between trains decreases faster in a reference frame of tracks, the distance between the trains is also bigger and in both frames of reference the trains will collide at the same time.