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Relative velocities in relativity

  1. Apr 18, 2003 #1

    I'm just trying to clean out some physics cobwebs in my head, as I was never much up on relativity; thanks in advance for accomodating me. Here's my question:

    Imagine two spacecraft widely separated, but travelling toward each other at significant fractions of c (observed relative to the same reference object, say a sun). What would observers on each spacecraft see in the following situations:

    a) each ship travels at v < 0.5c relative to the reference object
    b) each ship travels at v = 0.5c relative to the reference object
    c) each ship travels at v > 0.5c relative to the reference object

    My naive assumption in the case of (a) would be that observers on either ship would see the other ship approaching at 2(v), with light from the other ship blue-shifted in proportion to that speed. However, this doesn't seem to take into account dilation effects, so I'm not really sure of this answer.

    So, is the answer above correct, and furthermore, what happens when the ships' relative velocities sum to >= c in cases (b) and (c)? Finally, does anything change when the ships approach one another with velocities different from one other?

  2. jcsd
  3. Apr 18, 2003 #2


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    Your right in that your answer doesn't take relativistic effect into account when adding the velocities. They do not add up by the rule
    w = u+v but by

    w = (u+v)/(1+uv/c²)

    This is the correct formula for all additions of velocity. Note that when uv<<c (As in everyday experence), The formula gives an answer nearly identical with u+v, Which is why we still tend to use w = u+v for low velocity situations.
  4. Apr 18, 2003 #3
    Just come extra notes ...
    Note that as u->c and v->c the equation will reduce to :
    w = u+v/(1+c2/c2)
    w = u+v/2
    so if u=~v
    w =~ u =~ v
    And this is why the speed of any object (relative to any other object) will never pass the speed of light.

    Also note this equation is called "Lorentz Transformations"
  5. Apr 18, 2003 #4
    Cheers all, thanks for the excellent answers!
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