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Velocity addition formula

  1. May 28, 2009 #1
    Let's say there are two observers A and C. C is moving .5c with respect to A. If we introduce a third observer which we will call B and B is moving at .2c with respect to A and is moving in the same direction as C and is inbetween A and C, and we now apply the velocity addition formula we get that C is now moving at (.5/1.06) or .47c relative to A.
    So, does this mean that without B, C is moving at .5c relative to A and that with B, C is moving at .47 relative to A?
    Does just the mere presence of B change the velocity of C with respect to A?

    What am I missing?
     
  2. jcsd
  3. May 28, 2009 #2

    clem

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    Science Advisor

    You may be using a classical result that v_CA=v_CB+v_BA that is not true in relativity.
    B has no effect on v_CA.
     
  4. May 28, 2009 #3

    Doc Al

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    Staff: Mentor

    How did you apply the velocity addition formula? It seems that you have assumed that if the speed of B with respect to A is 0.2c, then the speed of C with respect to B must be 0.3c. Not so.
    Of course not.

    [oops... clem beat me too it]
     
  5. May 28, 2009 #4
    Yes. That is what I was assuming. So, does that mean that the velocity of C relative to B is such that when you apply the velocity addition formula you get a velocity of C relative to A of .5c?
     
  6. May 28, 2009 #5
    I'm a visual sort of guy, and there may be others out there, so I have attached a jpg of velocity addition for a slightly different scenario.

    The image is taken from the perspective of A, for whom B is moving at 0.25c and C is moving at 0.666c.

    Look at the red dashed lines, these are B's equivalents to A's blue dashed lines.

    Using those you can see that for B, C is moving at 0.5c and A is moving at (minus) 0.25c.

    [tex]v_{cB} = \frac{v_{cA} - v_{bA}}{1 + \frac{v_{bA}}{c}.\frac{v_{cA}}{c}}[/tex]

    [tex]v_{cB} = \frac{0.66c - 0.25c}{1 + 0.25 * 0.666}[/tex]

    [tex]v_{cB} = 0.5c[/tex]

    Alternatively,

    [tex]v_{cA} = \frac{v_{cB} - v_{bB}}{1 + \frac{v_{bB}}{c}.\frac{v_{cB}}{c}}[/tex]

    [tex]v_{cA} = \frac{0.66c - 0.5c}{1 + 0.5 * 0.666}[/tex]

    [tex]v_{cA} = 0.25c[/tex]

    cheers,

    neopolitan

    PS just to make it clear, va[\sub] on the diagram means the velocity of A. However, that velocity has to be relative to something (in the diagram which is taken from the perspective of A, it is 0m/s, that's why it is a vertical line). In the equations I have added a capital letter indicating who the velocity is relative to, so vaA[\sub] would be the velocity of A relative to A, and vcA[\sub] would be the velocity of C relative to A.
     

    Attached Files:

    Last edited: May 28, 2009
  7. May 29, 2009 #6

    Doc Al

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    Staff: Mentor

    Of course!

    You can use the velocity addition formula to figure out the speed of C relative to B. (It turns out to be c/3, not .3c.)
     
    Last edited: May 29, 2009
  8. Jun 3, 2009 #7
    To stretch it a bit further, the generalized velocity addition is neither commutative nor associative (see for instance Ungar).
     
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