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Derivation of Velocity-Addition Formula

  1. Jul 7, 2012 #1
    There was an old thread but for some reason it is now closed...

    So I'd have to restate this question:

    Why does any text book doesn't bother to derive the 3-vector-velocities addition formula using the more general 4-vector formalism?

    In detail, the Lorentz Transformation: U[itex]^{1'}[/itex]=γ(v)(U[itex]^{1}[/itex]-β(v)U[itex]^{0}[/itex])⇔γ(υ[itex]^{'}_{x}[/itex])υ[itex]^{'}_{x}[/itex]=γ(v)[γ(υ[itex]_{x}[/itex])(υ[itex]_{x}[/itex]-cβ(v))] (where υ[itex]_{x}[/itex] the velocity in Σ,υ[itex]^{'}_{x}[/itex] the velocity mesured in Σ', and v the relative velocity between the two systems) solved for υ[itex]^{'}_{x}[/itex] should grant υ[itex]^{'}_{x}[/itex]=(υ[itex]_{x}[/itex]-v)/(1-vυ[itex]_{x}[/itex]/c[itex]^{2}[/itex]) right?
    Last edited: Jul 7, 2012
  2. jcsd
  3. Jul 8, 2012 #2
    You're exactly right: If you know about the 4-velocity (or equivalently, the Energy-Momentum 4-vector), then you can derive the transformation law for velocities very easily.

    However, you can also do it the other way around: You derive the transformation law for velocities using the Lorentz transformations, and then you derive the relativistic form for momentum from that.
  4. Jul 8, 2012 #3


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    I wouldn't say this is correct, as you'd have to make assumptions about relativistic momentum in order to do this.
  5. Jul 8, 2012 #4
    Sure, you need to make assumptions, such as reducing to the Newtonian case as v/c --> 0, and being proportional to mass, and being in the same direction as the velocity, and so forth.
  6. Jul 9, 2012 #5
    At first glance the derivation seems straightforward but when put into practice I have some issues with the algebra:


    The above solved for u' would never yield the velocity-addition formula:frown:

    What am I missing?
  7. Jul 9, 2012 #6


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    Your last expression on the right should be simply (u-v). Then the algebra should (laboriously) work out - the rest of your derivation looks correct to me.

    [Edit: just worked it through - with the one correction, all the algebra does work out. This is your only mistake.]
  8. Jul 9, 2012 #7
    PAllen is correct. Now, write down the corresponding relationship for the time direction, and divide your x-direction equation by your t-direction equation. What do you get then?

  9. Jul 10, 2012 #8


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    Actually, this isn't necessary. Make the one fix to the equation in post #5, and the velocity addition formula follows by pure (messy) algebra. Easiest is just to plug in:

    u' = (u-v)/(1-uv/c^2)

    into the equation and see that it is a solution. To be complete, you then need to provide arguments that it is the only solution.
  10. Jul 10, 2012 #9

    I see it now. cβ(v) is of course v and not v/c...

    Working SR in 3D problems has always had some troublesome algebra though...
  11. Jul 10, 2012 #10
    There is absolutely no "messy" algebra in what I suggested. The velocity addition formula for the x-direction follows directly and immediately when you do as I suggested. You can also get the velocity addition formulas for the y- and z-directions directly by dividing the relationships in these directions by the relationship for the time direction. You can then substitute these results into the equation for the time direction to confirm that the solution satisfies the time equation identically.

  12. Jul 10, 2012 #11


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    True. However, my point was the equation in #5 (with the one correction) was informationally complete as it stands, and that seemed to be the direction the OP wanted to take. The tradeoff for working with only the x component relationship is more algebra to arrive at velocity addition formula.
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