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When is the 4-velocity useful?

  1. Aug 7, 2013 #1
    I'm a little confused about when I would use the 4-velocity for a particle on a world line. I got confused the other day when I was trying to derive the relativistic velocity addition formula. I thought I could write down a particle's 4-velocity in one frame, [itex]v^\mu=\gamma(c,v_x,0,0)[/itex], and then boost it into another frame, and I got something close to the velocity addition formula but not quite. And usually, when we say something is moving with some velocity in one frame, we are just talking about the classical velocity. So when would one actually use the 4-velocity? If anyone can also explain why it doesn't work to derive the velocity addition formula that would be very helpful!

    Thanks
     
  2. jcsd
  3. Aug 8, 2013 #2

    WannabeNewton

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    Hi there copernicus! Actually you can derive the velocity addition formula using the Lorentz transformation of the 4-velocity. Say we have a frame ##O## in which the 4-velocity has the components ##u = (u^t,u^x,u^y,u^z)##; let ##v^i = \frac{u^i}{u^t}## denote the components of the 3-velocity as measured with respect to ##O##. Consider a boost along the ##x##-axis of frame ##O## to a frame ##O'## with speed ##w##. The 4-velocity transforms as ##u^{\mu'} = \Lambda^{\mu'}_{\nu}u^{\nu}## so for example ##u^{t'} = \gamma u^{t}(1 - w v^{x}), u^{x'} = \gamma u^{t}( v^x-w )## hence ##v^{x'} = \frac{v^x-w}{1 - w v^{x}}## as usual. Similarly ##v^{y'} = \frac{v^{y}}{\gamma (1 - w v^{x})}## and ##v^{z'} = \frac{v^{z}}{\gamma (1 - w v^{x})}##.

    When you formulate SR through the framework of Minkowski space-time, you work with 4-vectors belonging to this space-time much like how you work with 3-vectors belonging to regular Euclidean 3-space when doing Newtonian mechanics.
     
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