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Relativistic momentum (Lorentz boost) low velocity limit

  1. Nov 17, 2015 #1

    If I have a momenta pμ=(E,px,py,pz) and transform it via lorentz boost in x-direction with velocity v I'll get for the new 0th component E′=γE+γvpx why is this in the limit of low velocities the same as transforming the energy by a galilei transformation with velocity v? For γvpx i get something like vpx+O(v³) and with a galilei transformation I'll have terms (with px=mu) like 1/2m(u−v)²=1/2mu²−vpx+1/2mv². So in the relativistic case I got the wrong sing for vpx+O(v³)and lost the 1/2mv²? Did I make a mistake?

    Thanks for help
  2. jcsd
  3. Nov 17, 2015 #2


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    Science Advisor
    Gold Member

    For the sign error, if if you have (u-v) in the Galilean case, you need E′=γE-γvpx for the corresponding relativistic case. The rest is just algebra, but I am not disposed to work it out right now.
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