Relativistic momentum (Lorentz boost) low velocity limit

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SUMMARY

The discussion centers on the transformation of momentum and energy using Lorentz boosts in the context of low velocities. The user queries why the relativistic transformation for energy, E′=γE+γvpx, appears to differ from the Galilean transformation, particularly in the sign of the term involving momentum. A participant clarifies that the correct relativistic transformation should be E′=γE-γvpx, highlighting a sign error in the user's understanding. The conversation emphasizes the algebraic manipulation required to reconcile the two frameworks.

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  • Understanding of Lorentz transformations in special relativity
  • Familiarity with Galilean transformations
  • Basic knowledge of momentum and energy concepts in physics
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Neutrinos02
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Hello,

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
Neutrinos
 
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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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