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We've already known that "p=γmv" . v is particle's velocity depends on our rest frame and mass too.

But shouldn't we observe particle's momentum by its displacement? for particle's view, displacement should be smaller than our observed displacement.

I read Halliday's book and it hasn't got any information.Just does say "use t = t_0.γ for particle's time dilation and you got relativistic momentum." we use lorentz transformation for time but not for x? Why? - The link that has been published by me is not an answer.It does just derivating momentum formula with relativistic mass as feynmann did it on his book.

Thanks for answers.

P.S.Forgive my syntax errors.I am not on my mother tongue.

Kadir Susuz
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  • Can you provide more context around the Halliday quote? Most treatments argue that the energy-momentum four-vector should transform under boosts like the time-space four-vector (though an introductory text, like the Halliday et al. that I know, might not given enough information to squeeze the fact into that single sentence). To start from non-relativistic momentum and modify it seems like a strange approach. – rob Apr 17 '19 at 23:09
  • https://physics.stackexchange.com/questions/190959/why-is-the-momentum-of-a-particle-gamma-mv

    this page has it.(But has no directly answer my question.) I've checked every single page on stack exchange but couldn't find an clear answer.

     – Kadir Susuz Apr 17 '19 at 23:20
  • Please [edit] your question to explain why the answers to the linked question aren't working for you. If I were to answer your question as posted here, I would basically write the same as here. Your edits will place your question in front of site users who will decide whether it should be re-opened. – rob Apr 17 '19 at 23:26
  • That proof is not clear. It just like Halliday's words.If you read my question, i am asking for why don't we use transformation for displacement but time(particle's displacement must be shorter than our observe. I did edit my post. – Kadir Susuz Apr 17 '19 at 23:35

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