How do you verify that v/c=pc/E? (1 Viewer)

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How do you verify that v/c=pc/E?
 

dextercioby

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HINT:

[tex] p=\gamma m_{0} v [/tex]

[tex] E=\gamma m_{0} c^{2} [/tex]

Daniel.
 

krab

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Your question is too short, and BTW only applies to free particles. I don't know what you are starting with. Or even whether you mean experimental verification. As Dextrecioby points out, if you already know expressions for p and E, the result is trivial. Another possible starting point is that you know only
[tex]E^2=p^2c^2-m_0^2c^4[/tex]
and Hamilton's equations. Then use H=E(p,x), and
[tex]v=dx/dt=\partial H/\partial p[/tex]
or the vector generalization.
 
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@@a
why does it only apply to free particles?
 

krab

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If a particle is in a potential field [itex]\Phi(x,y,z)[/itex], then
[tex](E-\Phi)^2=p^2c^2-m_0^2c^4[/tex]
so v/c is not equal to pc/E. One can define "kinetic energy" as [itex]K=E-\Phi(x,y,z)[/itex], then v/c=pc/K. That's just one example of non-free.
 
734
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i see... thank you very much!
 

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