How do you verify that v/c=pc/E?

[asdf1]

How do you verify that v/c=pc/E?

 

[dextercioby]

HINT:

$$p=\gamma m_{0} v$$

$$E=\gamma m_{0} c^{2}$$

Daniel.

 

[krab]

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
$$E^2=p^2c^2-m_0^2c^4$$
and Hamilton's equations. Then use H=E(p,x), and
$$v=dx/dt=\partial H/\partial p$$
or the vector generalization.

 

[asdf1]

@@a
why does it only apply to free particles?

 

[krab]

If a particle is in a potential field $\Phi(x,y,z)$, then
$$(E-\Phi)^2=p^2c^2-m_0^2c^4$$
so v/c is not equal to pc/E. One can define "kinetic energy" as $K=E-\Phi(x,y,z)$, then v/c=pc/K. That's just one example of non-free.

 

[asdf1]

i see... thank you very much!