(adsbygoogle = window.adsbygoogle || []).push({}); 4-velocity, the "mass shell" and potential energy

For a free relativistic particle we have the condition

[tex]E^2-\mathbf{p}^2c^2=m^2c^4[/tex].

putting [tex]p^0 = c^{-1}E[/tex] we may write this as

[tex]p^\mu p_\mu = m^2c^2[/tex]

And since [tex]p^\mu = m\frac{dx^\mu}{d\tau} = mu^\mu[/tex] we may express the same condition as

[tex]u^\mu u_\mu = c^2[/tex]

At least, this is true for afreeparticle. Is this condition affected by the presence of a potential?

In the presence of a potential [tex]A^\mu[/tex] the momentum-energy relation becomes

[tex](p^\mu-A^\mu)(p_\mu-A_\mu) = m^2c^2[/tex]

doesn't it? But then in that case, we are talking about the canonical momentum and we no longer have [tex]p^\mu = mu^\mu[/tex]. But is [tex]p^\mu = mu^\mu+A^\mu[/tex]? So that we still have [tex]u^\mu u_\mu = c^2[/tex]? Can't find this in my texts and just wanted someone to confirm I understand it correctly.

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# 4-velocity, the mass shell and potential energy

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