- #1
pellman
- 684
- 5
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 a free particle. 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.
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 a free particle. 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.