- #1
pellman
- 683
- 6
4-velocity, the "mass shell" and potential energy
For a free relativistic particle we have the condition
E^2-\mathbf{p}^2c^2=m^2c^4.
putting p^0 = c^{-1}E we may write this as
p^\mu p_\mu = m^2c^2
And since p^\mu = m\frac{dx^\mu}{d\tau} = mu^\mu we may express the same condition as
u^\mu u_\mu = c^2
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 A^\mu the momentum-energy relation becomes
(p^\mu-A^\mu)(p_\mu-A_\mu) = m^2c^2
doesn't it? But then in that case, we are talking about the canonical momentum and we no longer have p^\mu = mu^\mu. But is p^\mu = mu^\mu+A^\mu? So that we still have u^\mu u_\mu = c^2? 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
E^2-\mathbf{p}^2c^2=m^2c^4.
putting p^0 = c^{-1}E we may write this as
p^\mu p_\mu = m^2c^2
And since p^\mu = m\frac{dx^\mu}{d\tau} = mu^\mu we may express the same condition as
u^\mu u_\mu = c^2
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 A^\mu the momentum-energy relation becomes
(p^\mu-A^\mu)(p_\mu-A_\mu) = m^2c^2
doesn't it? But then in that case, we are talking about the canonical momentum and we no longer have p^\mu = mu^\mu. But is p^\mu = mu^\mu+A^\mu? So that we still have u^\mu u_\mu = c^2? Can't find this in my texts and just wanted someone to confirm I understand it correctly.
