4-velocity, the mass shell and potential energy

[pellman]

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.

 

[genneth]

Always, $$u^\mu u_\mu = c^2$$. Mechanical momentum is m times that. Canonical momentum is determined from what the Lagrangian of the system is.

 

[pmb_phy]

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.

If you look in Jackson's text Classical Electrodynamics you will find something very close.

Pete

 

[pellman]

Thanks, guys.