4-velocity length under a force

[pellman]

4-velocity "length" under a force

In relativity we have u^\mu u_\mu=c^2, which is just another way of saying that p^\mu p_\mu=m^2c^2 or E^2=m^2c^4+p^2c^2.

This is relatively (no pun!) easy to see for a free particle. But if we have a vector potential acting on the particle, is the above still true? Or do we have instead that

(p_\mu +qA_\mu)(p^\mu+qA^\mu)= constant

?

 

[pellman]

Ok. I think I have it. It is actually

(p_\mu -qA_\mu)(p^\mu-qA^\mu)= m^2c^2

but p here is the canonical conjugate momentum

p_\mu =u_\mu+qA_\mu

so when you put that in, you still get u_\mu u^\mu=c^2.

I'd still love to hear if you experts concur.

 

[Meir Achuz]

Ok. I think I have it. It is actually

(p_\mu -qA_\mu)(p^\mu-qA^\mu)= m^2c^2

but p here is the canonical conjugate momentum

p_\mu =u_\mu+qA_\mu

so when you put that in, you still get u_\mu u^\mu=c^2.

This is correct. That is how vector potentials affect the otion of particles.
For a 4-scalar potential S, you would let m-->m+S.

 

[pellman]

This is correct. That is how vector potentials affect the otion of particles.
For a 4-scalar potential S, you would let m-->m+S.

Really? that would be weird. This would be identical to situation in which the particle's mass depended on its position in spacetime--its actual rest mass, not its "relativistic mass".

 

[Meir Achuz]

No, m is still the invariant mass(which is equal to the rest mass). It is the energy E that changes with position.

 

[pellman]

No, m is still the invariant mass(which is equal to the rest mass). It is the energy E that changes with position.

Isn't energy the same as p^0? And isn't that modified already by the 0th component of the vector potential?

We're going off on a tangent here, but I'm curious.

 

[Meir Achuz]

Yes, E=p^0. With 4-vector A^\mu=(V,{\vec A}) potential
and scalar potential S, the equation is
(E-V)^2=({\vec p}-{\vec A})^2-(m+S)^2.