# 4-velocity length under a force

1. Jan 11, 2010

### 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

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

?

2. Jan 11, 2010

### pellman

Re: 4-velocity "length" under a force

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.

3. Jan 12, 2010

### clem

Re: 4-velocity "length" under a force

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.

4. Jan 13, 2010

### pellman

Re: 4-velocity "length" under a force

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".

5. Jan 13, 2010

### clem

Re: 4-velocity "length" under a force

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

6. Jan 15, 2010

### pellman

Re: 4-velocity "length" under a force

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.

7. Jan 15, 2010

### clem

Re: 4-velocity "length" under a force

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