Elecrodynamic gauge transform

1. Oct 16, 2009

evilcman

If we consider the following Lagrangian:
L = m * (dot x)^2 / 2 - e A dot x - e phi
with A the vectorpotential and phithe scalar potential, then
the Euler-Lagrange equations, reduce to the known formula
of Lorentz-force:
m ddot x = e dot x X B + e E
know, this equation is invariant under gauge transformations:
A->A + nabla F
phi -> phi - parcd phi / parcd t
Know this is great, but if we construct the Hamiltonian with the
usual Legendre transform we will get
H = (p+eA)^2 / 2m + e phi
My problem with this is, if a write the (Hamiltonian) equation of
motion now, it seems gauge dependent:
dot x = (p + eA)/m

Isn't this a problem? What's the resolution?

2. Oct 16, 2009

Bob_for_short

A Hamiltonian is always gauge-dependent. Remember the arbitrary constant C that can be added to the scalar potential? It is of the same nature.

3. Oct 16, 2009

evilcman

The gauge dependence of the Hamiltonian is not my problem. My problem is that, from these equations of motion, the velocity, which is something that should be mesurable, also seems to be gauge dependent...

4. Oct 16, 2009

Bob_for_short

No, the equation solutions are gauge invariant. Moreover, the equations themselves, expressed via EM filed tensions, are gauge invariant.

5. Oct 16, 2009

evilcman

I don't see why the solutions should be gauge independent when A(and not its curl or something) is explicitly in the equations of motion...

6. Oct 16, 2009

Bob_for_short

If you speak of velocity equation, it contains also p which is gauge-dependent. Write the equation for acceleration d2x/dt2. You will see that this equation will contain the electric and magnetic fields rather than potentials.

7. Oct 17, 2009

evilcman

You are right. And if I understand it correctly p is no longer a physical quantity, only p+eA is.

8. Oct 17, 2009

Bob_for_short

Yes, there are two momenta: a canonically conjugated p (a formal mathematical construction) and a kinetic P = p + eA (physical quantity).

Last edited: Oct 17, 2009