- #1
JustinLevy
- 895
- 1
Everytime I try to work out the Lagrangian for EM in different gauges, it gets messy really quick. Maybe there is some trick to simplify the process that I do not know, but either way I'd appreciate some suggestions.
Starting point:
For a point particle a (non-relativistic) Lagrangian that gives classical electrodynamics is:
[tex]\mathcal{L} = \frac{1}{2}m\dot{\vec{x}}^2 - q\phi + q\dot{\vec{x}} \cdot \vec{A} - \frac{1}{4\mu_0} \int F_{\mu\nu} F^{\mu\nu} \ d^3r[/tex]
The coordinates are [itex]x[/itex] and [itex]A^\mu[/itex], with the fields being a function of position.
I can obtain the two Maxwell source equations and the Lorentz force law with this (the remaining two Maxwell equations follow directly from the definition of the electric and magnetic fields in terms of the potential [itex]A^\mu[/itex]).
If instead of going all the way to Maxwell's equations, I just solve for the evolution of A^\mu, I get the source equations:
[tex]\partial_\nu F^{\mu \nu} = \mu_0 J^\mu[/tex]
Which is true in any gauge.
Question 1]
Due to the j.A term, the Lagrangian itself is not gauge invariant (right?) even though the evolution equations resulting from it may be.
So what "gauge" is this Lagrangian in? How do I determine it?
----
Now if I want to use as coordinates [itex]A^\mu[/itex] as a function of momentum space instead, I can write the following (deriving got a bit messy at times, so I won't show that):
[tex]\mathcal{L} = \frac{1}{2}m\dot{\vec{x}}^2 - q\int \phi(\vec{k})e^{+i\vec{k}\cdot\vec{x}} \frac{d^3k}{(2\pi)^3} + q\dot{\vec{x}} \cdot \int \vec{A}(\vec{k})e^{+i\vec{k}\cdot\vec{x}} \frac{d^3k}{(2\pi)^3} - \frac{1}{2} \int [ \frac{k^2}{\mu_0} \vec{A}(\vec{k})^2 - \epsilon_0 \dot{\vec{A}}(\vec{k})^2 -
\epsilon_0 k^2 \phi(\vec{k})^2 + \frac{1}{\mu_0} \dot{\phi}(\vec{k})^2
] \frac{d^3k}{(2\pi)^3}[/tex]
Now the coordinates are [itex]x[/itex] and [itex]A^\mu[/itex], with the fields being a function of momentum space.
You can derive:
[tex](-k^2 -\frac{1}{c^2}\frac{\partial^2}{\partial t^2})\phi(k) = -\rho(k)/\epsilon_0[/tex]
[tex](-k^2 -\frac{1}{c^2}\frac{\partial^2}{\partial t^2})\vec{A}(k) = -\mu_0 \vec{j}(k)[/tex]
Which are maxwell's equations in terms of the potentials in the Lorenz gauge, [itex]\partial_\mu A^\mu = 0[/itex]. (Correct?)
Question 2]
How did the evolution equations for [itex]A^\mu[/itex] somehow become gauge dependent now? Is there something I'm missing in the Lagrangian (probably the free field part)?
Question 3]
What does this look like in the Coulomb gauge? (Coulomb gauge, [tex]\partial_\mu A^\mu = \frac{1}{c^2}\frac{\partial}{\partial t}\phi[/tex])
All the most helpful sources I've found only do this for the completely free field, and so also add the constraint that [itex]\phi=0[/itex] which can't be enforced in the interacting case so screws things up for me here. When ever I try to work it out without that constraint the extra little piece of [itex]\phi[/itex] floating around gets annoying quick.
Thanks everyone.
Starting point:
For a point particle a (non-relativistic) Lagrangian that gives classical electrodynamics is:
[tex]\mathcal{L} = \frac{1}{2}m\dot{\vec{x}}^2 - q\phi + q\dot{\vec{x}} \cdot \vec{A} - \frac{1}{4\mu_0} \int F_{\mu\nu} F^{\mu\nu} \ d^3r[/tex]
The coordinates are [itex]x[/itex] and [itex]A^\mu[/itex], with the fields being a function of position.
I can obtain the two Maxwell source equations and the Lorentz force law with this (the remaining two Maxwell equations follow directly from the definition of the electric and magnetic fields in terms of the potential [itex]A^\mu[/itex]).
If instead of going all the way to Maxwell's equations, I just solve for the evolution of A^\mu, I get the source equations:
[tex]\partial_\nu F^{\mu \nu} = \mu_0 J^\mu[/tex]
Which is true in any gauge.
Question 1]
Due to the j.A term, the Lagrangian itself is not gauge invariant (right?) even though the evolution equations resulting from it may be.
So what "gauge" is this Lagrangian in? How do I determine it?
----
Now if I want to use as coordinates [itex]A^\mu[/itex] as a function of momentum space instead, I can write the following (deriving got a bit messy at times, so I won't show that):
[tex]\mathcal{L} = \frac{1}{2}m\dot{\vec{x}}^2 - q\int \phi(\vec{k})e^{+i\vec{k}\cdot\vec{x}} \frac{d^3k}{(2\pi)^3} + q\dot{\vec{x}} \cdot \int \vec{A}(\vec{k})e^{+i\vec{k}\cdot\vec{x}} \frac{d^3k}{(2\pi)^3} - \frac{1}{2} \int [ \frac{k^2}{\mu_0} \vec{A}(\vec{k})^2 - \epsilon_0 \dot{\vec{A}}(\vec{k})^2 -
\epsilon_0 k^2 \phi(\vec{k})^2 + \frac{1}{\mu_0} \dot{\phi}(\vec{k})^2
] \frac{d^3k}{(2\pi)^3}[/tex]
Now the coordinates are [itex]x[/itex] and [itex]A^\mu[/itex], with the fields being a function of momentum space.
You can derive:
[tex](-k^2 -\frac{1}{c^2}\frac{\partial^2}{\partial t^2})\phi(k) = -\rho(k)/\epsilon_0[/tex]
[tex](-k^2 -\frac{1}{c^2}\frac{\partial^2}{\partial t^2})\vec{A}(k) = -\mu_0 \vec{j}(k)[/tex]
Which are maxwell's equations in terms of the potentials in the Lorenz gauge, [itex]\partial_\mu A^\mu = 0[/itex]. (Correct?)
Question 2]
How did the evolution equations for [itex]A^\mu[/itex] somehow become gauge dependent now? Is there something I'm missing in the Lagrangian (probably the free field part)?
Question 3]
What does this look like in the Coulomb gauge? (Coulomb gauge, [tex]\partial_\mu A^\mu = \frac{1}{c^2}\frac{\partial}{\partial t}\phi[/tex])
All the most helpful sources I've found only do this for the completely free field, and so also add the constraint that [itex]\phi=0[/itex] which can't be enforced in the interacting case so screws things up for me here. When ever I try to work it out without that constraint the extra little piece of [itex]\phi[/itex] floating around gets annoying quick.
Thanks everyone.
Last edited: