0
$\begingroup$

I am studying classical electrodynamics and I have been introduced to the concept of gauge transformations and gauge fixing conditions. Right know I am trying to prove that some conditions are valid gauge fixing conditions.

I've been able to show that the Coulomb (which is solved in this Phys.SE question) and Weyl conditions are valid gauge fixing conditions, but I couldn't show that the Axial condition ($\vec{n} \cdot \vec{A} =0$) is valid too.

The only information that I get is that if $\vec{A}_{ax}$ verifies the axial condition, $\vec{A}_{no}$ doesn't verify the axial condition and $f$ is such that $\vec{A}_{ax} = \vec{A}_{no} + \vec{\nabla} f$, then ${A}_{no}^3 = - \vec{\nabla}f^3$ (where the superscript denotes the third component of the magnetic potential vector) but I haven't been able to proceed further from here.

I would like to get the proof of this fact or a reference where I can find it.

$\endgroup$
  • $\begingroup$ Can you clarify your notation (unless it's already clear to those skilled in the field). Should that be $\vec{n}\cdot\vec{A}$? What is $\vec{n}$? What is the $Ax$ subscript? What is the $no$ subscript? What is the cube of a vector? Are you cubing $f$ or its gradient? $\endgroup$ – garyp Oct 17 '15 at 23:01
  • $\begingroup$ @garyp ax and no are just two subscripts which I use to distinguish between a vector potential that fulfils the axial condition an one which doesn't. The 3 superscript is just the third coordinate of the vector. I use this notation because I'm used to work with tensors in the course that I'm taking and we consider $A^{\mu}=(\phi,A_x,A_y,A_z)$ but I don't know if this notation is usual or not. $\endgroup$ – A. A. Oct 17 '15 at 23:07
  • $\begingroup$ With the lower-case $a$ it made more sense. :) and the superscript. Thanks. $\endgroup$ – garyp Oct 17 '15 at 23:08
  • $\begingroup$ @Adolfo It is very uncommon to include the vector arrow, when extracting a component. That is, even when using a superscript, you would write the third component of $\vec A$ as $A^3$ not as $\vec A^3$ (side note: I like the notation $\vec x^3$ to mean $\vec x^3 = \vec x (\vec x \cdot \vec x)$). Also one would probably write $-\partial^3 f$ instead of $-\vec \nabla f^3$. Plus: Upper indices are really confusing when used with three dimensional vector notation (it is not confusing, when using 4-vector notation consistently). $\endgroup$ – Sebastian Riese Oct 17 '15 at 23:53
3
$\begingroup$

Let's call our four-vector potential $A_\mu$. This is related the the scalar and vector potentials by $A_\mu = \{\phi,A_x,A_y,A_z\}$.

Under a gauge transformation, \begin{equation} A_\mu \rightarrow A_\mu' = A_\mu + \partial_\mu \lambda \end{equation} What you are asking is, given a generic vector potential $A_\mu$, how can we choose a $\lambda$ such that $A_z' = 0$.

Well, plugging into our gauge transformation formula we find we want to solve \begin{equation} 0 = A_z + \partial_z \lambda \end{equation} which we can do easily \begin{equation} \lambda = \left(-\int dz A_z \right) + f(x,y,t) \end{equation} where $f(x,y,t)$ is an arbitrary function that does not depend on $z$.

The fact that we are left over with an $f$ means there is some residual gauge freedom. Indeed, axial gauge does not completely fix the gauge. So long as one compute gauge invariant quantities, this of course won't matter in the end.

In quantum theory, the residual gauge freedom requires us to be careful when setting up Hilbert space, if we apply the usual rules of canonical quantization we will find some unphysical states that we need to project out.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.