# Is the following gauge transformation possible?

1. Feb 21, 2010

### Marin

Hi there!

Few weeks ago I came upon the following problem:

Let B be a vector field derivable from a vector potential A (on a simply connected topological space, smooth enough and everything well established so that mathematicians do not have to care about), i.e. $$\vec B=rot \vec A=\vec\nabla\times\vec A$$.

Now, we know from Elecrodynamics that we could alter the vector potential A by some gradient field (guage transformation).

Further, assume $$\Delta\vec B\neq 0$$, i.e. B itself does not satisfy Laplace's equation.

The question is now, whether a scalar field $$\phi$$ exists, such that $$\vec A'= A+grad\phi$$ and $$\Delta\vec B=0$$?

The question is reasonable, since:

$$(\Delta\vec B)_i=\partial^2_l B_i=\partial^2_l\varepsilon_{ijk}\partial_j(\vec A+grad\phi)_k=\partial^2_l\varepsilon_{ijk}\partial_j(A_k+\partial_k\phi)=\varepsilon_{ijk}\partial_j\partial^2_l(A_k+\partial_k\phi)$$

although we know $$\partial^2_l A_k\neq 0$$, why shouldn't some $$\phi$$ exist such that $$\partial^2_l(A_k+\partial_k\phi)=0$$?!

(I tend to think that this is impossible, for the following reason: Let B be the magnetic field and A be the respective vector potential. We know that an EM-wave is a wave, where both the Electric and the Magnetic field obey the wave equation. Now if the former were true, I could always pick up that gauge for the A potential and make the magnetic wave equation trivial, which would be somekind of embarrassing, since I'm talking on my cell phone every day making use of the electroMAGNETIC waves :) - of course this is by no way a rigorous mathemacial disproof)

so I would be glad to here what you think of this :)

2. Feb 22, 2010

### torquil

Re: Is the following guage transformation possible?

First:

$$B := \nabla \times A$$

$$A' := A + \nabla \phi$$

$$B' := \nabla \times A'$$

Therefore

$$B' = \nabla \times (A + \nabla \phi) = B + \nabla \times \nabla \phi = B$$

This is the whole point of the gauge transformation. It is not able to change B.

So

$$\Delta B' = \Delta B \neq 0$$

Torquil

3. Feb 22, 2010

### Marin

Re: Is the following guage transformation possible?

what you've written is absolutely true..

but it is exactly the direct plug-in what I'm trying to avoid here, using the fact that I can change the order of derivatives according to Schwarz's thm.

It would be not affecting B itself, just its derivatives..?

4. Feb 22, 2010

### torquil

Re: Is the following guage transformation possible?

Firstly, the gauge transformation won't affect B in any way, e.g. as seen by the proof I gave that the function B' is the same function as B, and therefore all its derivatives are equal.

But I understand that you are interested in the resolution of the problem you posed in the context of your equation for A? Let me first repeat your equation A without components:

$$\Delta B' = \Delta (\nabla \times A') =\Delta (\nabla \times (A + \nabla\phi)) = \Delta B + \Delta (\nabla\times\nabla\phi) = \Delta B + \nabla\times (\Delta\nabla\phi)$$

Here I have done the same as you did in your component notation. I took the liberty of writing B instead of nabla x A in the last terms, just for neatness. Your question is why a $$\phi$$ cannot be chosen so that the last term makes the whole thing zero, i.e. to compensate for the non-zeroness of the Delta B term.

You can find a $$\phi$$ there that will not be killed by the Laplacian and on of the nablas, but the second nabla will kill it anyway... The quantity in the last parenthesis can be expressed as the gradient of a scalar, therefore it is killed by the second nabla. This is the case no matter what $$\phi$$ you use. Even if exchange the order of derivatives, the quantity that the (nabla X)
acts on can still be expressed as the gradient of a scalar.

Torquil

5. Feb 23, 2010

### Marin

sounds good to me :)

thanks for the 'enlightment' :)