# Gauge invariance of the vector potential

1. Jan 26, 2009

### muscaria

The vector potential can be expressed in the following way:

∇^2 Ay-∂/∂y (∇∙A)=-μJy

(Here only taking y components)

Vector A is not determined uniquely. We may add derivatives of an arbitrary function (gradient) to the components of A, and the magnetic field does not change (curl of gradient is zero). In other words: the magnetic field is invariant with regard to this calibration (just basic gauge invariance)
The arbitrary function is usually chosen so that we obtain the simplest form for the equation of A. The function is chosen so that ∇∙A=0 (lorentz condition) and we get

∇^2 A=-μJ

By doing this, don't we consider the divergence of A to be completely unphysical? The Lorentz condition is formulated purely as a mathematical convenience. Is it accepted mainly because we think that the magnetic field represents the whole physical effect? Shouldn't the divergence of A have a physical impact on the polarisation of the medium/vacuum? Maybe it could be observed in other non-transverse EM waves? Furthermore, the aharonov-bohm effect proves the physical impact of A even in a region with a vanishing B.
I think that there may be different types of waves (longitudinal) but that the classical theory would not account for them because it focuses on E and B as being the fundamental quantities, and disregarding the totallity of the effects of A and φ (electrostatic) potentials. Also for example the rate of change of φ with respect to time.
Has anyone ever come across this? Anyway, any kind of constructive criticism would be greatly appreciated.. Cheers

2. Jan 26, 2009

### Stingray

No, there's no observable consequence to choosing a particular gauge. You started off by (correctly) saying that the gauge doesn't affect the magnetic field. Even in the Ahonarov-Bohm effect, potentials are not measured directly. Instead, you're effectively observing their line integrals around loops. These are not affected by gauge transformations:
$$\oint ( A + \nabla \chi ) \cdot \mathrm{d} l = \oint A \cdot \mathrm{d} l .$$

I'm not sure what you mean with regards to longitudinal waves. These can exist in certain modifications to Maxwell's theory, but there is no experimental evidence for them. Gauge concepts certainly do change if you want to move away from Maxwell's equations.

3. Feb 2, 2009