# Coulomb and lorentz gauges.

ive just come across gauge transformations on maxwells equations, and mathematically its all pretty straightforward, however, i then come across choices of gauge, with the simplest being the coulomb gauge which is to set devirgence of A = 0 (A vector potential), youve also got the lorentz gauge.

i understand the maths operationals, however, im finding it difficult to attach any physical meaning to these statements.

can anyone explain this? particularly without the particle/QFT stuff that seems to swamp me wherever else i look.

cheers

## Answers and Replies

CompuChip
Science Advisor
Homework Helper
The point is there is no physical meaning. In the mathematical description of the theory, there exist certain substitutions (like $A \to A + \partial_\mu\xi$) which will alter the mathematics of the system, but eventually they will not affect the physical quantities that can be calculated from such a theory. Basically the reason is, that in the mathematical formulation certain degrees of freedom are included which the physical system does not have, and are therefore irrelevant for the physical properties. We can then use this freedom to make the mathematics easier to work with in a particular situation.

As an example, in electrodynamics the scalar potential determines the electric field by $\vec E = -\grad V$. The "real" quantity here is the electric field, and not the potential. This means any potential which gives the same electric field, is just as good. In this case, we can shift V by any constant we wish and still get the same electric field. This means that you and I can take different reference points for the potential (e.g. I might set it to zero at infinity, while you set it to zero at the origin of your coordinate system) and disagree on V. However, to say something about nature, we will both have to calculate and measure $\vec E$, and in both cases our results will agree.

Another example, you already mentioned, is the vector potential. It determines the magnetic field by $\vec B = \nabla \times A$. However, since "curl grad = 0", we can add to $\vec A$ the gradient of any scalar function. This will affect the potential $\vec A$, but not the physical quantity $\vec B$. In this case there is another nice point of view: note that $\vec A$ is a three dimensional vector (and in a relativistic description, it will have four components). However, the particles it is linked to, photons, only have two degrees of freedom (they can be polarized in only two directions). So if the vector potential is to describe actual photons, not all degrees of freedom in it should be independent, and indeed we can use our gauge freedom to fix some of these degrees of freedom and get something essentially two-dimensional.

thats brilliant, cheers. i suspected that it didnt have a direct physical meaning, so am i right in interpreting that you can choose A (and scalar potential) such that is suits a particular physical problem?

CompuChip
Science Advisor
Homework Helper
Yes, you can choose it anyway you want, as long as you don't affect the relevant physical quantities. One of the questions when considering a new theory, is always what exactly are the allowed gauge transformations. For electromagnetism, for example, it turns out that the most general one, which leaves $\vec E, \vec B, \rho, \vec J$ invariant, is the simultaneous transformation of
$$V \to V - \frac{d\xi(t, \vec x)}{dt}, \quad A \to A + \nabla\xi(t, \vec x),$$
and you are free to choose $\xi(t, \vec x)$ such that the mathematics for your particular problem becomes as easy as possible.

Yes, you can choose it anyway you want, as long as you don't affect the relevant physical quantities. One of the questions when considering a new theory, is always what exactly are the allowed gauge transformations. For electromagnetism, for example, it turns out that the most general one, which leaves $\vec E, \vec B, \rho, \vec J$ invariant, is the simultaneous transformation of
$$V \to V - \frac{d\xi(t, \vec x)}{dt}, \quad A \to A + \nabla\xi(t, \vec x),$$
and you are free to choose $\xi(t, \vec x)$ such that the mathematics for your particular problem becomes as easy as possible.

thanks for your help.

The point is there is no physical meaning. In the mathematical description of the theory, there exist certain substitutions (like $A \to A + \partial_\mu\xi$) which will alter the mathematics of the system, but eventually they will not affect the physical quantities that can be calculated from such a theory. Basically the reason is, that in the mathematical formulation certain degrees of freedom are included which the physical system does not have, and are therefore irrelevant for the physical properties. We can then use this freedom to make the mathematics easier to work with in a particular situation.

As an example, in electrodynamics the scalar potential determines the electric field by $\vec E = -\grad V$. The "real" quantity here is the electric field, and not the potential. This means any potential which gives the same electric field, is just as good. In this case, we can shift V by any constant we wish and still get the same electric field. This means that you and I can take different reference points for the potential (e.g. I might set it to zero at infinity, while you set it to zero at the origin of your coordinate system) and disagree on V. However, to say something about nature, we will both have to calculate and measure $\vec E$, and in both cases our results will agree.

Another example, you already mentioned, is the vector potential. It determines the magnetic field by $\vec B = \nabla \times A$. However, since "curl grad = 0", we can add to $\vec A$ the gradient of any scalar function. This will affect the potential $\vec A$, but not the physical quantity $\vec B$. In this case there is another nice point of view: note that $\vec A$ is a three dimensional vector (and in a relativistic description, it will have four components). However, the particles it is linked to, photons, only have two degrees of freedom (they can be polarized in only two directions). So if the vector potential is to describe actual photons, not all degrees of freedom in it should be independent, and indeed we can use our gauge freedom to fix some of these degrees of freedom and get something essentially two-dimensional.

Nicely done, CompuChip.

It seems that what you could call "physically real", is something that came be obtained through experimental measurement. What this might mean, though, could occupy an entire, and perhaps interesting, volume. This especially seems to pertain to the vector potential as well as quantum mechanical measurables.

There are a number of issues involving the measurable nature of the vector potential you might like to comment on.

First is the Bohm-Aharanov effect. That the magnetic potential is present is experimentally verified, but is the magnitude of the magnetic potential meaurable in general, or specifically in the arrangement where a long soleniod is placed between a double slit?

Second is the force obtained by acting a time invariant magnetic field on a moving conductor. Apparently, the magnetic field, local to the free charges, expressed by Faraday's law in vector form, together with the Lorentz force law, fails to predict any force upon the conductor.

It really isn't sufficient to say that the integral form of Faraday's law obtains the correct result: the vector form is a direct consequence of applying Stoke's theorem; and the local forces around a loop of wire conductor are not obtained from the integral solution, but only gives the total emf. But the local emf can be measured using distributed resistance around a loop of wire. In consequence, it's the Lorentz force law that appears to fail.

I don't know of any way out of this predicement but to look for a means by which the vector potential can be found responsible. Should a solution be found involving the vector potential,
is it the Dirac equation of the A field acting upon free electrons, and is an abolute or regaugable value of A obtained?

Third is Dirac's equation itself. Is the value of A in this equation gauge invariant?

Fourth, an interesting interpretation of the 4-vector potential is the five dimensional Kaluza-Klein unification of gravity and electromagetism. The potential is made equivalentto elements of a 5 dimensional metric where A_i = g_i4. In this case, A has an absolute value, as any other element of the metric. But, even at high enough energies, would the curvature in this dimension be measurable, and therefore A measurable?

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