# Is the electromagnetic 4-potential a Lorentz 4-vector in the Coulomb gauge?

In the course of Classical Electrodynamics I learned that in the Lorenz gauge it is easy to prove that the 4-potential $$A$$ is a 4-vector, that is its components $$A^\mu=(\phi/c,\vec{A})$$ transform as the component of a 4-vector under Lorentz transformation. The Lorenz gauge is written in a covariant form ($$\partial_\mu A^\mu=0$$) and so if I implement a generic Lorentz transformation, in the new reference frame I preserve the Lorenz gauge condition for $$A'$$.

In the case of Coulomb gauge in vacuum ($$div(\vec{A}))=0$$ and $$\phi=0$$) I read on Weinberg QFT that:

It's clear that the Coulomb gauge condition isn't Lorentz-invariant and so if I make a Lorentz-transformation in the new reference frame the Coulomb gauge isn't satisfied, but I don't understand how this can say that $$A$$ isn't a 4-vector. Furthermore, I don't understand how, even if $$A$$ isn't a 4-vector, the electromagnetic tensor $$F$$ remains a tensor. And again, if I impose the Coulomb gauge only in one (and not in all) reference frame, I can preserve the fact that $$A$$ is a tensor?

In the course of Classical Field Theory we construct the 4-potential form as a 1-form which function components are given by $$A_\mu$$ ($$A=A_\mu(x) dx^\mu$$) and the Professor says that this is sufficient to ensure that it is always a 1-form without any reference to gauge fixing. How this point of view is matching with the others?

I try to answer to these questions reading several books (QFT 1 Weinberg, Modern Electrodynamics Zangwill...), but I just created more confusion in my mind.

• The Coulomb gauge exhibits no gauge invariance and no causality. Gauge theory gets away with it by maintaining that the potential is not physical anyway. Feb 11 at 10:20

It's rather that the Coulomb gauge is not Lorentz invariant. If you fix the gauge in a given frame, then if you boost to another frame, the new 4-potential obtained by the Lorentz transformation does not verify the Coulomb gauge in the new frame so you still have a problem. This is the converse of Weinberg's argument. His argument is that if you fix the 4-potential in every frame by the Coulomb gauge, then the components of the 4-potential does not transform as a 4-vector. In any case, you have an incompatibility between Coulomb gauge and Lorentz transformation. This is to be contrasted with the Lorenz gauge which is compatible with the Lorentz transformation.

The fact that you can construct a Lorentz invariant quantity from an object that is not Lorentz invariant stems from gauge invariance. The the gauge freedom allows you to "cancel" out the contributions disrupting the Lorentz invariance.

Hope this helps.

but I don't understand how this can say that $$A$$ isn't a 4-vector

Because the four-tuple is defined, in all frames, in such a way that it obeys $$\nabla\cdot \mathbf A = 0, \varphi=0$$ (in all frames). This is possible, and prevents the four-tuples $$[\varphi/c,A_x,A_y,A_z]$$ in all frames from transforming between those frames according to the Lorentz transformation (and thus it can be quite misleading to denote it $$A^\mu$$ in this convention).

The components of the four-tuple (scalar potential, vector potential) have a lot of arbitrariness to them, and this allows us to define them in every frame obeying certain conditions, like the Coulomb gauge condition. This makes two four-tuples of potentials in two frames to be non-related via the Lorentz transformation, thus the four-tuple in this convention is not a four-vector.

Furthermore, I don't understand how, even if $$A$$ isn't a 4-vector, the electromagnetic tensor $$F$$ remains a tensor.

Because all possible choices of the four-tuple, 4-vector or not 4-vector, give the same field $$F$$. Since this field is an antisymmetric derivative of a 4-vector (when we use the convention where potentials are a 4-vector), the field $$F$$ has to be a 4-tensor. This is also required by the fact that the field determines proper acceleration of a test charged particle:

$$qF^{\mu\nu}u_\nu = m\frac{du^\mu}{d\tau}.$$

If $$F$$ was not a Lorentz 4-tensor, this equation could not hold in all frames.

And again, if I impose the Coulomb gauge only in one (and not in all) reference frame, I can preserve the fact that $$A$$ is a tensor?

Yes. We can take a 4-vector $$A^\mu$$, and do such a gauge transformation of it that makes it obey the Coulomb gauge condition in that frame. Then use the Lorentz transformation to find values of $$A'^\mu$$ from it in any other frame.