I Show 4-Potential is 4-Vector in Lorenz Gauge

[majinbeeb]

I am trying to show that the four-potential is a four-vector when working in the Lorenz gauge. In this gauge, we have $\Box A^{\mu} = 4\pi J^{\mu}$. If we perform a Lorentz transformation, we can show that $\Box A'^{\mu} = \Box \Lambda_{\nu}^{\mu} A^{\nu}$. From what I have seen, people have used this to conclude that $A'^{\mu} = \Lambda_{\nu}^{\mu} A^{\nu}$, but I don't see why this is necessarily the case. If we had any function $\xi$ such that $\Box \xi =0$, then we could have $A'^{\mu} = \Lambda_{\nu}^{\mu} A^{\nu} + \partial^{\mu} \xi$ instead, where $\partial^{\mu} \xi$ is not necessarily zero.

In a previous thread, someone invoked the quotient theorem. However the quotient theorem describes situations like if $B_{\mu} C^{\mu \nu}$ is a tensor whenever $B_{\mu}$ is a tensor, then $C^{\mu \nu}$ is a tensor. This does not apply to our case, as we only know that $S A^{\mu}$ is a tensor only when $S = \Box$; we do not have the universal quantification necessary to invoke the quotient theorem.

 

[PAllen]

You can always add a function (meeting appropriate requirements) to the vector potential. It thus only ever determined up to such functions. Thus, you can say the 4-potential can be treated as a 4-vector with no loss of generality. You can perversely choose not to, I guess.

 

[majinbeeb]

You can always add a function (meeting appropriate requirements) to the vector potential. It thus only ever determined up to such functions. Thus, you can say the 4-potential can be treated as a 4-vector with no loss of generality. You can perversely choose not to, I guess.

Thanks for the response. I do not find the idea of making a choice to set the function $\xi$ to zero too troubling, but making such a choice feels somewhat artificial (at least when making this choice directly). Would it be possible to exclude such a term for more physical reasons, such as boundary conditions at infinity? A solution to $\Box \xi = 0$ is a linear combination of plane waves propagating at the speed of light, and these do not vanish at infinity, so we should choose $\xi =0$ for this reason. Does this line of reasoning make sense/ seem plausible?

Edit: Boundary conditions at infinity are tricky in a relativistic framework, the new time coordinate $t'$ in particular is problematic. I was kind of hoping for a condition that, when taken together with the Lorenz gauge condition, would give uniqueness of the four-potential.

 

[vanhees71]

You can always add a function (meeting appropriate requirements) to the vector potential. It thus only ever determined up to such functions. Thus, you can say the 4-potential can be treated as a 4-vector with no loss of generality. You can perversely choose not to, I guess.

Why peversely? Often the Coulomb gauge is more convenient. Then you use a non-covariant four-potential. Of course, this is through the gauge dependence of the four-potential. The electromagnetic field is represented by the antisymmetric 2nd-rank field-strength tensor which is unanimously a Minkowski tensor.

Another very elegant possibility is to use the Riemann-Silberstein vector $\vec{\mathfrak{F}}=\vec{E}+\mathrm{i} \vec{B}$, where the proper orthochronous Lorentz transformations are represented by $\mathrm{SO}(3,\mathbb{C})$ matrices. The rotations are of course represented by the usual $\mathrm{SO}(3)$ subgroup, and the rotation-free boosts by rotations with imaginary angles (which take the physical meaning of $\mathrm{i}$ times rapidity).