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

• I
• majinbeeb
In summary, the four-potential can be treated as a four-vector in the Lorenz gauge, but it is not always determined up to functions. Another option is to use the Riemann-Silberstein vector, which has the advantage of being rotational invariant.

#### 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.

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
PAllen said:
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.

Last edited:
PAllen said:
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).