# Proof that four-vector potential is a valid four vector

The definition of $A^{\mu}$ that I've been given is as an object created from the scalar field $\phi$ and the vector field $\vec{A}$ i.e:

$$A^{\mu} = (\phi,\frac{\vec{A}}{c})$$ [in cgs units]

How do I formally prove that this is in fact a valid four vector, and not some jumble of objects?

(The Proof that four-potential is a four-vector question does not answer this - it posits some strange specific case.)

• it follows from the fact that $F^{\mu \nu}$ transforms as a second rank tensor, those follow from the Maxwell equations Sep 20 '17 at 21:20
• This type of question can't be answered generically. It depends on what assumptions you start from. Someone could choose a logical framework in which the four-vectorial nature of the potential is one of the postulates.
– user4552
Sep 20 '17 at 23:06

# Background: what are these potentials?

To see that this is a four-vector we must first understand where it comes from. We have the Maxwell equations, where I will use $\dot X = \partial_w X = c^{-1} \partial_t X$ to write as, \begin{align} \nabla\cdot E &= 4\pi\rho& \nabla\times E&= - \dot B\\ \nabla\cdot B &= 0 &\nabla\times B&= 4\pi c^{-1} J + \dot E, \end{align} and after using $\nabla\cdot B = 0$ to define a family of $A$ via $B = \nabla\times A$ and the consequent $\nabla \times (E + \dot A) = 0$ to define a family of $\varphi$ via $E = -\dot A - \nabla \varphi,$ we can derive that the remaining two equations state,\begin{align} -\nabla\cdot\dot A - \ddot\varphi &= 4\pi\rho\\ \nabla(\nabla\cdot A) - \nabla^2 A &= 4\pi c^{-1}J - \ddot A - \nabla\dot\varphi \end{align} which can be handily rewritten (by defining $\Box X =\ddot X -\nabla^2 X,$ and $\lambda = \dot\varphi + \nabla\cdot A$) as,\begin{align} \Box\varphi &= 4\pi\rho + \dot \lambda,\\ \Box A &= 4\pi c^{-1}J - \nabla\lambda. \end{align} Now we could have chosen a different set of potentials and still obtained the same fields: we know that we can add any $\nabla\chi$ to $A$ and preserve $B=\nabla\times A$ because the curl of a gradient is zero; looking at what's needed to preserve $E$ tells us only that we should simultaneously subtract $\dot\chi$ from $\varphi$ to preserve $E$. This does not affect the form of the last two equations directly, but it does map the value of $\lambda\mapsto \lambda - \Box \chi,$ and so we can essentially choose a different functional form for $\lambda,$ on the grounds that if we want, say, $\lambda = \dot\varphi$ (Coulomb gauge) we can take any other solution $(A, \varphi)$ and calculate its $\lambda$ and then solve $\Box\chi = \lambda - \dot\varphi$ to find a $\chi$ which gives us a different set of fields which has this functional form.

Of course the Lorenz gauge that we now assume is to solve for $A,\varphi$ such that $\lambda=0,$ yielding,\begin{align} \Box\varphi &= 4\pi \rho\\ \Box A_x &= 4\pi c^{-1}J_x\\ \Box A_y &= 4\pi c^{-1}J_y\\ \Box A_z &= 4\pi c^{-1}J_z \end{align}

I am going through all of this for three reasons:

1. You claimed your equation was valid in CGS units and I think you are mistaken,
2. To highlight that even when you find a solution it is not unique; other solutions $\Box\alpha = 0$ can be added to any component;
3. To assert that there is a clear order here: one starts from knowing the charge and current densities and one then solves for these fields, which can be used to derive the $E$ and $B$ fields if one wishes.

# Use the fact that $(\rho, J/c)$ is a four-vector.

The rough proof that $(\rho, J/c)$ is a four-vector involves imagining a static charge distribution first: a static charge distribution $(\rho_0, 0)$ becomes $(\gamma~\rho_0, -\gamma~ \rho_0 \vec \beta)$ under a Lorentz transform; this is easily seen to be a $(\rho, J/c)$ pair. But if this is a four-vector then a more complex arrangement is a four-vector, precisely because an arbitrary charge-current distribution can be arbitrarily-well approximated by a superposition of a bunch of little static charge-current distributions that have been boosted in various ways. If all of them individually transform appropriately over Lorentz transforms, then their sum must, too, because Lorentz transforms are linear.

If you agree with me that $(\rho, J/c)$ is a four-vector then this must also be the case for $(\varphi, A/c)$, also by a linearity argument, though it is a sufficiency and not a necessity (as it must be: the Lorenz gauge has not 100% nailed down the exact fields $\varphi$ and $A$, so there must be other fields that are not the Lorentz transform of these fields which also pass the Lorenz gauge and are valid).

Just examine the equations in the Lorenz gauge, after a Lorentz boost of the charge fields by $\beta$ in the $\hat z$ direction: \begin{align} \Box\phi' &= 4\pi ~\gamma~(\rho - \beta c^{-1}J_z)\\ \Box A_x' &= 4\pi c^{-1}~J_x\\ \Box A_y' &= 4\pi c^{-1}~J_y\\ \Box A_z' &= 4\pi c^{-1}~\gamma~(J_z - \beta c\rho) \end{align}It is clearly sufficient to solve these equations to find $A_x' = A_x$ and $A_y' = A_y.$ For $A_z'$ it is clearly sufficient to have $A_z' = \gamma~A_z - \gamma\beta\phi$ and for $\varphi' = \gamma~\varphi - \gamma\beta A_z.$

Thus: if you have solved $(\rho, J/c)$ for the fields $(\varphi, A)$ in the Lorenz gauge, then you can obtain a valid solution for the fields $(\varphi', A')$ that you would get from solving the same equations for the Lorentz-boosted $(\rho', J'/c)$ simply by Lorentz-transforming the fields $(\varphi, A)$ as a four-vector.

In other words: there are ways to do electromagnetism in which that pair is not a four-vector, but it does no harm to assume that it is.

• In my opinion the fact that the 4-dimensional quantity $\:\mathbf{J}=\left(c\rho,\mathbf{j}\right)\:$ is a Lorentz transformed 4-vector is not proved via the continuity equation. But, when applying the Lorentz transformation to the Maxwell equations and trying to leave their form unchanged (co-variant) we are inevitably constraint to define this in the new frame in such a way that finally it obeys the Lorentz transformation. In this definition it's necessary to use the Lorentz velocity 4-vector. Please correct me if I am wrong. Jun 9 '18 at 22:59
• ...for me not forget to remind you : you must swap $\:A^\mu, J^\mu\:$ in equation $\:A^\mu = \mu_0~\Box J^\mu$. Jun 9 '18 at 22:59
• Thank you for the comment about the latter. As for the former, I don't think I'm the right person to speak to this any more... I no longer am a big fan of the "how does it transform under coordinate transformation" physicsy view of what makes an $[a, b]$ tensor a tensor. What I would say is that there is probably a way, if one comes from that perspective, to short-circuit all of the debate... for small $\beta$ we know that we get back the nonrelativistic Galilean transform and I think one can argue that $\rho$ and $J$ mingle from there first. Jun 10 '18 at 2:35
• I disagree with the first paragraph. If $f$ is a function that satisfies $\square f=0$, then $\square (A^\mu +\delta^\mu_0 f)=\mu_0 J^\mu$. But we cannot conclude from this that $A^\mu +\delta^\mu_0 f$ is a vector -- indeed, it is not. Jun 10 '18 at 2:35
• @AccidentalFourierTransform Yeah this is why I'm no longer a big fan of that enterprise of doing things... once I got into some differential geometry it became much more clear that when you have something that's "obviously" not a tensor, like, say, $\Gamma^k_{~~ij}$, you actually need to be very careful about what's obvious because that entity does, in fact, represent a valid tensor; the objection is really that the tensor does not have that function of being a Christoffel symbol after a coordinate transform. I'd guess the same is true for your example? Jun 10 '18 at 2:42

I am answering this question for a homework so I'll try to share here what I am thinking:

Let us take for granted that at least $$J^\mu$$ is a 4-vector. Then, let's go back to Maxwell equations, which in the Lorentz gauge ($$\partial_\mu A^\mu=0$$) are (I'm doing GR so I'm using the metric $$\eta_{\mu\nu}=diag(-1,1,1,1)$$, and there are a couple of different signs with respect to the particle physics one, hopefully that should not interfere with the reasoning) $$\begin{equation} \Box A^\mu = -J^\mu \end{equation}$$ If $$J^\mu$$ is a 4-vector, it is the quantity $$\Box A^\mu$$ (and not just $$A^\mu$$) that must behave like a 4-vector.

As already pointed out by some previous answers, as $$\Box$$ is a Lorentz scalar, it follows that $$A^\mu$$ is a 4-vector, up to a function with null Dalembertian: if $$\tilde A^\mu(x)$$ is the 4-vector that satisfies $$\Box \tilde A^\mu=-J^\mu$$, and $$\chi^\mu$$ is any collection of functions which satisfy $$\Box \chi^\mu=0$$ for all $$\mu$$, then the quantity $$\begin{equation} A^\mu = \tilde A^\mu+\chi^\mu \end{equation}$$satisfies Maxwell's equations and the requirement that $$\Box A^\mu$$ be a 4-vector. This is not yet enough to state that $$A^\mu$$ is a 4-vector, but the additional constraint of the Lorentz gauge imposes $$\begin{equation} 0=\partial_\mu A^\mu=\partial_\mu\tilde A^\mu+\partial_\mu \chi^\mu\Rightarrow \partial_\mu\tilde A^\mu=-\partial_\mu \chi^\mu \end{equation}$$so again, the quantity $$\partial_\mu \chi^\mu$$ must behave like a scalar, which means that the set of functions $$\chi^\mu$$ has a pure 4-vector component $$\tilde\chi^\mu$$, and is defined up to a function $$h^\mu$$, not necessarily a 4-vector, but with null divergence: $$\begin{equation} \chi^\mu=\tilde \chi^\mu+h^\mu, \partial_\mu h^\mu=0 \end{equation}$$ Finally: $$\begin{equation} A^\mu=\underbrace{\tilde A^\mu}_{4-v}+\underbrace{\tilde \chi^\mu}_{4-v}+h^\mu \end{equation}$$Changing reference frame, $$\tilde A^\mu$$ and $$\tilde \chi^\mu$$ will transform as 4-vectors into $$\tilde A'^\mu$$ and $$\tilde \chi'^\mu$$, while $$h^\mu$$ in principle might transform in some $$h'^\mu$$ which is not $$\Lambda^\mu\,_\nu h^\nu$$, and which needs not satisfy $$\partial_\mu' h'^\mu=0$$. But thanks to the gauge invariance, we can always perform a redefinition of the field $$A^\mu$$ which cancels the nonzero 4-divergence: $$\begin{equation} A^\mu\rightarrow A'^\mu=\underbrace{\tilde A'^\mu}_{4-v}+\underbrace{\tilde \chi'^\mu}_{4-v}+h'^\mu+\partial'^\mu \Phi \end{equation}$$ $$\begin{equation} \partial_\mu h'^\mu+\partial_\mu\partial^\mu \Phi=0 \Rightarrow \Box\Phi =-\partial_\mu h'^\mu \end{equation}$$The function $$\Phi$$ can always be determined using the Green function for the Dalembertian. Everything on the left hand side is a scalar, so $$h'^\mu$$ can only be a 4-vector $$\tilde h'^\mu$$ plus a divergenceless part: $$\begin{equation} h'^\mu = \tilde h'^\mu+g'^\mu,\quad\partial_\mu' g'^\mu=0 \end{equation}$$So the field in this new frame is $$\begin{equation} A'^\mu=\underbrace{\tilde A'^\mu}_{4-v}+\underbrace{\tilde \chi'^\mu}_{4-v}+\underbrace{\tilde h'^\mu}_{4-v}+\underbrace{\partial'^\mu \Phi}_{4-v} +g'^\mu \end{equation}$$Comparing the two potentials in the two frames: $$\begin{equation} \begin{array}{cccccc} A^\mu & = & \underbrace{\tilde A^\mu+\tilde \chi^\mu}_{4-v} & + & \underbrace{h^\mu}_{\partial_\mu h^\mu=0} & \\ A'^\mu & = & \underbrace{\tilde A'^\mu+\tilde \chi'^\mu+\tilde h'^\mu+\partial'^\mu \Phi}_{4-v} & + & \underbrace{g'^\mu}_{\partial'_\mu g'^\mu=0}&\\ A'^\mu & = & \underbrace{\tilde A'^\mu+\tilde \chi'^\mu}_{4-v} & + & \underbrace{g'^\mu}_{\partial'_\mu g'^\mu=0}& \Rightarrow (\text{choose }\partial'^\mu \Phi=-h'^\mu) \end{array} \end{equation}$$Long story short: choosing suitably $$\Phi$$ we could cancel $$\tilde h'^\mu$$ out and at least make the 4-vector part of the two fields coincide. But we'll never be able to cut the non-vector, divergenceless part, out of the field, which then will remain globally a non-4-vector.

Imho, $$A^\mu$$ needs to be defined to be a 4-vector.

The four-vectors are defined by transformation laws. If the physical effects of $A^\mu$ in one frame are the same as physical effects of $\Lambda^\mu_\nu A^\nu$ in different frame - then one can say that $A$ is a four-vector.

Physical effects can be seen from the Maxwell equations, which have contravariant form when expressed in terms of the Field Strength tensor:

$$\partial_\mu F^{\mu\nu} = 0$$

Transformation law of $F$ implies four-vector transformation law of $A$.

See Electromagnetic Tensor for details.

• I would be very grateful to you if you show me the way by which the transformation law of the electromagnetic tensor $\:F^{\mu\nu}\:$ implies the four-vector Lorentz transformation law of the 4-potential $\:\mathbb A=\left(\mathbf A, \phi/c\right)=\left(\mathrm A_1, \mathrm A_2,\mathrm A_3,\mathrm A_4=\phi/c\right)\:$... (1)... Jun 13 '18 at 14:59
• ...(1)...For simplicity take the 1+1-dimensional case to show me how to prove the equations \begin{align} \mathrm A'_1 & = \gamma\left(\mathrm A_1-\dfrac{\upsilon}{c} \mathrm A_4\right) =\gamma\left(\mathrm A_1-\dfrac{\upsilon\phi}{c^{2}} \right) \tag{01a}\\ \mathrm A'_2 & = \mathrm A_2 \tag{01b}\\ \mathrm A'_3 & = \mathrm A_3 \tag{01c}\\ \mathrm A'_4 & = \gamma\left(\mathrm A_4-\dfrac{\upsilon}{c} \mathrm A_1\right) \quad \text{that is} \nonumber\\ \phi'& = \gamma\left(\phi-\upsilon \mathrm A_1 \right) \tag{01d} \end{align} Many thanks in advance. Jun 13 '18 at 15:05
• I downvoted your answer as incorrect. Jul 17 '18 at 5:39

@thesundayscientist. No, the Maxwell 4-potential $A^{\mu}$ is not a valid 4-vector field (i.e. transforming under the fundamental representation of $\mathcal{Lor} \left(3,1\right)$, or equivalently, under the $\left(1/2,1/2\right)$ representation of $\mbox{SL}(2,\mathbb{C})$), at least not in quantum field theory. The gauge symmetry spoils the correct transformation law under an element $\Lambda \in \mathcal{Lor} \left(3,1\right)$ of the quantized 4-potential field, as shown by S.Weinberg on pages 249-251 of the 1st volume of his famous treatise on QFT. However, the so-called Faraday tensor of electrodynamics $F_{\mu\nu}$ is a valid antisymmetric Lorentz tensor field, because the curl drops the gauge term. This is also spelled out by Weinberg, op.cit.

EDIT: There's a conflict of definitions here which is not addressed. A 4-vector field in classical field theory is [completely ignoring all diff.geom. aspects] a smooth assignment of a 4-tuple of functions to every point in spacetime, with the property that, under a change of inertial observers whose coordinates of their respective points are transformed by a restricted Lorentz transformation, they are transformed covariantly: $V^{\mu} (x) \longrightarrow V'^{\mu}(x') = \Lambda^{\mu}_{~\nu}V^{\nu}(x)$. Under this definition, you can say that the Maxwell 4-potential is a 4-vector. A 4-vector field in quantum field theory is defined as per Wightman axioms as: the 4-tuple of Fock-space operator-valued distributions on the flat Minkowski spacetime whose covariance under $\mbox{ISL}(2,\mathbb{C})$ is given by . What Wightman is saying is that the quantized electromagnetic potential - precisely due to the gauge invariance coming from the masslessness of the classical field - cannot be covariant under $\mbox{ISL}(2,\mathbb{C})$ in the sense of eqn. (4.1.82a) of Lopuszanski's book on axiomatical QFT

.

• The fact that it has gauge symmetry doesn't mean it's not a valid 4-vector. A 4-vector is defined by how it transforms not by its uniqueness Sep 23 '17 at 21:57
• That is what Weinberg is showing. A 4-vector field of zero mass is not covariant under a unitary representation of the subgroup of the universal covering group of the restricted Poincaré group isomorphic to the universal covering group of the restricted Lorentz group. So the Maxwell 4-potential is a not a 4-vector under $ISL(2,C)$. Sep 23 '17 at 22:04
• Maybe you should change the answer to just the edit part? It seems pretty clear the OP was talking about classical field theory.
– user12029
Sep 23 '17 at 23:23
• It's annoying when people answer simple questions with overly complicated answers just to show off here. Sep 24 '17 at 2:29
• I downvoted your answer because : "It's annoying when people answer simple questions with overly complicated answers..." Jul 17 '18 at 6:14