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Let \begin{equation} \boldsymbol{\Phi}=\Bigl(\dfrac{\phi}{c},\mathbf{A}\Bigr) \tag{01} \end{equation} the electromagnetic 4-potential. We know that if its 4-divergence is zero \begin{equation} \dfrac{1}{c^{2}}\dfrac{\partial \phi}{\partial t}\boldsymbol{+}\boldsymbol{\nabla}\boldsymbol{\cdot}\mathbf{A}=0 \quad \text{(the Lorenz condition)} \tag{02} \end{equation} then Maxwell's equations take the elegant form \begin{equation} \Box\boldsymbol{\Phi}=\mu_{0}\mathbf{J} \tag{03} \end{equation} where the so called d'Alembertian \begin{equation} \Box\equiv \dfrac{1}{c^{2}}\dfrac{\partial^{2} \hphantom{t}}{\partial t^{2}}\boldsymbol{-}\nabla^{2} \tag{04} \end{equation} and the 4-current \begin{equation} \mathbf{J}=(c\rho,\mathbf{j}) \tag{05} \end{equation} which has also its 4-divergence equal to zero \begin{equation} \dfrac{\partial \rho}{\partial t}\boldsymbol{+}\boldsymbol{\nabla}\boldsymbol{\cdot}\mathbf{j}=0 \quad \text{(the continuity equation)} \tag{06} \end{equation} and is a 4-vector.

The question is : under these conditions is the 4-potential a 4-vector ??? I ask for a proof or a reference (link,paper,textbook etc) with a proof.

EDIT

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$^\prime$Mainly Electromagnetism and Matter$^\prime$, The Feynman Lectures on Physics, Vol.II, The New Millenium Edition 2010.

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    $\begingroup$ With the conditions $\partial_\mu A^\mu=0$ and $\square A^\mu=j^\mu$ only? No: let $f(\vec x)$ be harmonic. Then $A'^\mu:=A^\mu+\delta^\mu_0 f$ satisfies $\partial_\mu A'^\mu=0$ and $\square A'^\mu=j^\mu$, but it is not a four-vector. $\endgroup$
    – AccidentalFourierTransform
    Jun 16, 2018 at 16:47
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    $\begingroup$ It's a bit annoying. Note that your equations do not specify how $A_\mu$ transforms, but they do specify how it transforms up to a gauge transformation. You can use that gauge freedom to get rid of the function $f$ that AccidentalFourierTransform talks about, recovering the four-vector transformation law. $\endgroup$
    – knzhou
    Jun 16, 2018 at 18:51
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    $\begingroup$ @knzhou, could you elaborate on that? Do you mean that one can choose the 4-tuple $\varphi,\mathbf A$ in each frame in such a way that it transforms as a four-vector, but it is not necessary to obey the wave and Lorenz equation? $\endgroup$
    – Ján Lalinský
    Jul 17, 2018 at 0:12
  • $\begingroup$ Before you ask how something transforms, you need to provide a definition that uniquely defines the quantity w.r.t to each frame (otherwise the question doesn't make sense). $\endgroup$
    – Filippo
    Oct 14, 2022 at 20:31

3 Answers 3

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under these conditions is the 4-potential a 4-vector ??? I ask for a proof or a reference (link,paper,textbook etc) with a proof.

No, not necessarily.

It is convenient to choose it as a 4-vector in relativistic theory. But Maxwell's equations together with the Lorenz constraint do not imply that 4-tuple $(\varphi/c,\mathbf A)$, must transform as four-vector.

The Lorenz condition

$$ \partial_t \varphi + \nabla \cdot \mathbf A = 0 ~~~\text{(in all inertial frames)} $$

and the wave equations

$$ \square^2\varphi = \rho/\epsilon_0 $$

$$ \square^2\mathbf A = \mu_0 \mathbf j $$

do not remove the arbitrariness of potentials completely: in one and the same frame $i$, they can still be changed using any scalar field $\chi_i$:

$$ \tilde{\mathbf A} = \mathbf A + \nabla \chi_i $$ $$ \tilde{\varphi} = \varphi - \partial_t \chi_i $$ provided it obeys the equation $$ \square^2\chi_i =0 $$ and is not constant. That equation has uncountable infinity of different solutions so we can assign one distinct solution to each inertial frame. Even if we begin with 4-tuple of potentials $(\varphi/c,\mathbf A)$ that transform as four-vector, if we then redefine the potentials in every frame using $\chi_i$ unique to that frame, there will be no simple relation between components of potentials in two inertial frames and thus they cannot be connected via the Lorentz transformation. This is impractical to do in this way, but the result - potentials in different frames not connected by the Lorentz transformation - is allowed by the above equations.

The potentials are artificial functions that we are free to define as we please as long as they give the actual electric and magnetic field via the usual formulae. It is possible to extend this definition so they are four-vectors and this is sometimes the most natural choice. For example, the well-known Lienard-Wiechert solution of the wave equations above, when used in all frames, gives electric and magnetic potentials that together transform as a four-vector.

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  • $\begingroup$ The Lienard-Wiechert potentials are the retarded parts. From the moment we remove the advanced parts the Lienard-Wiechert 4-potential couldn't be in any case a 4-vector. $\endgroup$
    – Frobenius
    Jun 16, 2018 at 22:19
  • $\begingroup$ I don't think that Lorentz transformations and gauge transformations interfere with each other. If $\Box A^\mu = J^\mu$ is Lorentz covariant, changing the guage to $\tilde{A}^\mu$ would not change the former's covariance. $\endgroup$
    – Oktay Doğangün
    Jun 16, 2018 at 22:43
  • $\begingroup$ @Frobenius I have never heard that before. When you look up in L&L where they discuss field of arbitrarily moving charge, they define its potentials as four-vector, as function of retarded radius four-vector and retarded four-velocity. I think their covariant formula gives the Lienard-Wiechert potentials. $\endgroup$
    – Ján Lalinský
    Jun 16, 2018 at 23:05
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    $\begingroup$ @OktayDoğangün, I did not mean to say those interfere. But the equation is Lorentz-covariant (the same in all inertial frames), even if the 4-tuple $(\varphi,\mathbf A)$ does not transform as a four-vector. $\endgroup$
    – Ján Lalinský
    Jun 16, 2018 at 23:11
  • $\begingroup$ Take a look in this document of mine : A Feynman Lectures EM Equation. Equations (4-2.26) and (4-2.27) give the Lienard-Wiechert potentials of an arbitrary moving charge. I don't think that they would compose a 4-vector. $\endgroup$
    – Frobenius
    Jun 16, 2018 at 23:29
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After searching in Web, in our PSE site and many books, textbooks, papers and so on I end up with this conclusion :

That the electromagnetic 4-potential $\:A^{\mu}(\mathbf{x},t)\:$ is a 4-vector is an assumption.

In $^{\prime}$Quantum Field Theory$^{\prime}$ by Itzykson C.-Zyber J., Edition 1980, we read (in $\S$1-1-2 Electromagnetic Field as an Infinite Dynamical System):

We assume $\:A^{\mu}(x)\:$ to transform as a four-vector field and the lagrangian as a scalar density in order for the action to be a Lorentz invariant.

Also Ben Crowell commented therein Proof that four-vector potential is a valid four vector:

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.- Ben Crowell Sep 20 '17 at 23:06

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    $\begingroup$ You can start with the four-vector nature of $A_\mu$ as a core postulate - but you absolutely do not need to, and there are plenty of alternative, physically-grounded sets of postulates in which the result is a theorem. There is no single 'best' set of such postulates, which is why you find so much hedging and a lack of clear, positive statements, but your answer's pretense that they do not exist is wrong. $\endgroup$
    – Emilio Pisanty
    Jul 16, 2018 at 19:05
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Yes the four potential $A^{\mu}=(\phi(\vec{x},t),\textbf{A}(\vec{x},t))$ is a four vector and it can be seen from the equation that it satisfies: \begin{align} \partial^{2}A^{\mu}=\frac{1}{c}J^{\mu} \end{align}

the $\partial^{2}$ operator is a scalar and $J^{\mu}$ is a Lorentz vector leading to $A^{\mu}$ being necessarily a four-vector itself.

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  • $\begingroup$ You mean that, because of the $j^\mu$ transform with a constante matrix of $so(3,1)$, and the $\partial^2$ operator conmute with it, it applies directly on the $A^\mu$, transforming it as a 4-vector? $\endgroup$
    – Iliado Odiseo
    Jun 16, 2018 at 19:08
  • $\begingroup$ Yes. It is after all an equality. All the objects are defined by the way that they transform. $J^{\mu}$ transforms under the $(\frac{1}{2},\frac{1}{2})$ representation of the Poincare group (vector representation), so that must be the case also for $A^{\mu}$. $\endgroup$
    – hal
    Jun 16, 2018 at 19:39

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