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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.

VoulKons
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3 Answers3

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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.

Ján Lalinský
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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

VoulKons
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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.

hal
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