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Since the photon has no mass, longitudinal electromagnetic waves cannot exist, and ordinary electromagnetic waves are always transversal waves. However, by speculating a bit we could introduce in the lagrangian density of the electromagnetic field a mass term for the photon, I understand that this is precisely what the Proca action does:

$$\mathcal{L}=-\frac{1}{2}(\partial_\mu B_\nu^*-\partial_\nu B_\mu^*)(\partial^\mu B^\nu-\partial^\nu B^\mu)+\frac{m^2 c^2}{\hbar^2}B_\nu^* B^\nu +\mathcal{L}_m, \quad \text{with }B^\mu = \left (\frac{\phi}{c}, \mathbf{A} \right)$$

being $\phi$ the electric potential and $\mathbf{A}$ the magnetic vector potential. My qüestion is: What would be the form of Maxwell's equations of this "electromagnetic" field formed by photons with mass (including sources)?

May guess is that the homogeneous equations remain the same:

$$\boldsymbol{\nabla}\times \boldsymbol{E} = -\frac{\partial \boldsymbol{B}}{\partial t}, \qquad \boldsymbol{\nabla}\cdot\boldsymbol{B} = 0$$

but the equations containing the sources are:

$$\boldsymbol{\nabla}\cdot \boldsymbol{E} + \frac{m^2c^2}{\hbar^2}\boldsymbol{A} = \frac{\rho}{\epsilon_0}, \qquad \boldsymbol{\nabla}\times\boldsymbol{B} + \frac{m^2c^2}{\hbar^2}\boldsymbol{A} = \mu_0\left( \boldsymbol{j} + \frac{\partial \boldsymbol{E}}{\partial t}\right)$$

Davius
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    Does this answer your question? Massive spin-1 field and Proca Lagrangian – Connor Behan Aug 23 '22 at 19:38
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    @ConnorBehan for the chargeless case, yes, but I am interestend in the full Maxwell equation including matter. – Davius Aug 23 '22 at 19:41
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    The Wikipedia article you linked to has Proca's generalization of Maxwell's equations, in covariant form, including charges and currents. It's the fourth equation in the section named "Equation". – Ghoster Aug 23 '22 at 19:47
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    @Ghoster yes, the source free equations. But, I am wondering how is the form of the full Maxwell equations with charged sources. Unfortunately, the Proca Action I gave is incomplete and does not include the sources (this this may be confusing). – Davius Aug 23 '22 at 19:49
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    No, the Proca equation with sources is in there too. It's the equation that starts with $c \mu_0 j^\nu$ on the left-hand side. $j^\nu$ is the source current four-density. – Michael Seifert Aug 24 '22 at 12:07

1 Answers1

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Maxwell homogeneous equations remain the same, because they are source-free:

$$\boldsymbol{\nabla}\times \boldsymbol{E} = -\frac{\partial \boldsymbol{B}}{\partial t} \tag{1}$$ $$ \boldsymbol{\nabla}\cdot\boldsymbol{B} = 0 \tag{2}$$

For Maxwell inhomogeneous equations, Proca field equation is: $$\partial_\nu F^{\lambda\nu} = \mu_0 J^\lambda- (\frac{m_\gamma c}{\hbar})^2 B^\lambda $$ and Maxwell field tensor is $$\begin{pmatrix} 0 & \frac{E^x}{c} & \frac{E^y}{c} & \frac{E^z}{c} \\ -\frac{E^x}{c} & 0 & B_z & -B_y \\ -\frac{E^y}{c} & -B_z & 0 & B_x \\ -\frac{E^z}{c} & B_y & -B_x & 0 \end{pmatrix}$$

Now for $\mu=0$ we have

$$\begin{aligned} \partial_\nu F^{0 \nu} &= \frac{\partial F^{00}}{\partial (ct)} + \frac{\partial F^{0i}}{\partial (x^i)} = \frac{\partial }{\partial x^i}(\frac{E^i}{c}) = \frac{1}{c} \nabla \cdot \mathbf{E} \\&= \mu_0 J^0 - (\frac{m_\gamma c}{\hbar})^2 B^0 = \mu_0 \rho c - (\frac{m_\gamma c}{\hbar})^2 \frac{\phi}{c} \end{aligned}$$

Or

$$\nabla \cdot \mathbf{E} + (\frac{m_\gamma c}{\hbar})^2 \phi =\frac{\rho}{\epsilon_0} \tag{3}$$

Also for $\mu=i$ we have

$$\begin{aligned} \partial_\nu F^{i \nu} &= \frac{\partial F^{i0}}{\partial (ct)} + \frac{\partial F^{ij}}{\partial (x^j)} = \frac{\partial }{\partial (ct)} (-\frac{E^i}{c}) + \frac{\partial }{\partial x^j}(\epsilon^{ijk}B_k) = (-\frac{1}{c^2} \frac{\partial}{\partial t} \mathbf{E} + \nabla × \mathbf{B})^i \\& = \mu_0 J^i - (\frac{m_\gamma c}{\hbar})^2 B^i = \mu_0 \mathbf{J}^i - (\frac{m_\gamma c}{\hbar})^2 \mathbf{A}^i \end{aligned}$$

Or $$\nabla × \mathbf{B} -\frac{1}{c^2} \frac{\partial}{\partial t} \mathbf{E} = \mu_0 \mathbf{J} - (\frac{m_\gamma c}{\hbar})^2 \mathbf{A} \tag{4}$$

Sancol.
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