I'm trying to figure out why the Jefimenko equations are not usually written in terms of 4-vectors. If they are fully relativistic, why not show it?

These are the equations:

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So, the electric and magnetic field are not part of a 4-vector. More like a 4-tensor. But $\rho$ and $\mathbf{J}$ are part of a 4-vector, so it would make sense to write the integrand of $\mathbf{E}$ and $\mathbf{B}$ as the tensor product of this 4-vector with something else.

The easiest thing is to start with the magnetic field. You can write the spatial part of the electro-magnetic tensor as $$ \dfrac{\mathbf{J}}{c}(\mathbf{r}',t_r)dV'\wedge\dfrac{\mathbf{r}-\mathbf{r}}{|\mathbf{r}-\mathbf{r}'|^3} + \dfrac{\partial}{\partial ct}\dfrac{\mathbf{J}}{c}(\mathbf{r}',t_r)dV'\wedge\dfrac{\mathbf{r}-\mathbf{r}}{|\mathbf{r}-\mathbf{r}'|^2} $$

which is the tensor product of 3-vectors so it's consistent. The problem arises when we consider the electric field components, which go to the mixed part of the electro-magnetic tensor.

If the spatial components go by the exterior product of two vectors, then the mixed components should go by this product $$ F_{0i} = a_0b_i - a_ib_0 $$

where $\mathbf{a}$ and $\mathbf{b}$ are the vectors involved in the spatial components of the tensor, and $a_0$, $b_0$ are the time components of the complete 4-vectors.

With this plan in mind, I identified two components of the electric field $$ \dfrac{\partial \rho}{\partial ct}(\mathbf{r}',t_r)dV'\dfrac{\mathbf{r}-\mathbf{r}'}{|\mathbf{r}-\mathbf{r}'|^2} - \dfrac{\partial}{\partial ct}\dfrac{\mathbf{J}}{c}(\mathbf{r}',t_r)dV'\dfrac{\mathbf{r}-\mathbf{r}'}{|\mathbf{r}-\mathbf{r}'|} $$

as coming from the exterior product of $\partial_{ct}(\rho,\mathbf{J}/c)$ and $\dfrac{dV'}{|\mathbf{r}-\mathbf{r}'|}\left(1,\dfrac{\mathbf{r}-\mathbf{r}'}{|\mathbf{r}-\mathbf{r}'|}\right)$, which I'm not sure are 4-vectors.

At this point, it would make total sense if there were two more components equal to $$ \rho(\mathbf{r}',t_r)dV'\dfrac{\mathbf{r}-\mathbf{r}'}{|\mathbf{r}-\mathbf{r}'|^3} - \dfrac{\mathbf{J}}{c}(\mathbf{r}',t_r)dV'\dfrac{1}{|\mathbf{r}-\mathbf{r}'|^2} $$

but alas, there is no $$ -\dfrac{1}{4\pi\epsilon_0}\int\left[\dfrac{1}{|\mathbf{r}-\mathbf{r}'|^2}\dfrac{\mathbf{J}}{c}(\mathbf{r}',t_r)\right]dV' $$

component of the electric field.

Am I deceiving myself? Is there any other way of writing Jefimenko's equations as a manifestly covariant tensor product?

  • $\begingroup$ your title doesn't seem to describe the question $\endgroup$
    – AXensen
    Sep 22, 2023 at 14:35

1 Answer 1


The main issue is that you want to write them as 3D integrals which break the covariance. It's easier to keep a Lorentz invariant from using 4D integrals.

I think that it's perhaps easier to rederive them keeping covariant notation. Your Maxwell equations are (setting $c=\mu_0=1$): $$ \begin{align} \epsilon_{\mu\nu\rho\sigma}F^{\mu\nu,\rho} &= 0 \\ F^{\mu\nu}_{,\nu} = \hat j^\mu \end{align} $$ which in Fourier space become: $$ \begin{align} \epsilon_{\mu\nu\rho\sigma}\hat F^{\mu\nu}ik^\rho &= 0 \\ \hat F^{\mu\nu}ik_\nu = \hat j^\mu \end{align} $$ You can invert these equations provided that you have conservation of charge: $$ \hat j^\mu k_\mu = 0 $$ and you should obtain (the only vectors you have are $k,j$ and you need to build an antisymmetric rank two tensor with them): $$ \hat F^{\mu\nu} = -i\frac{k^\nu \hat j^\mu-k^\mu \hat j^\nu}{k^2} $$ Thus, your Green's function is just: $$ \hat G^{\mu\nu}_\rho = -i\frac{k^\nu \delta^\mu_\rho-k^\mu \delta^\nu_\rho}{k^2} \\ \hat F^{\mu\nu} = \hat G^{\mu\nu}_\rho \hat j^\rho $$ Back in real space, you just need to calculate the Fourier transform: $$ \begin{align} G^{\mu\nu}_\rho &= \int -i\frac{k^\nu \delta^\mu_\rho-k^\mu \delta^\nu_\rho}{k^2} e^{ikx}\frac{d^4k}{(2\pi)^4} \\ &= -(\partial^\nu \delta^\mu_\rho-\partial^\mu \delta^\nu_\rho)\frac{\delta(t-r)}{4\pi r} \end{align} $$ and you recover Jefimenko's equations in covariant form as a 4D convolution by a Lorentz invariant kernel: $$ F^{\mu\nu} = G^{\mu\nu}_\rho \star j^\rho $$

Note that I used the causal kernel, but depending on your uses, other choices be may be more appropriate. In Fourier space, it all comes down to how you treat the poles $k^2=0$ (on shell).

Hope this helps.


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