# Constant currents do not interact with static charge. Why is that?

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:

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?

• your title doesn't seem to describe the question Sep 22, 2023 at 14:35

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).