0
$\begingroup$

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:

enter image description here enter image description here

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?

$\endgroup$
1
  • $\begingroup$ your title doesn't seem to describe the question $\endgroup$
    – AXensen
    Sep 22, 2023 at 14:35

1 Answer 1

1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.