So the vector potential $\mathbf{A}$ satisfies the following equation
$$ \begin{equation} \nabla^2\mathbf{A} - \frac{1}{c^2}\frac{\partial^2 \mathbf{A}}{\partial t^2} =-\mu_o\mathbf{J} + \frac{1}{c^2}\nabla \frac{\partial \Phi}{\partial t} \end{equation}$$

In Eq. (6.25) it says that the current density $\mathbf{J}$ can be decomposed as
$$\begin{equation} \mathbf{J} = \mathbf{J}_{l} + \mathbf{J}_{t} \end{equation} $$

where $\nabla \times \mathbf{J}_{l} = 0 $ and $\nabla \cdot \mathbf{J}_{t} = 0 $. He goes on by stating the following identities $$ \begin{equation} \nabla \times (\nabla \times \mathbf{J}) = \nabla(\nabla\cdot \mathbf{J})- \nabla^2\mathbf{J} \end{equation} $$ and $$ \begin{equation} \nabla^2\left(\frac{1}{|\mathbf{x}-\mathbf{x'|}}\right) = -4\pi\delta(\mathbf{x}-\mathbf{x'}). \end{equation} $$ He then states that expression of $\mathbf{J}_{l}$ and $\mathbf{J}_t$ can be obtained from $\mathbf{J}$ $$ \begin{equation} \mathbf{J}_{l} =-\frac{1}{4\pi} \nabla\int\frac{\nabla'\cdot \mathbf{J}}{|\mathbf{x}-\mathbf{x'}|} d^3\mathbf{x'} \end{equation} $$ $$ \begin{equation} \mathbf{J}_{t} =-\frac{1}{4\pi} \nabla \times \nabla \times \int\frac{\mathbf{J}}{|\mathbf{x}-\mathbf{x'}|} d^3\mathbf{x'}. \end{equation} $$ My question is how does he obtain the above two expressions for $\mathbf{J}_{t}$ and $\mathbf{J}_{l}$?


1 Answer 1


They are derived from the above identity, I’ll fill in the mathematical gaps. For $J_l$ think of electrostatics. Its already curl free, and you have:

$$ \nabla \cdot J_l= \nabla \cdot J $$

which is essentially Gauss’ law. You can solve this by convolving the charge with the Coulomb solution and taking the gradient which is the first equation.

For $J_t$ think of magnetostatics. Its already divergence free, and you have:

$$ \nabla \times J_t= \nabla \times J $$

which is Ampere’s law. This time, you use Biot-Savart and take the curl. Actually, this method gives you:

$$ J_t = \frac{1}{4\pi}\nabla \times \int \frac{\nabla’\times J}{|x-x’|}d^3x’ $$

which you can massage to your expression using integration by parts and taking the curl outside the integration.

Hope this helps and tell me if you find some mistakes.

  • $\begingroup$ Thank you very much. Of course, I will let you know if find mistakes. $\endgroup$
    – Fonon
    Apr 27, 2022 at 18:05

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