# Derivation of $\mathbf{J}_{l}$ and $\mathbf{J}_{t}$ in the Coulomb gauge in Jackson's EMT book

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

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

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

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.

• Thank you very much. Of course, I will let you know if find mistakes. Apr 27, 2022 at 18:05