# Derivation of $~\nabla^2\mathbf{H}=\sigma\mu{\partial\mathbf{H}\over\partial\mathrm{t}}+\epsilon\mu{\partial^2\mathbf{H}\over\partial\mathrm{t^2}}~$

The following equation(subset of Maxwell's equations of electromagnetic wave(s)) is said held in free space.

$$\underbrace{ \color{fuchsia}{\nabla^2\mathbf{H}_{}=\sigma\mu{\partial\mathbf{H}_{}\over\partial\mathrm{t}}+\epsilon\mu{\partial^2\mathbf{H}_{}\over\partial\mathrm{t^2}}} }_{\text{I want to derive this equation}}\tag{1}$$

It is said that above equation can be derived from the following equation but still I am in dought that this statement is true.

$$\nabla^2\mathbf{E}_{}=\sigma\mu{\partial\mathbf{E}_{}\over\partial\mathrm{t}}+\epsilon\mu{\partial^2\mathbf{E}_{}\over\partial\mathrm{t^2}}\tag{2}$$

$$\underbrace{\nabla\times\left(\nabla\times\mathbf{A}_{}\right)=\nabla\left(\nabla\cdot\mathbf A\right)-\nabla^2\mathbf A}_{\text{General formula for any vector}}\tag{3}$$

$$\begin{cases}\nabla\cdot\mathbf E=0\\\nabla\cdot\mathbf H=0\end{cases}~~\leftarrow~~ \text{Holds at freespace} \tag{4}$$

$$\nabla\times \left(\nabla\times \mathbf{E}_{} \right)= \nabla \left( \nabla\cdot \mathbf{E}_{} \right) -\nabla^2 \mathbf{E}_{}$$

$$\nabla\times \left(\nabla\times \mathbf{E}_{} \right)= -\nabla^2 \mathbf{E}_{}$$

$$~~\iff~~ \nabla^2 \mathbf{E}_{} = -\nabla\times \left(\nabla\times \mathbf{E}_{} \right)$$

$$= \underbrace{\sigma\mu{\partial\mathbf{E}_{}\over\partial\mathrm{t}}+\epsilon\mu{\partial^2\mathbf{E}_{}\over\partial\mathrm{t^2}}}_{\text{I've already derived it} }$$

Intuitively I think that it is dangerous to simply substituting $$~ \mathbf{H}_{} ~$$ for $$~ \mathbf{E}_{} ~$$ to obtain the first equation(pink-marked).

How eqn1 can be derived actually?

I don't know where you've seen this, but that source is wrong. This is the equation in a conducting medium, which means you have current $$\vec j = \sigma \vec E$$ (Ohm's law) so no vacuum. Furthermore, if $$\epsilon,\mu\neq\epsilon_0,\mu_0$$ this means you are also assuming an isotropic dielectric and magnetic medium. In general, if you have to appeal to $$\vec D,\vec H$$, it's because you are not in vacuum.

For the derivation of your formula, take the curl of Maxwell-Ampere's law, with the current replaced by Ohm's law, and substitute the curl of $$\vec E$$ by Faraday's law.

Hope this helps and tell me if you need more details.

$$\nabla × \vec{B} = \mu (\sigma \vec{E}) + \mu \epsilon \frac{\partial \vec{E} }{\partial t}$$

Take the curl and sub gauss law for magnetism

$$-\nabla^2 \vec{B} = \mu \sigma \nabla × \vec{E} + \mu \epsilon \frac{\partial \nabla × \vec{E} }{\partial t}$$

$$-\nabla^2 \vec{B} = -\mu \sigma \frac{\partial \vec{B}}{\partial t} - \mu \epsilon \frac{\partial^2 \vec{B}}{\partial t^2}$$

$$\nabla^2 \vec{B} = \mu \sigma \frac{\partial \vec{B}}{\partial t} + \mu \epsilon \frac{\partial^2 \vec{B}}{\partial t^2}$$

• I didn't know that curl operator is a linear operator. Apr 29, 2022 at 12:05
• I think it also best to use faradays law to solve for B, as that way you can identify the correct orientation and magnitude Apr 29, 2022 at 12:21