Component form of the fourth Maxwell's equation

Question

I am doing an introduction to Maxwell's equations, and it is said that originally the equations were in component form. Can anyone help with the derivation of the fourth equation? I have checked many sources, and I cannot find the derivation.

Answer 1

It is very clear from Maxwell's original work that he was writing component by component see 1865 A Dynamical Theory of the Electromagnetic Field in chapter III. See equations with letters (A to H).

Here is the example of equation (B):

$$\frac{\mathrm d H}{\mathrm d z } - \frac{\mathrm d G}{\mathrm d y }=\mu \alpha,$$ $$\frac{\mathrm d F}{\mathrm d x } - \frac{\mathrm d H}{\mathrm d z }=\mu \beta,$$ $$\frac{\mathrm d G}{\mathrm d y } - \frac{\mathrm d F}{\mathrm d x }=\mu \gamma.$$

which in modern notation is written as

$$\frac{\mathrm d A_y}{\mathrm d z } - \frac{\mathrm d A_z}{\mathrm d y }=\mu H_x,$$ $$\frac{\mathrm d A_z}{\mathrm d x } - \frac{\mathrm d A_x}{\mathrm d z }=\mu H_y,$$ $$\frac{\mathrm d A_x}{\mathrm d y } - \frac{\mathrm d A_y}{\mathrm d x }=\mu H_z.$$

where $\mathbf H$ is the magnetic force, $\mathbf A$ is the magnetic vector potential (Maxwell called it electromagnetic momentum) and $\mu$ is the permittivity. Equation (B) can be written using vector calculus as $\nabla \times \mathbf A = \mu \mathbf H=\mathbf B$ (here I introduce the magnetic field $\mathbf B$ ). This set of equations above is Gauss's law for magnetic fields, as the gradient of a curl is 0 i.e. $\nabla \cdot \mathbf B=\nabla \cdot (\nabla \times \mathbf A)=0$.

Maxwell equations in modern form vectorial form were introduced by Oliver Heaviside. Here there are Heaviside equations following Maxwell's order, extracted from Wikipedia own article of A Dynamical Theory of the Electromagnetic Field :

If by fourth equation you mean Ampère-Maxwell law, you just need to look at equation (A) and (C). Gauss' law is (G). And Faraday's law is hidden in (D), note that Lorentz' force is $\mathbf f=\mathbf E + \mu \mathbf v \times H$, so with (D) you can say that $\mathbf E=-\nabla \phi + \partial \mathbf A /\partial t$, take the curl and use (B).