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.

  • 2
    $\begingroup$ I am not a physicist. Could you check whether the following is what you are looking for: J.C. Maxwell, "VIII. A dynamical theory of the electromagnetic field," Philosophical transactions of the Royal Society of London, Vol. 155, 1865, pp. 459-512 (scan online, Google scan online) $\endgroup$
    – njuffa
    Aug 1 at 22:23

1 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 :

Heaviside's equations

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).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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