I have read, in many places, statements like this:

Heaviside was able to greatly simplify Maxwell's 20 equations in 20 variables, replacing them by four equations in two variables. Today we call these 'Maxwell's equations' forgetting that they are in fact 'Heaviside's equations'.

... but nowhere can I find exactly what these equations look like in the form Heaviside produced. There are many versions of the equations, most with more than just four.

(From this biographical article)

Please can anyone tell me exactly what are the original "four equations in two variables"?


1 Answer 1


This gives the four equations in the form Heaviside came up with: $$\varepsilon E = \rho$$ $$\nabla \times E = - \mu \frac{\partial H}{\partial t}$$ $$\nabla \cdot \mu H = 0$$ $$\nabla \times H = k E + \varepsilon \frac{\partial E}{\partial t}$$

where $E$ represents the electric field, $H$ represents the magnetic field, $\varepsilon$ is the permittivity, $\mu$ is the permeability, $\rho$ is the charge density, and $k$ is the conductivity.

Here, $\times$ and $\cdot$ denote the cross product and dot product. Also, $H$ is not the same as the perhaps more commonly used $B$.

Note: The first equation should properly be $\varepsilon \nabla \cdot E = \rho$, though the source leaves it out, probably due to a typo.

  • $\begingroup$ @HarryWeston Glad to help. $\endgroup$
    – HDE 226868
    Jun 14, 2015 at 18:46
  • 1
    $\begingroup$ It is important to note that in modern notation the magnetic field $B$ and the field $H$ are different entities: en.wikipedia.org/?title=Magnetic_field#The_H-field $\endgroup$
    – hjhjhj57
    Jun 14, 2015 at 19:02
  • $\begingroup$ @HDE226868 did you forget a divergence in the first equation? $\endgroup$
    – Danu
    Jun 14, 2015 at 19:44
  • $\begingroup$ @Danu Interestingly enough, I technically didn't, as that form is what is given. But there should be, in the correct form. $\endgroup$
    – HDE 226868
    Jun 14, 2015 at 19:46
  • $\begingroup$ @HDE226868 I think it's very probable that that is some kind of typo in the source. I think adding a note on this mistake (it is certainly the wrong equation!!) in your answer may be a good idea. I doubt that Heaviside wrote it down (or at least intended to write it down) like that. The correct equation would be $\nabla\cdot \varepsilon E=\rho$ $\endgroup$
    – Danu
    Jun 14, 2015 at 19:47

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