In Maxwell's theory, the interaction between two particles is transmitted through local variations of the electromagnetic field.

This propagation of the interaction through the electromagnetic field is possible via electromagnetic waves, which have a celerity \(c\).

To picture the interaction between two particles in the setting of a field theory, there is an image :

Two bottle caps A and B are floating on the water. They are initially steady.

A vertical oscillation of A creates oscillations of the water. The latter are transmitted locally in all directions until they reach the point B. Then point B is set in motion.

Maxwell's equations are true locally.

They express the relationship between the electromagnetic field and its sources :

• \(\mathrm{div} \vec{E} =\frac{\rho}{\epsilon_0}\) (Gauss' law for electricity)

• \(\mathrm{div} \vec{B} =0\) (Gauss' law for magnetism)

• \(\overrightarrow{\mathrm{rot}} \vec E = -\frac{\partial \vec B}{\partial t}\) (Faraday's law of induction)

• \(\overrightarrow{\mathrm{rot}} \vec B = \mu_0 \vec{\jmath} + \epsilon_0 \mu_0 \frac{\partial \vec E}{\partial t}\) (Ampere's law)

• Maxwell's equations and charge conservation :

Maxwell's equations hold the principle of charge conservation.

Indeed, if we apply the divergence to the equation about Ampere's law :

\(\mathrm{div} (\overrightarrow{\mathrm{rot}} \vec B) = 0 = \mu_0 \mathrm{div} \vec{\jmath}+\epsilon_0 \mu_0 \mathrm{div} (\frac{\partial \vec E}{\partial t}) = \mu_0(\mathrm{div} \vec{\jmath}+ \frac{\partial \rho}{\partial t})\)

Thus :

\(\mathrm{div} \vec{\jmath}+ \frac{\partial \rho}{\partial t} = 0\)

Because it is a consequence of Maxwell's equations, it is not necessary to add charge conservation to the statements of electromagnetic.

• Need for displacement current \(\vec{\jmath_D} = \varepsilon _0 \frac{{\partial \vec E}}{{\partial t}}\) :

For any kind of state, let's write :

\(\overrightarrow {rot} \;\vec B = \mu _0 (\vec j + \vec j_D )\)

Then :

\(div\;\vec j = \frac{1}{{\mu _0 }}div(\overrightarrow {rot} \vec B) - div\;\vec j_D = - div\;\vec j_D\)

Indeed : (see the lesson "Vector calculus")

\(div(\overrightarrow {rot} \vec B) = \vec \nabla .(\vec \nabla \wedge \vec B) = 0\)

Moreover, the charge conservation principle induces :

\(div\;\vec j = - div\;\vec j_D = - \frac{{\partial \rho }}{{\partial t}}\)

Gauss' law gives :

\(- div\;\vec j_D = - \frac{{\partial \rho }}{{\partial t}} = - \frac{\partial }{{\partial t}}\left( {\varepsilon _0 div\;\vec E} \right)\)

Thus :

\(div\left( {\vec j_D - \varepsilon _0 \frac{{\partial \vec E}}{{\partial t}}} \right)\; = 0\)

The most simple solution to this equation corresponds to the choice of a displacement current :

\(\vec{\jmath_D} = \varepsilon _0 \frac{{\partial \vec E}}{{\partial t}}\)