# Introduction to Mathematical Physics/Electromagnetism/Electromagnetic interaction

## Electromagnetic forces

Postulates of electromagnetism have to be completed by another postulate that deals with interactions:

Postulate:

In the case of a charged particle of charge $q$, Electromagnetic force applied to this particle is:

$f=qE+qv\wedge B$

where $qE$ is the electrical force (or Coulomb force) \index{Coulomb force} and $qv\wedge B$ is the Lorentz force. \index{Lorentz force}

This result can be generalized to continuous media using Poynting vector.\index{Poynting vector}

secenergemag

## Electromagnetic energy, Poynting vector

Previous postulate using forces can be replaced by a "dual" postulate that uses energies:

Postulate:

Consider a volume $V$. The vector $P=E\wedge H$ is called Poynting vector. It is postulated that flux of vector $P$ trough surface $S$ delimiting volume $V$, oriented by a entering normal is equal to the Electromagnetic power ${\mathcal P}$ given to this volume.

Using Green's theorem, ${\mathcal P}$ can be written as:

${\mathcal P}=\int\!\!\!\int E\wedge H n ds =-\int\!\!\!\int\!\!\!\int \mbox{ div } (E\wedge H)d\tau$

which yields, using Maxwell equations to:

${\mathcal P}=\int\!\!\!\int\!\!\!\int H\frac{\partial B}{\partial t}+E.j+E\frac{\partial D}{\partial t}d\tau$

Two last postulates are closely related. In fact we will show now that they basically say the same thing (even if Poynting vector form can be seen a bit more general).

Consider a point charge $q$ in a field $E$. Let us move this charge of $dr$. Previous postulated states that to this displacement corresponds a variation of internal energy:

$\delta U=\int E\delta D d\tau$

where $dD$ is the variation of $D$ induced by the charge displacement.

Theorem:

Internal energy variation is:

$\delta U=-f \delta r$

where $f$ is the electrical force applied to the charge.

Proof:

In the static case, $E$ field has conservative circulation ($\mbox{ rot } E=0$) so it derives from a potential. \medskip Let us write energy conservation equation:

$\delta U=\int E\delta D d\tau$

$\delta U=\int- \mbox{ grad } (V) \delta D d\tau$

$\delta U=\int V \mbox{ div }(\delta D) d\tau-\int \mbox{ div }(V\delta D) d\tau$

Flow associated to divergence of $V\delta D$ is zero in all the space, indeed $D$ decreases as $1/r^2$ and $V$ as $1/r$ and surface increases as $r^2$. So:

$\delta U=\int V \delta \rho d\tau$

Let us move charge of $\delta r$. Charge distribution goes from $q\delta(r)$ to $q\delta(r+dr)$ where $\delta(r)$ is Dirac distribution. We thus have $d \rho(r)=q(\delta(r+\delta r)-\delta(r))$. So:

,

$\delta U=\int\!\!\!\int\!\!\!\int qV(r)(\delta(r+\delta r)-\delta(r)) d\tau$

thus

$\delta U=qV(r+\delta r)-qV(r)$

$\delta U=\mbox{ grad }(qV)\delta r$

Variation is finally $\delta U=-fdr$. Moreover, we prooved that:

$f=-\mbox{ grad }(qV)=qE$