# 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 ${\displaystyle q}$, Electromagnetic force applied to this particle is:

${\displaystyle f=qE+qv\wedge B}$

where ${\displaystyle qE}$ is the electrical force (or Coulomb force) \index{Coulomb force} and ${\displaystyle qv\wedge B}$ 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 ${\displaystyle V}$. The vector ${\displaystyle P=E\wedge H}$ is called Poynting vector. It is postulated that flux of vector ${\displaystyle P}$ trough surface ${\displaystyle S}$ delimiting volume ${\displaystyle V}$, oriented by a entering normal is equal to the Electromagnetic power ${\displaystyle {\mathcal {P}}}$ given to this volume.

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

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

which yields, using Maxwell equations to:

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

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

${\displaystyle \delta U=\int E\delta Dd\tau }$

where ${\displaystyle dD}$ is the variation of ${\displaystyle D}$ induced by the charge displacement.

Theorem:

Internal energy variation is:

${\displaystyle \delta U=-f\delta r}$

where ${\displaystyle f}$ is the electrical force applied to the charge.

Proof:

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

${\displaystyle \delta U=\int E\delta Dd\tau }$

${\displaystyle \delta U=\int -{\mbox{ grad }}(V)\delta Dd\tau }$

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

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

${\displaystyle \delta U=\int V\delta \rho d\tau }$

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

,

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

thus

${\displaystyle \delta U=qV(r+\delta r)-qV(r)}$

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

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

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