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:

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

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:

$\delta U=\int E\delta Dd\tau$

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:

$\delta U=\int E\delta Dd\tau$

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

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

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:

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