Introduction to Mathematical Physics/Electromagnetism/Electromagnetic interaction

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Electromagnetic forces[edit]

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[edit]

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