Introduction to Mathematical Physics/Energy in continuous media/Electromagnetic energy

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Introduction[edit]

At section Electromagnetic energy, it has been postulated that the electromagnetic power given to a volume is the outgoing flow of the Poynting vector. \index{Poynting vector} If currents are zero, the energy density given to the system is:


dU=HdB+EdD

Multipolar distribution[edit]

It has been seen at section Electromagnetic interaction that energy for a volumic charge distribution \rho is \index{multipole}


U=\int \rho V d\tau

where V is the electrical potential. Here are the energy expression for common charge distributions:

  • for a point charge q, potential energy is: U=qV(0).
  • for a dipole \index{dipole} P_i potential energy is: U=\int V\mbox{ div }(P_i\delta)=\partial_i V.P_i.
  • for a quadripole Q_{i,j} potential energy is: U=\int    V(\partial_i\partial_jQ_{i,j}\delta)=\partial_i\partial_j V.Q_{i,j}.

Consider a physical system constituted by a set of point charges q_n located at r_n. Those charges can be for instance the electrons of an atom or a molecule. let us place this system in an external static electric field associated to an electrical potential U_e. Using linearity of Maxwell equations, potential U_t(r) felt at position r is the sum of external potential U_e(r) and potential U_c(r) created by the point charges. The expression of total potential energy of the system is:


U_t=\sum q_n (V_c(r_n)+V_e(r_n))

In an atom,\index{atom} term associated to V_c is supposed to be dominant because of the low small value of r_n-r_m. This term is used to compute atomic states. Second term is then considered as a perturbation. Let us look for the expression of the second term U_e=\sum q_n V_e(r_n). For that, let us expand potential around r=0 position:


U_e=\sum q_n V_e(r_n)=\sum q_n
(U(0) +x_i^n\partial_i(U)+
\frac{1}{2}x_i^nx_j^n\partial_i\partial_j(U)+\dots)

where x_i^n labels position vector of charge number n. This sum can be written as:


U_e=\sum q_n U(0)+\sum q_nx_i^n\partial_i(U)+\frac{1}{2}\sum
q_nx_i^nx_j^n\partial_i\partial_j(U)+\dots

the reader recognizes energies associated to multipoles.

Remark: In quantum mechanics, passage laws from classical to quantum mechanics allow to define tensorial operators (see chapter Groups) associated to multipolar momenta.

Field in matter[edit]

In vacuum electromagnetism, the following constitutive relation is exact:

eqmaxwvideE


D=\epsilon_0E

eqmaxwvideB


H=\frac{B}{\mu_0}

Those relations are included in Maxwell equations. Internal electrical energy variation is:


dU=EdD

or, by using a Legendre transform and choosing the thermodynamical variable E:


dF=DdE

We propose to treat here the problem of the modelization of the function D(E). In other words, we look for the medium constitutive relation. This problem can be treated in two different ways. The first way is to propose {\it a priori} a relation D(E) depending on the physical phenomena to describe. For instance, experimental measurements show that D is proportional to E. So the constitutive relation adopted is:

D=\chi E

Another point of view consist in starting from a microscopic level, that is to modelize the material as a charge distribution is vacuum. Maxwell equations in vacuum eqmaxwvideE and eqmaxwvideB can then be used to get a macroscopic model. Let us illustrate the first point of view by some examples:

Example:

If one impose a relation of the following type:


D_i=\epsilon_{ij}E_j

then medium is called dielectric .\index{dielectric} The expression of the energy is:


F=F_0+\epsilon_{ij}E_iE_j

Example:

In the linear response theory \index{linear response}, D_i at time t is supposed to depend not only on the values of E at the same time t, but also on values of E at times anteriors. This dependence is assumed to be linear:


D_i(t)=\epsilon_{ij}*E_j

where * means time convolution.

Example:

To treat the optical activity [ph:elect:LandauEle], a tensor \index{optical activity} a_{ijk} such that:


D_i=\epsilon_{ij}E_j+a_{ijk}E_{j,k}

is introduced. Not that this law is still linear but that D_i depends on the gradient of E_i.

The second point of view is now illustrated by the following two examples:

Example:

A simple model for the susceptibility: \index{susceptibility} An elementary electric dipole located at r_0 can be modelized (see section Modelization of charge) by a charge distribution \mbox{ div } (p\delta (r_0)). Consider a uniform distribution of N such dipoles in a volume V, dipoles being at position r_i. Function \rho that modelizes this charge distribution is:

\rho=\sum_V \mbox{ div } (p_i\delta (r_i))

As the divergence operator is linear, it can also be written:

\rho=\mbox{ div } \sum_V (p_i\delta (r_i))

Consider the vector:

eqmoyP


P(r)=\lim_{d\tau\rightarrow 0}\frac{\sum_{d\tau}p_i}{d\tau}

This vector P is called polarization vector\index{polarisation}. The evaluation of this vector P is illustrated by figure figpolar.

figpolar

Polarization vector at point r is the limit of the ratio of the sum of elementary dipolar moments contained in the box d\tau over the volume d\tau</math> as it tends towards zero.}

Maxwell--gauss equation in vacuum


\mbox{ div } \epsilon_0E=\rho

can be written as:


\mbox{ div } (\epsilon_0E-P)=0

We thus have related the microscopic properties of the material (the p's) to the macroscopic description of the material (by vector D=\epsilon_0E-P). We have now to provide a microscopic model for p. Several models can be proposed. A material can be constituted by small dipoles all oriented in the same direction. Other materials, like oil, are constituted by molecules carrying a small dipole, their orientation being random when there is no E field. But when there exist an non zero E field, those molecules tend to orient their moment along the electric field lines. The mean P of the p_i's given by equation eqmoyP that is zero when E is zero (due to the random orientation of the moments) becomes non zero in presence of a non zero E. A simple model can be proposed without entering into the details of a quantum description. It consist in saying that P is proportional to E:

P=\chi  E

where \chi is the polarisability of the medium. In this case relation:

D=\epsilon_0E-P

becomes:

D=(\epsilon_0+\chi )E

Example: A second model of susceptibility: Consider the Vlasov equation (see equation eqvlasov and reference [ph:physt:Diu89]. Function f is the mean density of particles and n_0 represents the density of the positively charged background.

vlasdie

\frac{\partial f}{\partial t}+{v}\partial_x f+\frac{F}{m}\partial_v f= 0

let us assume that the force undergone by the particles is the electric force:

\vec{F}=-eE(x,t)

Maxwell equations are reduced here to:

eqmaxsystpart

\mbox{ div } E=\rho

where electrical charge \rho(x,t) is the charge induced by the fluctuations of the electrons around the neutral equilibrium state:

\rho=-e\int f(x,v,t)dv+en_0

Let us linearize this equation system with respect to the following equilibrium position:

\begin{matrix}
f(x,v,t)&=&f^0(v)+f^1(x,v,t)\\
F(x,v,t)&=&0+F_1
\end{matrix}

As the system is globally electrically neutral:


\int f^0(v) = n_0

By a x and t Fourier transform of equations vlasdie and eqmaxsystpart one has:

\begin{matrix}
\epsilon_0 ik \hat{E_1}&=&-e\int \hat{f_1} dv\\
-i\omega \hat{f_1}+ivk\hat{f_1}&=&e\frac{\hat{E_1}}{m} \frac{\partial
\hat{f^0}}{\partial v} 
\end{matrix}

Eliminating \hat{f_1} from the previous system, we obtain:


ik\hat{E_1}(\epsilon_0 - \frac{e^2}{km}\int
\frac{1}{vk-\omega}\frac{\partial \hat f^0}{\partial v}dv)=0

The first term of the previous equation can be considered as the divergence of a vector that we note D which is D=\epsilon*E_1, where * is the convolution in x and t:

eqmaxconvol

\mbox{ div }(\epsilon *E_1)=0

Vector D is called electrical displacement. \epsilon is the susceptibility of the medium. Maxwell equations eqmaxsystpart describing a system of charges in vacuum has thus been transformed to equation eqmaxconvol that described the field in matter. Previous equation provides \epsilon(k,\omega):


\epsilon (k,\omega)=(\epsilon_0 - \frac{e^2}{km}\int
\frac{1}{vk-\omega}\frac{\partial \hat{f}^0}{\partial v}dv)