Maxwell's equation in microscopic formulation and speed of EM-waves

Join the discussion
Registration is free. Start your own thread to ask a follow-up.
3 replies · 2K views
Homework Helper
Insights Author
Messages
6,002
Reaction score
2,629
Starting from the microscopic form of Maxwell's equations and following standard mathematical procedure outlined in
Inhomogeneous electromagnetic wave equation - Wikipedia
we can have as end result the following equations:
$$(\nabla^2-\frac{1}{c^2}\frac{\partial^2}{\partial t^2})\mathbf{E}=\frac{\nabla\rho}{\epsilon_0}+\mu_0\frac{\partial \mathbf{J}}{\partial t}$$
$$(\nabla^2-\frac{1}{c^2}\frac{\partial^2}{\partial t^2})\mathbf{B}=\mu_0\nabla\times\mathbf{J}$$

These two equations seem to imply that the speed of the waves is always c (due to the ##\frac{1}{c^2}## term appearing in front of the second time derivative in the left hand side). On the right hand side of course the ##\rho## and ##\mathbf{J}## are not the free charge and current density but rather the total ##\rho=\rho_{free}+\rho_{bound}## and ##\mathbf{J}=\mathbf{J_{free}}+\mathbf{J_{bound}}##. But what's the catch here, how can the exact nature of the right hand side (that depends on the materials used) can affect the left hand side and the speed of the waves?
 
Physics news on Phys.org
Of course for em. waves in a medium the phase velocity changes as we well know from optics (##c_{\text{med}}=c/n## with ##n## the refractive index, but that's of course only an approximation).

You have to take into account that ##\vec{J}_{\text{bound}}## is a function(al) of the fields themselves. So you have to first find a self-consistent equation for the field and these currents. One model is to assume a homogeneous distribution bound elastically (small external fields) and subject to friction when in motion. For a detailed discussion, see Sommerfeld, Lectures on Theoretical Physics, vol. 4 (optics) or Jackson.
 
  • Like
Likes   Reactions: Delta2
vanhees71 said:
Of course for em. waves in a medium the phase velocity changes as we well know from optics (##c_{\text{med}}=c/n## with ##n## the refractive index, but that's of course only an approximation).

You have to take into account that ##\vec{J}_{\text{bound}}## is a function(al) of the fields themselves. So you have to first find a self-consistent equation for the field and these currents. One model is to assume a homogeneous distribution bound elastically (small external fields) and subject to friction when in motion. For a detailed discussion, see Sommerfeld, Lectures on Theoretical Physics, vol. 4 (optics) or Jackson.
Thanks @vanhees71 . This seems to be the only successful resolution of this, that the current density (bound and possibly free) is a function of the fields, like for example when ##\mathbf{J}=\sigma\mathbf{E}##. Thanks again.
 
  • Like
Likes   Reactions: vanhees71
This is of course for a conductor.

For a bound electron the equation of motion reads
$$m \ddot{\vec{x}}+m \gamma \dot{\vec{x}} + m \omega_0^2 \vec{x}=q \vec{E}(t,\vec{x}),$$
where ##\vec{E}## is the total electric field at the position of the particle and ##q=-e## the charge of electron. Now let ##N## be the number density of bound electrons. Then the polarization of the medium is
$$\vec{P}=q N \vec{x}.$$
So multiplying the equation of motion with ##q N## leads to
$$\ddot{\vec{P}}+\omega_0 \gamma \dot{\vec{P}} + \omega_0^2 \vec{P}=\omega_{\text{P}}^2 \vec{E}$$
with the plasma frequency
$$\omega_{\text{P}}=\sqrt{\frac{N q^2}{m}}.$$
Now assume a harmonic time dependence for all fields and assume further that the frequency is much lower than the relaxation time of the medium, you can assume that
$$\vec{P}(\omega,\vec{x})=[\epsilon(\omega,\vec{x})-1] \vec{E}(\omega,\vec{x}),$$
which is in linear response approximation (valid for fields \vec{E} small against the binding fields of the electrons) and ##\epsilon(\omega,\vec{x})## is the usual permittivity of the medium. Plugging this into the equation of motion and the Maxwell equations,
$$\vec{\nabla} \times \vec{E}=-\frac{1}{c} \partial_t \vec{B}, \quad \vec{\nabla} \cdot{\vec{B}}=0, \quad \vec{\nabla} \cdot \vec{E}=0, \quad \vec{\nabla} \times \vec{B}=\frac{1}{c} (\partial_t \vec{E}+\partial_t \vec{P}).$$
leads to
$$\epsilon(\omega,\vec{x}) = 1+\frac{\omega_{\text{P}}^2}{\omega_0^2-\omega^2-\mathrm{i} \gamma \omega}.$$
Plugging all this into the equation for ##\vec{E}## leads to a propagator ##G## such that
$$\vec{P}(t,\vec{x})=\int_{\mathbb{R}} \mathrm{d} t' G(t-t',\vec{x}) \vec{E}(t',\vec{x})$$
with
$$G(t)=\int_{\mathbb{R}} \mathrm{d} \omega \frac{1}{2 \pi} [\epsilon(\omega,\vec{x})-1].$$
and you can calculate the propagation of a wave entering from the vacuum into the medium by using the appropriate boundary conditions (a la Fresnel's equations with ##n(\omega,\vec{x})=\sqrt{\epsilon(\omega,\vec{x})}##).

The assumption of a practically instantaneously relaxing medium leads to dispersion only in frequency and time. Under more general conditions, e.g., for a plasma you also have to take into account spatial dispersion. For details see Landau&Lifshitz vol. 8.
 
  • Like
Likes   Reactions: Delta2