Due to electron's intrinsic property of spin and electric charge, a magnetic dipole moment is generated. It is easy to visualise the magnetic dipole moment as the orbital motion of the electron resulting in a current loop around the atom. The magnetic dipole moment is defined as $$ \textbf{m}=I\textbf{A}, $$ where $ I $ is the current generated by the electron's motion, and $ \textbf{A} $ is the vector defining the area enclosed by the current loop. Hence, magnetisation is defined as $$ \textbf{M}=N\textbf{m}, $$ where $ N $ is the number density of atoms in the dielectric. Magnetisation is defined by the Right-Hand Rule.

Current Densities

Show Definitions Consider a small element of a dielectric material. Taking a layer of this element which is an atom thick, we can see that the net current is generated along the surface in a direction determined by the orientation of the current loops (look at the diagram); within the layer, the adjacent current loops act to oppose each other and cancel out. Hence, for the dielectric element, a current, $ I_{element} $ is generated along the faces which are perpendicular to the direction of the magnetic moment of the element $ \textbf{m}_{element} $. $$ m_{element}=MlA=I_{block}A $$ where $ lA $ is the volume of the element. Hence, Surface magnetisation current density $ J_{sm}, $ which is defined as current per unit width of the surface, is $$ J_{sm}=\frac{I_{block}}{L}=\frac{ML}{L}=M $$ As a vector, $ J_{sm} $ is defined as $$ \textbf{J}_{sm}=\textbf{M} \times \hat{\textbf{n}} $$ For uniform $ \textbf{M}, $ meaning that $ \textbf{M} $ is that same for each element, when multiple elements are placed adjacent to each other, the current travels along the outer faces. The current along the inner shared faces cancel out.

The Magnetisation current density $ \textbf{J}_{m} $ is $$ \textbf{J}_{m}=\nabla \times \textbf{M}. $$ Returning to polarisation, polarisation current density can be determined by considering the time variation of $ \textbf{P}: $ $$ \textbf{J}_{p}=\frac{\partial \textbf{P}}{\partial t}=Nq\frac{\partial \textbf{d}}{\partial t}=Nq\textbf{v}. $$ This suggest that there is a variation in the dipole separation over time. Hence, the bound charges move, which produces a current density.

Visualisation Guide

The visualisation on the right allows you to construct an arbitrary shape using small dielectric element. Press the directional buttons, the first six buttons, to add elements in specific direction. Note that you cannot overlap the elements. The surface arrows shows $ \textbf{J}_{sm}. $ The arrow head is represented by a red line. The "RESET!" button resets the number of elements to one, and "SHOW THE NORMAL VECTOR!" shows the normals vectors of each of the faces, represented by a black line. Finally, you can rotate the direction of uniform magnetization, based in spherical polar coordinate system, by using the sliders. Note the z-axis. Enjoy!

A conductor contains mobile free electrons. The positively charged ion left behind is usually fixed so we investigate only the behavior of the free electrons in an electric field.

The Drude Model

The Drude Model studies the motion of electrons in material by assuming that there are only two forces on the electrons,

1. electric force from applied electric field, $ \mathbf{F}_E $

2. force arises from the collisions among the electrons and between electrons and ions, $ \mathbf{F}_c $

so the coulombic forces between electrons and ions are neglected.

By Newton's second law of motion, $$ m \mathbf{a} = \mathbf{F}_{total} = \mathbf{F}_E + \mathbf{F}_c $$ So, $$ m_e \frac{\partial \mathbf{v}_e}{\partial t} = -e \mathbf{E} - \frac{m_e \mathbf{v}_e}{\tau_c} $$ where $ \tau_c $ is the electron collision time.

We consider only oscillatory electric fields (or alternating current) to test the material so $ E_{applied} \propto e^{-i \omega t} $ or $ \mathbf{E} = \mathbf{E}_0 e^{-i \omega t} $, where $ \omega $ is the angular frequency of the field and $ \mathbf{E}_0 $ is real. In this case, the solution to the equation is such that $ \mathbf{v}_e \propto e^{-i \omega t} $.

Now we obtain $$ -i \omega t m_e \mathbf{v}_e = -e \mathbf{E} - \frac{m_e \mathbf{v}_e}{\tau_c} $$ so $$ \mathbf{v}_e = -\frac{e}{m_e} \frac{\tau_c}{1-i \omega \tau_c} $$ Recall that the conduction current density is given by $$ \mathbf{J}_c = N_i e \mathbf{v}_i - N_e e \mathbf{v}_e $$ where N is the number density, subscript $ i $ is for ions and $ e $ for electrons. In general, $ \mathbf{v}_e \gg \mathbf{v}_i $ given the same initial conditions for ions and electrons because ions are much heavier than electrons. We assume that the first term tends to zero so $$ \mathbf{J}_c = - N_e e \mathbf{v}_e $$ Taking the result for $\mathbf{v}_e$ from above, we get $$ \mathbf{J}_c = \frac{\sigma \mathbf{E}}{1 - i \omega \tau_c} $$ where $\sigma = \frac{N_e e^2 \tau_c}{m_e}$. We can rewrite $\mathbf{J}_c$ in terms of complex exponential form: $$ \mathbf{J}_c = \frac{\sigma}{\sqrt{1+\omega^2 \tau^2}} e^{i\tan^{-1} {(\omega \tau})} \mathbf{E} $$ The phase shift of $ \mathbf{J}_c $ to $ \mathbf{E} $ is given by $ \tan^{-1} {(\omega \tau}) $. An alternative way of looking at the phase shift between $ \mathbf{J}_c $ and $ \mathbf{E} $ is that $ \mathbf{J}_c $ is analogous to the velocity of the electrons and $ \mathbf{E} $ is analogous to the acceleration of the electrons. (In fact, it is just a matter of scaling.)

This phase shift comes out of the fact that this is effectively a damped driven harmonic oscillation. The source of the driving force is the applied electric field while the source of the damping force is the collisions between electrons or electrons and ions.

In a damped driven harmonic oscillation, the phase shift of the steady state is due to the amount of time for the transient phase to decay away.

Charge Rearrangement Time

Given some arbitrary electric field, the free electrons would rearrange themselves until the total electric field inside the conductor is 0. This arrangement happens within the charge rearrangement time $ \tau^{\star} = \frac{\epsilon}{\sigma} $, where $ \epsilon = \epsilon_0\epsilon_r $, $ \epsilon_0 $ is the permittivity of free space, $ \epsilon_r $ and $ \sigma $ are the relative permittivity and the conductivity of the material respectively. A physical insight to $ \tau^{\star} $ is that $ \epsilon $ is 'a measurement of how much the material dampens the electric field felt by the charges' while $ \sigma $ is 'a measurement as to how ready is the charge to move given some electric field'. So the charge rearrangement time $ \tau^{\star} $ is a characteristic of the conductor.

The derivation, based on the Drude Model is given below: $$ \mathbf{J}_c = \frac{N_e e^2 \tau_c}{m_e \sqrt{1+\omega^2 \tau^2}} e^{i\tan^{-1} {(\omega \tau})} \mathbf{E}_{applied} $$ where $ \mathbf{E}_{applied} $ is the applied electric field. The phase shift of $ \mathbf{J}_c $ to $ \mathbf{E}_{applied} $ is given by $ \tan^{-1} {(\omega \tau}) $.

We consider only oscillatory electric fields (or alternating current) to test the material so $ \mathbf{E}_{applied} = \mathbf{E}_0 e^{-i \omega t} $, where $ \omega $ is the angular frequency of the field and $ \mathbf{E}_0 $ is real.

In a conductor, the electron collision time $ \tau_c $ is much smaller than the 'field timescale' $ T = \frac{2\pi}{\omega}, $ i.e. $ \tau_c \ll T $ or $ \frac{\tau_c}{T} \ll 1 $ so $ \omega \tau_c \ll 1. $ Taking the limits $ \omega \tau_c + 1 \rightarrow 1 $ and $ \tan^{-1} {(\omega \tau}) \rightarrow 0, $ $$ \mathbf{J}_c = \sigma \mathbf{E} $$ This is Ohm's Law.

Approximating the material to be linear, meaning that charges respond to electric fields by producing a current that is a linear function of the electric field they feel, we can use Ohm's law: $$ \mathbf{J}_c = \sigma \mathbf{E} $$ where $ \sigma $ is a constant (not a function position because we take the material to be homogeneous) scalar (not a tensor because we take the material to be isotropic).

Conservation of free charge requires $$ \frac{\partial \rho_f}{\partial t} = - \nabla \cdot \mathbf{J}_c $$ Substituting Ohm's Law for the electric field, $$ \frac{\partial \rho_f}{\partial t} = - \sigma \nabla \cdot \mathbf{E} $$ $$ = - \sigma \frac{\rho_f}{\epsilon} = - \frac{\sigma}{\epsilon} \rho_f $$ So, $$ \rho_f = \rho_{f, initial} e^{-\frac{t}{\tau^{\star}}} $$ where $ \tau^{\star} = \frac{\epsilon}{\sigma}. $

Good or Poor Conductor ?

Not to be confused with electrical conductivity, the conductivity we are talking about here is how responsive the material is to any changes in the electric field applied to it. We consider only oscillatory electric fields (or alternating current) to test the material so $ E_{applied} \propto e^{-i \omega t}, $ where $ \omega = \frac{2\pi}{T} $ is the angular frequency of the field.

Now we ask the question: Are the free electrons rearranging themselves fast enough to 'cancel out' $ \mathbf{E}_{applied} $? In other words, we want to know if the timescale of electron movement is small enough relative to the timescale of the electric field changing in order for $ \mathbf{E}_{applied} $ to be cancelled out. And so an important quantity for us is the ratio of timescales $ \frac{\tau^{\star}}{T}. $ That ratio turns out to be $ \frac{\tau^{\star}}{T} = \frac{1}{2\pi} \frac{\mid \mathbf{J}_d \mid}{\mid \mathbf{J}_c \mid}, $ or $ \omega \tau^{\star} = \frac{\mid \mathbf{J}_d \mid}{\mid \mathbf{J}_c \mid} $, which is shown later.

If the free electrons take much less time than the wave period to rearrange, i.e. $ \omega \tau^{\star} \ll 1, $ then we expect the material to be a good conductor, as they can move quickly enough to cancel out the applied field. The 'good conductor' criterion depends on $ \omega $ and $ \tau^{\star} $ so it depends on both the material and the applied electric field.

The conduction current density is $$ \mathbf{J}_c = \sigma \mathbf{E}_{applied}, $$ (This is the same from the Ohm's law approximation as above)

$ \mathbf{J}_c $ is a current density due to electrons experiencing a non-zero electric field. Notice $ \sigma $ is a property of the material, which relates the electric field to the observed conduction current.

The displacement current density is $$ \mathbf{J}_d = \frac{\partial \mathbf{D}}{\partial t} = \frac{\partial }{\partial t} (\epsilon \mathbf{E}_{applied}) $$ $ \mathbf{J}_d $ is an apparent current density due to the rate of change of the electric field. Notice $ \epsilon $ is a property of the material, which relates the electric field to the observed displacement current. Notice that both the conduction current density and the displacement current density are linearly proportional to the amplitude of the applied electric field (if the field amplitude is doubled, the rate of change of the field value is also doubled). In this case, $ E_{applied} \propto e^{-i \omega t}, $ and so $$ \mathbf{J}_d = -i \omega \epsilon \mathbf{E}_{applied} $$ We only care about how quickly the field is changing and the material properties in order to classify a good or a poor conductor, not about the value of the field itself. We can achieve such a quantity by dividing the magnitudes of the two currents, as both are only linearly proportional to the electric field amplitude. $$ \frac{\mid \mathbf{J}_d \mid}{\mid \mathbf{J}_c \mid} = \frac{\omega \epsilon}{\sigma} $$ Using the charge rearrangement time $ \tau^{\star} = \frac{\epsilon}{\sigma}, $ we see $$ \frac{\omega \epsilon}{\sigma} = \omega \tau^{\star} = 2\pi \frac{\tau^{\star}}{T} $$ which is the ratio of electron movement timescale to the field timescale (period). And so $$ \frac{\omega \epsilon}{\sigma} $$ dictates whether the material is a good conductor for an EM field of that frequency.

Skin Effect

When an alternating current passes through a conductor, the alternating electric field produces an alternating magnetic field. This magnetic field in turn creates an electric field that opposes the original alternating current, known as the "back emf". The back emf is the strongest at the centre of a conductor, pushing the current to the surface of the conductor.

The current diffusion equation is given by $$ \nabla^2 \mathbf{J}_c = \mu_0 \sigma \frac{\partial \mathbf{J}_c}{\partial t} $$ where $\mathbf{J}_c$ is the conduction current density in the conductor. The LHS of the above above equation gives the gradient field of the flow of current density in the conductor. The current density of an alternating current in a cylindrical wire can be written as $\mathbf{J}_c = \widetilde{J}_c (r) e^{-i \omega t} \mathbf{\hat{z}}$, which gives $$ \nabla^2 \widetilde{J}_c = -\kappa ^2 \widetilde{J}_c $$ where $ \kappa^2 = i \omega \mu_0 \sigma. $ This is a Helmholtz Equation, where $\kappa$ provides a measure of the wavenumber. Hence, $$ \kappa = \sqrt{\mu_0 \omega \sigma} \, i^{\frac{1}{2}} = \frac{1+i}{\delta} $$ where $ \delta = \sqrt{ \frac{2}{\mu_0 \sigma \omega} } $ is the skin depth, a current diffusion length scale. This means that $ J_c $ reduces by a factor of $ \frac{1}{e} $ after travelling for one $ \delta. $

Let $ a $ be the radius of a wire. If the AC frequency $ ( \omega ) $ is low, $ \delta = \sqrt{ \frac{2}{\mu_0 \sigma \omega} } \gg a $ and $ \mathbf{J}_c $ is roughly uniform throughout the wire. However, if $ \omega $ is high, $ \delta \ll a $ and $ \mathbf{J}_c $ is confined within the layer of width $ \approx \delta $ on the surface on the wire.

We shall only talk about collisionless plasmas in oscillatory electric fields.

Based on the Drude Model, $$ \mathbf{J}_c = \frac{N_e e^2 \tau_c}{m_e \sqrt{1+\omega^2 \tau^2}} e^{i\tan^{-1} {(\omega \tau})} \mathbf{E}_{applied} $$ where $ \mathbf{E}_{applied} $ is the applied electric field. The phase shift of $ \mathbf{J}_c $ to $ \mathbf{E}_{applied} $ is given by $ \tan^{-1} {(\omega \tau}). $

We consider only oscillatory electric fields (or alternating current) to test the material so $ \mathbf{E}_{applied} = \mathbf{E}_0 e^{-i \omega t}, $ where $ \omega $ is the angular frequency of the field and $ \mathbf{E}_0. $ Taking the limits $ \omega \tau_c + 1 \rightarrow \omega \tau_c $ and $ \tan^{-1} {(\omega \tau}) \rightarrow \frac{\pi}{2}, $ $$ \mathbf{J}_c \approx \frac{N_e e^2 \tau_c}{m_e} e^{i \frac{\pi}{2}} \mathbf{E}_{applied} $$ $ E^{real} $ and $ J^{real}_c $ are out of phase by $ \frac{\pi}{2}. $

Plasmas Oscillation

Consider a cloud of neutral plasma left alone to reach its equilibrium distribution in the absence of external electromagnetic field. What happens if we displace one of the electrons in the plasma slightly? (Hint: the keywords are equilibrium and displace slightly.)

The electron should be sitting at the bottom of the potential well initially. The small displacement will result in simple harmonic motion of the electron. Since plasma is a cloud of free electrons and ions, this small displacement actually causes the entire cloud to oscillate hence the collective behaviour of plasma oscillations.

It turns out that plasma oscillates at the plasma frequency: $$ \omega_p = \sqrt{\frac{N_e e^2}{m_e \epsilon_0}} $$ where $ N_e $ is the number density of electrons, $ e $ is the fundamental charge and $ \epsilon_0 $ is the permittivity of free space. Notice how the plasma frequency here depends only on the characteristics of electron distribution. This is due to the assumption that the electrons moving much faster than the positive ions in a neutral plasma.

How do we justify this assumption? Consider the lightest element, hydrogen. Ignoring the heavier isotopes, the simplest hydrogen atom consists of a proton and an electron. The mass of a proton is roughly 1800 times greater than that of an electron. Given the same initial conditions, an electron would be accelerated much more than a proton in an electric field that is not too strong. Therefore, in any plasmas, the positive ions are always much heavier than the electrons and so the movement of the positive ions can be ignored given that the electric field is not too strong.

Derivation: Displacing an electron in a neutral plasma means that the number density of electron, $ N_e $ depends on the position and time. Let $ N_e = N_0 + N_1 (x, t), $ where $ N_0 $ is the initial number density of electron and $ x $ is the direction of perturbation. Generally, the free charge density $ \rho_f $ is given by $ \rho_f = N_i e - N_e e, $ where $ N_i $ is the number density of positive ions. In this case, $ \rho_f = N_i e - (N_0 + N_1(x, t))e $ but $ N_i e = N_0 e $ for a neutral plasma so $$ \rho_f = -N_1(x, t)e $$ Also, in general, the conduction current density $ \mathbf{J}_c $ is given by $$ \mathbf{J}_c = N_i e \mathbf{v}_e - N_e e \mathbf{v}_e $$ Based on the assumption of small perturbation of electron such that the movements of positive ions can be ignored, the first term vanishes so $ \mathbf{J}_c = - N_e e \mathbf{v}_e = -(N_0 + N_1(x, t)) e \mathbf{v}_e. $ Also based on the assumption of small perturbation, $ N_0 \gg N_1, $ so $$ \mathbf{J}_c = - N_0 e \mathbf{v}_e $$ In a closed system, we require conservation of free charge, $$ \frac{\partial \rho_f}{\partial t} + \nabla \cdot \mathbf{J}_c = 0 $$ so, $$ -e\frac{\partial N_1}{\partial t} - N_0 e \nabla \cdot \mathbf{v}_e = 0 $$ Consider only in the direction of perturbation, $$ -e\frac{\partial N_1}{\partial t} - N_0 e \frac{\partial v_{e, x}}{\partial x} = 0 $$ We can rewrite this as $$ \frac{\partial v_{e, x}}{\partial x} = -\frac{1}{N_0} \frac{\partial N_1}{\partial t} $$ Assuming that $$ \frac{\partial^2 v_{e, x}}{\partial x \partial t} = \frac{\partial^2 v_{e, x}}{\partial t \partial x} $$ We get (1): $$ \frac{\partial^2 v_{e, x}}{\partial x \partial t} = -\frac{1}{N_0} \frac{\partial^2 N_1}{\partial t^2} \tag{1} $$ By Gauss' Law for electrostatics, $$ \nabla \cdot \mathbf{E} = \frac{\rho}{\epsilon_0} $$ where $ \rho = \rho_f + \rho_p, $ $ \rho_f $ is the free charge density while $ \rho_p $ is the polarisation charge density. In plasmas, $ \rho_p = 0 $ so $$ \nabla \cdot \mathbf{E} = \frac{\rho_f}{\epsilon_0} $$ From the result above, $ \rho_f = -N_1(x, t)e $ so $$ \nabla \cdot \mathbf{E} = -\frac{N_1 e}{\epsilon_0} $$ Consider only in the direction of perturbation, $$ \frac{\partial E_x}{\partial x} = -\frac{N_1 e}{\epsilon_0} $$ Now, by Newton's second law, $$ m_e \frac{\partial \mathbf{v}_e}{\partial t} = -e \mathbf{E} $$ Consider only in the direction of perturbation, $$ m_e \frac{\partial v_{e, x}}{\partial t} = -e E_x $$ So $$ \frac{\partial^2 v_{e, x}}{\partial x \partial t} = - \frac{e}{m_e} \frac{\partial E_x}{\partial x} $$ From the result for $ \frac{\partial E_x}{\partial x} $ above, we obtain (2): $$ \frac{\partial^2 v_{e, x}}{\partial x \partial t} = - \frac{e}{m_e} \cdot -\frac{N_1 e}{\epsilon_0} = \frac{e^2}{m_e \epsilon_0} N_1 \tag{2} $$ Equating (1) and (2) and rearranging the result, $$ \frac{\partial^2 N_1}{\partial t^2} + \frac{N_0 e^2}{m_e \epsilon_0} N_1 = 0 $$ This is the equation of simple harmonic motion for plasma, with a natural frequency of $$ \omega_p = \sqrt{\frac{N_0 e^2}{m_e \epsilon_0}} $$

The Key Idea

Consider a magnetic material that stretches to infinity. Placing it in a uniform magnetic field $ \mathbf{B}_{applied} $ will induce a magnetisation field $ \mathbf{M} $ in it. This will give rise to a total magnetic field $ \mathbf{B}_{total}. $

In general, $ \nabla \times \mathbf{M} = \mathbf{J}_m, $ where $ \mathbf{J}_m $ is the magnetisation current density. However, $ \nabla \cdot \mathbf{M} $ depends on the material.

Define the magnetising field, $ \mathbf{H} $ as $ \mathbf{H} = \frac{1}{\mu_0}\mathbf{B}_{total} - \mathbf{M}, $ where $ \mu_0 $ is the permeability of free space. In general, $ \nabla \cdot \mathbf{H} = - \nabla \cdot \mathbf{M} $ and $ \nabla \times \mathbf{H} = \mathbf{J}_c + \frac{\partial \mathbf{D}}{\partial t}, $ where $ \mathbf{J}_c $ is the conduction current density. (Note: $ \frac{\partial \mathbf{D}}{\partial t} = \mathbf{J}_d $ - displacement current density.)

In a magnetic material, $ \mathbf{M} = \chi_m \mathbf{H}, $ where $ \chi_m $ is the magnetic susceptibility of the material. So, $$ \mathbf{H} = \frac{1}{\mu_0\mu_r}\mathbf{B}_{total}, $$ where $ \mu_r = 1 + \chi_m. $

Note: The units of the B-field are different to those of the M and H fields, hence field line density in this visualisation only provides a qualitative representation of field strength

The Subtlety

You might notice that the total magnetic field remaining constant when $ \mu_r $ is varied. The logic behind is similar to that of D field remains constant with varying $ \epsilon_r. $ The change in total magnetic field is complemented by the change in magnetisation field to keep the total magnetic field constant.

Now let us state the Maxwell's equations (differential) in matter: $$ \nabla \cdot \mathbf{D} = \rho_f $$ $$ \nabla \cdot \mathbf{B} = 0 $$ $$ \nabla \times \mathbf{E} = -\frac{\partial \mathbf{B}}{\partial t} $$ $$ \nabla \times \mathbf{H} = \mathbf{J}_c + \frac{\partial \mathbf{D}}{\partial t} $$ We shall use these equations to study dielectrics, conductors and plasmas in the following visualisations.

In conductors Maxwell's equations are the same as in non-conducting media, except for the curl of B, which becomes: $$ \nabla \times \mathbf{B} = \mu \epsilon \frac{\partial \mathbf{E}}{\partial t} + \mu \sigma \mathbf{E}. $$ This reduces to the equation for nonconducting media in the case when $ \sigma = 0. $ From this we can obtain plane wave solutions: $$ \tilde{\textbf{E}}(z,t) = \tilde{\textbf{E}}_0 e^{i(\tilde{k}z-\omega t)}, $$ $$ \tilde{\textbf{B}}(z,t) = \tilde{\textbf{B}}_0 e^{i(\tilde{k}z-\omega t)} $$ Where $ \tilde{k} $ is the complex wave number, given by: $$ \tilde{k}^2 = \mu \epsilon \omega^2 + i \mu \sigma \omega. $$ The imaginary part of $ \tilde{k} $ results in attenuation inside the conductor, as seen previously, but this time the attenuation strength is dependent on the frequency of the wave and sigma; it is also significantly stronger than attenuation in dielectrics. The attenuation is due to the fact that, in a conductor, electrons move in such a way as to oppose any electric field. In a good conductor the electrons can move faster to oppose the electric field and with a faster frequency of EM wave, it is easier for the electrons to oppose the electric field.

Furthermore, since $ \tilde{k} $ can be expressed as $ \tilde{k} = Ke^{i\phi} $ and $ \tilde{\textbf{B}}_0 = \frac{Ke^{i\phi}}{ \omega} \tilde{\textbf{E}}_0. $

the B field is phase shifted, from E, by $ \phi $ inside the conductor.

A dispersion relation relates a wave's angular frequency $ \omega $ to its wavevector $ k $.

The dispersion relation for a plasma is given by $$ k = \frac{\omega}{c}\Big(1 - \frac{\omega_p^2}{\omega^2}\Big)^{\frac{1}{2}}, $$ where $ c $ is the speed of light, the plasma frequency $ \omega_p = \sqrt{\frac{N_e e^2}{m_e \varepsilon_0}} $, $ N_e $ is the plasma free electron number density, e is the electron charge, $ m_e $ is the mass of an electron and $ \varepsilon_0 $ is the vacuum permittivity.

Try varying the $ N_e $ slider to in order to change $ \omega_p $. Move the $ \omega $ slider to change the frequency of the wave. Notice that when $ \omega $ is less than $ \omega_p $, the wavevector $ k $ becomes imaginary, and the wave is completely attenuated by the plasma.

The dispersion relation for a vacuum is also shown in the graph. As the number density of electrons in the plasma $ N_e $ is decreased, it becomes more like a vacuum, so it makes sense that the plasma dispersion relation should tend to the vacuum relation. Another way of thinking about it is that increasing the frequency of the wave relative to the plasma frequency (the frequency of oscillation of the electrons within the plasma) means that the electrons cannot react quickly enough to affect the wave, so the effect of passing through the plasma becomes negligible.

Derivation

To derive this dispersion relation, we start with Maxwell's equations. $$ \nabla\cdot\boldsymbol{E} = \frac{\rho_f}{\varepsilon} $$ $$ \nabla\cdot\boldsymbol{B} = 0 $$ $$ \nabla \times \boldsymbol{E} = -\frac{\partial \boldsymbol{B}}{\partial t} $$ $$ \nabla \times \boldsymbol{B} = \mu \boldsymbol{J}_c + \mu \varepsilon \frac{\partial \boldsymbol{E}}{\partial t} $$, where $ \boldsymbol{B} $ is the magnetic field, $ \boldsymbol{E} $ is the electric field, $ \mu $ is the permeability, $ \varepsilon $ is the permittivity, $ \rho_f $ is the free charge density and $ J_c $ is the conduction current density.

For a plane wave, $ \nabla $ becomes $ i\boldsymbol{k} $ and $ \frac{\partial}{\partial t} $ becomes $ -i\omega. $ Therefore

Maxwell's equations become: $$ \boldsymbol{k}\cdot\boldsymbol{E} = -i\frac{\rho_f}{\varepsilon} $$ $$ \boldsymbol{k}\cdot\boldsymbol{B} = 0 $$ $$ \boldsymbol{k} \times \boldsymbol{E} = \omega \boldsymbol{B} $$ $$ \boldsymbol{k} \times \boldsymbol{B} = -i \mu_0 \boldsymbol{J}_c - \mu_0 \varepsilon \omega \boldsymbol{E} $$ In a plasma we assume that:

• the free charge density is non zero because charge is free to move in the plasma, $\rho_f \neq 0$.

• $\boldsymbol{J}_c = i \frac{N_e e^2}{m_e \omega}\boldsymbol{E}$

• $\varepsilon = \varepsilon_0$

• a plasma has no D or H fields, since a plasma is unable to be polarised.

We can now take the fourth equation, the Ampere-Maxwell equation, and substitute in our assumptions: \begin{align} \boldsymbol{k} \times \boldsymbol{B} & = -i\mu_0\Big(i\frac{N_e e^2}{m_e \omega}\boldsymbol{E}\Big) - \varepsilon_0 \mu_0 \omega \boldsymbol{E} \\ & = \varepsilon_0 \mu_0 \omega \Big(\frac{N_e e^2}{m_e \varepsilon_0 \omega^2} -1 \Big)\boldsymbol{E} \\ & = \frac{\omega}{c^2}\Big(\frac{\omega_p^2}{\omega^2} -1 \Big)\boldsymbol{E} \end{align}
where $ \omega_p = \sqrt{\frac{N_e e^2}{m_e e^2}}. $ $$ \boldsymbol{k} \times (\boldsymbol{k} \times \boldsymbol{B}) = \frac{\omega}{c^2}\Big(\frac{\omega_p^2}{\omega^2} - 1\Big)\boldsymbol{k}\times\boldsymbol{E} $$ Using an identity to rewrite the vector triple product on the left, we get $$ \boldsymbol{k}(\boldsymbol{k} \cdot \boldsymbol{B}) - \boldsymbol{B}(\boldsymbol{k} \cdot \boldsymbol{k})= \frac{\omega}{c^2}\Big(\frac{\omega_p^2}{\omega^2} - 1\Big)\boldsymbol{k}\times\boldsymbol{E}. $$ The first term on the LHS becomes zero because $ \boldsymbol{B} $ is perpendicular to $ \boldsymbol{k}, $ so this becomes $$ -k^2\boldsymbol{B} = \frac{\omega}{c^2}\Big(\frac{\omega_p^2}{\omega^2} - 1\Big)\omega\boldsymbol{B} = \Big(\frac{\omega_p^2 - \omega^2}{c^2}\boldsymbol{B}\Big) $$ $$ \Big(k^2 + \frac{\omega_p^2 - \omega^2}{c^2}\Big)\boldsymbol{B} = 0 $$ Finally, we can rearrange to get our dispersion relation for a plasma. $$ k^2 = \frac{\omega^2 - \omega_p^2}{c^2} $$ $$ k = \frac{\omega}{c}\Big(1 - \frac{\omega_p^2}{\omega^2}\Big)^\frac{1}{2} $$

When an electromagnetic wave is incident on a dielectric boundary, some of the wave is reflected and some is transmitted. When the medium containing the incident wave has a greater refractive index than the other medium, there exists a critical angle $ sin^{-1}(\frac{n2}{n1}). $ If the wave is incident at greater than the critical angle, then classically, total internal reflection takes place.

But in reality, through a process that is analogous to quantum tunnelling, some of the wave does enter the second medium. This wave propagates in a direction parallel to the boundary, but decays in intensity as you move perpendicularly away from the boundary. This is an evanescent wave. (The refractive index of the material is a function of intrinsic properties of the material and the frequency of the electromagnetic wave, but we will not describe those formulae here.)

Similar to tunnelling between two potential wells in quantum mechanics, let's say there existed a material 2 of refractive index $ n_2, $ but we introduced a small gap inside of refractive index $ n_1. $ When a wave travelling in medium 2 strikes medium 1 at greater than the critical angle, total internal reflection takes place, and the transmitted wave in medium 1 decays as you move away from the boundary. But once this decaying wave hits medium 2 again, it continues propagating but with a smaller amplitude governed by how much it decayed.

The underlying maths of quantum tunnelling and the phenomenon of evanescent waves is exactly the same.

Visualisation Guide

The wave is propagating from left to right. The sliders can be varied to observe the behaviour of the incident, reflected, their combination, and the transmitted waves. Doing this, Snell's Law can be observed.

To observe the phenomenon of evanescent waves, the angular frequency ratio slider should be varied until $ n2 > n1, $ and then the incident angle slider should be taken to an angle greater than the critical angle.

This visualisation aims to explain a somewhat surprising property of waves incident on a dielectric boundary. More specifically, it will show that for any non-unity reflection coefficient, all of the energy of the incident wave can be made to transmit through the dielectric.

To understand this, consider a wave normally incident on a dielectric of thickness $ d. $ Recall that a wave can both transmit through and be reflected by a dielectric boundary. Here we label the reflection coefficient and transmission coefficient as $ r $ and $ t $ respectively.

Note that the magnitude of the reflection and transmission coefficients are the same going both from a vacuum to the dielectric and vice versa. There is, however, a $ \pi $ phase change on reflection.

It therefore follows that for any non-unity reflection coefficient, we can have an infinite number of reflections within the dielectric, with some transmission at each stage.


It can then be shown that the sum of the energies of an infinite series of transmitted waves is given by $$ E_{tot} = \frac{t'te^{i\phi}}{1-r'^2e^{2i\phi}}E^i, $$ where $ \phi=k_2d, $ $ k_2 $ is the wave vector in the dielectric and $ E^i $ is the energy of the incident wave. Note also that the prime denotes the coefficient from the dielectric to the vacuum rather than from the vacuum to the dielectric. Show Proof We can now define the total energy transmission coefficient as $ T_{layer}=|E_{tot}|^2/|E^i|^2. $ Substituting our earlier expression for $ E_{tot} $ we obtain $$ T_{layer}=\frac{(tt')^2}{1+r'^4-2r'^2\cos2\phi}. $$ We also note that the reflection coefficient $ R=r^2=r'^2, $ thus we have the final expression. $$ T_{layer}=\left(1+\frac{4R}{(1-R)^2}\sin^2\phi\right). $$ Here we can see that regardless of the value of $ R, $ if the thickness of the dielectric is an integer multiple of half the wavelength, ie. $ d=\frac{n\lambda}{2}, $ $ \sin\phi $ will always be 0, thus all of the wave will transmit through the dielectric.

For a more physical understanding as to why this is consider a reflection coefficient of 0.99. Under such conditions, 99% of the incident wave will be reflected at every boundary.

However, due to the $ \pi $ phase changes upon reflection, over an infinite number of reflections, the waves on the incident side of the dielectric sum to zero while the waves on the transmitted side add to the original amplitude. Thus, the net result is the entirety of the incident wave being transmitted through the dielectric.