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Plasma effects


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Plasma effects

Electrodynamics January 2019


So far we have discussed generation and absorption of radiation in vacuum, that is to say that electrons emit and absorb radiation as though there are no other charged particles around. The transport of radiation was similarly treated under vacuum conditions. Here we wish to understand the collective effect of the underlying plasma on the generation and transport of radiation.

It is worth emphasizing that there is nothing fundamentally different about the pres- ence of plasma. Maxwell’s equations are remain and are always valid. Plasma effects are merely a secondary effect. Because electrons are accelerated by an external EM-field, they in turn radiate. Plasma effects is an implicit study of the effects of this secondary radiation. One way to understand our philosophy for this lecture is to make an analogy with the concept of buoyancy.You know that buoyant force is not a fundamental force, but rather a handy way to include the effect of the media in some circumstances. In the same way, we will use the di-electric constant or the refractive index of a medium to un- derstand the effect it has on propagation of electro-magnetic (EM hereafter) waves.

Linear response of a medium

Before delving into specific cases, we will state some general properties of linear media (defined below) and what effect they have on EM fields. We start with Maxwell’s equa- tions in real and Fourier notation. We will be using the latter mostly. The Fourier notation assumes that all quantities vary as exp(i(k·rωt))


∇ ×E = −1 c


∂t ; ik×E =iω

c B (1)

∇ ×B = 1 c


∂t +

c j ; ik×B = −iω


c j (2)

∇ ·E =4πρ ; ik·E =4πρ (3)

∇ ·B =0 ; ik·B=0 (4)

Note that we have explicitly included the charge density ρ and current density j in the equations since the EM waves are not propagating in free-space but in some medium.

We will also add to this, the continuity equations which basically says that charge is con- served. So the amount of charge leaving some volume element per unit time must be equal to the (negative) of the time derivative of the charge density.

∇ ·j = −∂ρ

∂t ; ik·j=iωρ (5)

To proceed further, we need to relate the EM field to ts effect on the medium. In other words, we need a relationship between E, B and j, ρ. This is eventually obtained by solving the equations of motion of charged particles in a varying EM field. We will do so later. First, we obtain a general set of linear solutions. By linear we mean that we can write j = σ.E. Notice that the R.H.S is a tensor product, that is, if E and j are 3 dimensional vectors, then σ is a 3 by 3 matrix. Here σ is called the tensor conductivity of the medium. It is a property of the medium because it relates how the medium responds to external forcing. That is, it relates the field E to the current it drives in the medium j.

If the response of the medium is isotropic, then we do not need a tensor notation and we can simply write j=σE. In this case the x component of the electric field only influences the current flowing in the x direction and so on.

There are immediate consequences of this linear relationship on the Maxwell’s equa- tions. We get

ik×E = iω

c B (unchanged) (6)

ik×B = −iω c

1− σ

.E (7)


1− σ

.E = 0 (8)

ik·B = 0 (9)

Notice how the response of the medium has been absorbed into the equations such that they look similar to Maxwell’s equations in vacuum. The only substitution is E →

.E in two of the equations where  is the dielectric constant and is related to the conduc- tivity tensor as


σ (10)


Let us see what kind of wave solutions we get from this substitution. The wave solu- tion is obtained by taking the curl of the first equation and substituting the second. In the Fourier notation this is equivalent to taking the cross product of the first equation with k and then substituting the second. We get

k× (k×E) = −ω


c2.E (11)

We can use the vector triple product identity, A× (B×C) = B(A·C) −C(A·B) on the L.H.S to get k× (k×E) =k(k·E) −k2E, where we have used k2 =k·k. The wave equation is

k(k·E) −k2E+ω


c2 .E =0 (12)

This is a general matrix equation in that the L.H.S can be written as a 3 by 3 matrix multiplying a 3 by 1 vector electric field. Of course the trivial solution that always satis- fies the equation is E =0, but we are interested in non-trivial solutions which only exist when the determinant of the matrix multiplying the vector electric field is zero. Equating the determinant to zero gives us the properties of the waves the media can support and hence the effect of the media on wave propagation.

Waves in an isotropic medium

Without loss of generality let us take k to be along the z-axis. We can do this in an isotropic medium because there is no preferred direction. So we have k =zˆk, where ˆz is the unit vector along the z axis. Then the wave equation is



c2 .E =0 (13)

Let us take the simplest case, where the media is isotropic. This is true for example, in an un-magnetized plasma. There is no preferred direction in such a medium and hence its response should be isotropic. In such a case,  is no longer a tensor. It is a basically a scalar. Then the wave equation becomes.

k2Ez−k2E+ ω


c2 eE =0 (14)

In matrix form, we can write

k2+ ωc22e 0 0 0 −k2+ωc22e 0

0 0 ωc22e

 Ex




 0 0 0

 (15)

Equating the determinant of the transform matrix to zero, we get


ω2 c2e−k2



c2e =0 (16)

There are two possible solutions to this equation which are usually referred to as dis- persion relations. The first solution is eω2 =c2k2, and the second solution is e=0.

If we substitute the first solution into the matrix equation, then we get Ez =0. Hence there is no electric field in the direction of wave propagation. In other words, the solution is a transverse EM wave. The phase velocity of the wave can be found from its dispersion relationship

vph = ω k = √c

e (17)

The group velocity can be again found from the dispersion relation. But since e in may in general depend on ω, we cannot determine the group velocity, ∂ω/∂k without specifying this variation.

The second solution, when substituted into the matrix equation gives Ex = Ey = 0.

This is therefore a longitudinal wave! This is an important difference between free space propagation (where only transverse solutions are allowed) and propagation in a dielectric medium, where longitudinal waves are allowed when e =0. Note that the solution e=0 may seem strange to those of you who have experience with optics where the refractive index η = √

e is not zero for typical materials like glass, water etc. But plasma as a different ball game, as we shall soon see. They support solutions where e =0!

Cold isotropic plasma

By cold we mean that we will neglect the thermal motion of plasma particles and by isotropic we mean that there is no preferred direction in the plasma. We will further as- sume that the plasma is homogeneous.

So far we have only concerned ourselves with deriving the general properties of any linear medium (one where the induced current density is linearly proportional to the electric field; j = σ.E). We have not derived this actual linear relationship which is a property of the media— plasma in our present case. To determine this relationship, we have to understand the microscopic response of the media to EM fields.

To compute the current induced in a plasma by an electromagnetic wave, we must begin with the equation of motion of a charged particle

m ˙v =q E+v

c ×B

(18) where the dot in ˙v should be understood as a time derivative. The L.H.S is basically mass times acceleration (or the force) and the R.H.S is the Lorentz force on a charged particle.


For electrons, we have q = −e. Furthermore, if the electron is non-relativistic, then v/c 1, and the second term can be safely neglected (note than E and B have compara- ble magnitudes in an EM-wave). We therefore have

me˙v = −eE; −iωmev = −eE (19)

The current density is j = −nev where n is the electron density in the plasma. This yields us the linear relationship that we are after:

j = − ne



E (20)

The conductivity is therefore

σ = − ne


iωme (21)

Notice that the conductivity is isotropic and there is no need for a full tensor formulation.

The dielectric constant is

e =1−



ω2me (22)

The refractive index approaches zero as the wave frequency approaches the plasma frequency, ωpgiven by

ωp = s


me (23)

The dielectric constant is written in terms of the plasma frequency as

e =1−ωp ω


(24) With this equation for the dielectric constant, the dispersion relationship for transverse waves becomes

ω2 =ω2p+k2c2 (25) We can now find the group and phase velocities of transverse EM waves as

vph = ω k =c

s ω2

ω2ω2p (26)


vgr = ∂ω

∂k =c s


ω2 (27)


Temporal dispersion

Notice how the dielectric constant (and hence the phase and group velocities) depend on the frequency of the EM-wave. This dependence is called temporal dispersion. One manifestation of this dispersion is the different speed at which EM waves at different frequencies travel. The time of flight of an EM signal is given by

t = Z ds

vgr (28)

where ds is the length-element along the ray. Substituting for the group velocity from above we get

t= Z


c 1−ω




(29) When ω ωp, we can Taylor expand the square root to get

t= 1 c


ds+1 c


ds ω2p

2 (30)

The first terms on the RHS is the time of light in vacuum. The additional delay due to the medium is then

∆t= 1 c


dsω2p ω2 = 1

c Z


2 (31)

In a homogeneous medium, the integral of the plasma density gives the column den- sity, N. The extra time-delay is then

∆t= 2πNq


cmω2 (32)

The inverse-square relationship between the arrival time and wave frequency is rou- tinely exploited in astronomy to measure the integrated column of free electrons from the source to the observer.

In astronomically convenient units, the additional time delay due to cold plasma dis- persion is

∆t=4.15 ms


pc cm3

 ν 1 GHz



Cold magnetized plasma

Let is now retain the cold-plasma assumption but relax the isotropic assumption. The most common astrophysical scenario where this happens is when there is a static external magnetic field present which breaks the underlying isotropy of the plasma.


Let the external magnetic field be B0. We will assume that B0  B, that is, the external magnetic field is much larger than the magnetic field of the EM-wave. In such a case, we can no longer assume that vB0/c  E in our equation for the Lorentz force on the electron. Retaining the magnetic part of the Lorentz force then gives

m ˙v =q E+ v

c ×B0

; −imωv=q E+v

c ×B0

(34) Using j=qnv we get

j= iq


E+j×B0 qnc

(35) To simplify the algebra, we will assume that B0 is along the z axis. We will further express everything in terms of the plasma frequency ω2p = 4πnq2/m and the cyclotron frequencyΩ=qB0/(mc).

After some tedious algebra, we get the linear response of the medium as

j =σ.E; σ =



ω iΩ 0

iΩ ω 0 0 0 ω2ω2

 (36)

One thing we immediate notice is that unlike the previous situation of isotropic plasma, our dielectric tensor has non-diagonal entries. This make the algebra quite complicated.

In this case, we can avoid this complication by “diagonalizing” this matrix. In other words, the natural modes in the medium are not aligned with our axes of choice (x, y z in this case). Often by choosing the right axes we can not just simplify the algebra but get a deeper understanding of what is going on. In the previous case of an isotropic medium, the electrons were moving along a line in simple harmonic motion due to the presence of a harmonic field. This rectilinear motion of the electrons was amenable to a description in Cartesian co-ordinates. In the presence of the magnetic field however, the electrons are forced into circular motion with an axis aligned with the magnetic field. Hence we will be better off by choosing left and right hand circular motion as our natural basis in the x, y plane instead of the usual Cartesian basis. This can be done via the transformations

vL =vx+ivy ; EL =Ex+iEy (37) vR =vx−ivy ; ER =Ex−iEy (38) (39) You can check that the new basis are indeed orthogonal to one another.

The conductivity tensor now becomes

σ = i


ω 0 0

0 ω


ω+ 0

0 0 ω





The dielectric tensor is  =1−4π/()σ. The non-zero terms in the tensor are

eLL = 1− ω


ω(ω) (41)

eRR = 1− ω


ω(ω+) (42)

ezz = 1−ω


ω2 (43)


Parallel transport

Let us start with the case of parallel transport; that is a wave that is traveling along the magnetic field (z axis in our case). We have kL =kR =0 and kz =k.

The wave equation can be written in matrix form as

k2+ωc22eLL 0 0 0 −k2+ ωc22eRR 0

0 0 ωc22ezz


 EL



 0 0 0

 (45)

Again non-zero solutions exist when one the following three cases is satisfied

Case(i) k2 =ω2eLL/c2. This is called the Left-hand wave solution because ER = Ez = 0

Case(ii) k2 =ω2eRR/c2. This is called the Right-hand wave solution because EL =Ez =0 Case(iii) ezz =0. This is the usual longitudinal wave solution as before (EL =ER =Ex = Ey =0)

The left-hand and right-hand circular waves have different dielectric constants (and hence different phase velocities) in a magnetized medium. This leads to a well known effects called the Faraday effect.

Faraday rotation

Notice that the left and right hand wave solution have different dispersion relations. They have different phase and group velocities. Hence left hand and right hand circularly polarized waves travel with different phase and group velocities in a magnetized media.

This is popularly known as the Faraday effect. Any linearly polarized wave can be written as a unique linear combination of left and right hand polarized waves. When there is a phase shift between left and right hand waves imparted due to their different phase velocities in a magnetized medium, the net effect is the rotation of the plane of linear polarization.


We can appreciate the magnitude of this effect by computing the differential phase of the left and right hand waves when they travel a length s through a magnetized medium.

∆φ= Z

ds(kL −kR) (46)

From the dispersion relations, we get

∆φ= Z

dsw c

 s

1− ω

2p2 1−Ω/ω


1− ω

2p2 1+Ω/ω

 (47)

Let us now assume that ω  Ω. Then we can put 1/(1−ω/Ω) ≈ 1+ω/Ω and 1/(1+ω/Ω) ≈ 1ω/Ω. Let us also assume that ωωp. This then allows us to Taylor expand the square-root sign. These two assumptions are collectively referred to as the high-frequency solutions. We get

∆φ= −

R dsΩω2p

2 (48)

Notice that the phase different varies as the inverse square of the wave frequency. This is an important characteristic of the Faraday effect.

Finally, we can compute the effect in the plane of linear polarization. We have Ex = (EL +ER)/2 and Ey = (EL−ER)/(2i). The factor of two in our definition makes the phase angle between Exand Eyvary at half the rate at which the phase angle between EL

and ERvaries. Hence the angle by which the plane of linear polarization rotates is given by

∆φxy = −

R dsΩω2p

2cω2 (49)

Putting in q = −e and the expressions for the plasma and cyclotron frequency, we get

∆φxy = 2πe


m2ec2ω2 (50)

where N = R

nds is the column density of electrons along the ray path. Notice again the inverse square relationship between the angle by which the plane of polarization is rotated and the frequency of the EM wave. This effect is routinely exploited in astronomy to compute the magnetic field in the media between the observed and the source.

What happens when the wave vector is not aligned with the magnetic field? We will state the result for Faraday rotation in the high-frequency limit without proof. All we have to do is replace the magnetic field strength by its component along the direction of propagation. So we make the substitution B0→ B0cos θ where θ is the angle between the wave vector and the static magnetic field.

The angle by which the plane of linear polarization is rotated is therefore


∆φxy = 2πe

3NB0cos θ

m2ec2ω2 ; ω Ω, ω ωp (51) In astrophysically convenient units we have

∆φxy =0.073 rad

 B0cos θ 106Gauss


pc cm3

 ν 1 GHz





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