Scattering Processes and the Electromagnetic Wave Picture

So far we have worked mainly with the macroscopic description of radiation. Light traveled along rays, and the main equation was the transfer equation along a path length \(s\):

We used this equation many times, but we did not really calculate the interaction coefficients from first principles. We introduced optical depth as the accumulated extinction along the path,

and in the random-walk picture we also wrote the extinction coefficient in the form

Here \(n\) is the number density, or more generally an occupation number of the particles or states that interact with the radiation. For electron scattering it can be the number density of free electrons. For atomic lines it can be the occupation number of a particular atomic level. The quantity \(\sigma_\nu\) is the cross-section for the interaction.

So the next question is unavoidable: how do we obtain \(\sigma_\nu\)? If we know the cross-section and the number density, then we know \(\alpha_\nu\), and then we can calculate optical depths and solve the transfer problem.

This chapter starts that calculation for scattering processes. The examples coming next are Thomson scattering, Rayleigh scattering, and eventually line scattering. For these ideas, the electromagnetic-wave description of light is the natural starting point.

The reason is scale. In the ray picture we imagine a bundle of light moving through matter. But when light interacts with an atom or an electron, the interaction occurs on an atomic scale. A typical atomic length scale is of order an angstrom. Optical light has wavelength of order \(5000\) angstrom. So for optical light,

That is exactly the regime where the geometrical ray idea breaks down. We can still use the macroscopic transfer equation later, but the coefficients inside it must be calculated using the microscopic wave description.

Picture an incoming electromagnetic wave hitting an electron. The electron is driven by the wave, starts to oscillate, and then emits radiation. In scattering, the emitted radiation is the scattered light.

The cross-section is defined by comparing two quantities:

This definition has the correct units. Power has units of energy per time. Flux has units of energy per time per area. Power divided by flux is therefore an area. That effective area is the cross-section.

This is directly connected to the luminosity and flux language we already used. Power is energy per unit time, so in astrophysics it is often just luminosity. A luminosity can be written as a surface integral of flux:

For a spherical star with radial flux \(F_r\), this becomes

The microscopic scattering problem uses the same physical idea, but the flux is now the electromagnetic-wave flux, and the emitting object is an accelerated charge rather than a whole star.

We use Gaussian/CGS units in this chapter. These units are common in older radiative-transfer and astrophysics texts, and they have one convenient feature for plane waves: the electric and magnetic fields have the same magnitude.

Maxwell's equations are

Here \(\rho\) is charge density and \(\mathbf j\) is current density. For moving charges,

If we set \(\rho=0\) and \(\mathbf j=0\), we get the vacuum case. Then Maxwell's equations imply a wave equation for both \(\mathbf E\) and \(\mathbf B\):

This tells us that electromagnetic disturbances propagate with the speed \(c\). This is exactly the wave picture of light.

Now introduce the Lorentz force on a charge \(q\):

For the scattering calculations in this introductory treatment we take the non-relativistic limit,

Then the magnetic part of the Lorentz force is smaller by a factor \(v/c\), so the force is approximately

In most of the examples we imagine the charge to be an electron. We often write the charge magnitude as \(e\), keeping in mind that the electron charge itself is negative. For the emitted power, the sign will not matter because the result contains \(e^2\).

The mechanical work done on the charge is force dotted with velocity. In the non-relativistic limit this becomes

For a continuous charge distribution, \(q\mathbf v\) becomes the current density \(\mathbf j\). So the work done per unit volume is

This is the term that tells us how the electromagnetic field exchanges energy with matter.

The rate at which the electromagnetic field does work on matter is

If this term is positive, electromagnetic energy is being converted into mechanical or thermal energy of the charges. If it is negative, the charges are giving energy back to the electromagnetic field.

Using Maxwell's fourth equation,

we can solve for \(\mathbf j\) and dot with \(\mathbf E\):

After using the vector identity

and also using Faraday's law, this becomes Poynting's theorem:

This is the differential form of Poynting's theorem. If we integrate it over a volume and use Gauss's divergence theorem, the divergence term becomes a surface flux through the boundary. That is the integral conservation statement: the energy inside a volume changes because energy flows through the surface and because fields do work on charges.

This has the same conservation-law shape we used before:

Therefore we identify the electromagnetic energy density as

For a plane electromagnetic wave in Gaussian units, \(|\mathbf E|=|\mathbf B|\), so

The electromagnetic flux is the Poynting vector:

As a sanity check, take a plane wave. Since \(|\mathbf E|=|\mathbf B|\),

That is exactly what a flux should look like: speed times energy density.

A plane wave travels in one direction. So in the specific-intensity language, it is represented by intensity concentrated into one direction:

The radiation energy density is

For this one-direction plane wave, the angular integral just picks out \(I_0\), so

Now write the electric field of the wave as

The wave oscillates extremely fast, so when we compare it to the macroscopic intensity description, we average over one cycle. Since the average of \(\cos^2\) over a full cycle is \(1/2\),

Therefore the time-averaged electromagnetic energy density is

Matching the two descriptions gives

The same result appears from the flux side. In the free-streaming limit,

So a single plane wave has the same basic relation we used before: flux equals speed times energy density. This is the bridge between the microscopic wave picture and the macroscopic transfer picture.

We now need the power emitted by the electron. The power is the surface integral of the electromagnetic flux:

First consider a static charge. Its electric field is Coulomb's law:

If we put this into the power integral, \(E_r^2\) falls like \(1/r^4\), and after multiplying by the surface area factor \(r^2\), the contribution still falls like \(1/r^2\). So the static Coulomb field is not the radiation that reaches us from far away. It dies away too fast.

The important part appears when the charge accelerates. Imagine an electron initially at rest. Its electric field lines are radial. Then we give the electron a small kick for a short time \(\Delta t\). A later time \(t\) afterward, the information about that kick has only traveled a distance \(ct\). Outside that radius, the field still looks like the old field. Inside, the field knows about the new position. The field lines must connect, so a thin shell of distorted field lines moves outward.

The shell thickness is \(c\Delta t\). The sideways displacement of the charge, projected perpendicular to the line of sight, is \(\delta v\,t\sin\theta\). Therefore

Since the acceleration is \(a=\delta v/\Delta t\), and since the disturbance observed at radius \(r\) was emitted a time \(t=r/c\) earlier,

Using \(E_r=e/r^2\), the transverse electric field is

This is the crucial point. The static Coulomb field falls as \(1/r^2\), but the radiation field from the accelerated charge falls only as \(1/r\). That is why radiation can travel to large distances.

Now put this radiating field into the power integral:

Substitute \(E_\theta=e a\sin\theta/(c^2r)\):

The \(r^2\) from the area cancels the \(1/r^2\) from \(E_\theta^2\), so the emitted power is independent of how far away we place the sphere. That is exactly what we want for radiation.

Now evaluate the angular integral:

Therefore

This is the Larmor formula. It gives the total power radiated by a non-relativistic accelerated charge.

Dipole radiation pattern

The radiation is not emitted equally in all directions. The differential power is

Along the acceleration direction, \(\theta=0\), so \(\sin^2\theta=0\). There is no radiated power in that direction. Perpendicular to the acceleration, \(\theta=\pi/2\), so the emission is maximum. This produces the characteristic dipole pattern.

Many charges and the dipole moment

The Larmor formula was derived for one charge, but it can be generalized to a collection of charges. Define the electric dipole moment

Taking two time derivatives gives

Then the emitted power can be written in the compact form

This is why the result is often described as dipole radiation. Accelerating charges change the dipole moment, and that changing dipole moment radiates.

This same Larmor formula will be used for scattering. An incoming wave accelerates an electron; the accelerated electron radiates; that re-radiated light is the scattered radiation. The same idea will also reappear later in free-free emission: an electron is deflected by a proton, the deflection is an acceleration, and the accelerated charge emits radiation.

The next scattering topics use this foundation. Thomson scattering is the free-electron limit, and it will let us estimate why stars cannot become arbitrarily luminous before radiation force becomes important. Rayleigh scattering explains why shorter wavelengths scatter more strongly, which is the basic reason the sky is blue. Line scattering needs more quantum mechanics, but the classical driven-and-damped oscillator picture already gives useful intuition about why spectral lines are narrow and powerful, and why a line profile has a characteristic shape.