====== 5. Bremsstrahlung radiation ======
A charge particle passing by another another charged particle accelerates and radiates. This radiation is called Bremsstrahlung because in German //brem// means //to brake// and //sstrahlung// means //radiation//.

In order to understand this radiation, let us think of a plasma of ions with the following assumptions:

  - The gas is monatomic and in thermal equilibrium, so it can be described using [[en:Maxwell-Boltzmann distribution]]. For such a gas the temperature is related to the average translational kinetic energy as $ 3kT/2 = (mv^2/2)_{av}$ where $k$ is Boltzmann constant and $m$ and $v$ are the mass and thermal speed of an individual atom.
  - The plasma is made of completely ionized Hydrogen, i. e. electrons and protons. If energy is more than 13.6 eV, H will be mostly ionized and the gas will be made of single electrons and protons. This energy gives a minimum temperature of $10^5$ K. The **degree of ionization**, calculated using the [[en:Saha equation]], and particle densities will determine the actual required temperature, but generally low-density Hydrogen gas becomes completely ionized at $\sim 20$ kK.
  - The plasma is made of electrons with charge $-e$ and ions with charge $+Ze$ where $Z$ is the atomic number ($Z=1$ for H). The electrons and ions are in thermal equilibrium, their average kinetic energies are equal.
  - Electrons are $1836$ times lighter than protons and, hence, move $40$ times faster. So the protons are stationary and the electrons are moving.
  - For slowly moving particles ($v<<c$) and for //near// collisions, number of photons exchanged is large and, hence, quantum and relativistic effects are negligible. In this **classical approximation**, photons radiated from electrons have energies much smaller than the kinetic energies of the electrons.
  - The **free-free radiation** has a **continuum spectrum**. Because of low density, the plasma is optically thin, i. e. its [[en:optical depth]] $\tau<<1$.

===== Electromagnetic radiation =====
A charge moving with constant velocity $v<<c$ has electric-filed ($E$) lines around it. The instantaneous pattern of the E-lines will be isotropic. If the charge accelerates, waves will propagate along the lines similar to waves propagating along a stretched string due to oscillation of one end of the string.

{{ :en:6.1.png?nolink&650 |}}

For a positive charge, $\mathbf{E}$ has direction opposite to $a\sin\theta$ and $E\propto q/r$. Energy flux is $\propto E^2\propto r^{-2}$. For a negative charge direction of $\mathbf{E}$ and $a\sin\theta$ is the same. The electric field

$$ \mathbf{E}(r,t) = E\hat{n} = \frac{qa(t')\sin\theta}{4\pi\epsilon_0 c^2 r} \hat{n} = \frac{q}{4\pi\epsilon_0c^2r} (\mathbf{a}\times\hat{k}) $$

where $\epsilon_0$ is the //permittivity of free space//, $\hat{k}$ gives the direction of propagation and $t'=t-r/c$ is an earlier time.

Figure (a) shows the electric and magnetic field vectors and figure (b) the dipole radiation pattern. In (b), the acceleration is vertical and the wave propagates in a doughnut-shaped toroidal pattern outward from the charge. $E=0$ at the poles and maximum at $\theta=90^\circ$.

The flux is given by the **Poynting vector**

$$\mathbf{F}_p = \frac{1}{\mu_0} (\mathbf{E}\times\mathbf{B})$$

where $\mu_0$ is the //permeability of free space// and $B=E/c$ is the magnitude of the magnetic field. The speed of such a wave

$$ c = \frac{1}{\epsilon_0\mu_0} = 3\times 10^8 \ \text{m/s}$$

which can be derived from the wave-equation solution of Maxwell's equations. The magnitude of the Poynting vector thus becomes

$$F_p = \epsilon_0 c E^2 = \frac{q^2a^2\sin^2\theta}{(4\pi)^2\epsilon_0 c^3 r^2}. $$

If we integrate this function over the surface of a sphere at a distance $r$ from the charge, we will get the total power radiated in all directions as

$$ P = \int F_p r^2 d\Omega = \int\int F_pr^2 \sin\theta \ d\theta d\phi = \frac{1}{6\pi\epsilon_0} \frac{q^2a^2}{c^3} $$

which is called **Larmor's formula**. It gives power in W.

==== Energy and frequency ====
The Coulomb force between two charges

$$ \mathbf{F} = \frac{q_1 q_2}{4\pi\epsilon_0 r^2} \hat{r}. $$

The acceleration of an electron because of the force from an ion of charge $Ze$ is

$$ \mathbf{a} = \frac{\mathbf{F}}{m} = -\frac{1}{4\pi\epsilon_0} \frac{Ze^2}{mr^2} \hat{r}. $$

{{ :courses:ap:6.2.png?nolink&600 |}}

The **impact parameter** $b$ is defined as the projected distance of the closest approach as shown above. Replace $r$ with $b$ to get the maximum acceleration

$$ a_{max} \approx \frac{1}{4\pi\epsilon_0} \frac{Ze^2}{mb^2}. $$

Collision time $\tau_b=b/v$ if $v$ is the speed of an electron. Now the total energy emitted can be found by integrating $P(t)$ from Larmor's formula over the total collision time as

$$ Q(b,v) = \int_{-\infty}^{\infty} P(t) dt = \frac{e^2}{6\pi\epsilon_0c^3} \int_{-\infty}^{\infty} a^2dt. $$

Two simplifications: the integration does not have to up to infinity because Coulomb force is vanishingly small at large distances and the acceleration can be considered constant near maximum value. So

$$ Q(b,v) \approx \frac{e^2}{6\pi\epsilon_0c^3} a_{max}^2 \tau_b \approx \left(\frac{1}{4\pi\epsilon_0}\right)^3 \frac{2}{3} \frac{Z^2e^6}{c^3m^2b^3v} $$

which gives the total energy radiated in all directions by a single electron of speed $v$ as it passes an ion of charge $Ze$ with impact parameter $b$.

The acceleration changes only once and, hence, we get only a pulse of $E$ shown in the figure above.

The angular frequency of the pulse $\omega\approx 1/\tau_b = v/b$ and, hence, the characteristic frequency of the radiation

$$ \nu = \frac{\omega}{2\pi} = \frac{v}{2\pi b}. $$

This is the frequency at which most power is emitted and also the maximum frequency. The relation between impact parameter and frequency

$$ b = \frac{v}{2\pi\nu} \Rightarrow db = \frac{v}{2\pi\nu^2}d\nu. $$