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:

1. The gas is monatomic and in thermal equilibrium, so it can be described using 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.
2. 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 Saha equation, and particle densities will determine the actual required temperature, but generally low-density Hydrogen gas becomes completely ionized at $\sim 20$ kK.
3. 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.
4. Electrons are $1836$ times lighter than protons and, hence, move $40$ times faster. So the protons are stationary and the electrons are moving.
5. 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.
6. The free-free radiation has a continuum spectrum. Because of low density, the plasma is optically thin, i. e. its optical depth $\tau<<1$.

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.

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.

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

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

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

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