Thermal bremsstrahlung

How are the X-rays produced in this hot gas? Since the gas temperature is very high, most elements are highly ionised, and so the gas is really a hot plasma. The negatively charged electrons in this plasma experience a force when they pass close to an ion, and an accelerating charge emits electro-magnetic dipole radiation. This is the radiation that we observe. Since the electron looses energy in the encounter, it will slow down. Hence the name thermal bremsstrahlung^6.3. And as the electron slows down, the plasma also looses energy, and so the gas cools.^6.4 The electron is not bound to any particular ion throughout the interaction. This type of radiation is therefore also called free-free radiation. This is all you need to know about this, but if you want more details, here they come.

Let's try to obtain an expression for the radiated power. The aim of this derivation is to show that the power emitted is $\propto n_e n_i T^{1/2}$, where $n_e$ and $n_i$ are the electron and ion densities, and $T$ is the temperature. You could have guessed the first bit, since we're talking about an encounter between two particles. So the exercise is how to get the $T^{1/2}\propto v$ dependence.

If the deflection angle is small, we can simple neglect it, and say that the electron moves at constant speed $v$ along a straight line with impact parameter $b$. The acceleration is then

$$\hbox{$\bf a$}(t) = {Ze^2\over m_e d^3}\hbox{$\bf d$}= {Ze^2\over m_e(b^2+v^2t^2)}{\hbox{$\bf d$}/d} ,$$ (6.1)

where $Z$ is the charge of the ion in units of the electron charge $e$, and $\hbox{$\bf d$}$ is the distance between electron and ion. $t=0$ is the time of closest approach, and so the encounter lasts from $t\rightarrow -\infty$ to $t\rightarrow +\infty$.

The dipole electric field at distance $r$ from the electron is

$$\vert{\bf E}(t)\vert = {e a\over c^2 r}\sin(\theta)={Z e^3\sin(\theta)\over m_e c^2 r (b^2+v^2t^2)} .$$ (6.2)

This is the electric field as function of time $t$, but we want it as a function of frequency $\nu$. The trick to obtain $E(\nu)$ is by Fourier transforming $E(t)$,

$$E(\nu) \equiv \int_{-\infty}^{\infty} E(t)\exp(2\pi i\nu t) dt$$

$$= {Ze^3\sin(\theta)\over m_ec^2 r}\int_{-\infty}^{\infty} {\exp(2\pi i\nu t)\over b^2+v^2t^2} dt$$

$$= {Ze^3\sin(\theta)\over m_ec^2 r}{1\over b v} \int_{-\infty}^{\infty} {\exp(2\pi ix(\nu b/v))\over (1+x^2)} dx$$

$$= {Ze^3\sin(\theta)\over m_ec^3b^2v^2} {\pi\over bv} \exp(-2\pi\nu b/v) .$$ (6.3)

Along the way, I've introduced the variable $x=vt/b$. The integral $\int\exp(2\pi x\alpha)/(1+x^2) dx=$ some function of $\alpha$ is a little tricky to perform, and I don't expect you to be able to do it. What is important is to note that this change of variables gives rise to a factor $1/bv$ .

Now we're done, because it means that the power radiated by this single electron per unit frequency is

$${dW\over d\nu}\propto \int E^2(\nu) d\Omega\propto {1\over (b v)^2}\exp(-4\pi\nu b/v) .$$ (6.4)

So the spectrum is independent of $\nu$ for low frequencies $\nu\ll v/b$, and cuts off exponentially for large $\nu$. Now the number of encounters with impact parameter between $b$ and $b+db$, in time $dt$ is $n_e 2\pi b db vdt$, and so this gives the power as

$${dW\over d\nu dt dV}\propto n_e n_i v^{-1} \exp(-4\pi\nu b/v)\propto n_e n_i T^{-1/2} \exp(-4\pi\nu b/v) .$$ (6.5)

So the origin of the $1/v$ is that the energy radiated per encounter $\propto 1/v^2$ whereas as the rate of encounters is $\propto v$.

To obtain the final result, we still need to integrate over the impact parameter $b$, and average over all impact velocities $v$ (for example by assuming a Maxwell-Boltzmann distribution for $v$). The result is that the total power emitted is

$${\hbox{\rm power }} \propto n_e n_i T^{1/2} .$$ (6.6)

Notice that the total power is much more dependent on the density of the gas ($\propto$ density squared), than on its temperature ( $\propto T^{1/2}$)