radiated power

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

$$\hbox{$\bf a$}(t) = {Ze^2\over m_e d^3}\hbox{$\bf d$}= {Ze^2\over m_e(b^2+v^2t^2)}{\hbox{$\bf d$}/d}\,,$$ (7.1)

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

$$\vert{\bf E}(t)\vert = {e\,a\over c^2\,r}\sin(\theta)={Z\,e^3\sin(\theta)\over m_e\,c^2\,r\,(b^2+v^2t^2)}\,.$$ (7.2)

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 $E(t)$,

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

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

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

$$= {Ze^3\sin(\theta)\over m_ec^3b^2v^2}\,{\pi\over bv}\,\exp(-2\pi\nu\,b/v)\,.$$ (7.3)

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

$${dW\over d\nu}\propto \int E^2(\nu)\,d\Omega\propto {1\over (b\,v)^2}\exp(-4\pi\nu\,b/v)\,.$$ (7.4)

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

$${dW\over d\nu\,dt\,dV}\propto n_e\,n_i\,v^{-1}\,\exp(-4\pi\nu\,b/v)\propto n_e\,n_i\,T^{-1/2}\,\exp(-4\pi\nu\,b/v)\,.$$ (7.5)

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

$${\hbox{\rm power }} \propto n_e\,n_i\,T^{1/2}\,.$$ (7.6)

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