# Radiation Lecture 11

<latex> \documentclass[11pt]{article} \def\inv#1Template:1 \over \def\ddtTemplate:D \over dt \def\mean#1{\left\langle #1\right\rangle} \def\sigot{\sigma_{12}} \def\sigto{\sigma_{21}} \def\eval#1{\big|_{#1}} \def\tr{\nabla} \def\dce{\vec\tr\times\vec E} \def\dcb{\vec\tr\times\vec B} \def\wz{\omega_0} \def\ef{\vec E} \def\ato{{A_{21}}} \def\bto{{B_{21}}} \def\bot{{B_{12}}} \def\bfieldTemplate:\vec B \def\apTemplate:A^\prime \def\xp{{x^{\prime}}} \def\yp{{y^{\prime}}} \def\zp{{z^{\prime}}} \def\tp{{t^{\prime}}} \def\upxTemplate:U x^\prime \def\upyTemplate:U y^\prime \def\e#1{\cdot10^{#1}} \def\hf{\frac12} \usepackage{fullpage} \usepackage{amsmath} \usepackage{eufrak} \begin{document} \subsection*{ Bremsstrahlung (braking radiation)}

Bremsstrahlung is the continuum of emission from a plasma caused by
the deflection of charged particles off of one another. The most
important deflection which we will talk about is of an $e^-$ by
a positive nucleus. Recall that the power radiated by an accelerated
$e^-$ is:
$$P={2\over3}{e^2a^2\over c^3}$$
In this case, the acceleration is from the electrical force of a
nucleus, so $a\sim{Ze^2\over b^2m_e}$, where b is the distance of
closest approach between the $e^-$ and the nucleus. Thus:
$$P={2\over3}{Z^2e^6\over b^4m_e^2c^3}$$
The energy released by this encounter is given by $E\sim P\Delta t$, where
$\Delta t$ is about how long the $e^-$ is within order b of the nucleus. Thus
$\Delta t\sim{2b\over v}$, where v is the velocity of the electron. Therefore:
$$E\sim{4\over3}{Z^2e^6\over m_e^2c^3b^4}{b\over v}$$
From the point of view of the $e^-$, the force of the $\ef$ of the nucleus
starts out pulling the $e^-$ almost directly forward, and ends up pulling
the $e^-$ nearly backward. If we graph the portion of this force which
pulls the $e^-$ sideways, we have something that looks a lot like the
upper half of a sine wave of period $2\Delta t$.
This enables us to relate $d\nu$ to
$db$, which will be useful to us in a minute:
$$d\nu\sim{v\over4b}db$$
Now let's consider $e^-$'s going in a ring around a nucleus at radius
b. The ``fraction* of an electron in a section db of that ring is *
$2\pi b\,db\,n_ev$, where $n_e$ is the number of electrons in the ring.
Thus, the power radiated by that section of ring is:
$$dP\sim E\cdot2\pi b\,db\,n_ev$$
Using our relation between $d\nu$ and db, and our expression for E:
$${dP\over d\nu}\sim{32\pi\over3}{Z^2e^6n_e\over m_e^2c^3v}$$
So the power per frequency interval is independent of distance.\par
Now let's assume a Maxwellian
velocity distribution. We'll define a $\nu_{fix}$ such that
$h\nu_{fix}\sim\hf mv_{min}^2$, where $v_{min}$ is the minimum velocity
required to keep an $e^-$ moving unbound around a nucleus. Then
the average total power released over all frequencies is:
$$\begin{aligned}\meanTemplate:P\over\nu&=\int_{v_{min}}^\infty{{dP\over d\nu}4\pi
\left({m_e\over2\pi kT}\right)^{3\over2}e^{-m_ev^2\over2kT}v^2dv}\\
&={64\sqrt{\pi}\over3\sqrt{2}}{Z^2e^6n_e\over m_e^{3\over2}c^3(kT)^\hf}
e^{-h\nu\over kT}\\ \end{aligned}$$
\def\jnff{j_{\nu,ff}}
We now define $\jnff$ to be the volume emissivity for free-free interactions.
That is, $\jnff$ measures the power radiated by plasma, per volume, into
a solid angle $d\Omega$. We can calculate the ``per $d\Omega$* because this*
radiation is isotropic:
$$\boxed{\jnff={16\over3\sqrt{2\pi}}{Z^2e^6\over m_e^{3\over2}c^3(kT)^\hf}
n_en_pe^{-h\nu\over kT}}$$
where $n_p$ is the \# density of ions. This is the expression for
Thermal Bremsstrahlung. Note that the definition of $\jnff$ in Rybicki \&
Lightman has an additional factor of ${\pi\over\sqrt{3}}\_g_{ff}(v,T)$, which
is a quantum mechanics correction factor of order unity. It's called the
``Gaunt factor*. Compare $\nu\jnff$ to the power emitted by a blackbody:*
they both peak at $4kT\over h$, and for small $\nu$, they both go as
$\nu^3$.

\subsection*{ Inverse Bremsstrahlung}

An $e^-$ can also absorb a photon and become more energetic and ``free*. We *
define the coefficient for thermal free-free absorption as:
\def\anff{\alpha_{\nu,ff}}
$$\boxed{\anff\equiv{\jnff\over B_\nu}}$$
We can express the opacity as:
\def\knff{K_{\nu,ff}}
$$\boxed{\knff={\anff\over\rho}\propto{n_en_p\nu^{-3}(e^{-h\nu\over kT}-1)\over
\rho\sqrt{T}e^{h\nu\over kT}}}$$
where $\rho$ is the total density. Since most photons have $h\nu\sim kT$,
$$\begin{aligned}\knff&\propto{n_en_p\over\rho}{T^{-3}\over\sqrt{T}}\\
&\propto\rho T^{-3.5}\\ \end{aligned}$$
See how we got back to Kramer's opacity!\par
It is important to remember the assumptions we made to get here:
\begin{itemize}
\item Maxwellian velocity distribution, which should be valid since the
collision time scale is very small, so the system should relax to a
Maxwellian distribution quickly.
\item Non-relativistic, so $T\le{m_ec^2\over k}\sim7\e9K$, which
is an okay assumption for most plasmas.\par
Some examples of Thermal Bremsstrahlung are HII regions, and the diffuse
IGM (which contains nuclei and H).
\end{itemize}

\end{document}
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