Radiation Lecture 11

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<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} <\latex>