Thermal Bremsstrahlung

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Bremsstrahlung is the continuum of emission from a plasma caused by the deflection of charged particles off of one another. It is a common source of X-ray emission. The deflection we will discuss, which is the most important, is of an $e^{-}$ by a positive nucleus, where we assume the nucleus to be stationary.

Recall that the power radiated by an accelerated $e^{-}$ is (see Larmor Formula):

$P={2 \over 3}{e^{2}a^{2} \over c^{3}}\,\!$ In this case, the acceleration is due to the electric force of a nucleus, so $a\sim {Ze^{2} \over b^{2}m_{e}}$ , where $b$ is the distance of closest approach between the $e^{-}$ and the nucleus, known as the impact parameter. Thus:

$P={2 \over 3}{Z^{2}e^{6} \over b^{4}m_{e}^{2}c^{3}}\,\!$ The energy released by this encounter is given by $E\,\sim \,P\Delta t$ , where $\Delta t$ is the time of interaction. We can approximate the interaction to occur roughly over a distance $b$ to either side of the nucleus, and so, if the $e^{-}$ is traveling at a speed $v$ , then $\Delta t\sim {2b \over v}$ . Therefore:

$E\sim {4 \over 3}{Z^{2}e^{6} \over m_{e}^{2}c^{3}b^{4}}{b \over v}\,\!$ From the $e^{-}$ ’s point of view, the force of the nucleus’s ${\vec {E}}$ is initially pulling the $e^{-}$ almost directly forward, and ends up pulling the $e^{-}$ nearly completely backwards. At the point of closest approach, the $e^{-}$ is not being pulled forwards or backwards, the electric force is solely perpendicular to the original direction of motion. We can therefore see that if we plot the parallel component of the force, we end up with something that looks a lot like a sine wave of period $2\Delta t$ . Parallel component of the Electric Force from the electron’s point of view.

From earlier, we know though that $\Delta t\sim {2b \over v}$ . Multiplying this by 2 in order to get a period, and inverting it to get a frequency, we have $\nu \sim {v \over 4b}$ . This enables us to relate $d\nu$ to $db$ , which will be useful to us in a minute:

$d\nu \sim {-{v \over 4b^{2}}}db\,\!$ Now let’s consider a nucleus in a sea of electrons.

The electrons of interest to us are those a distance $b$ from the nucleus. This fraction of electrons is given by the number density of electrons multiplied by the area of the ring: $2\pi b\,db\,n_{e}$ . Thus, the power radiated by that ring is:

$dP\sim E\cdot 2\pi b\,db\,n_{e}v\,\!$ where we need to include the $e^{-}$ velocity to account for the increased interaction rate for fast moving electrons. Using our relation between $d\nu$ and $db$ , and our expression for $E$ we arrive at:

${dP \over d\nu }\sim {-{32\pi \over 3}}{Z^{2}e^{6}n_{e} \over m_{e}^{2}c^{3}v}\,\!$ So the power per frequency interval is independent of distance!

The remaining term left to define is the velocity. We have seen that in a plasma, the collision rates between electrons are high enough to relax into a Maxwellian velocity distribution fairly quickly:

$f(v)=\left({m_{e} \over 2\pi kT}\right)^{3 \over 2}4\pi v^{2}e^{-{m_{e}v^{2} \over 2kT}}\,\!$ However, we are considering free-free emission, and so we must define a minimum velocity below which the electron would otherwise get captured by the nucleus. An $e^{-}$ with this minimum velocity will have a kinetic energy of order the energy radiated by the acceleration, which is the energy of a photon: $h\nu \sim {\frac {1}{2}}mv_{min}^{2}$ . Then the average total power released over all velocities due to the interaction with one nucleus is:

{\begin{aligned}\left\langle {dP \over d\nu }\right\rangle &=\int _{v_{min}}^{\infty }{{dP \over d\nu }4\pi \left({m_{e} \over 2\pi kT}\right)^{3 \over 2}e^{-{m_{e}v^{2} \over 2kT}}v^{2}dv}\\&={64{\sqrt {\pi }} \over 3{\sqrt {2}}}{Z^{2}e^{6}n_{e} \over m_{e}^{3 \over 2}c^{3}(kT)^{\frac {1}{2}}}e^{-{h\nu \over kT}}\\\end{aligned}}\,\! where the $h\nu$ in the exponential is from the lower velocity bound which we just defined. We now define $j_{\nu ,ff}$ to be the volume emissivity for free-free interactions. That is, $j_{\nu ,ff}$ measures the power radiated by plasma, per volume, into a solid angle $d\Omega$ . We can calculate the “per $d\Omega$ ” because the radiation is isotropic:

${j_{\nu ,ff}={16 \over 3{\sqrt {2\pi }}}{Z^{2}e^{6} \over m_{e}^{3 \over 2}c^{3}(kT)^{\frac {1}{2}}}n_{e}n_{p}e^{-{h\nu \over kT}}}\,\!$ where we have to include the # density of ions, $n_{p}$ , as well to account for the contribution from multiple nuclei. This is the expression for Thermal Bremsstrahlung. Note that the definition of $j_{\nu ,ff}$ in Rybicki & Lightman has an additional factor of ${\pi \over {\sqrt {3}}}\,{\bar {g}}_{ff}(v,T)$ , which is a quantum mechanical correction factor of order unity. It’s called the “Gaunt factor”.

Inverse Bremsstrahlung

We now consider the inverse effect, an $e^{-}$ can also absorb a photon and become more energetic and “free”.

We define the coefficient for thermal free-free absorption as:

$\alpha _{\nu ,ff}\equiv {j_{\nu ,ff} \over B_{\nu }}\,\!$ where we use the Planck function because we know that in the optically thick case the specific intensity, $I_{\nu }$ , becomes the source function, $S_{\nu }={j_{\nu ,ff} \over \alpha _{\nu ,ff}}$ , and in the case of optically thick thermal radiation, the source function becomes the Planck function. And so, just as the expression for $j_{\nu ,ff}$ was the definition of Thermal Bremsstrahlung, the expression for $\alpha _{\nu ,ff}$ is the definition for Inverse Bremsstrahlung.

${\alpha _{\nu ,ff}={8 \over 3{\sqrt {2\pi }}}{Z^{2}e^{6} \over m_{e}^{3 \over 2}c(kT)^{\frac {1}{2}}}{n_{e}n_{p} \over h\nu ^{3}}\left({1-e^{-{h\nu \over kT}}}\right)}\,\!$ where we can once again include a factor of ${\pi \over {\sqrt {3}}}\,{\bar {g}}_{ff}(v,T)$ to obtain the same result as Rybicki & Lightman. In the regime where $h\nu \gg kT$ the exponential term becomes negligible and the absorptivity will go as $\,\nu ^{-3}$ . The $h\nu \ll kT$ regime is the Rayleigh-Jeans regime and the expression for absorptivity goes as $\,\nu ^{-2}$ , where we have lost a factor of ${\,1/\nu }$ due to the taylor expansion of the exponential.