Past exam of the radiative processes and high energy astrophysics course of the University of Oxford / 2024 / 3

Thermal bremsstrahlung is radiation from the acceleration of electrons in the Coulomb force of ions. At fixed temperature, the encounter rate per electron is proportional to ion number density $n_i$, and the number of radiating electrons per unit volume is $n_e$. Therefore $j_{\nu }\propto n_e{n}_i$ in a dilute, optically thin plasma. This is a binary-encounter rate, not a one-particle emission process.

For a fixed encounter geometry, the ionic electric field and hence the electron acceleration are proportional to ionic charge $Ze$. The Larmor formula is quadratic in acceleration, giving $j_{\nu }\propto Z^2$ at fixed species number densities. The dimensionless Gaunt factor $g(\nu ,T)$ corrects the simple classical result for the detailed quantum scattering and radiation calculation, after averaging over the thermal electron distribution. It depends on frequency and temperature and, in a more complete notation, ionic charge.

For $h\nu \ll k_{B}T$, expand the exponential in the thermal free-free emissivity:The energy emitted per unit frequency is therefore approximately flat, with the weak Gaunt factor dependence superposed. At these low photon energies, most of the thermal electrons can participate in emission; there is no exponentially small population threshold.

For $h\nu >k_{B}T$, the spectrum turns down approximately as $e^{-h\nu /(k_{B}T)}$, apart from the Gaunt factor. Conservation of energy requires an emitting electron to supply at least $h\nu$. The high-energy tail of its Maxwell-Boltzmann distribution is exponentially rare, which produces this cutoff.

“Flat” refers to $j_{\nu }$, an energy spectrum per unit frequency. The emitted photon number spectrum instead contains $1/(h\nu )$, while energy per logarithmic frequency interval is $\nu j_{\nu }$. At sufficiently low frequency the plasma can also become optically thick to free-free absorption; the given optically thin emissivity is then not the escaping spectrum.

Neglect electron mass and nuclear binding corrections when converting mass density $\rho$ into hydrogen and helium number densities. With $x=x_H$,The thermal free-free emissivity adds over ion species with weights $Z_i^2$. ConsequentlyWith each Gaunt factor set to one, the mixture hasAt fixed $n_e$, pure hydrogen gives the composition factor $1$, and pure helium gives $2$. The latter check follows independently from $n_{\mathrm{He}}=n_e/2$ and $Z^2=4$; simply replacing $Z$ by an average charge would lose this weighting.

The coefficient $j_{\nu }$ is power per unit volume, frequency and solid angle. For uniform, optically thin, isotropic emission from $V=4\pi R^3/3$, the spectral luminosity is $L_{\nu }=4\pi V{j}_{\nu }$. With photon energy $\epsilon =h\nu$, conservation of energy and the change of spectral variable giveThusThe $4\pi$ outside $V$ integrates over emitted solid angle; $1/\epsilon$ changes radiated energy to a photon count, and $1/h$ is the Jacobian determinant from frequency to energy.

With $h$ in $\mathrm{J}\mathrm{s}$ and $\epsilon$ in joules, this formula produces ${\mathrm{J}}^{-1}{\mathrm{s}}^{-1}$. To compare with counts per $\mathrm{keV}$, multiply by $1.602176634\times 10^{-16}J/keV$. Equivalently, the conversion cancels the same joule-per-$\mathrm{keV}$ factor in $\epsilon$: one can use the numerical $\epsilon$ in $\mathrm{keV}$ in the $1/\epsilon$ factor, retain $h$ in $\mathrm{J}\mathrm{s}$, and express $k_{B}T$ in $\mathrm{keV}$ in the exponential. The radius in the volume must still be converted to metres because $j_0$ is per ${\mathrm{m}}^3$.

In the thermal Sunyaev-Zeldovich effect, hot intracluster electrons transfer energy to cosmic microwave background photons by inverse Compton scattering. Their thermal energy is much larger than the CMB photon energy; in the nonrelativistic limit the mean fractional photon energy gain per scattering is about $4k_{B}T/(m_e{c}^2)$. Multiplication by the incremental Thomson scattering optical depth $d\tau ={\sigma }_T{n}_{e}dx$ explains why $y=\int [k_{B}T/(m_e{c}^2)]d\tau$ measures the electron pressure integrated along the line of sight. The result is a spectral distortion, with a low-frequency decrement and a high-frequency increment, rather than a positive additional emission component at every frequency.

For a uniform sphere, a line of sight at impact parameter $b$ has path length $2\sqrt{R^2-b^2}$. ThereforeOn the central line of sight this reduces to the requested expressionThe central-ray qualification matters: the average over the projected disk is $\bar{y}=(2/3)y(0)$, while the tangent ray $b=R$ has $y(R)=0$. Thus the printed formula does not hold for an arbitrary sightline. Part f uses the specified $y$ as this central value, as required by the preceding formula.

Write $\Theta =k_{B}T$ in $\mathrm{keV}$. The photon count ratio in d eliminates radius, electron number density and the common normalization:The extra ${\epsilon }_2/{\epsilon }_1$ is essential for a photon number spectrum. HenceThe measured central Compton y parameter and the supplied conversion coefficient then determine a column combination:To determine the two remaining quantities separately, define $\hat{R}=R/(1\mathrm{Mpc})$ and $\hat{n}=n_e/(1\mathrm{c}{\mathrm{m}}^{-3})$. Use $1\mathrm{Mpc}=3.08567758\times 10^{22}\mathrm{m}$, $h=6.62607015\times 10^{-34}\mathrm{J}\mathrm{s}$, $j_0=4.885\times 10^{-51}\mathrm{W}{\mathrm{m}}^{-3}\mathrm{H}{\mathrm{z}}^{-1}\mathrm{s}{\mathrm{r}}^{-1}$, $T_0=1.977\times 10^9\mathrm{K}$, and $x_H=0.76$ in d. With the spectral unit conversion explained there, the $0.2\mathrm{keV}$ count luminosity isExplicitly the numerical coefficient isSince $\hat{R}\hat{n}=0.00200087$, substitution gives ${\hat{R}}^3{\hat{n}}^2=\hat{R}(0.00200087{)}^2$, soRounding consistently with the supplied coefficient $0.004$, the cluster has $T\simeq 10^7\mathrm{K}$, $R\simeq 1\mathrm{Mpc}$ and $n_e\simeq 2\times 10^{-3}\mathrm{c}{\mathrm{m}}^{-3}$. Substituting the unrounded values into both X-ray count luminosities gives $2.8800\times 10^{54}$ and $4.4000\times 10^{50}\mathrm{ke}{\mathrm{V}}^{-1}{\mathrm{s}}^{-1}$, and the central Compton y parameter returns $1.3800\times 10^{-5}$. If the observed $y$ were a projected-disk average, its central value would be $3/2$ times larger and this same inversion would give $R$ smaller by $4/9$ and $n_e$ larger by $27/8$; specifying the sightline is therefore part of the inference.