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.

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.

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 ${\mathrm{m}}^3$.

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.