Rob Jeffries found this document for me that describes how the X-Ray emissions from galaxy clusters are used to calculate the mass of the cluster. I'm unable to follow the steps to calculating the gas mass. I get the part where the central electron density is calculated, but I'm not sure how to go from there to a mass within a radius, $r$.

Specifically, in section (2) there is a formula for the surface brightness profile that appears to be pretty standard: $$ S(r) = S_{0} \left(1 + \left({\frac{r}{r_c}}\right)^2\right)^{(0.5 - 3 \beta)} $$ The authors then go on to calculate the central electron density $$ n_{0} = 2.89 \times 10^{-3} \ h_{50}^{1/2} \ cm^{-3} $$ I can't make the connection from the central electron density to a function that provides the mass at a given radius, r. The authors conclude the hot gas mass is $$ 5.1 \times 10^{14} \ h_{50}^{-5/2} \ M_{\odot} $$ but I don't see how they get from A to B.

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    $\begingroup$ It will probably be useful to copy the portion of the paper that you are not understanding (and be clear/direct about what it is they're saying that you're not getting), than to leave us to read a 4-page paper and guess what it is that you're missing. $\endgroup$ – Kyle Kanos Oct 7 '15 at 12:42
  • $\begingroup$ @DonaldRoyAirey - If the emissions are x-rays, then the next question is whether those emissions are from some form of bremsstrahlung, synchrotron, or just regular thermal blackbody emissions (my guess is one of the first two). If it is bremsstrahlung or synchrotron, I think you can relate the power to the density and local background parameters. That would be my guess as a good starting point. $\endgroup$ – honeste_vivere Oct 7 '15 at 13:03
  • $\begingroup$ bremsstrahlung - I've got a starting point, thanks, I need someone to fill in what apparently is an implicit step to the authors but I can't find the background on it. $\endgroup$ – Donald Airey Oct 7 '15 at 13:05

The paper by Briel et al uses the formulae in Henry & Henriksen (1986): they start with the spatial electron number density (eq. (2) in Henry & Henriksen) $$ n_e(r) = n_0\left(1 + \left(\frac{r}{a}\right)^2\right)^{-3\beta/2}. $$ The cluster gas produces thermal bremsstrahlung, which has an emissivity of the form $$ \epsilon_\text{br}(r) \sim n^2_e(r)\,g(E,T)\,(kT)^{-1/2}\,e^{-E/kT}, $$ where Henry & Henriksen use $$ g(E,T) = 0.9(E/kT)^{-0.3}. $$ The surface brightness is then the total emissivity along the line of sight $z$ and within a certain energy range $[E_1,E_2]$ $$ \begin{align} S(R) &= \int_{E_1}^{E_2}\text{d}E\int_{-\infty}^{+\infty} \epsilon_\text{br}(r)\,\text{d}z \sim I_1I_2, \end{align} $$ with the integrals $$ \begin{align} I_1&= (kT)^{-1/2}\int_{E_1}^{E_2}(E/kT)^{-0.3}\,e^{-E/kT}\text{d}E\\ &= \sqrt{kT}\,\bigg[\gamma(0.7,E_2/kT) - \gamma(0.7,E_1/kT)\bigg] \end{align} $$ and $$ \begin{align} I_2 &= \int_{-\infty}^{+\infty}n^2_0\left(1 + \left(\frac{R}{a}\right)^2+ \left(\frac{z}{a}\right)^2 \right)^{-3\beta}\,\text{d}z\\ &= an^2_0\left(1 + \left(\frac{R}{a}\right)^2\right)^{-3\beta+0.5} \int_{-\infty}^{+\infty}(1+u^2)^{-3\beta}\,\text{d}u\\ &= \sqrt{\pi}\frac{\Gamma(3\beta-1/2)}{\Gamma(3\beta)}an^2_0\left(1 + \left(\frac{R}{a}\right)^2\right)^{-3\beta+0.5}, \end{align} $$ where $r^2 = R^2 + z^2$ and $R$ is the projected radius on the plane of the sky; the exact formula for $S(r)$ is given by eq. (3) in Henry & Henriksen. So this is where the formula for the surface brightness comes from in Briel et al: $$ S(R) = S_0\left(1 + \left(\frac{R}{a}\right)^2\right)^{-3\beta+0.5}. $$ Unfortunately, in their notation they use $r$ instead of $R$ for the projected radius.

The total mass inside a radius $r_b$ then follows from the spatial electron density: $$ M(r_b) = 4\pi\,m_\text{A}\int_0^{r_b}n_e(r)\,r^2\text{d}r = 4\pi\,m_\text{A}n_0\int_0^{r_b}\left(1 + \left(\frac{r}{a}\right)^2\right)^{-3\beta/2}\,r^2\text{d}r, $$ where $m_\text{A}$ is the average mass of an atom in the cluster gas. To calculate $m_\text{A}$, we can assume that the cluster gas is made up entirely of hydrogen and helium atoms. We need to be careful though: each hydrogen atom corresponds with one electron, but each helium atom corresponds with two electrons. Therefore, if $\rho$ is the total mass density, $$ \rho = n_em_\text{A}= (n_\text{H} + 2n_\text{He})m_\text{A}, $$ where $n_\text{H}$ and $n_\text{He}$ are the number densities of hydrogen and helium atoms, respectively. Now, let's call $X$ the mass fraction of hydrogen atoms. Then $$ X\rho = n_\text{H}m_\text{H},\qquad (1-X)\rho = n_\text{He}m_\text{He} \approx 4n_\text{He}m_\text{H}, $$ where we used in the last line the fact that the mass of a helium atom is about 4 times the mass of a hydrogen atom. Putting it all together, we find $$ m_\text{A} = \frac{\rho\, m_\text{H}}{(n_\text{H} + 2n_\text{He})m_\text{H}}\approx \frac{2\rho\, m_\text{H}}{2X\rho + (1-X)\rho} = \frac{2m_\text{H}}{1+X}. $$ I don't know what value of $X$ is used by Briel et al, but a common value is $X = 0.768$ (see e.g. eqs. (16) & (17) in Wu et al, 1999). Since $$ m_\text{H} = 1.67\times 10^{-27}\,\text{kg} = 8.42\times 10^{-58}\,\text{M}_\odot, $$ we obtain $$ m_\text{A} = 9.52\times 10^{-58}\,\text{M}_\odot. $$ The other values are listed in Briel et al: $$ \begin{align} n_0 &= 2.89\times 10^{-3}\,h_{50}^{1/2}\,\text{cm}^{-3} = 8.49\times 10^{70}\,h_{50}^{1/2}\,\text{Mpc}^{-3},\\ r_b &= 5\,h_{50}^{-1}\,\text{Mpc},\\ a &= 0.42\,h_{50}^{-1}\,\text{Mpc},\\ \beta &= 0.75, \end{align} $$ which indeed gives $$ M(r_b) = 5.1\times 10^{14}\,h_{50}^{-5/2}\,\text{M}_\odot. $$

Update: the relation between angles and intrinsic size in Coma:

At the time this article was written, its redshift was measured to be $z=0.0235$ (Sarazin et al, 1982). All distances were also given in terms of $h_{50}$, a dimensionless constant defined as $$ H_0 = 50h_{50}\,\text{km}\,\text{s}^{-1}\,\text{Mpc}^{-1}. $$ In other words, a value of $h_{50}=1$ corresponds with a Hubble constant of $H_0 = 50\,\text{km}\,\text{s}^{-1}\,\text{Mpc}^{-1}$. For a modern-day value $H_0 = 68\,\text{km}\,\text{s}^{-1}\,\text{Mpc}^{-1}$, you get $h_{50}=68/50=1.36$.

From Hubble's Law, we get the co-moving distance to Coma: $$ D_c \approx \frac{cz}{50h_{50}\,\text{km}\,\text{s}^{-1}\,\text{Mpc}^{-1}} = 141h_{50}^{-1}\,\text{Mpc}. $$ But to convert angles into intrinsic sizes, we need to use the angular diameter distance, which is a small cosmological correction: $$ D_A = \frac{D_c}{1+z} = 138h_{50}^{-1}\,\text{Mpc}. $$ Therefore, $$ a = 10.5' = 0.00305\,\text{rad} \rightarrow aD_A = 0.42h_{50}^{-1}\,\text{Mpc}, $$ which is given in the first paragraph on page L33 in Briel et al.

  • $\begingroup$ Thank you very much for the derivation. It's far more detailed than the papers I found which seemed to take these steps as implicit. Anyway, we agree on the formula for mass and, approximately, the number for the density. My point is that this formula describes a solid. If you double the radius (10 Mpc) you triple the mass (6.8 e14 M⊙). What kind of a gas works like that? $\endgroup$ – Donald Airey Oct 12 '15 at 1:02
  • $\begingroup$ Actually, no, we don't agree on the numbers. How did you get 0.42 Mpc for a core radius? At 99 Mpc, that number should be about 0.22 Mpc (for 10.5 arcmin). Would you mind showing me how you calculated this value? $\endgroup$ – Donald Airey Oct 12 '15 at 1:42
  • $\begingroup$ @DonaldRoyAirey The value 0.42 Mpc is given in the first paragraph on page L33 in Briel et al. I've added an explicit calculation to my post. $\endgroup$ – Pulsar Oct 12 '15 at 9:43
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    $\begingroup$ @DonaldRoyAirey The density profile is only valid within a certain radius, where the gas is isothermal and in thermal equilibrium. At the outer edges of the cluster, this will no longer be the case, and the density will drop off rapidly. So you cannot extrapolate this density profile to larger radii, beyond what has been observed. $\endgroup$ – Pulsar Oct 12 '15 at 9:49
  • $\begingroup$ It's the density that determines the boundary of the cluster. We define the boundary of the cluster when the density has dropped below 200 times the critical density of the universe. According to this function, it never gets to this point and extends forever like a solid. I was hoping you could answer my question: what kind of a gas acts like this? $\endgroup$ – Donald Airey Oct 12 '15 at 11:47

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