Hydrostatic equilibrium

An important application of X-ray observations is to probe the gravitational potential of the galaxy. Indeed if the system is in equilibrium, then the gravitational pull will be balanced by the gradient of the pressure in the hot gas,

$$\nabla\Phi=-\nabla p/\rho .$$ (6.7)

(This should look familiar from hydrostatic equilibrium in stars!) By modelling the spectrum as function of position, we can determine $T$ and $\rho$ separately, and hence the rhs of Eq.(6.7). From that, one can reconstruct the gravitational potential, and hence infer the mass distribution.

The temperature $T$ is observed to remain approximately constant, in which case we can obtain an approximate expression for the density profile. The pressure is

$$p={\hbox{${\rm k}_{\sc\rm B}$} T\over \mu \hbox{$m_{\sc\rm p}$}} \rho .$$ (6.8)

If we take $T$ constant then the rhs of Eq. (6.7) becomes $-{\hbox{${\rm k}_{\sc\rm B}$} T\over \mu \hbox{$m_{\sc\rm p}$}}\nabla\ln(\rho)$. The lhs of this equation is just the gravitational force, $\nabla\Phi=GM(<r)/r^2$, with $M(<r)$ the mass interior to $R$. Combining these two, we obtain after some rearranging

$$r^2 {d\ln(\rho)\over dr}={G\mu \hbox{$m_{\sc\rm p}$}\over\hbox{${\rm k}_{\sc\rm B}$} T}M(<r) .$$ (6.9)

Taking the derivative of both sides with respect to $r$, gives

$${d\over dr} \left(r^2 {d\ln(\rho)\over dr}\right) = -{4\pi G\mu \hbox{$m_{\sc\rm p}$}\over \hbox{${\rm k}_{\sc\rm B}$} T} r^2\rho ,$$ (6.10)

since $dM(<r)/dr=4\pi r^2 \rho$. Try a solution of the form $\rho=C r^b$, where $C$ and $b$ are constants. Substituting this, the lhs of Eq. (6.10) is $b$, and the rhs is $-(4\pi G\mu\hbox{$m_{\sc\rm p}$}/\hbox{${\rm k}_{\sc\rm B}$} T) C r^{b+2}$. If this is to be a solution, then $b=-2$, and

$$\rho(r) = {\hbox{${\rm k}_{\sc\rm B}$} T\over 2\pi G\mu\hbox{$m_{\sc\rm p}$}} r^{-2} .$$ (6.11)

This would be the density profile if the gas were to furnish the gravitational potential required to keep it in hydrostatic equilibrium. However, in addition to the gas, there is also the gravitational potential due to stars. And so we should have used $dM/dr=4\pi r^2(\rho_{\rm gas}+\rho_{stars})$. By estimating the mass in stars, we can correct our density distribution to what it should be, in the presence of stars. Since this extra gravity increases the rhs of Eq. (6.9), the density $\rho_{\rm gas}$ will need to increase as well.

Given the corrected density profile, we can compute the expected X-ray emissivity, since that depends on the density $\propto\rho_{\rm gas}^2$. The emissivity is much higher than expected, and hence we infer the presence of dark matter in ellipticals.