The Schwarzschild radius $R_s$ is the distance from a black hole such that for $r < R_s$, the escape velocity exceeds the speed of light $c$. Since nothing in the universe can go faster than $c$, nothing which passes over this surface (an event horizon) can ever return: it is permanently captured by the gravitational pull of the black hole.

$R_s$ is therefore defined by the point where the force due to gravity, $\frac{GMm}{R_s}$, is equal to the kinetic energy of a photon: $\frac{1}{2}mc^2$. Rearranging for $R_s$ gives $$ \begin{equation*} R_s = \frac{2GM}{c^2}. \tag{1} \end{equation*} $$ In solar mass units, $R_s \approx 3\cdot \frac{M}{M_{\odot}}$km. Beyond the event horizon, general relativity breaks down and nobody knows what happens. It's possible that the black hole collapses down to a zero-dimensional point, in which case its volume is $0$ and hence its density is undefined; it is also possible that some unknown force or mechanism counteracts the gravitational collapse, and so its density is finite but very large. Since nothing which crosses the Schwarzschild radius can ever again escape, we will simplify matters and say that this is the radius which defines the black hole's volume. ^[1] Under this assumption, we find the density $\rho$ by substituing (1) into $\rho = \frac{M}{V} = \frac{M}{\frac{4}{3} \pi R^3}$: $$ \begin{align*} \rho_{BH} &= \frac{M}{\frac{4}{3} \pi R_s^3} \\ &= \frac{M}{\frac{4}{3} \pi \left(\frac{2GM}{c^2}\right)^3} \\ &= \frac{3c^6}{32 \pi G M^2} \\ &\approx 1.85 \cdot 10^{19} \frac{1}{(M/M_{\odot})^2} \end{align*} $$ (in the last line we have converted to solar mass units). So $\rho \propto M^{-2}$, i.e. the more massive the black hole, the less dense it is.

This is a surprising result. Say we take some substance of uniform density - a lump of clay - and double it so that $V_{f}=2V_{i}$ (subscripts designate final and initial). We have doubled both the volume and the mass, so $\rho_i = \rho_f$. However, doubling the volume implies $$\begin{align*} \frac{4}{3} \pi r_{f}^3 & = 2 \cdot \frac{4}{3} \pi r_{i}^3 \\ \implies r_f & = \sqrt[3]{2}~r_i. \end{align*} $$ So, adding an equal sized lump of clay doubled mass and volume, but increases the radius (we form the clay into a sphere) by a factor of only $\sqrt[3]{2} \approx 1.26$.

For a black hole, on the other hand, equation (1) implies that $V \propto M^3$. Similarly $R_s \propto M$, and so doubling the mass doubles the radius. Instead of density remaning constant as we add mass of the same density like we would expect, we have seen that it decreases as $M^{-2}$. Doubling the mass of a black hole implies that $p_f=\frac{1}{4}p_i$.

This means, for example, that Sagittarius A* - the supermassive black hole at the centre of the Milky Way with mass $M_{Sgr A^*} \approx 4.3*10^6M_{\odot}$ - is about $2 \cdot 10^{13}$ times less dense than a stellar mass black hole. If we had a 4.3 billion solar mass black hole, it would only be as dense as water. Apparently the most massive known black hole is the evocatively named S5 0014+813, weighing in at 40 billion solar masses. This has density $\rho \approx 0.01~$kg/m$^3$ - less dense than the atmosphere at an altitude of 30km.

^{[1] As you might have guessed, we have already glossed over quite a bit of complexity. The first point is that the black hole volume is not invariant across different reference frames. Different observers might measure different volumes (note, however, that the surface area is invariant). This is complicated by the fact that for $r < R_s$ the radial Schwarzschild coordinate $r$ becomes timelike, and the time coordinate $t$ becomes spacelike. Then to find the volume we must integrate with respect to $t$, and because black holes exist (classically) for infinite time, we find that they have infinite volume. So, to summarise the major assumptions:}

• We have chosen a coordinate system where the volume is $V = \frac{4}{3} \pi R_s^3$, with $R_s$ given by (1).
• We go on to calculate some kind of mean density, on the assumption that the mass distribution is uniform across the spherical volume defined by $R_s$. This is untrue, but an interesting question nonetheless. In fact what we are calculating is the minimum volume of some substance of uniform density required for it to collapse into a black hole.

A final point is that we are obviously using Newtonian (non-relativistic) arguments. We get the right answer, but I'm not really sure why. This is probably some kind of happy coincidence; black holes are decidedly non-Newtonian.↩

Let's calculate some upper bounds on the pressure and temperature inside a solar mass star. See chapters 1 and 2 of Stellar Structure and Evolution by Kippenhahn and Weigert for more details.

Consider a thin spherical mass shell of thickness $dr$. The mass per unit area of the shell is $\rho dr$ The force per unit area due to the pressure gradient over the shell, i.e. between $r$ and $r+dr$, is $P_i - P_e = \frac{\partial P} {\partial r} dr$, where $P_i$ and $P_e$ are the pressures at the interior and outer boundary of the mass shell respectively. For hydrostatic equilibrium, we have that the forces due to gravity and pressure balance: per unit volume, $$ \begin{align} \frac {\partial P} {\partial r} & = -g \rho \nonumber \\ & = -\frac {Gm}{r^2} \rho \end{align} $$ where g is the gravitational acceleration and G is the gravitational constant, $G = 6.673 \times 10^{-8}$ cm$^3$ g$^{-1}$ s$^{-2}$. Since the mass enclosed by a shell is a monotonic function of r (with $m=0$ at the centre of the star, and $m=M$ the total mass at the surface), we can transform $ r \mapsto m $. We have $$ \begin{align*} m & = \frac{4}{3} \pi r^3 \rho \\ \implies \frac {\partial m} {\partial r} & = 4 \pi r^2 \rho \\ \end{align*} $$ Hence (1) becomes $$ \begin{equation} \frac {\partial P} {\partial m} = - \frac { Gm } {4 \pi r^4 } \end{equation} $$ We can approximate these quantities by taking mean values: $ \frac {\partial P} {\partial m} \approx \frac { P_0 - P_c}{M} $ where $P_0 = 0$ and $P_c$ are the pressures at the surface and at the centre of the star respectively; $r \approx \frac{1}{2}R$ and $m \approx \frac{1}{2}M$. We end up with $$ \begin{equation} P_c \approx \frac { 2GM^2 } {\pi R^4 } \end{equation} $$ To relate this to the temperature, we need the ideal gas law: $PV = n \overline{R} T$, with pressure P , volume V, number of moles in the volume n, the gas constant $\overline{R} = 8.3145$ J/mol$\cdot$K (c.f. $R$, the stellar radius), and temperature T. It is convenient to define the gas constant per g instead of per mol, introducing the dimensionless mean molecular weight $\mu$. We then write the ideal gas law as: $$ \begin{equation} \rho = \frac { \mu } { \mathfrak{R} } \frac{P}{T} \end{equation} $$ where the gothic $\mathfrak{R} = 8.3145$ J/g$\cdot$K is the gas constant defined per gram. (To see that this transformation is correct, you could perform dimensional analysis on the two formulations of the gas law given here: $\frac{P}{T} = \frac{nR}{V} = \frac {\rho \mathfrak{R}}{\mu}$).

We're almost ready to calculate some numbers. Take the mean density $\overline{\rho} = \frac {3M}{4 \pi R^3}$. Then, evaluated at the centre, (4) gives $$\begin{align} T_c & = \frac {P_c}{\rho_c} \frac {\mu}{\mathfrak{R}} \nonumber \\ & = P_c \frac {\mu}{\mathfrak{R}} \frac {\overline{\rho}}{\rho_c} \frac {4 \pi R^3}{3M} \nonumber \\ & \approx \frac{8GM}{3R} \frac{\mu}{\mathfrak{R}}\frac{\overline{\rho}}{\rho_c} \end{align} $$ where we have multiplied by $1 = \overline{\rho}~\frac {4 \pi R^3}{3M}$ in the second line and substituted (3) in the last line. We now have equations (3) for pressure and (5) for temperature at the steller centre. Substituting the mass $M_{\odot}=1.989 \times 10^{33}$ g and radius $R_{\odot} = 6.96 \times 10^{10}$ cm of the sun, we get $$\begin{equation*} P_c \approx 7 \times 10^{15}\text{ erg/cm}^2 \end{equation*}. $$ For ionised hydrogen in the stellar interior, we take $\mu = 0.5$: neglecting electron mass, there is half an atomic mass unit per free particle. Since density increases monotonically with $r$, $\frac {\overline{\rho}} {\rho_c} < 1$. (5) then gives an upper bound on the temperature: $$\begin{equation*} T_c < 3 \times 10^7\text{ K} \end{equation*} $$ According to wikipedia, the temperature at the centre of the sun is indeed $\approx 1.57 \times 10^7$ K.

So, about 16 million degrees. Pretty hot.

The Friedmann equation describes the expansion of the universe: $$ \begin{equation} \left(\frac{\dot{a}(t)}{a(t)}\right)^2 = \frac{8 \pi G}{3} - \frac{kc^2}{a(t)^2}, \end{equation}\tag{1}$$ where $$k = -\frac{2U}{mc^2r^2}.$$

$G$ is the gravitational constant and $c$ is the speed of light; the other terms will be defined as we progress. We know from the redshifting of distant galaxies that the universe is expanding. Imagine a grid placed over all of space, such that as space expands the grid is stretched. Let's call this grid the comoving coordinate system, and the distance between points on the grid the comoving distance. Note that the expansion of space has no effect on the comoving distance, because our coordinate system expands with it.

By contrast, expansion causes the proper distance between any two points to increase with time. The coordinate $r'$ in this proper frame is defined by the product of the comoving coordinate $r$ and the scale factor $a(t)$: $$\begin{equation} r'(t) = a(t) r \end{equation}\tag{2}$$ The scale factor is a measure of the expansion of space, quantifying the amount by which the comoving coordinate system has been stretched.

Since we'll need it later, let's take the derivative. By the chain rule, $$ \begin{align*} \dot{r}' &= \frac{da(t)}{dt}\cdot r + a(t) \frac{dr}{dt} \\ &= \frac{da(t)}{dt}\cdot r \end{align*} $$

The second term has dropped out because comoving coordinates are constant in time. We can now derive the Friedmann equation from Newtonian arguments (the full derivation uses Einstein's field equations). Assume that the universe is homogeneous and isotropic with density $\rho$, and consider a sphere of (proper) radius $r'$. The total mass contained in the sphere is $M= \frac{4}{3} \pi r'^3 \rho $. The gravitational potential energy of a test particle of mass $m$ located on the sphere is then $$ \begin{align*} V_g & = - \frac{GMm}{r'} \\ & = - \frac {4 \pi G r'^2 \rho m}{3} \end{align*} $$ We know from Newton's shell theorem that we can ignore the effect of any mass outside the sphere. The kinetic energy of the test particle is $E_k = \frac{1}{2}m \dot{r}'^2$. Hence, the total energy is $$ \begin{align*} U & = V_g + E_k \\ \implies \frac{2}{ma^2r^2} \cdot U & = \frac{2}{ma^2r^2} \left[ -\frac{4 \pi G \rho (ar)^2m}{3} + \frac {1}{2}m(\dot{a}r)^2 \right] \\ \implies \left(\frac{\dot{a}}{a} \right)^2 & = \frac {8 \pi G \rho}{3} + \frac {2U}{ma^2r^2} \end{align*} $$ where we have expressed $r'$ in the comoving reference frame using (2) and its time derivative. We now let $k = -\frac{2U}{mc^2r^2}$ and obtain the Friedmann equation: $$ \begin{equation*} \left(\frac{\dot{a}(t)}{a(t)}\right)^2 = \frac{8 \pi G}{3} - \frac{kc^2}{a(t)^2}. \end{equation*} $$ Maybe I'll say more about the physical interpretation of this equation (and in particular the significance of $k$) another time.