It Almost Seems Too Convenient: The Virial Theorem

a) Consider a spherical distribution of particles, each with a mass $m_i$ and a total (collective) mass $\displaystyle\sum\limits_{i}^N m_i = M$, and a total (collective) radius $R$. Convince yourself that the total potential energy $U$ is of the form\[U=-\frac{GM^2}{R}\]

Let's first examine what it is we mean by a "spherical distribution of particles." It doesn't actually matter how densely packed all of these particles are--this could be a globular cluster, or it could be a single star. The key is that all of the component parts--whether they're stars or atoms--are evenly distributed across a spherical volume.

Pictured: A spherical distribution of particles (source).

In order to find the total potential energy of the sphere, we're going to integrate out from the center, summing the potential energies of spherical shells. Gravitational potential energy is defined as $U = -\frac{GMm}{r}$, where in the case of a spherical shell, $M$ is the mass enclosed by the shell, $m$ is the mass of the shell, and $r$ is the radius of the shell. Our first step, then, should be to define each of these terms for a given shell at radius $r_i$.

$G$ is constant, which makes our lives a bit easier. $M_i$ is going to be the volume enclosed by the spherical shell ($4/3 \pi r_i^3$) multiplied by the density of the distribution, which we can describe by dividing its total mass, $M$, by its total volume, $4/3 \pi R^3$ (where $R$ is the total radius). However, for simplicity's sake, we'll just plug $\rho$ into the equation for $\frac{M}{4/3 \pi R^3}$, giving us $M_i = 4/3 \pi r_i^3 \rho$.

Next we need to look at the mass $m_i$ of the spherical shell with radius $r_i$. If we were to integrate by mass, this quantity would simply be $dm$. However, since ultimately we're going to integrate by radius, we need to define it further. We'll take the usual approach when integrating by summing spherical shells and describe the volume of a given shell as its surface area times its width, or $4 \pi r_i^2 \times dr$. Multiplying this by density once again gives us $m_i = \rho 4 \pi r_i^2 dr$. Putting everything together now gives us an equation for the potential energy $U_i$ of a given spherical shell.\[U_i = -\frac{G \times 4/3 \pi r_i^3 \rho \times \rho 4 \pi r_i^2 dr}{r_i} = -\frac{16 \pi ^2}{3}(G r_i^4 \rho^2 dr)\] Now we're going to want to integrate, in order to go from the potential energy of a single shell to the overall potential energy of the distribution--as such, it makes more sense to describe $U_i$ as $dU$ on the left side of the equation. On the right, we're integrating radii from 0 out to $R$, so everything else is a constant that we can pull out. This gives us \[\int dU = -\frac{16 \pi^2 G \rho^2}{3} \int_0^R r^4dr\] The left integral just comes out to $U$, while the right integral comes out to $1/5 R^5$. Including all of the constants, then, we have \[U = -\frac{16 \pi^2 G \rho^2 R^5}{15}\] The last step is to recall that we defined $\rho$ as $\frac{M}{4/3 \pi R^3}$ and substitute back in for it. Once we do so, this mess of algebra quickly collapses into exactly what we were looking for.\[U = -\frac{16 \pi^2 GR^5M^2}{15(\frac{16}{9})\pi^2R^6} = -\frac{9GM^2}{15R} = -\frac{3GM^2}{5R}\] Exactly the form we wanted, with a coefficient of $3/5$ thrown in for good measure. This makes a great deal of sense, since most of the particles in the distribution are closer to the outside (imagine the number of particles in a spherical shell at the edge compared to the number in a spherical shell near the center).

b) If the average speed of a star in a cluster of thousands of stars is $<v>$, give an expression for the total mass of the cluster.

Star clusters are pretty cool objects. They range from tightly bound globular clusters in the galactic halo, composed of hundreds of thousands of ancient stars, to less numerous, looser associations of bright young stars. Of the former, the most famous is probably Messier 13, the Great Globular Cluster in Hercules, while of the latter the most well-known is almost certainly the Pleiades, a collection of brilliant blue stars in Taurus.

The Great Globular Cluster (source) and the Pleiades (source).

In order to figure out the mass of our cluster, we're going to use the Virial Theorem to which the title of this post refers. The Virial Theorem is both deceptively simple and incredibly useful. All it states is that the kinetic energy of a collection of matter bound by a force that follows an inverse-square law (gravity is a good, and applicable, example of such a force) is equal to half of its potential energy (multiplied by -1). In math-ese, this means that\[K = -1/2U\]. Now we need to apply this to the problem at hand. We know the potential energy, since we found that in the last part: $-\frac{3GM^2}{R}$, where $M$ and $R$ are the mass and radius of the cluster, respectively. The total kinetic energy, meanwhile, is the number of stars (which we'll call $N$) multiplied by the average mass of a given star, $m$ and the average velocity of a given star--$v$ (all multiplied by 1/2). All of this put together and plugged into the Virial Theorem gives us \[-\frac{3GM^2}{10R} = \frac{Nmv^2}{2}\] Now it's just a matter of solving for $M$--which is substantially easier when we note that $N \times m = M$, allowing us to cancel an $M$ from each side. A few algebraic steps give us a final expression for $M$.\[M=\frac{5Rv^2}{3G}\]

c) The cluster M80 has an angular diameter about 10 arcminutes and resides about $10^4$ parsecs from the Sun. If the average velocity of a star in M80 is $<v> \approx 10 \textrm{km s}^{-1}$, Approximately how much mass, in solar masses ($M_\odot$), does the cluster contain?

Our first step here is to figure out the radius of the cluster, which we can do using its angular diameter and distance by solving the very familiar small angle approximation $\theta=2R/D$ for $R$, where $\theta$ is the cluster's angular diameter, $D$ is its distance, and $R$ is its radius. We can then plug in $\frac{\theta D}{2}$ in for R in the equation we derived for $M$ in the last part, giving us $M = \frac{5\frac{\theta D}{2}v^2}{3G}$.

The cluster M80 in Scorpius. M80 is notable for containing numerous blue stragglers, one of which is easily visible just to the left side of the main body of the cluster (source).

From here, it's just a matter of inserting values for each of the constants and solving out (remembering to convert $10$ arcminutes into $3 \times 10^{-3}$ radians, $10^4$ parsecs into $3 \times 10^22$ cm, and $10$ km/s into $10^6$ km/s). This gives us \[M = \frac{5\frac{3 \times 10^{-3} \times 3 \times 10^{22}}{2} (10^6)^2}{3G} = 10^{39} /: /: \textrm{g} = 5 \times 10^5 \: \: M_\odot\] According to Wikipedia, this is dead on--score one for the Virial Theorem.

Credit goes out once again to Anne Madoff and Scott Zhuge for working through these tricky problems with me.