Energy in space is anything but one-dimensional

We are bounded in a nutshell of Infinite space: Blog Post #17, Worksheet # 6.1, Problem #3: Energy in space is anything but one-dimensional

3. One of the most useful equations in astronomy is an extremely simple relationship known as the Virial Theorem. It can be used to derive Kepler’s Third Law, measure the mass of a cluster of stars, or the temperature and brightness of a newly-formed planet. The Virial Theorem applies to a system of particles in equilibrium that are bound by a force that is defined by an inverse central-force law \(F \propto 1/r^{\alpha} \) It relates the kinetic (or thermal) energy of a system, K, to the potential energy, U, giving \[ K = - \frac{1}{2} U \]

(a) Consider a spherical distribution of N particles, each with a mass m. The distribution has total mass M and total radius R. Convince yourself that the total potential energy, U, is approximately \[ U \approx - \frac{GM^2 }{ R} \]  You can derive or look up the actual numerical constant out front. But in general in astronomy, you don’t need this prefactor, which is of order unity.

(b) Now let’s figure out what K is equal to. Consider a bound spherical distribution of N particles (perhaps stars in a globular cluster), each of mass m, and each moving with a velocity of \(v_i\) with respect to the center of mass. If these stars are far away in space, their individual velocity vectors are very difficult to measure directly. Generally, it is much easier to measure the scatter around the mean velocity if the system along our line of sight, the velocity scatter \(\sigma^2\) . Show that the kinetic energy of the system is: \[ K = N \frac{3}{2} m \sigma^2 \]  

(c) Use the Virial Theorem to show that the total mass of, say, a globular cluster of radius R and stellar velocity dispersion \(\sigma\) is (to some prefactor of order unity): \[M \approx \frac{\sigma^2 R}{G} . \]

(a) A simple relationship which continues to help coalesce information and difficult problems into understandable proportions, the Virial Theorem utilizes the correlation of potential and kinetic energies to describe the entirety of a large stellar system, such as that of a solar system or an entire galaxy. In this particular case, the Virial Theorem can be used to arrive at the velocity scatter of a galaxy as well as the relation this holds to mass.

First, we must find the relationship that gravitational potential energy has to the kinetic energy present in the system. Therefore, we begin with the basic Newtonian equation for the force of gravity from an object onto any other: \[ F_{Grav} = \frac{G (M_1 \cdot M_2)}{R^2} , \] (where G is the gravitational constant, the M’s are the objects’ masses, and R is the distance between them). This can be supplanted into the gravitational potential energy equation: \[U_{Grav} = - m\cdot g \cdot R ,  \]  \[U_{Grav} = - F_{Grav} \cdot R ,  \] \[U_{Grav} = - \frac{G (M_1 \cdot M_2) }{R^2} \cdot R ,  \] \[U_{Grav} \approx - \frac{G (M^2) }{R} , \] which is what the equation asks us to find and prove.

(b) In observational astronomy, finding the velocity of any one, singular, small point, is both incredibly difficult, nigh impossible, as well as inefficient. It is much more useful to measure the combined velocity scatter of a system and use this value interchangeably in velocity equation to describe the total velocity of a large enough system. Now then, the general equation for kinetic energy is: \[ K = \frac{1}{2} m_{Total} v^2 , \] and we know the total mass of a galaxy can be approximated with the amount of particles (objects) (N) times the median mass of each individual particle (m). Thus the equation becomes:  \[ K = \frac{1}{2} N m  v^2 , \] and now we introduce our definition of \(v\) in a large system: \[ K = \frac{1}{2} N m \sigma^2 , \] but this is still not the equation we are looking for. The coefficient of 3 is still missing, but what do we know exists in 3 dimensions (as far as we know): space! Not Space-Time, but rather just distance measurements, which corresponds to velocity vectors that exist in 3 directions/dimensions. Thus, the equation has to be added to the other two physical dimensions to make the entire energy equation: \[K_{Total} = K_x + K_y + K_z\] \[ K_{Total} = \frac{1}{2} N m \sigma^2  + \frac{1}{2} N m \sigma^2 + \frac{1}{2} N m \sigma^2 , \] \[ K_{Total} =\frac{3}{2} N m \sigma^2 .\]

(c) Therefore, taking both the equations we have just established (and in the case of K using it as only one dimensions since gravity is in one direction), we can now look for the relationship between speed and mass. Starting with \[ K = - \frac{1}{2} U , \] and substituting, \[ \frac{1}{2} N m \sigma^2 \approx - \frac{1}{2} \left(-\frac{G (M^2) }{R} \right) , \] and simplifying a bit, \[ M_{Total} \sigma^2 \approx \frac{G (M^2) }{R} , \] cancel out the M, \[ \sigma^2 \approx \frac{G (M) }{R} , \] and rearrange: \[ M_{Total} \approx \frac{ \sigma^2 R }{G} , \] the original equation we were looking for.