# Virial Theorem in Stellar Dynamics

1. Oct 7, 2008

### cepheid

Staff Emeritus
After reading in Longair's Galaxy Formation through the derivation of the virial theorem in the context of a dynamical system in equilibrium consisting of "point masses" interacting only through gravity, I proceeded to try to understand his comments on how the theorem can be applied in order to determine the mass of a galaxy. In these comments, he makes a claim:

The total kinetic energy T is equal to one half the total mass of the system M times the velocity dispersion <v^2>. Now, based on the definition of velocity dispersion I was able to find (in another book, mind you), it is the root mean square of the velocities:

$$\langle v^2 \rangle = \frac{1}{\sqrt{N}}(\sum_i{v_i^2})^{\frac{1}{2}}$$

Even if I assume in this context that <v^2> is actually that *without* the power of 1/2 (so that things will sort of make sense), his claim still basically amounts to:

$$2T = M\langle v^2 \rangle$$

$$\sum_i m_iv_i^2 = (\sum_i m_i)(\frac{1}{N}\sum_i{v_i^2})$$

This claim is clearly false. So what's going on? With what justification can he say that the total kinetic energy is just one half the total mass times the velocity dispersion squared?

Last edited: Oct 7, 2008
2. Oct 8, 2008

### cepheid

Staff Emeritus
no takers, I guess...

3. Oct 9, 2008

### Arch2008

I unfortunately don’t have the software to type mathematical expressions, but here goes:
We know that kinetic energy is one half the mass times the velocity squared, or

KE= (mv^2)/2

Longair is saying then that the total kinetic energy of a galaxy must be one half the total mass of the system times the “velocity dispersion”:

T= (M <v^2>)/2

This seems to me to be pretty intuitive. Solving for M then gives:

M= 2T / <v^2>

This gives perhaps the approximate mass of a galaxy from its total kinetic energy and derived velocity dispersion. So the clearly false part is…

4. Oct 9, 2008

### Kurdt

Staff Emeritus
Is it not just the average of the squared velocities?

5. Oct 9, 2008

### Arch2008

Well, what did Longair mean by velocity dispersion?

6. Oct 9, 2008

### Kurdt

Staff Emeritus
In my book by E Battaner, that same notation the author says is the "mean squared velocity".

Last edited: Oct 9, 2008
7. Oct 9, 2008

### Arch2008

Maybe, "there's the rub". We'll have to see what cepheid has to say.

8. Oct 9, 2008

### Arch2008

http://nedwww.ipac.caltech.edu/level5/Glossary/Glossary_V.html

"Velocity Dispersion The spread of a velocity distribution - that is, how stars move relative to one another. Technically, the velocity dispersion is the standard deviation of the velocity distribution. Stars with similar velocities have a small velocity dispersion, whereas stars with wildly different velocities have a large velocity dispersion. [C95] "

I found this, which is hardly a conclusive definition. It’s perhaps a term that can be molded as needed according to the author.

9. Oct 9, 2008

### cepheid

Staff Emeritus
Isn't that what I wrote in my first post? The LaTeX is showing up for me anyway? Maybe my definition of mean squared velocity is wrong. I interpreted it as (for N particles):

1/N*(sum over i of (v_i)^2 )

My main problem with it was the particulars of the algebra. Using my expressions, at least, you end up with the claim reducing to

sum of products = product of sums (???)

10. Oct 9, 2008

### cepheid

Staff Emeritus
I also talked to my prof (instructor, not supervisor) about velocity dispersion (although I didn't ask him this question in all its detail). He said that it is indeed a measure of the width of the velocity distribution (in a galaxy or galaxy cluster etc). However, there is no clear cut way of defining it. If your distribution is a nice gaussian then obviously you have a meaningful way of defining that width. But typically you have some sort of messy histogram with a bunch of outliers that you may have to cut out. In other words, picking some portion of the distribution that is centred on the mean in such a way that will give you the most reasonable estimate of the mass. Apparently, finding the best way to do this is part of the art of observing and depends upon the specific data in question. That is what I got out of what he was saying anyway.

11. Oct 10, 2008