Virial Theorem derivation

[Piano man]

Currently working through a derivation of the Virial Theorem relating average internal pressure to gravitational potential energy.

So I've got to
-3\int^V_0 Pdv=-\int^M_0 \frac{Gm}{r}dm

which is meant to give

3 \langle P \rangle V=-E_{grav}

But if I'm right in saying that E_{grav} is \frac{GM^2}{r} then the above integral on the rhs gives an extra factor of 1/2.

What am I missing?

[zhermes]

Are you sure the 'dm' should be referring to the same 'm' as the one in the integrand?
I.e. maybe it should me dm' ("dee-em prime")?

[Piano man]

That would give the right answer yes, but I don't see any justification for it.
The 'm' refers to the gravitational mass within radius r, and dm has already been substituted in for 4\pi r^2\rho dr

So do you see any reason for using dm' other than the fact that it works?

[facenian]

could you state the problem precisely?. The virial theorem I know relates relates mean values in time of kinetic energy and potencial energy

[Piano man]

This is the part of the derivation before you introduce kinetic energy.

Starting with \frac{dP}{dr}=-\frac{Gm\rho}{r^2}

Multiplying both sides by 4\pi r^3 and integrating over the entire radius of the star:
\int^R_0 4\pi r^3\frac{dP}{dr}dr=-\int^R_0\frac{Gm}{r}4\pi r^2\rho dr

Integrating the left by parts, and subbing dm=4\pi r^2\rho dr on the right:

\left[4\pi r^3P\right]^R_0-3\int^R_04\pi r^2Pdr=-\int^M_0\frac{Gm}{r}dm

The first term on the left is 0 since P(R)=0, ie, pressure at the surface.

Sub in volume of spherical shell dv=4\pi r^2dr and you get the original equation in the first post:

-3\int^V_0Pdv=-\int^M_0\frac{Gm}{r}dm


From that you can relate the average pressure to the gravitational energy, and also to the thermal energy and mash things up a bit to get the familiar Virial theorem.
But for now, I'm still wondering where the factor of 1/2 has gone. Any ideas?
And thanks for your contributions so far! :D

[facenian]

I think that is correct when you assert that
E_{grav}=-\int^M_0 \frac{Gm}{r}dm
but don't thing is correct to say that
E_{grav}=\frac{GM^2}{r}

[Piano man]

Thank you - I was beginning to start thinking something similar, because I found somewhere referring to
E_{grav}=\alpha\frac{GM^2}{R}

so I'm guessing the alpha accounts for the 1/2.
It would be interesting to know what other values it could take...

Thank you very much for your help.

[facenian]

The way I understand it is that the integration can not be calculated without knowing m as a funtion of r so the factor 1/2 is not correct. You simply must realize that the integral is the correct expresion for E_grav.

[Piano man]

Okay that's logical. Thank you.