Where Does the Missing Factor in the Virial Theorem Derivation Come From?

[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.