Before looking up the formula for the gravitational binding energy of a uniform sphere, I simply figured that the general formula for binding energy of an arbitrarily-shaped mass distribution would be $\left\langle V,\rho\right\rangle $, where $V$ is the potential as a function of space due to the distribution, $\rho$ is the density distribution as a function of space, and $\left\langle ,\right\rangle $ is the inner product (ie, integral over all space).
Going ahead for the special case of a uniform sphere of density $\rho$ and radius $R$, I used the well-known result for the potential inside of a sphere of uniform density,
$$V(r)=\frac{2}{3}\pi G\rho(r^{2}-3R^{2})\mbox{ for }r\leq R.
$$
I then computed the inner product,
$$\left\langle V,\rho\right\rangle =\int_{0}^{R}r^{2}dr\int_{0}^{\pi}\mbox{sin}(\theta)d\theta\int_{0}^{2\pi}d\phi V(r)\rho=-\frac{32}{15}G\pi^{2}R^{5}\rho^{2}=-\frac{6GM^{2}}{5R},
$$
which is twice the correct result, $\frac{-3GM^{2}}{5R}$. (The last equality follows from $\rho=\frac{M}{4/3\pi R^{3}}$).
I fully understand the geometric proof that the binding energy of a sphere is $\frac{-3GM^{2}}{5R}$ that proceeds by successively moving shells in from infinity, but was a bit confused when the inner product approach gave an extra factor of 2. My question is as follows:
- Is the extra factor of two in $\left\langle V,\rho\right\rangle$ due to double-counting the interaction energies?
- If so, is the correct formula for the binding energy of an arbitrary mass distribution $U=\frac{1}{2}\left\langle V,\rho\right\rangle$?
I am trying to estimate the total energy released when two planets which are just barely touching each other collapse to form a single large sphere, and in the course of the derivation this issue sprung up. Any help explaining the extra factor of 2 would be greatly appreciated!
Best Answer
You are right that the factor of $\frac{1}{2}$ is from double-counting. To see this let's assume we have two objects $A$ and $B$, with densities $\rho_A$ and $\rho_B$ so that the total density is $\rho$. The potential $V$ is a sum of the potential $V_A$ from object $A$ and $V_B$ from object $B$. Now let's evaluate the energy using your formula. We get $$E = \langle \rho , V \rangle = \langle \rho_A + \rho_B , V_A + V_B \rangle = \\ \langle \rho_A , V_A \rangle+\langle \rho_A , V_B \rangle+\langle \rho_B , V_A \rangle+\langle \rho_B , V_B \rangle$$
Now $\langle \rho_A , V_A \rangle$ and $\langle \rho_B , V_B \rangle$ are the gravitational self energies for objects $A$ and $B$ (according to your formula anyway, actually there should be a factor of one half). Also, the terms $\langle \rho_A , V_B \rangle$ and $\langle \rho_B , V_A \rangle$ are equal, basically because the $\nabla^{2}$ operator is symmetric.
Now we ask what happens when we start with object $A$ and bring object $B$ in from infinity? No work is done on object $A$ since it does not move. The work done on object $B$ is $\langle \rho_B , V_A \rangle$. However, your formula predicts twice this change in energy, because you have both $\langle \rho_A , V_B \rangle$ and $\langle \rho_B , V_A \rangle$. This is exactly the manifestation of the double counting. This problem is solved by the factor of $\frac{1}{2}$.