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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!

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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 $\frac{1}{2}$.

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  • $\begingroup$ Actually, the factor of 1/2 came out when I was computing the self-energy of a single sphere, not two interacting ones. But basically, you're saying that the formula for gravitational self-energy is always 1/2*<V,rho>? Thanks for replying BTW. $\endgroup$ – DumpsterDoofus Oct 8 '13 at 0:17
  • $\begingroup$ I was trying to convince you that if you don't have a factor of one half you are double counting. Your single sphere can be broken up into two half spheres you know. Another way of saying it is to image the work required to increase the charge from $0$ to $\rho_0$. The energy is $E=\int dE = \int \langle V, d \rho \rangle = \int \langle \nabla^{-2} \rho, d \rho \rangle = \int_0^1 \langle \nabla^{-2} \alpha \rho_0, d \alpha \rho_0 \rangle = \langle \nabla^{-2} \rho_0 , \rho_0 \rangle \int_0^1 \alpha d \alpha = \frac{1}{2} \langle V , \rho_0 \rangle$ $\endgroup$ – Brian Moths Oct 8 '13 at 0:29
  • $\begingroup$ Interesting. I think I get it now. And when you say the $\nabla^{2}$ operator is symmetric, do you mean it is a self-adjoint operator, and thus we can shuttle it from one side of the inner product to the other to obtain equivalence of the cross-terms? $\endgroup$ – DumpsterDoofus Oct 8 '13 at 0:44
  • $\begingroup$ ya that is what I meant $\endgroup$ – Brian Moths Oct 8 '13 at 1:07

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