# Electric field energy

Is it true that $\int_V \rho \phi$ is the equation for the energy of a charge distribution in an external potential and $\frac{1}{2} \int_V \rho \phi$ is the equation for the energy of a charge distribution in its own potential?

I never noticed this difference before and there are not many reference that explicitely touch this topic, so I thought that I could ask this.

• In short, yes, you are correct. Also, see physics.stackexchange.com/q/79970. NowIGetToLearnWhatAHeadIs explained why the factor of $\frac{1}{2}$ comes about in his answer. May 12, 2014 at 20:24
• thank you, this is actually something I was not aware of before. May 12, 2014 at 20:26
• In his comments, he explains a rather clever way of seeing where the factor of $\frac{1}{2}$ comes from. He does $E=\int dE = \int \langle V, d \rho \rangle = \int \langle \nabla^{-2} \rho, d \rho \rangle = \int_0^1 \langle \nabla^{-2} \alpha \rho_0, d \alpha \rho_0 \rangle = \langle \nabla^{-2} \rho_0 , \rho_0 \rangle \int_0^1 \alpha d \alpha = \frac{1}{2} \langle V , \rho_0 \rangle.$ May 12, 2014 at 20:31
• A more visual way of seeing it is to imagine you discretize $\rho$ into a bunch of little cubic chunks. The total energy becomes the sum over pairs of cubes, $E=\sum_{i=1}^N\sum_{j=1}^{i-1}\frac{m_im_j}{4\pi\epsilon_0| \mathbf{r}_i-\mathbf{r}_j|}=\frac{1}{2}\sum_{i=1}^N\sum_{j=1,j\neq i}^{N}\frac{m_im_j}{4\pi\epsilon_0|\mathbf{r}_i-\mathbf{r}_j|}$ which in the limit of infinitely small cubes becomes $\frac{1}{2}\int_V\int_V\frac{1}{4\pi\epsilon_0}\frac{\rho(\mathbf{r}_1)\rho( \mathbf{r}_2 )}{|\mathbf{r}_1-\mathbf{r}_2|}\,d\mathbf{r}_1d\mathbf{r}_2=\frac{1}{2} \int_V \rho\phi$. May 12, 2014 at 20:39
• So the factor of $\frac{1}{2}$ comes about because you need to avoid double-counting when you do a sum over pairs of charges. May 12, 2014 at 20:41

There are really two questions here. First, why is there a factor of $$\frac{1}{2}$$ in the self-energy expression $$E=\frac{1}{2}\int\rho(\mathbf{r})\phi(\mathbf{r})\,d\mathbf{r}=\frac{1}{2}\langle\rho,\phi\rangle$$? Second, why is there no factor of $$\frac{1}{2}$$ in the expression for energy in an external potential?

## Self-energy of $$\rho$$

"NowIGetToLearnWhatAHeadIs" answered this in a related question, but there are two arguments you can give. The first was one which he provided in a comment, namely that $$E=\int dE = \int \langle V, d \rho \rangle = \int \langle \nabla^{-2} \rho, d \rho \rangle = \int_0^1 \langle \nabla^{-2} \alpha \rho_0, d \alpha \rho_0 \rangle \\= \langle \nabla^{-2} \rho_0 , \rho_0 \rangle \int_0^1 \alpha d \alpha = \frac{1}{2} \langle V , \rho_0 \rangle.$$ It's a clever argument, and somewhat subtle; the idea is to gradually "materialize" the charge distribution $$\rho$$ bit by bit, sort of like a ghost which slowly gains solid form.

A second, more direct way is to imagine discretizing space into lots of little cubes. The total energy then becomes the sum of the pairwise contributions, $$E=\sum_{i=1}^N\sum_{j=1}^{i-1}\frac{m_im_j}{4\pi\epsilon_0| \mathbf{r}_i-\mathbf{r}_j|}=\frac{1}{2}\sum_{i=1}^N\sum_{j=1,j\neq i}^{N}\frac{m_im_j}{4\pi\epsilon_0|\mathbf{r}_i-\mathbf{r}_j|}$$ which in the limit of infinitely small cubes becomes $$E=\frac{1}{2}\int_V\int_V\frac{1}{4\pi\epsilon_0}\frac{\rho(\mathbf{r}_1)\rho( \mathbf{r}_2 )}{|\mathbf{r}_1-\mathbf{r}_2|}\,d\mathbf{r}_1d\mathbf{r}_2=\frac{1}{2} \int_V \rho(\mathbf{r})\phi(\mathbf{r})\,d\mathbf{r}=\frac{1}{2}\langle\rho,\phi\rangle.$$

So the factor of $$\frac{1}{2}$$ comes about because you need to avoid double-counting when you do a sum over pairs of charges.

## Energy in an external potential

This is hopefully more straightforward: since $$\phi$$ is energy per charge, if you put a point charge $$q$$ there, it will take an energy of $$q\phi$$. And if you instead put a big complicated charge distribution $$\rho$$ there, then you just add up the energies: $$E=\int\rho(\mathbf{r})\phi(\mathbf{r})\,d\mathbf{r}=\langle\rho,\phi\rangle.$$