- #1

- 29

- 2

## Summary:

- Why does Jackson's "Energy Type Functionals" seem to double count the energy?

In section 1.12 Variational Approach to the Solution of the Laplace and Poisson Equations, Jackson mentions that in electrostatics, we can consider "energy type functionals". He gives, for Dirichlet Boundary Conditions,

$$I[\psi]=\frac{1}{2}\int_{V}\nabla\psi\cdot\nabla\psi d^3x-\int_{V}g\psi d^3 x$$

After variation, he then finds the equations of motion,

$$\nabla^2\psi=-g$$

This would intuitively lead me to make the identification of ##\psi=\phi## as the scalar potential, and ##g=\frac{\rho}{\varepsilon_0}## to be the charge density. However, substituting this into the functional gives

$$I[\psi]=\frac{1}{2}\int_{V}|\vec{E}|^2d^3x-\frac{1}{\varepsilon_0}\int_{V}\rho\phi d^3x$$

The first term is the electrostatic energy, and the second is the electrostatic energy too. Doesn't this double count the energy? While I understand that if energy is minimized, twice the energy is minimized too, why

$$I[\psi]=\frac{1}{2}\int_{V}|\vec{E}|^2d^3x$$

$$I[\psi]=\frac{1}{2}\int_{V}\nabla\psi\cdot\nabla\psi d^3x-\int_{V}g\psi d^3 x$$

After variation, he then finds the equations of motion,

$$\nabla^2\psi=-g$$

This would intuitively lead me to make the identification of ##\psi=\phi## as the scalar potential, and ##g=\frac{\rho}{\varepsilon_0}## to be the charge density. However, substituting this into the functional gives

$$I[\psi]=\frac{1}{2}\int_{V}|\vec{E}|^2d^3x-\frac{1}{\varepsilon_0}\int_{V}\rho\phi d^3x$$

The first term is the electrostatic energy, and the second is the electrostatic energy too. Doesn't this double count the energy? While I understand that if energy is minimized, twice the energy is minimized too, why

*must*we double count the energy in order to get the correct equations of motion? Why can't we just minimize this functional?$$I[\psi]=\frac{1}{2}\int_{V}|\vec{E}|^2d^3x$$