"Energy Type Functionals" in Jackson

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

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vanhees71
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Why should the functional be the field energy to begin with? It's just a functional letting you derive the electrostatic field equations given the source ##g=\rho/\epsilon_0##.

To get the field energy (as well as momentum and angular momentum) from variational principles you need to analyze the action functional and apply Noether's theorem to the relativistic space-time symmetries (temporal and spatial translations, rotations, Lorentz boosts, in short the proper orthochronous Poincare group).
 
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samalkhaiat
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[tex]I[\psi]=\frac{1}{2}\int_{V}|\vec{E}|^2d^3x-\frac{1}{\varepsilon_0}\int_{V}\rho\phi d^3x[/tex]
The first term is the electrostatic energy, and the second is the electrostatic energy too.
No, the second term is not the electrostatic energy. The electrostatic energy is given by [tex]\mathcal{E} = \frac{1}{2}\int d^{3}x \ \rho (x) \ \varphi (x).[/tex] If you use [tex]\rho = \frac{1}{4 \pi} \nabla \cdot \vec{E} = - \frac{1}{4 \pi} \nabla^{2}\varphi ,[/tex] and integrate by parts, you find [tex]\frac{1}{2}\int d^{3}x \ \rho (x) \ \varphi (x) = \frac{1}{8 \pi} \int d^{3}x \ \nabla \varphi \cdot \nabla \varphi = \frac{1}{8 \pi} \int d^{3}x \ \vec{E} \cdot \vec{E} .[/tex] So, if you rearrange the terms, you find [tex]\mathcal{E} - \int d^{3}x \ \rho (x) \varphi (x) = \int d^{3}x \left( \frac{1}{8 \pi} \nabla \varphi \cdot \nabla \varphi - \rho \ \varphi \right) = -\frac{1}{2} \int d^{3}x \ \rho (x) \varphi (x) .[/tex] Clearly, the expression in the middle (which is the [itex]I[\varphi][/itex] of Jackson) is not double the electrostatic energy.
Now, the meaningful question (which I will answer) is: where did Jackson get his functional from? Well, he knew that [itex]\nabla^{2}\varphi + 4 \pi \rho = 0[/itex]. So, he needs to construct [itex]A[\varphi][/itex] such that [tex]\frac{\delta A}{\delta \varphi} = - \left( \nabla^{2}\varphi + 4 \pi \rho \right) .[/tex] The minus sign is there for convenience and allows us to interpret [itex]A[\varphi][/itex] as “energy functional”. Assuming that the functional [itex]A[/itex] exists, we consider the functional [itex]A[\psi_{\lambda}][/itex] evaluated along the straight line “path” [tex]\psi_{\lambda}(x) = \lambda \varphi (x) + (1 - \lambda ) \psi_{0}(x) , \ \ \ \ \ 0 \leq \lambda \leq 1 ,[/tex] that begins at the point [itex]\psi_{0}(x) = 0[/itex] and end at [itex]\psi_{1}(x) = \varphi (x)[/itex]. The derivative of the functional [itex]A[\psi_{\lambda}][/itex] with respect to [itex]\lambda[/itex] is [tex]\frac{d}{d\lambda}A[\psi_{\lambda}] = \int d^{3}x \frac{\delta A}{\delta \psi_{\lambda}} \frac{\partial \psi_{\lambda}}{\partial \lambda} .[/tex] Now, using [tex]\frac{\delta A}{\delta \psi_{\lambda}} = - \left( \nabla^{2} \psi_{\lambda} + 4 \pi \rho \right), \ \ \ \ \frac{\partial \psi_{\lambda}(x)}{\partial \lambda} = \varphi (x),[/tex] we find
[tex]\frac{d}{d\lambda}A[\psi_{\lambda}] = - \int d^{3}x \left( \lambda \ \nabla^{2}\varphi + 4 \pi \rho (x) \right) \varphi (x) .[/tex] Integrating from [itex]\lambda = 0[/itex] to [itex]\lambda = 1[/itex], we find [tex]A[\psi_{1}] \equiv A[\varphi] = \int d^{3}x \left( - \frac{1}{2} \varphi \nabla^{2}\varphi - 4 \pi \rho (x) \varphi (x) \right) + \mbox{Const.} \ ,[/tex] where the constant term is [itex]A[\psi_{0}][/itex]. Finally, integrating the first term by parts gives us the wanted “energy functional” (up to additive constant) [tex]A[\varphi] = \int d^{3}x \left( \frac{1}{2} \nabla \varphi \cdot \nabla \varphi - 4 \pi \ \rho (x) \varphi (x) \right) .[/tex]
 
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