Let's consider this boundary-value problem:

$$\begin{cases} -\Delta V = \rho & \rm{in}\ \Omega \\ V=0 & \rm{on}\ \partial \Omega \end{cases}.$$

We know that this problem has a variational formulation: its solutions are critical points of the energy functional

$$I[u]=\int_{\Omega} \left( \frac{1}{2}\lvert \nabla u \rvert^2 - u\rho \right)\, dx,$$

(cfr. Evans Partial differential equations 2nd ed., §2.2.5).

Now let us look at one (of the many) possible physical interpretations of this problem, the electrostatic one. If units are chosen so that $\varepsilon_0=1$, this equation describes the electric potential $V$ in a region $\Omega$ filled with a charge distribution $\rho$ and whose boundary is grounded.

It would be nice if the energy functional introduced above coincided with the total energy of this physical system. Unfortunately, it seems to me not to be so. In fact, as I read in Feynman's Lectures on physics, vol.II, §8-5, the total energy of this system consists of two additive parts: one is

$$\frac{1}{2}\int_{\Omega} \lvert \nabla V \rvert^2\, dx$$

and is due to the electric field; the other is

$$\frac{1}{2}\int_{\Omega} V\rho\, dx$$

and is due to the charge distribution.

This induces me to think that the energy functional should be

$$U[u]=\int_{\Omega} \left(\frac{1}{2}\lvert \nabla u \rvert^2 + \frac{1}{2}u\rho \right)\, dx,$$ instead of the correct $$I[u]=\int_{\Omega} \left( \frac{1}{2}\lvert \nabla u \rvert^2 - u\rho \right)\, dx.$$ Why am I wrong?

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    $\begingroup$ As for the sign: if you instead consider $J[u] := I[-u]$, the first term doesn't change, while the second one gets a plus sign. The critical points of the functionals $J$ and $I$ also agree (up to sign, of course). As for the factor 1/2, see joriki's post. $\endgroup$ – Gerben May 18 '11 at 11:08
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    $\begingroup$ I have recently read in this paper (Berestycki- Lions - Nonlinear scalar field equations-existence of ground states, part I) some useful remarks on the (ab)use of language of mathematical literature around the terms "energy" and "action". I think that those remarks completely settle the present question. $\endgroup$ – Giuseppe Negro May 26 '16 at 15:53

I think this has to do with what you treat as the dynamic variables. In your first formulation, $\rho$ was given and fixed, and you wanted to calculate $V$ by varying only $V$, not $\rho$. In the second formulation, $V$ is just a quantity derived from $\rho$, and this is the energy suitable for obtaining $\rho$ by varying $\rho$, with $V$ just a shorthand for a certain linear transformation of $\rho$. A third possibility is to consider $V$ as given and find the motion of a charge distribution in this potential -- here, again, there would not be a factor of $\frac{1}{2}$. This is the case, for instance, if we calculate the motion of the electron in the potential of a hydrogen nucleus, or of the Earth in the gravitational potential of the Sun.

Mathematically speaking, we need a factor of $\frac{1}{2}$ in front of quadratic terms (where $V\rho$ is quadratic in $\rho$ if $V$ is considered as a derived quantity derived from the dynamical variable $\rho$) but not in front of linear terms. Physically speaking, the factor of $\frac{1}{2}$ avoids double-counting when the energy is regarded as the interaction energy of a charge distribution with itself.

Now you may ask: But surely there is some well-defined energy and we can't just pick and choose the factors according to expedience? That's true, but consider again the hydrogen atom: If we consider the potential of the nucleus as given, then we're not counting the potential energy of the nucleus in the field of the electron in the integral over $\rho V$. But the entire energy is there and has to be accounted for in the integral, hence the factor $1$. On the other hand, in treating a helium atom, where both electrons are treated as dynamic and their interaction energy enters into the Hamiltonian, we do include a factor of $\frac{1}{2}$ to avoid double-counting the energy in the double integral over $\rho_1\rho_2/r_{12}$.

  • $\begingroup$ +1: Maybe it is worth to add that a Legendre transform brings you from the form where $V$ is a "dynamical" variable to the form where $\rho$ is the "dynamical" variable. $\endgroup$ – Fabian May 18 '11 at 11:44

In the Lagrangian formulation of classical mechanics, the functional that we extremise to derive the Euler-Lagrange equation of motion is the Lagrangian, defined as

$$L[u] = T[u] - V[u]$$

where $T[u]$ is the kinetic part and $V[u]$ is the potential part. Contrast this with the Hamiltonian $H=T+V$ from which we derive Hamilton's equations of motion (these are equivalent to the Euler-Lagrange equations, and the two are related by Legendre transform). That should explain the minus sign discrepancy.

I am not sure about the factor of 1/2. Is it possible that we are simply defining $\rho_{\mathrm{Feynman}} = 2\rho_{\mathrm{Evans}}$?

  • $\begingroup$ don't you have a problem with $L$ not being conserved? In the theory of ODE's/PDE's, you explicitly construct an energy functional to have something that stays constant along flow lines (of the solution). $\endgroup$ – Gerben May 18 '11 at 11:10
  • $\begingroup$ $L$ is not the energy functional, it is the Lagrangian. If you want something corresponding to the total energy of the system you need to be looking at the Hamiltonian $H$. $\endgroup$ – Chris Taylor May 18 '11 at 11:17
  • $\begingroup$ I'm not sure this is it, Chris. This one we have here is not an action functional, but an energy functional. It's not that we have two different kinds of energy s.t. one transforms into the other, as kinetic and potential energy do in mechanical systems. I'm more inclined to think that the error lies in the fact that I am not quantifying energy correctly, as Luboš Motl and joriki point out. $\endgroup$ – Giuseppe Negro May 18 '11 at 15:39

The energy functional has to agree with the total energy and there can't be any ambiguity or "freedom of conventions" that would add the factors of two, among other things. After all, the total energy/mass is a subject to conservation laws and it determines both inertial of the system as well as the strength of its gravitational field.

If you integrated Feynman's two $\int E^2/2$ and $\int \rho V/2$ contributions over the whole 3-space, you would surely be double-counting the energy. The correct energy is just one of them, as given by the Evans formula. In the localized case, the two expressions may be shown to be equal.

You may also use Feynman's formula but you must be careful to use the $\int E^2/2$ exclusively for the part of the electric fields that are "unrelated" to localized charged sources.

  • $\begingroup$ Double-counting the energy... In fact if we integrate by parts $$\int_{\Omega} \frac{E^2}{2}\, dx$$ we get $$\frac{1}{2}\int_{\Omega} V \rho \, dx.$$ So you say I must take into account one or the other energy term and not the sum of the two... In the end, this must be the same suggestion joriki gives above. I need to think about this a little bit. In the meantime, thank you. $\endgroup$ – Giuseppe Negro May 18 '11 at 15:34
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    $\begingroup$ I still don't get it. In Evans' formula $$\int_{\Omega} \frac{1}{2}\lvert \nabla V \rvert^2 - V\rho\, dx$$ it looks as that $-V\rho$ is a corrector term, that is: total energy at a point is $\frac{1}{2}\lvert \nabla V \rvert^2$ diminished by $V\rho$. Why do we diminish energy like that? Do you mind explaining this, please? $\endgroup$ – Giuseppe Negro May 18 '11 at 16:42

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