I am reading the Feynman Lectures on Physics about the Principle of Least Action. During this lecture (around 3/4th of the page), Dr. Feynman formulated the electrostatics problem as:

\begin{equation*} U\star=\frac{\epsilon}{2}\int(\nabla{\phi})^2\,dV- \int\rho\phi\,dV, \end{equation*}

From one of his previous lectures, I understand that the first integral refers to the energy at a certain location due to electric field (eqn. 8.35).

As for the second integral, it should refer to the potential energy gained by bringing charges to their current locations (eqn. 8.28).

However, eqn. 8.28 had a 1/2 factor to account for double counting the forces between a pair of charges.

Question: Why does this factor disappear from the formulation of the electrostatics Principle of Least Action problem?

Equations Mentioned

Eqn 8.28 - $$U = \frac {1}{2} \int \rho \phi d V \tag{8.28}$$

This equation can be interpreted as follows. The potential energy of the charge ρdVρdV is the product of this charge and the potential at the same point. The total energy is therefore the integral over ϕρdVϕρdV. But there is again the factor 1212. It is still required because we are counting energies twice. The mutual energies of two charges is the charge of one times the potential at it due to the other. Or, it can be taken as the second charge times the potential at it from the first.

Eqn 8.35 - $$U = \frac {\epsilon_0}{2} \int\limits_{all space} (\nabla \phi) * (\nabla \phi) d V = \frac {\epsilon_0}{2} \int\limits_{all space} E * E d V\tag{8.35}$$

We see that it is possible for us to represent the energy of any charge distribution as being the integral over an energy density located in the field.

  • $\begingroup$ Please tell us what equation 8.28 is, and what it is represents. $\endgroup$
    – garyp
    Commented Aug 3, 2016 at 14:51
  • $\begingroup$ Welcome to Physics.SE Kiat! Please include the relevant equations so that we don't have to read through his previous lecture to find it. $\endgroup$
    – auden
    Commented Aug 3, 2016 at 14:54
  • $\begingroup$ @garyp, I added the equations, and copyed over the relevant sentences from his link. Does that help? $\endgroup$
    – auden
    Commented Aug 3, 2016 at 15:03
  • $\begingroup$ @garyp I apologize, I was trying to insert the links earlier but I don't have enough reputation for more than 2 links. Thanks for the help @heather! $\endgroup$
    – Kiat
    Commented Aug 3, 2016 at 17:11

1 Answer 1


The quantity $\:U^{\boldsymbol{*}}\:$ of the unnumbered equation, page 19-10 of the textbook(1) \begin{equation} U^{\boldsymbol{*}}=\dfrac{\epsilon_{o}}{2} \int\left(\boldsymbol{\nabla}\phi \right)^{2}dV- \int \rho\phi dV \tag{001} \end{equation} is a quantiy of the electrostatic field like the "action integral". Also it's the volume integral of the Lagrangian density $\:\mathcal{L}_{em}\:$ of the field, which in general is(2) \begin{equation} \boxed{\: \mathcal{L}_{em}\:=\:\epsilon_{0}\cdot\dfrac{\left|\!\left|\mathbf{E}\right|\!\right|^{2}-c^{2}\left|\!\left|\mathbf{B}\right|\!\right|^{2}}{2}-\rho \phi + \mathbf{j} \boldsymbol{\cdot} \mathbf{A} \:} \tag{002} \end{equation} where \begin{align} \left\Vert\mathbf{E}\right\Vert^{2} & = \left\Vert - \boldsymbol{\nabla}\phi -\dfrac{\partial \mathbf{A}}{\partial t}\right\Vert^{2} = \left\Vert \mathbf{\dot{A}}\right\Vert^{2}+\Vert \boldsymbol{\nabla}\phi \Vert^{2}+2\left(\boldsymbol{\nabla}\phi \boldsymbol{\cdot} \mathbf{\dot{A}}\right) \tag{003a}\\ & \nonumber\\ \left\Vert\mathbf{B}\right\Vert^{2} & = \left\Vert\boldsymbol{\nabla} \boldsymbol{\times} \mathbf{A}\right\Vert^{2}=\sum^{k=3}_{k=1}\left[\Vert \boldsymbol{\nabla}\mathrm{A}_{k}\Vert^{2}-\dfrac{\partial \mathbf{A}}{\partial x_{k}}\boldsymbol{\cdot} \boldsymbol{\nabla}\mathrm{A}_{k}\right] \tag{003b} \end{align}

For the electrostatic field \begin{equation} \mathcal{L}_{em}\:=\:\epsilon_{0}\cdot\dfrac{\left|\!\left|\mathbf{E}\right|\!\right|^{2}}{2}-\rho \phi =\:\epsilon_{0}\cdot\dfrac{\left(\boldsymbol{\nabla}\phi \right)^{2}}{2}-\rho \phi \tag{004} \end{equation} and (001) comes from \begin{equation} U^{\boldsymbol{*}}= \int \mathcal{L}_{em} dV \tag{005} \end{equation}

This function being minimum, that is when its variation is zero \begin{equation} \Delta U^{\boldsymbol{*}}=0 \tag{006} \end{equation} yields the well known equation for electrostatics \begin{equation} \nabla^{2}\phi = -\dfrac{\rho}{\epsilon_{0}} \tag{007} \end{equation} what Feynman shows in Chapter 19. The Principle of Least Action of the textbook.(1)

So, $\:U^{\boldsymbol{*}}\:$ is NOT the volume integral of the electrostatic energy $\:U\:$ stored in the field, equations (8.28) and (8.35) \begin{equation} U= \dfrac{1}{2} \int \rho\phi dV \tag{8.28-textbook} \end{equation}

\begin{equation} U= \dfrac{\epsilon_{o}}{2} \int\limits_{\text{all space}}\left(\boldsymbol{\nabla}\phi \right)\boldsymbol{\cdot}\left(\boldsymbol{\nabla}\phi \right)dV=\dfrac{\epsilon_{o}}{2} \int\limits_{\text{all space}}\mathbf{E}\boldsymbol{\cdot}\mathbf{E}dV \tag{8.35-textbook} \end{equation}

Our misunderstanding here is similar to this :

To confuse the Lagrangian of a point particle under the gravity force \begin{equation} L=T-V=\dfrac{1}{2}mv^{2}-mgh \tag{008} \end{equation} with its total energy \begin{equation} E=T+V=\dfrac{1}{2}mv^{2}+mgh \tag{009} \end{equation}

(1) The Feynman Lectures on Physics - Volume II : Mainly Electromagnetism and Matter , New Millenium Edition 2010.

(2) See my answer here Deriving Lagrangian density for electromagnetic field

  • 1
    $\begingroup$ Thank you for your detailed explanation! I now understand that we are actually integrating a "Lagrangian" (Dr. Feynman mentioned it later in this lecture). However, the story about Mr. Bader at the beginning of this lecture seemed to suggest that we were integrating the difference between kinetic and potential energy. Do I need to read up on Lagrangian density theory to understand why the integral terms in Mr. Bader's example had an easy physical meaning while the electrostatic case produced a non-intuitive integral? $\endgroup$
    – Kiat
    Commented Aug 5, 2016 at 21:53

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