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Let $\Omega$ be a region in $\mathbb{R}^2$ with $f:\Omega \to \mathbb{R}$ a smooth function. Why is the quantity, $$ \tfrac{1}{2} \iint_{\Omega} \|\nabla f\|^2 $$ Called the "energy" of $f$? I am sure it comes from physics but I do not know why.

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The term energy comes from electrostatics. The energy density of electric field $\mathbf E$ is $\frac12 \varepsilon_0\mathbf E^2$; to get the total energy of the field we integrate that.

The derivation of $\frac12 \varepsilon_0\mathbf E^2$ is given in the Wikipedia article Electric potential energy. But I'll outline the steps here: the potential energy of a charge in an electric field (normalized to be zero at infinity) is proportional to the product of charge amount and the potential $\Phi$. For distributed charges, the charge density is the divergence of $\mathbf E$. So the potential energy is $$\frac12 \varepsilon_0 \int \Phi \nabla \mathbf E$$ which after integration by parts (using $\nabla \Phi = -\mathbf E$) becomes $$\frac12 \varepsilon_0 \int \mathbf E\cdot \mathbf E = \frac12 \varepsilon_0 \int |\nabla \Phi|^2 $$

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    $\begingroup$ Can you recommend an article, or webpage, or short book, that explains all of these terms from physics, such as, "electrostatic", "potential energy", "charge", ect, to someone who knows very little physics? $\endgroup$ – Nicolas Bourbaki Aug 9 '14 at 22:26
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    $\begingroup$ @ILoveMr.Paul Maybe Khan Academy videos on E&M? $\endgroup$ – user147263 Aug 9 '14 at 22:29
  • $\begingroup$ Any chance for assistance here: math.stackexchange.com/questions/1066495/… $\endgroup$ – Royi Dec 14 '14 at 6:54
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Let us imagine that the graph of $f$ in $\mathbb{R}^3$ represents some sheet of uniform elastic rubber. If we study the idealized case, we find that the force in the horizontal direction is $\Delta f$. (This is reasonable, as when we study Laplace's equation we find that harmonic functions, those which satisfy $\Delta f =0$ satisfy a property where $f(x)$ is equal to the average of value of $f$ on any circle about $f(x)$, so the rubber will be at an equilibrium). Thus the integral given captures the potential energy in the rubber pulling on itself (you get a factor it squared so that kinetic and potential energy interact correctly).

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    $\begingroup$ Not-physics guy here. How is the sheet exerting force? Is the sheet actually the $xy$-plane stretched to fill out the graph of $f$, so it "wants" to go back to its original shape? Otherwise interesting answer. $\endgroup$ – blue Aug 9 '14 at 23:05
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    $\begingroup$ Yeah, sorry, it's meant to have originally been the xy-plane, and distortion is assumed to be vertical. I think there may also be a "small distortion" assumption that allows us to cutoff a Taylor series. $\endgroup$ – JHance Aug 10 '14 at 14:57

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