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Physicists sometimes get the lagrangian $$\mathcal{L}=-\frac{1}{2}\mbox dA \wedge \star \mbox dA - A \wedge \star J$$ define a functional given by $$S(A)=\int_{N_4} \mathcal{L}= \int_{N_4}-\frac{1}{2}\mbox dA \wedge \star \mbox dA - A \wedge \star J$$ and calculate variation of such functional $$\delta S(A)(\delta A)=\delta \int_{N_4} \mathcal L = \int_{N_4} \delta \mathcal{L}=\int_{N_4}-\frac{1}{2}\delta(\mbox dA\wedge \star \mbox dA)-\delta A\wedge \star J$$ etc. and they derive Maxwell's equation $d\star F=-\star J$. The functional is well defined - it's simply an integral of a 4-differential form over an oriented manifold (submanifold). But how is the variation defined? If $M$ is Riemannian manifold we've got Riemannian metric and induced inner product $(\cdot, \cdot)$ given by $$(\omega, \eta):=\int_M \left<\omega, \eta \right>\mathrm{dvol}$$ Hence we've got induced norm of a differential form and can define variation of a functional and of a differential form as its Frechet derivative. Alas, in physics, manifold is psudoriemannian manifold hence $$(\omega, \eta)=\int_M \left<\omega, \eta \right>\mathrm{dvol}$$ is no longer an inner product. We have no norm of a differential form, and without norm we cannot define variations as Frechet derivatives. Moreover! We cannot even define a local extremum of a functional.

Can you tell me how is defined variation of such functional, and a differential form? Does it even make sence what physicists are doing?

P Lrc
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1 Answers1

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To cut a long story short, you don't use a Fréchet derivative but a Gateaux derivative, which doesn't run into any of the issues you bring up.

Anyhow, if $A$ is a fixed $1$-form on $N$, then for any $1$-form $\delta A$ of compact support in the interior of $N^4$, you can still define the variation $\delta S(A)(\delta A)$ of $S$ about $A$ in the direction of $\delta A$ by the Gateaux derivative $$ \delta S(A)(\delta A) = \left.\frac{d}{d\epsilon}\right\rvert_{\epsilon=0}S(A+\epsilon\delta A) = \lim_{\epsilon \to 0}\frac{S(A+\epsilon\delta A) - S(A)}{\epsilon}. $$ So, in this case, since $$ L(A+\epsilon\delta A) = -\frac{1}{2}d(A+\epsilon \delta A) \wedge \star d(A+\epsilon \delta A) -(A + \epsilon \delta A) \wedge \star J\\ = L(A) + \epsilon\left( -\tfrac{1}{2}dA \wedge \star d(\delta A) - \tfrac{1}{2}d(\delta A) \wedge \star dA -\delta A \wedge \star J\right) + O(\epsilon^2)\\ = L(A) + \epsilon\left(-d(\delta A) \wedge \star dA - \delta A \wedge \star J \right) + O(\epsilon^2)\\ = L(A) + \epsilon\left( -\delta A \wedge \left(d\!\star\!dA - \star J \right) -d\left(\delta A \wedge \star dA\right) \right) + O(\epsilon^2) $$ it follows that $$ \delta S(A)(\delta A) = \int_{N^4} \delta A \wedge \left(-d\!\star\!dA - \star J\right) - \int_{N^4} d(\delta A \wedge \star dA)\\ = \int_{N^4} \delta A \wedge \left(-d\!\star\!dA - \star J\right) = \int_{N^4} g\left(\delta A,-\Delta A - J\right)\,dvol, $$ where the Laplacian $\Delta$ on $1$-forms is defined (in Lorentzian signature) by $\Delta = \star d \! \star \! d$. Since the bilinear form $\langle A,B \rangle := \int_{N^4} g(A,B)\,dvol$ is still non-degenerate even in the Lorentzian case, it still follows that $\delta S(A)(\delta A) = 0$ for all suitable variations $\delta A$ if and only if the Euler–Lagrange equation $\Delta A = -J$ holds.

Branimir Ćaćić
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  • Thank you very much for your precise answer!
    1. I realize Gateaux derivative of such functional still makes scense. But I rejected it since Gateaux derivate of a differential form still doesn't make sense whereas physicists seem not to bother themselves with it and calculate variations of differential forms (Walter Thirring for instance).
    2. Without a norm in linear space of differential forms we cannot define a local extremum of functional $F: \Omega^k(M) \to \mathbb{R}$. I understand we don't fuss about it?
     – P Lrc Dec 27 '17 at 21:10
  • I understand notation $O(\epsilon^2)$ is connventional since differential forms are not real-valued functions. But inserted under an integral they become.
    • – P Lrc Dec 27 '17 at 21:11
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    In the literature on global analysis on Lorentzian manifolds, it seems to be common practice to just pick an auxiliary Riemannian metric and connection on $T^\ast M$ and topologise $\Omega^k(M)$ accordingly, since, for instance, when $M$ is closed, different choices of Riemannian metric and connection will give rise to equivalent $C^m$ seminorms, and hence to equivalent Fréchet topologies. So, it looks like you can develop variational calculus on Lorentzian manifolds using more-or-less the same kind of functional-analytic framework as on Riemannian manifolds. – Branimir Ćaćić Dec 27 '17 at 22:17
  • Very interesting! Could you recommend me some textbooks on calculus of variations on manifolds (pseudoriemannian manifolds) which approach this issue with mathematical accuracy? Thank you for your assistance and happy new year! :) – P Lrc Dec 29 '17 at 16:31