# Calculus of variations and string theory

In Polchinski's String theory book, Vol 1., in chapter 1, p. 18, he is deriving the Lagrangian in the light cone gauge (that's not necessary to know in order to answer this question), and he gets

$$L~=~L_{ok}+\int_{0}^{l}d\sigma \gamma_{\tau\sigma}\partial_{\sigma}Y^- ,\tag{1.3.11}$$

where

$$Y^-(\tau,\sigma)~=~X^-(\tau,\sigma)-x^-(\tau) \tag{1.3.12b}$$ and

$$x^-(\tau)~=~\frac{1}{l}\int_{0}^{l}d\sigma X^-(\tau,\sigma),\tag{1.3.12a}$$ so

$$\langle Y^-\rangle_{\sigma}~=~0 .$$

Now he makes the variation of the Lagrangian respect to $Y^-$ :

$$\delta L_{Y^-}~=~\int_{0}^{l}d\sigma \gamma_{\tau\sigma}\delta(\partial_{\sigma}Y^-)=0$$

than he says that the variation yields $$\partial^2_{\sigma}\gamma_{\tau\sigma}~=~0$$

where the extra $\partial_{\sigma}$ is due to the fact that $$\langle Y^-\rangle_{\sigma}~=~\int_{0}^{l}d\sigma Y^-(\tau,\sigma)~=~0$$

I don't get why this fact should lead to an extra derivative, can anyone help? I "suspect" it has something to do with the fact that $Y^-$ is not positive definite (?)

The extra derivative in Polchinski comes from the following version of the Fundamental Lemma of Calculus of Variation (FLCV):

$$\tag{1} \left[ \forall g : ~~\int_a^b\! dx~ g(x) ~=~0 \quad\Rightarrow \quad \int_a^b\! dx ~f(x) g(x) ~=~0\right]\quad\Rightarrow \quad f^{\prime}~=~0.\quad$$

FLCV (1) states in words:

If it is true that for all functions $g$ with zero average that the integral $\int_a^b\! dx ~f(x) g(x) =0$ vanishes, then $f$ is a constant.

(Here we will for simplicity assume in what follows that $f$ and $g$ are sufficiently smooth functions, e.g. $f\in C^{1}[a,b]$. The mathematically minded reader is encouraged to try to weaken these assumptions.) The standard FLCV reads

$$\tag{2} \left[ \forall g : \quad \int_a^b\! dx ~f(x) g(x) ~=~0\right]\quad\Rightarrow \quad f~=~0.\quad$$

Actually, the following FLCV (3) holds as well

$$\tag{3} \left[ \forall g : ~~g(a)~=~0~=~g(b) \quad\Rightarrow \quad \int_a^b\! dx ~f(x) g(x) ~=~0\right]\quad\Rightarrow \quad f~=~0,\quad$$

because $f$ is continuous. Let us prove FLCV (1) using FLCV (3). To this end, define the antiderivative

$$\tag{4} G(x)~:=~\int_a^x\! dx^{\prime}~ g(x^{\prime}).$$

Then we can reformulate FLCV (1) as

$$\tag{5} \left[ \forall G : ~~G(a)~=~0~=~G(b) \quad\Rightarrow \quad \int_a^b\! dx ~f(x) G^{\prime}(x) ~=~0\right]\quad\Rightarrow \quad f^{\prime}~=~0.\quad$$

If we integrate (5) by parts, this becomes exactly FLCV (3). So FLCV (1) holds.

• In (3), how does the vanishing of $g$ at the endpoints guarantee that the integral vanishes? I can think of many counterexamples... – Ryan Unger Mar 26 '15 at 19:40
• @0celo7: Thanks for the feedback. To make sure we are on the same page, which counterexample do you have in mind? – Qmechanic Mar 26 '15 at 19:55
• @Qmechanic Maybe I'm not understanding your arrows properly. If I take $g(x)=\sin(x)$, then for $a=0$ and $b=\pi$ we have $g(a)=0=g(b)$ but the integral does not necessarily vanish. Follow-up question: what are the brackets for? – Ryan Unger Mar 26 '15 at 19:57
• @0celo7: Note the universal quantification. It has to hold for all functions $g$, not just the sine function. – Qmechanic Mar 26 '15 at 20:00
• @Qmechanic How can it be true for all functions which vanish at the endpoints if it is not true for a subset of functions which vanish at the endpoints? – Ryan Unger Mar 26 '15 at 20:02