# variation of an action functional

I would like to confirm my answer since it is in direct contradiction from my book on physics. So suppose we have an action functional $$S(\phi) = \int_M \phi(\Delta + m^2)\phi dV,$$ where $M$ is a manifold and $\phi$ is in $C_c^\infty(U)$, the space of compactly supported smooth functions in an open set $U$.

So I get the variation $$S(\phi+t\psi) = S(\phi) + t(\int_M \psi(\Delta + m^2)\phi dV + \int_M \phi(\Delta + m^2)\psi dV) + t^2 S(\psi),$$ hence getting the functional derivative $$\int_M \psi(\Delta + m^2)\phi dV + \int_M \phi(\Delta + m^2)\psi dV.$$

But then the book claims that the Euler Lagrange equation is $(\Delta + m^2)\phi=0$. Is there some condition on $\psi$ that I don't know about or did I misunderstand some of the notations involved?

• Can the person who downvoted please explain his/her reason(s)? – user591553 Sep 9 '18 at 17:16
• I didn't downvote, but I'd like to know where $\rho$ comes from, and if it is related to $\psi$ It looks to me like you might want $S(\phi + t\psi)$ on the left. Cheers! – Robert Lewis Sep 9 '18 at 17:25
• Maybe the downvote was cast because the relationship between $\rho$ and $\psi$ is not clear? – Robert Lewis Sep 9 '18 at 17:29
• Thanks for the fix! – user591553 Sep 9 '18 at 17:35
• That functional is uncommon. A more normal form for the equation $\Delta \phi + m^2\phi^2 = 0$ would be something like $$S(\phi) = \frac12 \int_M ((\nabla\phi)^2 - m^2\phi^2) \, dV$$ – md2perpe Sep 10 '18 at 17:07

I wish to show that, under reasonable hypotheses,

$$\displaystyle \int_m \psi(\nabla^2 + m^2) \phi \; dV = \int_m \phi(\nabla^2 + m^2) \psi \; dV. \tag 0$$

First of all observe that

$$\displaystyle \int_M \psi (\nabla^2 + m^2) \phi \; dV = \int_M \psi \nabla^2 \phi \; dV + \int_M m^2 \psi \phi \; dV, \tag 1$$

and

$$\displaystyle \int_M \phi (\nabla^2 + m^2) \psi \; dV = \int_M \phi \nabla^2 \psi \; dV + \int_M m^2 \phi \psi \; dV; \tag 2$$

since evidently

$$\displaystyle \int_M m^2 \psi \phi \; dV = \int_M m^2 \phi \psi \; dV, \tag 3$$

we only need show that

$$\displaystyle \int_M \psi \nabla^2 \phi \; dV = \displaystyle \int_M \phi \nabla^2 \psi \; dV; \tag 4$$

we have the vector identity

$$\nabla \cdot (fX) = \nabla f \cdot X + f \nabla \cdot X \tag 5$$

for functions $$f$$ and vector fields $$X$$ (see here); if we set $$f = \psi$$ and $$X = \nabla \phi$$ this yields

$$\nabla \cdot (\psi \nabla \phi) = \nabla \psi \cdot \nabla \phi + \psi \nabla \cdot \nabla \phi = \nabla \psi \cdot \nabla \phi + \psi \nabla^2 \phi, \tag 6$$

or

$$\psi \nabla^2 \phi = \nabla \cdot (\psi \nabla \phi) - \nabla \psi \cdot \nabla \phi, \tag 7$$

which we may integrate over $$M$$:

$$\displaystyle \int_M \psi \nabla^2 \phi \; dV = \int_M \nabla \cdot (\psi \nabla \phi) \; dV - \int_M \nabla \psi \cdot \nabla \phi \; dV. \tag 8$$

We wish to apply the divergence theorem to the first integral on the right of (8); in order to so this, we must pause and carefully consider the regions in $$M$$ on which $$\psi, \phi \ne 0$$; that is, the support of of $$\psi$$ and $$\phi$$. We are given that $$\phi$$ (and I assume $$\psi$$) are in $$C_c^\infty(U)$$,

$$\phi, \psi \in C_c^\infty(U), \tag 9$$

$$U \subset M$$ open; it follows that the compact set

$$K = \text{supp} \; \phi \cup \text{supp} \; \psi \subset U, \tag{10}$$

and we require the existence of an open $$\Omega \subset U$$ with $$\bar \Omega$$ compact, $$K \subset \Omega$$, and

$$\partial \Omega = \bar \Omega \setminus \Omega \tag{11}$$

a smooth (at least $$\mathcal C^1$$), orientable, compact hypersurface in $$M$$. These conditions may seem somewhat arbitrary and or strange, but what I have tried to do in imposing them is simply ensure that we may perform, on the manifold $$M$$, the kind of calculations around the divergence theorem that we typically do in $$\Bbb R^n$$; e.g., assuming for the moment that in fact

$$M = \Bbb R^n, \tag{12}$$

then we see that we may take

$$\Omega = B(R, 0) = \{ x \in \Bbb R^n \mid \vert x \vert < R \}, \tag{13}$$

that is, $$\Omega$$ is the open ball of radius $$R$$ centered at the origin; then

$$\bar \Omega = \bar B(R, 0) = \{ x \in \Bbb R^n \mid \vert x \vert \le R \}, \tag{14}$$

and

$$\partial \Omega = \bar \Omega \setminus \Omega = S(R, 0) = \{ x \in \Bbb R^n \mid \vert x \vert = R \} \tag{15}$$

is the sphere of radius $$R$$ about the origin. Since $$S(R, 0)$$ is compact and has a well-defined outward pointing normal field $$\mathbf n$$, we may, for $$R$$ large enough, use the divergence theorem applied to the first integral on the right of (8):

$$\displaystyle \int_{\Bbb R^n} \nabla \cdot (\psi \nabla \phi) \; dV = \int_{\bar B(R, 0)} \nabla \cdot (\psi \nabla \phi) \; dV = \int_{S(R, 0)} \psi \nabla \phi \cdot \mathbf n \; dA, \tag{16}$$

where $$dA$$ is the volume measure on $$S(R, 0)$$; now by virtue of the hypothesis that both $$\psi, \phi \in C_c^\infty(\Bbb R^n)$$, we see that for $$R$$ sufficiently large,

$$\displaystyle \int_{S(R, 0)} \psi \nabla \phi \cdot \mathbf n \; dA = 0, \tag{17}$$

and so by (8) we have

$$\displaystyle \int_{\Bbb R^n} \psi \nabla^2 \phi \; dV = \int_{\Bbb R^n} \nabla \cdot (\psi \nabla \phi) \; dV - \int_{\Bbb R^n} \nabla \psi \cdot \nabla \phi \; dV$$ $$= \displaystyle \int_{S(R, 0)} \nabla \cdot (\psi \nabla \phi) \; dV - \int_{\Bbb R^n} \nabla \psi \cdot \nabla \phi \; dV = - \int_{\Bbb R^n} \nabla \psi \cdot \nabla \phi \; dV. \tag{18}$$

From this perspective, we see that they hypotheses we have placed upon $$\Omega \subset M$$ have been devised to allow a calculation similar to (18) to proceed on/within $$M$$; with such $$\Omega$$ as described above ca. equations (10)-(11) we may write

$$\displaystyle \int_M \nabla \cdot (\psi \nabla \phi) \; dV = \int_{\bar \Omega} \nabla \cdot (\psi \nabla \phi) \; dV = \int_{\partial {\bar \Omega}} \psi \nabla \phi \cdot \mathbf n \; dA = 0, \tag{19}$$

where again $$\mathbf n$$ is the outward-pointing normal on $$\partial{\bar \Omega}$$; we have used the fact that $$\phi, \psi = 0$$ on $$\partial{\bar \Omega}$$ in establishing this equation. In the light of (19) we see that (8) becomes

$$\displaystyle \int_M \psi \nabla^2 \phi \; dV = -\int_M \nabla \psi \cdot \nabla \phi \; dV; \tag{20}$$

we may reverse the roles of $$\phi$$, $$\psi$$ in (20) and arrive at

$$\displaystyle \int_M \phi \nabla^2 \psi \; dV = -\int_M \nabla \phi \cdot \nabla \psi \; dV; \tag{21}$$

by virtue of (20) and (21) in collusion we have

$$\displaystyle \int_M \psi \nabla^2 \phi \; dV = \int_M \phi \nabla^2 \psi \, dV \tag{22}$$

and thus, combining (3) and (22) we have

$$S(\phi + t\psi) = S(\phi) + 2t \displaystyle \int_M \psi(\nabla^2 + m^2)\phi \; dV + t^2 S(\psi); \tag{23}$$

we see that $$S(\phi)$$ is stationary precisely when

$$\displaystyle \int_M \psi(\nabla^2 + m^2)\phi \; dV = 0, \; \forall \psi \in C_c^\infty(M), \tag{24}$$

which shows the Euler-Lagrange equation must be

$$(\nabla^2 + m^2)\phi = 0. \tag{25}$$