# Second Variation of an Action: Part I

So your advisor asks you, “what are the background fluctuations for your action?” If you don’t know what they are talking about, you’ve come to the right place. In the meantime, just smile and wave guys $–$ just smile and wave.

Over this three part post series, you will realize that finding the background fluctuations of an action amounts to determining the the second variation of an action. It might not be clear right now, but we will build to it.

Suppose you have an action of one variable with the canonical form $$\mathcal{S}[\phi] = \int\text{d}^4x\mathcal{L}(\phi(x),\partial_\mu\phi(x))$$ To obtain the second variation of an action, we will have to state the definition of a functional’s derivative.

Functional Derivative: Given a manifold, $\mathcal{M}$ representing smooth functions $\phi$, accompanied with relevant boundary conditions, and a functional $\mathcal{F}:\mathcal{M}\to\mathbb{R}$, a functional derivative of $\mathcal{F}[\phi]$ is obtained through $$\mathcal{F}[\phi+\delta\phi]=\mathcal{F}[\phi]+\int\prod_{n\ge 1}(\text{d}^dx)^n\frac{\delta^n\mathcal{F}[\phi]}{(\delta\phi)^n}(\delta\phi)^n$$ with $n\in\mathbb{N}.$

While this may not seem useful at first, this is essentially the only definition one needs to perform the second variation of an action. The action, according to classical mechanics, is simply a functional and may be expanded accordingly: $$\mathcal{S}[\phi+\delta\phi] = \mathcal{S}[\phi] +\int\text{d}^dx \frac{\delta\mathcal{S}[\phi(z)]}{\delta\phi(x)}\delta\phi(x) + \int\text{d}^dx\int\text{d}^dy\frac{\delta^2\mathcal{S}[\phi(z)]}{\delta\phi(x)\delta\phi(y)}\delta\phi(y)\delta\phi(x)+\dots$$ While we are effectively done, and could easily introduce a Lagrangian, we can first employ the Principle of Least Action:

Principle of Least Action: The path taken by a system between two times is one for which the action is stationary to first order, i.e. $\delta S=0$ to first order.

Generally, the principle dictates that the extrema of the action in functional space vanishes, but one typically takes the first order variation to be a minima and sets it to zero accordingly. With that in mind, we have $$\delta \mathcal{S}= \mathcal{S}[\phi+\delta\phi]-\mathcal{S}[\phi]= \int\text{d}^dx\int\text{d}^dy\frac{\delta^2\mathcal{S}[\phi(z)]}{\delta\phi(x)\delta\phi(y)}\delta\phi(y)\delta\phi(x)+\dots\ \ .$$ For an example, consider the Klein-Gordon Lagrangian given by, $$\mathcal{L}(\phi,\partial_\mu\phi)=\frac{1}{2}\partial_\mu\phi\partial^\mu\phi-\frac{1}{2}m^2\phi^2.$$ Then the variation of the action to second order becomes, in $\text{d}=4$, \begin{align} \delta\mathcal{S}&=\int\text{d}^4x\int\text{d}^4y\int\text{d}^4z\frac{\delta\mathcal{L}}{\delta\phi(x)\phi(y)}\delta\phi(y)\delta\phi(x)\\ &= \int\text{d}^4x\int\text{d}^4y\frac{\delta}{\delta\phi(x)}\left[\int\text{d}^4z\frac{\delta\mathcal{L}(\phi(z),\partial_\mu\phi(z))}{\delta\phi(y)}\right]\delta\phi(y)\delta\phi(x)\\ &=\int\text{d}^4x\int\text{d}^4y\frac{\delta}{\delta\phi(x)}\left[\int\text{d}^4z\left(\frac{\partial\mathcal{L}}{\partial\phi(z)}\frac{\delta\phi(z)}{\delta\phi(x)} +\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi(z))}\frac{\delta(\partial_\mu\phi(z))}{\delta\phi(y)} \right) \right]\delta\phi(y)\delta\phi(x). \end{align} As the variational derivative and the partial derivative act on different objects, they may be commuted past one another. This allows us to express the above line as \begin{align} \delta\mathcal{S} &=\int\text{d}^4x\int\text{d}^4y\frac{\delta}{\delta\phi(x)}\left[\int\text{d}^4z\left(\frac{\partial\mathcal{L}}{\partial\phi(z)}\frac{\delta\phi(z)}{\delta\phi(x)} +\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi(z))}\partial_\mu\left(\frac{\delta\phi(z)}{\delta\phi(y)}\right) \right) \right]\delta\phi(y)\delta\phi(x). \\ \end{align} With the definition  $$\frac{\delta\phi(z)}{\delta\phi(y)}=\delta^d(z-y)$$ with $d$ being the dimension of the delta function, we have \begin{align} \delta\mathcal{S} &=\int\text{d}^4x\int\text{d}^4y\frac{\delta}{\delta\phi(x)}\left[\int\text{d}^4z\left(\frac{\partial\mathcal{L}}{\partial\phi(z)}\delta^4(z-y) + \frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi(z))}\partial_\mu\left(\delta^4(z-y)\right) \right) \right]\delta\phi(y)\delta\phi(x). \end{align} We may integrate the second term by parts; \begin{align} \int\text{d}^4z\partial_\mu\left[\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi(z))}\delta^4(z-y)\right]=\int\text{d}^4z\partial_\mu\left[\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi(z))}\right]\delta^4(z-y)+\int\text{d}^4z\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi(z))}\partial_\mu\left[\delta^4(z-y)\right]. \end{align} The term on the left hand side is a total divergence. By the divergence theorem of an ordinary vector valued function with a volume measure, such as $\int\text{d}^4z$, we integrate the argument on the boundary of the volume. As the boundary of a four dimensional spacetime is a three$-$dimensional hypersurface with positive curvature, we assert that the boundary is strictly a spatial boundary. Thus our manifold with a Lorentzian metric, $(\mathcal{M},\eta_{\mu\nu})$, has a boundary with a hypersurface metric, $(\partial\mathcal{M},h_{ij})$ where $\mu,\nu=0,1,2,3$ and $i,j=1,2,3.$ Given these conditions, the divergence theorem may be stated as:

Divergence Theorem: Let $(\mathcal{M},\eta_{\mu\nu})$ be a manifold with a hypersurface boundary, $(\partial\mathcal{M},h_{ij})$. Then the volume integral of a total divergence is related to the boundary by $$\int_\mathcal{M}\text{d}^nx\sqrt{|\det(\eta_{\mu\nu})|}\nabla_\mu X^\mu = \int_{\partial\mathcal{M}}\text{d}^{n-1}x\sqrt{|\det(h_{ij})|}n_i X^i$$ with $n_i$ being the unit normal to the boundary.

Given the above theorem the total divergence term becomes \begin{align} \int_{\mathcal{M}}\text{d}^4z\ \partial_\mu\left[\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi(z))}\delta^4(z-y)\right] &= \int_{\partial\mathcal{M}}\text{d}^3z\ n_i\left(\frac{\partial\mathcal{L}}{\partial(\partial_i\phi(z))}\delta^3(z-y)\right)\\ &= n_i\left(\frac{\partial\mathcal{L}}{\partial(\partial_i\phi(z))}\right)\bigg|_{\partial\mathcal{M}} = 0. \end{align} This last line is a common assumption, that being that the fields, and their derivatives, are only relevant locally. The reasoning for this assumption is that we can take the distance of the boundary from the point being evaluated to be as large as necessary to cause the fields to vanish. If one does not like this explanation, an actual reason to why we are allowed to make this assumption may be found in Compact Support. With these simplifications, the variation in the action becomes \begin{align} \delta\mathcal{S} &=\int\text{d}^4x\int\text{d}^4y\frac{\delta}{\delta\phi(x)}\left[\int\text{d}^4z\left(\frac{\partial\mathcal{L}}{\partial\phi(z)}\delta^4(y-x) – \partial_\mu\left(\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi(z))}\right)\delta^4(y-x) \right) \right]\delta\phi(y)\delta\phi(x)\\ &=\int\text{d}^4x\int\text{d}^4y\frac{\delta}{\delta\phi(x)}\left[\int\text{d}^4z\left(\frac{\partial\mathcal{L}}{\partial\phi(z)} – \partial_\mu\left(\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi(z))}\right)\right)\delta^4(z-y)\right]\delta\phi(y)\delta\phi(x)\\ &=\int\text{d}^4x\int\text{d}^4y\frac{\delta}{\delta\phi(x)}\left[\left(\frac{\partial\mathcal{L}}{\partial\phi(y)} – \partial_\mu\left(\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi(y))}\right)\right) \right]\delta\phi(y)\delta\phi(x). \end{align} This is an optimal point to introduce the form of the Klein$-$Gordon Lagrangian: \begin{align} \delta\mathcal{S} &= \int\text{d}^4x\int\text{d}^4y\frac{\delta}{\delta\phi(x)}\left[ -m^2\phi(y)-\partial_\mu(\partial^{\mu}\phi(z)) \right]\delta\phi(y)\delta\phi(x)\\ &=\int\text{d}^4x\int\text{d}^4y\left[-m^2\frac{\delta\phi(y)}{\delta\phi(x)}-\partial_\mu\left(\partial^{\mu}\left(\frac{\delta\phi(y)}{\delta\phi(x)}\right)\right)\right]\delta\phi(y)\delta\phi(x)\\ &=\int\text{d}^4x\int\text{d}^4y\left[\delta\phi(y)\left(-m^2-\partial_\mu\partial^\mu\right)\delta^{4}(y-x)\right]\delta\phi(x). \end{align} While we are more or less finished finding the second variation, the physics of the situation is better understood when we transfer the $\partial_\mu\partial^\mu$ operation onto the variation of the fields instead of the delta function. Before we embark on this procedure, note that the boundary conditions for the fields also apply to the variation of the fields: $$\phi\bigg|_{\partial\mathcal{M}}\to0,\ \partial_\mu\phi\bigg|_{\partial\mathcal{M}}\to0,\ \delta\phi\bigg|_{\partial\mathcal{M}}\to0, \text{ and } \partial_\mu(\delta\phi)\bigg|_{\partial\mathcal{M}}\to0$$ Then the term in question may be found from integrating by parts twice and applying the relevant boundary conditions. This type of integration is strictly not legal. Since the delta function is not a continuous function, it cannot be differentiable either. Only in the context of integration does the conditions of a function make sense for the delta function. When integrating by parts, we are instead relying on the fact that $\phi$ is smooth and has compact support in an open set of $\mathcal{M}$ in order to transfer the derivatives onto $\delta\phi$:Then the term in question may be found from integrating by parts twice and applying the relevant boundary conditions. This type of integration is strictly not legal. Since the delta function is not a continuous function, it cannot be differentiable either. Only in the context of integration does the conditions of a function make sense for the delta function. When integrating by parts, we are instead relying on the fact that $\phi$ is smooth and has compact support in an open set of $\mathcal{M}$ in order to transfer the derivatives onto $\delta\phi$: \begin{align} &\int_{\mathcal{M}}\text{d}^4y\delta\phi(y)\partial_\mu\partial^\mu(\delta^4(y-x)) = -\int_\mathcal{M}\text{d}^4y\partial_\mu(\delta\phi(y))\partial^\mu(\delta^4(y-x)) = +\int_\mathcal{M}\text{d}^4y\partial^\mu\partial_\mu(\delta\phi)\delta^4(y-x) \end{align} Loosely speaking, the total divergence term on the left hand side underwent a procedure similar to integration by parts. Basically integrate once, transfer the derivative from the delta function to the variation, use compact support to kill the boundary integral. Then repeat the process to transfer the second derivative on the delta function to the variation of $\phi.$ Putting all of this together, we find \begin{align} \delta\mathcal{S} &=\int\text{d}^4x\int\text{d}^4y\ \delta\phi(x)\left[\delta\phi(y)\left(-m^2-\partial_\mu\partial^\mu\right)\delta^{4}(y-x)\right]\\ &=\int\text{d}^4x\int\text{d}^4y\ \delta\phi(x) \left[\delta^{4}(y-x)\left(-m^2-\partial_\mu\partial^\mu\right)\delta\phi(y)\right] \\ &=\int\text{d}^4x\ \delta\phi(x)(-m^2-\partial_\mu\partial^\mu)\delta\phi(x), \end{align} where in the last line, we have used up the delta function to kill an integral. This is it! We’re finished. The last bit is to interpret the result form the physical standpoint. If we call $\delta\phi(x)=\chi(x)/\sqrt{2}$, we find \begin{align} \delta\mathcal{S} &= \frac{1}{2}\int\text{d}^4x(-\chi(x) m^2\chi(x)-\chi(x)\partial_\mu\partial^\mu\chi(x))=\frac{1}{2}\int\text{d}^4x(-m^2\chi^2(x)+\partial_\mu\chi(x)\partial^\mu\chi(x))\\ &=\int\text{d}^4x\mathcal{L}^{(1)}(\chi(x),\partial_\mu\chi(x)), \end{align} where $\mathcal{L}^{(1)}(\chi(x),\partial_\mu\chi(x))$ has the form $$\mathcal{L}^{(1)}(\chi(x),\partial_\mu\chi(x)) = \frac{1}{2}(\partial_\mu\chi(x)\partial^\mu\chi(x)-m^2\chi^2(x)).$$ This is a remarkable form for this mysterious $\mathcal{L}^{(1)}$ as it looks exactly like the Klein-Gordon Lagrangian with a field $\chi(x)$ instead of $\phi(z).$ I call these functions Lagrangians of Type 1, or $\mathcal{L}^{(1)}$ for short. $\mathcal{L}^{(1)}$ are the associated energy densities describing the fluctuations of a scalar field.
Setting $\delta\mathcal{S}=\mathcal{S}^{(1)}$, we have an Action of Type 1, given by $$\mathcal{S}^{(1)}=\int\text{d}^4x\mathcal{L}^{(1)}(\chi(x),\partial_\mu\chi(x)),$$ which we could extremize and demand that the Principle of Least Action holds for Type 1 actions as well. In doing so, we would obtain the Euler Lagrange equations  $$\frac{\delta\mathcal{S}^{(1)}}{\delta\chi(x)} = 0 =\frac{\partial\mathcal{L}^{(1)}}{\partial\chi(x)} – \partial_\mu\left(\frac{\partial\mathcal{L}^{(1)}}{\partial(\partial_\mu\chi(x))} \right).$$

These above equations describe what we have been looking for; the equations of motion for the background, i.e. the background fluctuations.

## 2 thoughts on “Second Variation of an Action: Part I”

1. Andy Knapp says:

The term
$$\oint_{\partial\partial\mathcal{M}} \delta^2(y-x)\delta\phi(y)\vec{n} \cdot \text{d}^2\vec{y}$$
vanishes by some exterior-derivative powered abstract nonsense, and therefore you’ve shown the vector field $X^\mu$ is divergence-free. I’m not sure that’s what you wanted to show…

$$\int_{\partial\partial\mathcal{M}} \omega = \oint_{\partial\mathcal{M}}d\omega = \oint_{\mathcal{M}}dd\omega = 0$$

1. Awesome comment. I’ve changed it now to reflect this oversight.

For context: I attempted to use the divergence theorem twice to remove the Laplacian on the delta function onto the scalar variation $\delta\phi$. That was wrong.

Actual solution: Integrate by parts and don’t forget Calc III.