I have seen many potential abuse of notation that prevents me from clearly understanding variational methods in QFT and GR that I want to get this settled once and for all. This may be a bit long but I think it pays to put everything in one place.

Functional derivative in QFT

Suppose I want to obtain an equation of motion. If I follow standard definition (e.g. Wikipedia, which does give standard expression as far as I remember), given an action for a field theory of the form

$$S[\Phi] = \int d^4x\,\mathcal{L}[\Phi,\partial_\mu\Phi]$$ where $\Phi$ is a particular field we are interested in. I will set the variation of the action $\delta S=0$. Now, this variation is formally defined as \begin{align} \delta S := \int d^4x\, \frac{\delta S}{\delta \Phi}\delta\Phi \end{align} and we formally define the quantity $\delta S/\delta\Phi$ to be the functional derivative of $S$ with respect to $\Phi$ (there may be rigorous alternative/interpretation using Frechet derivative which I am not familiar of so I appreciate if anyone can clarify this).

Now, the expression on RHS of $\delta S$ is meaningless unless I know what is $\delta \Phi$ and the functional derivative $\delta S/\delta \Phi$. This is settled by using some appropriate space of test functions, which for asymptotically flat spacetimes would be the space of functions that vanish on the boundary $\partial M$ of the manifold $M$ (e.g. compactly supported functions on $M$, denoted $C^\infty_c(M)$). If $h\in C^\infty_c(M)$, we have \begin{align} \int d^4x\,\frac{\delta S}{\delta \Phi}h = \lim_{\epsilon\to 0}\frac{S[\Phi+\epsilon h]-S[\Phi]}{\epsilon} = \left.\frac{d}{d\epsilon}\right|_{\epsilon=0}S[\Phi+\epsilon h]\,, \end{align} and what we usually call $\delta \Phi$ is in fact $\epsilon h$, which is consistent with the name "variation of $\Phi$". The above expression also furnishes a definition of how to take functional derivative of any functional. The standard Euler-Lagrange equation for field theory is then obtained by saying that $\delta S=0$ for all variations $\delta\Phi$ that vanish on the boundary, which then implies that \begin{align} 0 = \frac{\delta S}{\delta\Phi} = \frac{\partial \mathcal{L}}{\partial\Phi}- \partial_{\mu}\frac{\partial \mathcal{L}}{\partial(\partial_\mu\Phi)}\,. \end{align} While it may be obvious to some, it should be stressed that $\partial \mathcal{L}/\partial \Phi$ is not a function but a functional of $\Phi,\partial_\mu\Phi$: just note that $\Phi=\Phi(x^\mu)$. This is followed e.g. QFT texts by Blundell, implicitly by Peskin, and many other places.

If we follow Weinberg's QFT route, he instead works with Lagrangian: \begin{align} L = \int d^3x\,\mathcal{L}(\Phi,\partial_\mu\Phi)\,,\hspace{0.5cm} S= \int d x^0 L\,, \end{align} and then show that the same Euler-Lagrange equation is obtained when $\delta L = 0$. You can check in Weinberg's textbook that the steps used are exactly the same as what I outlined using the actions $S$ except that he chose to work with $L$, the usual Lagrangian (not the Lagrangian density) instead of the full action $S$.

Q1: why can we do these two different variations $\delta S=0$ and $\delta L=0$ and get the same answer? Clearly there is some connection between $\delta S$ and $\delta L$, but my problem stems from this issue: it looks to me that the variation $\delta\Phi$ looks different in these two cases, since one is variation under $d^4x$ and the other is in $d^3x$: effectively, the test function $\delta\Phi\equiv \epsilon h$ for $\delta L$ case only need to care about spatial integral, while $\delta S$ requires spacetime integral. Either the two mean the same thing or some subtle thing I missed render them equal in the end.

UPDATE 1: I think I might have figured out Q1 (or at least partially). It has to do with the fact that Weinberg had to split the Euler-Lagrange for spatial derivative and time derivatives, so he treated $\partial_j\Phi$ and $\dot{\Phi}$ separately (see discussions around his Eq. (7.2.1-7.2.7) or so). I certainly could use some clarification/confirmation.

Functional derivative in GR

In GR, there is a situation in which you want to work with canonical formalism which leads you to understand surface charges and conserved quantities similar to the above. The usual difference, however, is that the method formally differential forms to make things work. You don't work with Lagrangian density but Lagrangian 4-form $\mathbf{L}$ (see e.g. Iyer-Wald formalism or advanced lecture notes on GR by Compere here, among many others). In here, $\mathbf{L} = L\,d^4x$ so $L$ is really a Lagrangian density as we usually know in QFT. For convenience let's focus on Compere's notes (which is quite clean and well-written) However, in these contexts, the variation of $\mathbf{L}$ is the one that gives the equation of motion, and they formally define \begin{align} \frac{\delta L}{\delta \Phi}:=\frac{\partial L}{\partial \Phi}-\partial_\mu\frac{\partial L}{\partial (\partial_\mu\Phi)}\,. \end{align}

As far as I know, in these contexts where one works with Lagrangian 4-form and symplectic formalism, the calculation is rigorous (modulo doing hardcore analysis), i.e. there is no handwaving and whatsoever but the definitions here are to me inconsistent with QFT one I wrote above: after all, in these two papers/notes $L$ is Lagrangian density and hence it would, by replacing $L$ with $\mathcal{L}$ to match QFT version, mean that Euler-Lagrange equation is \begin{align} 0=\frac{\delta \mathcal L}{\delta \Phi} \neq \frac{\delta S}{\delta \Phi}\,,\frac{\delta L_{\text{Weinberg}}}{\delta \Phi}\,. \end{align} Note also that in this formalism, the definition of conserved stress-tensor also follows from variation of the Lagrangian 4-form with respect to infinitesimal diffeomorphism generated by vector $\xi^\mu$, i.e. \begin{align} \delta_\xi\mathbf{L} = \text{d}(...)\Longrightarrow \partial_\mu T^{\mu\nu} = 0\,. \end{align} where $\text{d}(...)$ is exterior derivative of some 3-form (i.e. RHS is an exact 4-form).

Q2: is this abuse of notation, inconsistency, or is there something I am fundamentally missing here?

Out of all people, I find it hard to believe that Wald/Compere (and many others I cannot remember) commit abuse of notation of this kind (if at all), so either I miss something trivial or there is something going on that I don't understand.


1 Answer 1

  1. The main point is (as OP already mentions) that while the action $$S[\Phi]~=~\int\!dt~L[\Phi(\cdot,t),\dot{\Phi}(\cdot,t),t]~=~ \int\! d^4x~{\cal L}(\Phi(x),\partial\Phi(x),x) \tag{A}$$ is a functional of $\Phi$, the Lagrangian $$L[\Phi(\cdot,t),\dot{\Phi}(\cdot,t),t]~=~ \int\! d^3{\bf x}~{\cal L}(\Phi({\bf x},t),\dot{\Phi}({\bf x},t),\nabla\Phi({\bf x},t),{\bf x},t)\tag{B}$$ at some instant $t$ is a functional of two independent fields $\Phi(\cdot,t)$ and $\dot{\Phi}(\cdot,t)$, cf. my Phys.SE answers here and here. The Lagrangian density ${\cal L}$ is a (density-valued) function of its arguments.

  2. On one hand, for a variationally defined functional derivative (FD) $$ \frac{\delta S[\Phi]}{\delta\Phi(x)}\tag{C} $$ to exist, appropriate boundary conditions (BCs) are necessary.

    On the other hand, Compere, Iyer & Wald consider 'same-spacetime' FDs $$ \frac{\delta {\cal L}(\Phi(x),\partial\Phi(x),\ldots,x)}{\delta\Phi(x)},\tag{D} $$ defined via their (possibly higher-order) Euler-Lagrange (EL) expressions, where BCs are irrelevant, cf. my Phys.SE answers here, here & here. (The only requirement is that ${\cal L}$ should be a sufficiently smooth function. We stress that the notation (D) becomes meaningless if interpreted as a variationally defined FD.) There is a parallel 'same-spacetime' story for the Lagrangian 4-form $${\bf L}~=~d^4x~{\cal L}.\tag{E}$$

  • $\begingroup$ Thanks, that's illuminating. I still have some doubts; for what you called "algebraically defined" FDs, since you said they are tied to "same-spacetime" FDs (not sure if it's formalized elsewhere), does that mean they are simply abusing notation? What I understood from your other posts seem to be that there is a way to consistently and rigorously define those despite being ill-defined in principle (due to $\delta(0)$ thing when you do FD). $\endgroup$ Commented Aug 25, 2019 at 22:34
  • $\begingroup$ I retired the terminology "algebraically defined" FD and updated the answer. $\endgroup$
    – Qmechanic
    Commented Aug 26, 2019 at 8:25
  • $\begingroup$ Thanks, I think more or less I get what you meant. The key was the fact that the BCs are irrelevant in (D). $\endgroup$ Commented Aug 28, 2019 at 17:26
  • $\begingroup$ $\uparrow$ Exactly. $\endgroup$
    – Qmechanic
    Commented Aug 28, 2019 at 17:31

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