# Tag Info

30

Typically: $\rm d$ denotes the total derivative (sometimes called the exact differential):$$\frac{{\rm d}}{{\rm d}t}f(x,t)=\frac{\partial f}{\partial t}+\frac{\partial f}{\partial x}\frac{{\rm d}x}{{\rm d}t}$$This is also sometimes denoted via $$\frac{Df}{Dt},\,D_tf$$ $\partial$ represents the partial derivative (derivative of $f(x,y)$ with respect to $x$ ...

11

If the functional derivative $$\tag{1} \frac{\delta F[\phi]}{\delta\phi^{\alpha}(x)}$$ exists (wrt. to a certain choice of boundary conditions), it obeys infinitesimally $$\tag{2}\delta F ~:=~ F[\phi+\delta\phi]- F[\phi] ~=~\int_M \!dx\sum_{\alpha\in J} \frac{\delta F[\phi]}{\delta\phi^{\alpha}(x)}\delta\phi^{\alpha}(x).$$ OP's functional integral ...

10

Regardless of the context and the meaning of the symbols, both sides of the equation have perfectly the same units: they are dimensionless. The integral has units $js$ as you write, using your notation, but the functional derivative has the compensating units $1/(js)$ so the units cancel. To see that dimension of the functional derivative is $1/(js)$, one ...

8

It is not. The correct identity is $$\frac{\delta}{\delta \Phi(y)} \Phi (x) = \delta(x-y)$$ where the derivative is the functional derivative. If $F : D(F)\ni \Phi \mapsto F(\Phi)\in \mathbb C$ is a function from a space of functions $D(F)$ to $\mathbb C$, the functional derivative of $F$, if it exists is the distribution $\frac{\delta F}{\delta \Phi}$ ...

8

I) It is worthwhile mentioning that there exists a basic approach well-suited to physics applications (where we usually assume locality) that avoids multiplying two distributions together. The idea is that the two inputs $F$ and $G$ in the Poisson bracket (PB) $$\tag{1}\{F,G\} ~=~ \int_M \!dx \left( \frac{\delta F}{\delta \phi(x)}\frac{\delta G}{\delta ... 7 Contrary to your claim near the end of your question, I claim that the time-derivative of the field is being treated as an "independent" argument of the Lagrangian. I'll try to convince you of this by showing you how this independence leads to everything working out the way you think it should. Some of the key points are at the end, so please read all the ... 6 OP considers the 'same-time' functional derivative (FD)$$\tag{1} \frac{\delta f(t)}{\delta x(t)}~:=~\frac{\partial f(t)}{\partial x(t)} - \frac{d}{dt} \frac{\partial f(t)}{\partial \dot{x}(t)} +\ldots. $$Here f(t) is shorthand for the function f(x(t), \dot{x}(t), \ldots;t). Although the 'same-time' FD (1) can be notationally useful, it has various ... 5 In general functional derivatives obey chain and product rules. If the concept troubles you you can always think of a function as a vector with an infinity of coordinates. Then a functional derivative is just a partial derivative. If F[h] is a functional of the function h(x). You can think of this as$$ h \to \vec{h} = \left(h(x_1), h(x_2), ..., ...

5

Comment to the question (v2): P&S is using the notation of a 'same-spacetime' functional derivative. To illustrate this notation, let us for simplicity stay within first variations, and leave it to the reader to generalize to higher-order variations. I) First of all, functional/variational derivatives should not be confused with partial derivatives. In ...

5

Whenever I have troubles with functional derivative things, I just do the replacement of a continuous variable $x$ into a discrete index $i$. If I'm not mistaken this is what they call a "DeWitt notation". The hand waiving idea is that you can think of a functional $F[f(x)]$ as of a "ordinary function" of many variables ...

5

A "functional" may also be a map of a function to another function. Hence in this case, $L:\mathcal{F}\times\mathcal{F}\rightarrow\mathcal{F};(\Psi(t),\dot\Psi(t))\mapsto L[\Psi(t),\dot\Psi(t)]$, $L$ maps functions in a function space $\mathcal{F}$ into $\mathcal{F}$ (this assumes that $\Psi(t)$ is well enough controlled, and that $L$ is not too exotic, for ...

4

This answer can be view as a supplement to joshphysics' correct answer, possibly stressing slightly different things and using slightly different words. Before defining functional/variational derivatives in Lagrangian formalism, it is crucial to understand exactly which variables are independent of each other and which are not? In other words, which ...

4

Have a look first at several chapters in Stone and Goldbart, "Mathematics for Physics" (the free preprint is here) before entering into more specific books. I think you may want to see chapters 1, and parts of 2 and 9. You may find some parts of what you want in classic books of the "Comprehensive Mathematical Methods for Physics" type, but they don't ...

4

The term functional is used in at least two different meanings. One meaning is in the mathematical topic of functional analysis, where one in particular studies linear functionals. This meaning is not relevant for the discussion at page 299 in Ref. 1. Another meaning is in the topics of calculus of variations and (classical) field theory. This is the ...

4

The expression : $[d\phi(x)] \frac{\delta F}{\delta \phi(x)}$ could be interpreted as a formal $dF(\phi)$ : $$\int [d\phi(x)] \frac{\delta F}{\delta \phi(x)} \sim \int \frac {\partial F}{\partial \phi_i} d\phi_i \sim \int dF(\phi) =F(+\infty) - F(-\infty)$$ So the left hand side of the expression is zero only for identical boudary conditions, for instance ...

4

Yes. Here, we are dealing with functional derivatives and these satisfy the chain rule and the product rule, which is really an important reason why it can be called a derivative to begin with. Important note: The definition that you give for the functional derivative is not the standard one, and does not satisfy its usual properties (as shown by ...

4

The physicist's derivative notation denotes the components of a Frechet derivative in the direction of the delta-function supported at $y$. This is one of those places where the habit of denoting the function $f$ by its value $f(x)$ gets confusing. It's somewhat clearer if you write $\delta_y$ for the delta function at $y$, and $$\frac{\delta F}{\delta ... 4 First, I want to say that different people use different notation and I welcome any comments. I also feel as if I am about to enter a minefield. Here the answer is made up with examples of use of d, \partial and \delta. I would say for d that dV \over dx would be the total derivative in one dimension for V(x) where the potential V is a ... 4 Right, one is only supposed to put the sources J=0 to zero after the very last J-differentiation has been performed. Figuratively speaking, short of writing out the calculation in full detail: Some of the Js downstairs can "couple" to the Js upstairs in the exponential. 3 As suggested by Frederic, you should integrate the$$ \int d^n x \: \alpha^a(x) \cdot \nabla_a \delta \Psi term (with whatever \alpha should be) by parts using the Ostrogradsky-Gauss theorem and the boundary condition \delta \Psi (\infty) \rightarrow 0. But Euler and Lagrange have already done it for you, so you can just use the formula for an ... 3 Basically you are missing a term. Think of Lagrangian mechanics. If q(t) is your generalized coordinate, and you make the change q(t) \to q(t) + \delta q(t), then remember that the change in the action is \delta S = \int_{t_i}^{t_f} \left(\partial_q L - \partial_t \partial_\dot{q} L \right) \delta q. Thus you can basically think of the change in the ... 3 It seems that OP is pondering the following. What happens in a field theory [in OP's case: GR] if spacetime M has a non-empty boundary \partial M\neq \emptyset, and we don't impose pertinent (e.g. Dirichlet) boundary conditions (BC) on the fields \phi^{\alpha}(x) [in OP's case: the metric tensor g_{\mu\nu}(x)]? I) Firstly, it should stressed ... 3 Let's consider a single scalar field for simplicity. The following step is a misapplication of the functional derivative: \begin{align} \delta Z(J) = \frac{\delta Z}{\delta\phi(x)}\delta\phi(x) \end{align} By definition, one can only take the functional derivative of a functional F with respect to \phi if F is a functional of \phi. The functional ... 3 I) The mentioned integral \color{Red}{\int \!\mathrm{d}^4x} should really be there. If we define the action as\tag{1}S_J[\phi]~:=~S[\phi]+\int \!\mathrm{d}^4x~ J_a(x) \phi^a(x),$$then the infinitesimal variation of the action reads$$\tag{2}\delta S_J~=~ \color{Red}{\int \!\mathrm{d}^4x} ~\frac{\delta S_J}{\delta \phi^a(x)} ~\delta \phi^a(x).$$... 3 To formally show the Schwinger-Dyson equations, use the fact the$$\int [d\phi]\frac{\delta}{\delta \phi^{\alpha}(x)} \exp\left[\frac{i}{\hbar}\left(S[\phi]+\int \!d^nx^{\prime}~J_{\alpha}(x^{\prime})\phi^{\alpha}(x^{\prime}) \right)\right] ~=~0,$$cf. this Phys.SE post. Without specifying the action S[\phi] and field content \phi^{\alpha} further, it ... 3 Since R[A] is gauge invariant, the variation of R[A] is zero when A^a_{\mu} undergoes the infinitesimal gauge transformation A^a_{\mu}\rightarrow A^a_{\mu} + \epsilon (D_{\mu}\alpha)^a where \alpha^a is any Lie algebra valued field and \epsilon is an infinitesimal parameter. The variation of R[A] under this gauge transformation is$$0=\delta R ...

3

One way to see that considering the dependence of $\dot{x}$ on $x$ is problematic is as follows: $x(t)$ maps a real number $t$ to another real number $x$. So $\dot{x}=dx/dt$ is the derivative of that map, meaning we take $$\lim_{\Delta t \to 0} \frac{x(t+\Delta t) - x(t)}{\Delta t}$$ So we can see that $dx/dt$ is itself another map from a real number $t$ ...

3

Functional derivative: $F_d[{\bf p}] = \int \mathrm d\boldsymbol{r}\ f( \boldsymbol{r}, {\bf p}(\boldsymbol{r}), \nabla\cdot {\bf p}(\boldsymbol{r}) )$ $\frac{\delta F_d}{\delta {\bf p}(\boldsymbol{r})} := \frac{\partial f}{\partial {\bf p}} - \nabla \cdot \frac{\partial f}{\partial\nabla\cdot {\bf p}},$ although something's weird with your signs. Your ...

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