15
votes
Accepted
Does the "Euler-Lagrange operator" $(\gamma,L) \mapsto (\partial_x L)\circ d\gamma- (\partial_v L\circ d\gamma)'$ have some geometric interpretation?
I am going to use the $\infty$-jet formalism (one could also work with finite order jets but that's actually more complicated), and consider higher order variational problems as well as the case of ...
- 11.6k
13
votes
Variations wrt. Time Derivatives
This is a fairly common misconception.
The action functional $S$ eats a function $q$ and spits out the following number:
$$S[q] := \int_{t_1}^{t_2} L\big(q(t),\dot q(t), t\big)\ dt$$
where the ...
- 71.5k
9
votes
Accepted
Why does the integral symbol disappear when applying a functional derivative?
Define functional
$$ G[g]~:=~\int \!d^4x ~\sqrt{-g(x)}.\tag{0}$$
Method 1:
$$\begin{align} \int \!d^4x ~\color{red}{\frac{\delta G[g]}{\delta g_{\mu\nu}(x)}} \delta g_{\mu\nu}(x)
~=~&\delta G[g]\...
- 213k
8
votes
Accepted
Definition of the Lagrangian in (classical) field theory
Perhaps it would be more pedagogical to use the notation $q$ and $v$ as the notation for the fields $\psi$ and $\dot{\psi}$, respectively, because they are independent fields: $\mathbb{R}^{n+1}\to\...
- 213k
8
votes
Accepted
Computing functional derivative of exchange-correlation functional
In my opinion the notation is a bit weird. So let us first define $$\epsilon_n := \epsilon \circ n :x \mapsto \epsilon(n(x))$$ for some function $\epsilon$ and
$$E[n] := \int \mathrm dx\, n(x) \, \...
- 8,442
8
votes
Accepted
Fourier transform of a functional derivative
Note that if a functional $S[\phi]$ is local in position space with variable $x$, it is typically not local in wavevector space with Fourier transformed variable $k$, where$^1$
$$\begin{align}\...
- 213k
8
votes
Accepted
Functional derivative under a path integral sign
For what it's worth, to gain some intuition, consider the following crude discretization of OP's path integral:
$$\frac{\partial}{\partial\chi^k}\underbrace{\left[\prod_{i\in I} \int d\phi^{i} e^{-{\...
- 213k
7
votes
Accepted
How to calculate functional derivative correctly?
Once nice way to calculate functional derivatives is to use the concept of the Gateaux derivative as follows:
$$\frac{d}{d\epsilon}S[\phi+\epsilon \eta]\bigg|_{\epsilon=0} = \int d^4x\frac{\delta S}{\...
- 71.5k
7
votes
How to calculate functional derivative correctly?
Here is a second way to see the correct result for taking the functional derivative of the spacetime derivative of the field, which I hope will be helpful.
Recall that the definition of the functional ...
- 2,073
6
votes
Accepted
Variational derivative of function with respect to its derivative
The definition of the functional derivative of a functional $I[g]$ is the distribution $\frac{\delta I}{\delta g}(\tau)$ such that
$$\left\langle \frac{\delta I}{\delta g}, h\right\rangle := \frac{d}{...
- 78.1k
6
votes
Accepted
Why does Fermat's principle (optics) not apply to all paths?
If I have a ball rolling on some type of crazy set of hills and valleys, where is the ball going to want to sit? You might say immediately "at the bottom of a valley, of course!" But let's rephrase ...
- 8,582
6
votes
Accepted
Functional derivative commutes with total derivative
OP is essentially asking the following.
Why the total time derivative
$$ \frac{d}{dt}~=~\frac{\partial}{\partial t} + \sum_{m=0}^{\infty}\sum_{i=1}^n q^{i(m+1)}(t)\frac{\partial}{\partial q^{i(m)}(t)}...
- 213k
6
votes
Accepted
Notation of derivatives in field theory
There are two notions which tend to be a bit confused in Lagrangian mechanics for fields.
First, there is the functional derivative. As the name implies, this is a derivative for functionals, such ...
- 16.7k
6
votes
Fourier transform of a functional derivative
Without paying much attention to mathematical hypotheses (however they can be fixed), your idea is correct in view of the following "proof".
$$E[\hat{x}(q)]= E\left[ \frac{1}{(2\pi)^{n/2}}\...
- 78.1k
5
votes
Accepted
Help with taking derivative of Lagrangian scalar model of graviton
You have to be careful with the way you've defined your Lagrangian. You're treating the second derivative of $h$, $\Box h$, as if it were independent of both $h$ and $\partial_\mu h$. If you integrate ...
- 690
5
votes
Accepted
Confusion about dimensions of a functional, its functional derivative and its variation
No, $F$ and the infinitesimal variation $\delta F$ have the same dimension. And $\phi$ and the infinitesimal variation $\delta \phi$ have the same dimension. But by definition of the functional/...
- 213k
5
votes
Functional derivative or Euler-Lagrange?
Briefly, the Euler-Lagrange (EL) equation
$$ \frac{\delta S}{\delta q(t)}~=~0$$
states that the functional derivative of the action $S[q]$ vanishes.
- 213k
5
votes
Accepted
How would one calculate the inverse of Dirac delta function?
To expand on @mikestone's answer, the required result is not$$\frac{\delta\phi(x)}{\delta\phi(y)}\frac{\delta\phi(y)}{\delta\phi(x)}=1,$$but$$\int_{\Bbb R^3}\frac{\delta\phi(x)}{\delta\phi(y)}\frac{\...
- 25.4k
5
votes
Accepted
Equation 13.20 of Peskin & Schroeder
The object $\Gamma[\phi]$ (I'll drop the subscript to keep notation more simple) is a real number. We'd like to expand it in a Taylor series with respect to its dependence on the field variable $\phi$....
- 23k
5
votes
How to Relate the Functional Derivative to Infinitesimal Change in Noether's Theorem
One way to do a calculation similar to OP's without a notion of infinitesimal is to consider a one-parameter deformation of the field $\phi^\epsilon$ (with $\phi = \phi^0$) and define $\delta$ to mean ...
- 6,119
5
votes
Accepted
Leibniz rule and Nakahara's definition for functional derivatives with respect to Grassmann variables
The definition given by Nakahara is correct. Specifically:
$$
\frac{\delta G[\psi(x)]}{\delta \psi(y)} \\
= \frac{1}{\epsilon}\{G[\psi(x) + \epsilon \delta(x-y)] - G[\psi(x)]\} \\
= \frac{1}{\epsilon}...
- 4,833
4
votes
Variation of a field action
I think its best to first remember the finite-dimensional case and then we generalize.
Consider a function of a single variable $f(\phi)$. Note here that $\phi$ is a real number, NOT a function. I am ...
- 27.7k
4
votes
Why is this "the" functional of Laplace's equation?
The Lagrangian density of the electromagnetic field is
$${\cal L} ~=~-\frac{1}{4\mu_0}F_{\mu\nu}F^{\mu\nu}\pm J^{\mu}A_{\mu} ~=~\frac{\varepsilon_0E^2}{2} - \frac{B^2}{2\mu_0} -\rho \phi + {\bf J}\...
- 213k
4
votes
Derivatives in Euler-Lagrange for fields
Varying $\phi$ satisfies $$\delta\partial_\lambda\phi=\partial_\lambda\delta\phi,\,\delta\partial^\lambda\phi=\partial^\lambda\delta\phi=\eta^{\lambda \mu}\delta\partial_\mu\phi$$ so $\frac{\partial\...
- 25.4k
4
votes
Accepted
Second variation of a functional
The rule that defines functional derivatives is
$$\frac{\delta \rho(\mathbf{r})}{\delta \rho(\mathbf{r}')} = \delta(\mathbf{r}-\mathbf{r}').$$
Note that the right hand side is the Dirac delta function....
- 22.8k
4
votes
Accepted
Functional derivative and variation of action $S$ vs Lagrangian $L$ vs Lagrangian density $\mathcal{L}$ vs Lagrangian 4-form $\mathbf{L}$
The main point is (as OP already mentions) that while the action
$$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}$$ is a ...
- 213k
4
votes
Functional Derivative
The answer can simply be seen by applying Taylor series:
$$
g[x] = g[x_0] + g'[x_0] \cdot (x-x_0) + ...
$$
In your case we should take $x_0 = f(x)$ and $x=f(x)+\epsilon \delta(x - x_0)$.
You are ...
- 1,815
4
votes
Accepted
I am stuck in the derivation of Schwinger-Dyson equation for 1-point Function in Schwartz's QFT book
Let us clarify these two issues first. The expansion of the exponential is:
$$e^x = 1 + x + \mathcal{O}(x^2)\tag{1}\label{eq:exponential}$$
Second let us remind ourselves about integration by parts:
$$...
- 4,478
4
votes
Variations wrt. Time Derivatives
Here is the cheat sheet:
On one hand, in a partial differentiation the variable $x,\dot{x},\ddot{x},\ldots$ are independent. E.g. the partial derivatives $\frac{\partial \dot{x}}{\partial x}=0$ and $\...
- 213k
4
votes
Deriving Klein-Gordon from Hamilton's equations for fields using functional derivatives
You are most likely confused because methods from functional calculus are often obscured by hacks to make the calculations easier to understands but it also hides a lot of the intuition. So starting ...
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