# Tag Info

## Hot answers tagged variational-calculus

153

The V shape makes sense for an ideal chain with all of its mass concentrated at the midpoint (with the rest of the chain being massless). But now consider a chain with its mass distributed over 3 points, e.g. equal masses at the quarter-way points and the midpoint. The quarter point masses will pull the V out of shape, introducing their own bends into the ...

71

Sorry to put in my contribution so late in the proceedings, but I don't think anyone has yet presented this simple argument based on forces… Suppose the chain did hang in a V shape. Consider a small portion P of the chain. It is acted upon by a downward force, W, of gravity and by forces tangential to the chain from neighbouring parts of the chain. Now draw ...

64

The V has a higher gravitational potential energy than the catenary does. To see this, consider pulling down at the center of the catenary very hard. As you pull down, the chain deforms into a V. But you have to do work to pull it down, which means energy is going into the system and its center of mass is moving up. Hence the catenary has a lower center of ...

62

TL;DR: Yes, it is just a short-cut. The main point is that the complexified map $$\tag{A} \begin{pmatrix} \phi \\ \phi^{*} \end{pmatrix} ~=~ \begin{pmatrix} 1 & i\\ 1 &-i \end{pmatrix} \begin{pmatrix} \phi_1 \\ \phi_2 \end{pmatrix}$$ is a bijective map :$\mathbb{C}^2 \to \mathbb{C}^2$. Notation in this answer: In this answer, let $\phi,\phi^{*}... 50 When I asked my undergrad analytic mechanics professor "what does it mean for a rotation to be infinitesimal?" after he hand-wavily presented this topic in class, he answered "it means it's really small." At that point, I just walked away. Later that day I emailed my TA who set me straight by pointing me to a book on Lie theory. Fortunately, I don't ... 19 It does follow from calculus. Here's the standard way this is treated (I'm not going to be explicit about mathematical details such as smoothness assumptions here). Definition of$\delta q$. Given a parametrized path$q:t\mapsto q(t)$, we consider a deformation of the path which we call$\hat q:(t, \epsilon)\mapsto \hat q(t,\epsilon)$satisfying$\hat q(t,...

18

The Lagrangian density for a Dirac field is $$\mathcal{L} = i\bar\psi\gamma^\mu\partial_\mu\psi -m \bar\psi\psi$$ The Euler-Lagrange equation reads $$\frac{\partial\mathcal{L}}{\partial\psi} - \frac{\partial}{\partial x^\mu}\left[\frac{\partial\mathcal{L}}{\partial(\partial_\mu\psi)}\right] = 0$$ We treat $\psi$ and $\bar\psi$ as independent dynamical ...

18

We do not treat $\dot q$ as an independent variable in the derivation of the Euler-Lagrange equations. The rough answer is that $q$ and $\dot q$ are independent as inputs to the Lagrangian, but become linked once we specify a path through configuration space - I expand on this in points 5 and 6. I'll be quite formal in what follows, but perhaps the ...

16

I would like to make a comment, which may clarify and simplify the things a little bit. In complex analysis [see e.g. Introduction to Complex Analysis" by B.V. Shabat] by definition derivatives over the complex variables $z$ and $\bar z$ are given by: $$\mbox{def:} \quad \partial_z \equiv \frac{1}{2} \left(\partial_{\rm a} - i \partial_{b}\right) \... 14 Since the metric g_{\mu\nu}=g_{\nu\mu} is symmetric, we must demand that$$\begin{align} \delta g_{\mu\nu}~=~&\delta g_{\nu\mu}\cr~=~&\frac{1}{2}\left(\delta g_{\mu\nu}+\delta g_{\nu\mu}\right)\cr~=~&\frac{1}{2}\left( \delta_{\mu}^{\alpha}\delta_{\nu}^{\beta} + \delta_{\nu}^{\alpha}\delta_{\mu}^{\beta}\right)\delta g_{\alpha\beta},\end{align}\...

14

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 multiple independent variables. Jet spaces: Let's fix some notation. I will consider a fibred manifold $\pi:Y\rightarrow X$, and I will suppose that $\pi$ is the ...

12

The Wikipedia quote appears to be lifted from this Solid State Physics text by CTI Reviews, and then plastered all over the web. The text does not give any citation of Lanczos, however. Here is the only passage in Lanczos's 300 pages long Variational Principles of Mechanics that contains the word "self-adjoint": "Schroedinger, on the other hand, ...

12

The Euler Lagrange equation is a differential equation resulting from the search for the extremum of a functional: this extremum is given by the first variation only. This is similar to the condition for finding a point where a function $f$ is extremum: the condition $df/dx=0$ is on the first derivative only. In both cases, one does not seek to ...

12

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 Lagrangian $L: \mathbb R^3 \rightarrow \mathbb R$ is just a function of three variables. One might have, for example, $$L(a,b,c) = \frac{1}{2} mb^2 - \frac{1}{2} m\... 11 Of course @QMechanic's answer is correct. i would like to show a very simple reason why this is so (and also point to possible generalisations) First of all, any complex number z=a+bi, is 2-dimensional and each part (the real part a or the imaginary part bi) can be completely independent of each other. As a result a complex number can represent in ... 11 Let us here for simplicity consider point mechanics with generalized positions q^k(t) defined on a time interval [t_i,t_f]\subseteq \mathbb{R}. The generalization to field theory with fields \phi^{\alpha}(x) defined on a spacetime region \Omega\subseteq\mathbb{R}^D is straightforward.^1 Given an (off-shell) action functional$$ I[q]~=~\int_{t_i}^{...

11

This is a common point of confusion when one gets started in Lagrangian mechanics. The important thing to notice is that we are taking partial derivatives, not full derivatives with respect to $x$. From the point of view of our partial derivatives, $x$ and $\dot{x}$ are completely separate variables with no relation to each other. To give you a bit of ...

11

Perhaps the following comment is helpful: If $M=M^{\prime}\times M^{\prime\prime}$ is a product manifold then the exterior differential $\mathrm{d}=\mathrm{d}^{\prime}+\mathrm{d}^{\prime\prime}$ on $M$ is a sum of the exterior differentials on $M^{\prime}$ and $M^{\prime\prime}$. We can assign degree $(1,0)$ to $\mathrm{d}^{\prime}$ and degree $(0,1)$ to $\... 10 Suppose now the Lagrangian$L$is a function of$y(x), y'(x)$and also$y''(x)$, i.e. it contains second derivatives w/r to the parameter$x. It is straightforward to adapt the usual procedure to this case: write \begin{align} Y(x,\epsilon)=y(x)+\epsilon\,\eta(x) \end{align} for an otherwise arbitrary function\eta$. We then have the parametrized ... 10 No, one of the partial derivative symbol$\partial_{\mu}$in OP's equation (2) is not correct if it is supposed to mean partial derivatives. The correct Euler-Lagrange (EL) equations read $$\tag{2'} 0~\approx~\frac{\delta S}{\delta \phi^{\alpha}} ~=~\frac{\partial {\cal L}}{\partial \phi^{\alpha}} - \sum_{\mu} \color{Red}{\frac{ d}{dx^{\mu}}} \frac{\... 9 I) The point of Ref. 1 is similar to why the generalized positions q^j and the generalized velocities \dot{q}^j in the Lagrangian L(q,\dot{q},t) are independent variables, see e.g. this Phys.SE post. A less confusing notation would probably be to denote the generalized velocities v^j instead of \dot{q}^j. Ref. 1 is referring to the non-commutative ... 9 Theorem: let L be a homogeneous function of degree k; then the on-shell lagrangian is a total derivative. Proof: according to the Euler's homogeneous function theorem,$$ k\ L(q,\dot q)=q\frac{\partial L}{\partial q}+\dot q\frac{\partial L}{\partial \dot q}\tag{1} $$On the other hand, because of the Euler-Lagrange equations,$$ (1)=q\frac{\partial L}{\... 9 First, let's make sure we understand the notion of the total derivative in the particle case: The Lagrangian itself is a real-valued function$L(q,\dot{q},t)$, where$q$and$\dot{q}$are treated as independent variables, cf. this question or this answer of mine. When we speak of a "total" derivative in the context of the Euler-Lagrangian equations, we ... 9 Define functional $$G[g]~:=~\int \!d^4x ~\sqrt{-g(x)}.\tag{0}$$ Method 1: $$\int \!d^4x ~\color{red}{\frac{\delta G[g]}{\delta g_{\mu\nu}(x)}} \delta g_{\mu\nu}(x)~=~\delta G[g]~\stackrel{(0)}{=}~ \int \!d^4x ~\color{red}{\frac{\partial\sqrt{-g(x)}}{\partial g_{\mu\nu}(x)}} \delta g_{\mu\nu}(x). \tag{1}$$ Method 2: $$\color{red}{\frac{\delta G[g]}{\delta ... 8 OP considers the 'same-time' functional derivative (FD)$$\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. $$Here f(t) is shorthand for the function f(x(t), \dot{x}(t), \ldots;t). Although the 'same-time' FD (1) can be notationally useful, it has various ... 8 This has nothing to do with "bras" or "kets" and more with the elementary observation that a complex number has two real degrees of freedom, and that derivatives are with respect to one real degree of freedom. The \frac{\partial}{\partial\phi} and \frac{\partial}{\partial\phi^\ast} are the Wirtinger derivatives, which in particular fulfill \frac{\... 8 Calculus of variations is used to find maxima and minima of functionals. The usual functions f(x) are from some set of numbers to another. For example,$$f: \mathbf{R} \to \mathbf{C}$$Functionals F[f] are from a a vector space (usually of functions) to real numbers:$$F: \mathcal V \to \mathbf R$$Basically they are mathematical quantities ... 8 The Lagrangian density is a function. Consider the following examples:$$ A[f]=\int_0^1\mathrm dx\ f(x) $$and$$ B(f(x))=f(x) $$It is clear that A is a functional, because for example$$ A[\sin]=1-\cos1=0.45\in\mathbb R $$is a number, while B is a function, because$$ B(\sin)=\sin$$is not a number, but a function. In your notation,$L\$ is a ...

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