149

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 ...


68

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 ...


65

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 ...


52

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^{*}...


41

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 ...


18

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,...


14

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 ...


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 ...


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 ...


10

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) \...


9

Just like in normal differentiation, you can get yourself confused if you call all your variables the same thing. So let's write things carefully: $$ J[\rho] = \frac 12\int \frac{\rho(r)\rho(r')}{|r - r'|} dr\,dr' $$ and the functional derivative will use a different dummy variable: $$ \frac {\delta J}{\delta \rho (x)} = \frac{\delta}{\delta\rho(...


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

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 ...


9

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}^{...


8

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 ...


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

Simplest is to go with \begin{align} \frac{df}{dt}&= \frac{\partial f}{\partial q}\dot q + \frac{\partial f}{\partial t}\, ,\\ \frac{d}{dt}\left(\frac{\partial }{\partial \dot{q}}\frac{df}{dt}\right)&= \frac{d}{dt}\left(\frac{\partial f}{\partial q} \right)\\ &= \frac{\partial }{\partial q}\left( \frac{df}{dt}\right) \end{align} You can deal with ...


7

Let's do what Heidar says and write it with indices, and identify the Lagrangian. $$ L=\frac{1}{2}(\vec{\nabla}\times \vec{A})^2 = \frac{1}{2}\epsilon_{ijk}\partial_j A_k \epsilon_{ilm}\partial_l A_m $$ where, if you haven't heard of it yet, you pretend there is a summation symbol for each repeated index. Then since there are no bare $A_i$ sitting by ...


7

Edit: Note that I am doing only the first variation, and I am not doing each and every step, mainly those pertinent in understanding how the general shape equation is determined. If you want to see the full derivation, you will need to understand the Geometric Mathematic Primer discussed in Sections 2 and 3 of the book. Geometric Methods in the Elastic ...


7

The Lagrangian provided is given by, $$L=\sqrt{t+\dot{y}^2t}$$ Clearly, $\partial L/\partial y = 0$. We compute the second term in the Euler-Lagrange equations: $$\frac{d}{dt}\frac{\partial L}{\partial \dot{y}} = \frac{d}{dt} \left( \frac{t\dot{y}}{\sqrt{t+\dot{y}^2 t}} \right) = \frac{t(\dot{y} + \dot{y}^3 + 2t\ddot{y})}{2[t(1+\dot{y}^2)]^{3/2}} = 0$$ ...


7

Within the framework of differential geometry, the construction goes as follows. Let there be given a 4-dimensional orientable spacetime manifold $M$. A field is a section $$\phi~\in~\Gamma(E) \tag{1}$$ in a bundle $(E,\pi,M)$. The Lagrangian$^1$ 4-form $\mathbb{L}$ is a bundle map $$ \begin{array}{rcl} J^1(M,E) &\stackrel{\mathbb{L}}{\longrightarrow}&...


7

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 ...


6

Your confusion really just comes down to understanding the notation that is widely used for partial derivatives. For simplicity, I'll restrict the discussion to a system with one coordinate degree of freedom $x$. In this case, the Lagrangian is a real valued function of two real variables which we suggestively label by the symbols $x$ and $\dot x$. ...


6

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 $F(f_{-N},\cdots,f_0,f_1,f_2,\cdots,...


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