154 votes
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Why is the shape of a hanging chain not a "V"?

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 ...
PM 2Ring's user avatar
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71 votes

Why is the shape of a hanging chain not a "V"?

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 ...
Philip Wood's user avatar
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64 votes

Why is the shape of a hanging chain not a "V"?

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 ...
knzhou's user avatar
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21 votes

What's the variation of the Christoffel symbols with respect to the metric?

$ \Gamma^{a}_{bc} = \cfrac{1}{2}g^{ad}(\partial_{b}g_{dc} + \partial_{c}g_{bd} - \partial_{d}g_{bc}) \Rightarrow $ $ δ\Gamma^{a}_{bc} = \cfrac{1}{2}δg^{ad}(\partial_{b}g_{dc} + \partial_{c}g_{bd} - \...
Noone's user avatar
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19 votes
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What's the variation of the Christoffel symbols with respect to the metric?

The difference between two connections is a tensor, so $\delta\Gamma$ is a tensor. Evaluate your variational formula in Riemannian normal coordinates at some arbitrary point $x_0$. Since the metric ...
Bence Racskó's user avatar
18 votes
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The dynamical variables in Lagrangian formalism

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 ...
J. Murray's user avatar
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15 votes
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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 ...
Bence Racskó's user avatar
14 votes

Why treat complex scalar field and its complex conjugate as two different fields?

(This post is an answer to the marked as duplicate question there : Independent fields and the Lagrange Density of Schrodinger equation) The Lagrangian Density of the Schr$\ddot{\bf o}$dinger equation ...
Frobenius's user avatar
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14 votes

Why is that in the action principle, the Taylor's series is limited to the first order?

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 ...
ZeroTheHero's user avatar
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13 votes
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When is numerical value of Lagrangian evaluated on-shell a full differential?

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{\...
AccidentalFourierTransform's user avatar
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"Any physical law which can be expressed as a variational principle describes a self-adjoint operator"

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 ...
Conifold's user avatar
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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 ...
J. Murray's user avatar
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13 votes
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Variation of metric tensor

$g_{\mu\lambda}g^{\lambda\nu}=\delta_\mu^\nu$ is constant, so its variation is zero. You can also write this equation as $g g^{-1} = 1$, where $1$ is the identity tensor. The variation of $g^{-1}$ is ...
Andrew's user avatar
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12 votes

Derivation of Euler-Lagrange equations for Lagrangian with dependence on second order derivatives

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 ...
ZeroTheHero's user avatar
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12 votes
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Is the shorthand $ \partial_{\mu} $ strictly a partial derivative in field theory?

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'} ...
Qmechanic's user avatar
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11 votes
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Why does the partial derivative with respect to $x$ of a function depending only on $\dot{x}$ vanish?

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 $x$. ...
Mason's user avatar
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11 votes
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Inverse metric in linearised gravity

Hint: Ansatz (3) is not necessary. To derive eq. (4) from eqs. (1), (2) & (5), use instead that for an infinitesimal variation $$ gg^{-1}~=~{\bf 1}\qquad\Rightarrow\qquad \delta (g^{-1})~=~- g^{-...
Qmechanic's user avatar
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11 votes

Is the shorthand $ \partial_{\mu} $ strictly a partial derivative in field theory?

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 ...
ACuriousMind's user avatar
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11 votes
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Anticommutation of variation $\delta$ and differential $d$

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 $M$...
Qmechanic's user avatar
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9 votes
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Is the Lagrangian density a functional or a function?

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]=...
AccidentalFourierTransform's user avatar
9 votes
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How to deal with explicit time dependence of the Lagrangian?

The Euler-Lagrange equations in the case of explicitly time dependent Lagrangian are the same as the ones without explicit time dependence. You are getting mixed up with what is getting varied and how ...
Futurologist's user avatar
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9 votes
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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]\...
Qmechanic's user avatar
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9 votes

About variation of Ricci tensor

This confusion can be settled with a much simpler example. Let $$ F(x) = \frac{d}{dx}f(x)\,, $$ Now $$ \delta F(x) = \frac{d}{dx}\delta f(x)\,. $$ With your reasoning one would say $$ \frac{\delta F}{\...
MannyC's user avatar
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9 votes

Confusion in derivation of Euler-Lagrange equations

If $\eta(t_i) = \eta(t_f) = 0$, $$ m \dot{x}(t_f) \eta(t_f) - m \dot{x}(t_i) \eta(t_i) = 0 \cdot (m \dot{x}(t_f) - m \dot{x}(t_i)) = 0 $$ no matter what are the values of $\dot{x}$ at the endpoints, ...
Adrien Martina's user avatar
8 votes
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Why can the bra and ket be varied independently?

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 ...
ACuriousMind's user avatar
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8 votes

Use partial or covariant derivatives when deriving equations of a field theory?

I) Assuming that the variational problem for the action $S=\int \! d^nx~{\cal L}$ is well-posed (with appropriate boundary conditions), the field-theoretic Euler-Lagrange (EL) equations read in ...
Qmechanic's user avatar
  • 202k
8 votes

Why we do calculus of variation instead of finding maxima or miniama of function?

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, $$f: \mathbf{R} \to \mathbf{C}$$ ...
valerio's user avatar
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8 votes

Inverse metric in linearised gravity

I am late, but I'll answer anyway. In linearized gravity, $g_{\mu \nu} = \eta_{\mu \nu} + h_{\mu \nu}$. We wish to find the expression for $g^{\mu \nu}$ up to linear order, which in general, must be ...
Avantgarde's user avatar
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8 votes

Is the Lagrangian density a functional or a function?

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}...
Qmechanic's user avatar
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