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25

I reproduce a blog post I wrote some time ago: We tend to not use higher derivative theories. It turns out that there is a very good reason for this, but that reason is rarely discussed in textbooks. We will take, for concreteness, $L\left(q,\dot q, \ddot q\right)$, a Lagrangian which depends on the 2nd derivative in an essential manner. ...

19

Could someone please convince me that there is something natural about the choice of the Lagrangian formulation... If I ask a high school physics student, "I am swinging a ball on a string around my head in a circle. The string is cut. Which way does the ball go?", they will probably tell me that the ball goes straight out - along the direction the ...

17

The Hamiltonian H and Lagrangian L which are rather abstract constructions in classical mechanics get a very simple interpretation in relativistic quantum mechanics. Both are proportional to the number of phase changes per unit of time. The Hamiltonian runs over the time axis (the vertical axis in the drawing) while the Lagrangian runs over the trajectory of ...

16

The notes from week 1 of John Baez's course in Lagrangian mechanics (http://math.ucr.edu/home/baez/classical/#lagrangian) give some insight into the motivations for action principles. The idea is that least action might be considered an extension of the principle of virtual work. When an object is in equilibrium, it takes zero work to make an arbitrary ...

11

There is also Feynman's approach, i.e. least action is true classically just because it is true quantum mechanically, and classical physics is best considered as an approximation to the underlying quantum approach. See http://www.worldscibooks.com/physics/5852.html or http://www.eftaylor.com/pub/call_action.html . Basically, the whole thing is ...

10

In physics, it is often implicitly assumed that the Lagrangian $L=L(q^i,v^i,t)$ depends smoothly on the (generalized) positions $q^i$, velocities $v^i$, and time $t$, i.e. that the Lagrangian $L$ is a differentiable function. Let us now assume that the Lagrangian is of the form $$L~=~\ell(v^2),\qquad\qquad v~:=~|\vec{v}|,\qquad\qquad(1)$$ where $\ell$ is ...

10

Dear Ondřeji, a good question but a part of the answer is that your equation for the fluid is underdetermined. It treats $p,\rho$ as independent variables. But the physical system only knows how to behave if you also substitute some equation of state, i.e. a function $p=p(\rho)$ or $p=p(\rho,\vec v)$. Note that your Ansatz for the stress-energy tensor ...

10

Yes, the invariance of the action follows from special relativity – and special relativity is right (not only) because it is experimentally verified. All the equations of motion may be derived from the condition $\delta S = 0$, the action is stationary (which usually means it has the minimum value on the allowed trajectory/history among all ...

9

We are considering a transformation, which may transform the field variables $\phi^{\alpha}(x)$ and which may transform the space-time points $x^{\mu}$. The transformation in turn apply to The action $S$. The Euler-Lagrange equations = the equations of motion (EOM). A solution of EOM. If any of the items 1-3 are invariant under the transformation, we ...

9

Quantum systems are essentially defined by their symmetries. For example, in QFT's you expect all terms not forbidden by the symmetries of the problem to appear in the Lagrangian, with irrelevant operators suppressed by large scales, etc. So I think your first step in this approach would be to write down the most general 2D QFT respecting the 2D Diff and ...

8

As that lovely article linked by dfan says the virial theorem comes from varying the action $S[x]$ by $x\rightarrow(1+\epsilon)x$ $$\frac{1}{T}\delta S = \frac{1}{T}\epsilon\int_{0}^{T} dt\{m\dot{x}^2 -x\frac{\partial V}{\partial x}\}$$ This is a variation of the action and therefore must vanish up to some boundary terms if $x$ is a solution of the ...

7

As you can see from the image below, you want the variation of the action integral to be a minimum, therefore $\displaystyle \frac{\delta S}{\delta q}$ must be $0$. Otherwise, you are not taking the true path between $q_{t_{1}}$ and $q_{t_{2}}$ but a slightly longer path. However, even following $\delta S=0$, as you know, you might end up with another ...

7

Excellent question, and one that I've never really found a completely satisfactory answer for. But consider this: in elementary classical mechanics, one of the fundamental laws is Newton's second law, $\mathbf{F} = m\mathbf{a}$, which relates the force on an object to the object's acceleration. Now, most forces are exerted by one particular object on another ...

7

We vary the action $$\delta \int {Ldt} = \delta \int {\int {\Lambda \left( {A_\nu ,\partial _\mu A_\nu } \right)d^3 xdt = 0} }$$ ${\Lambda \left( {A_\nu ,\partial _\mu A_\nu } \right)}$ is the density of lagrangian of the system. So: $$\int {\int {\left( {\frac{{\partial \Lambda }}{{\partial A_\nu }}\delta A_\nu + \frac{{\partial \Lambda ... 7 OP wrote: As far as I can tell, from here it's a matter of playing around until you get a Lagrangian that produces the equations of motion you want. Too often, as a student, one is only shown how to derive Newton's 2nd law from Euler-Lagrange equations by postulating some particular Lagrangian L. If one believes that Newton's laws are more natural ... 7 The action functional and Hamilton's principal function are two different mathematical objects related to the same physical quantity. The action along a trajectory \gamma:[t_1,t_2]\rightarrow Q is given by$$ S[\gamma] = \int_{t_1}^{t_2}L(\gamma(t'),\dot\gamma(t'),t')dt' $$whereas the pricipal function is the solution of the Hamilton-Jacobi equation$$ ...

7

Building an action: If you know the field content (which I assume means you know the gauge group and reps of all the fields) then: Write down every term that is Lorentz scalar (so combinations like $\partial_\mu A^\mu$, $\bar{\psi}\gamma^\mu \partial_\mu\psi$ allowed but not things like $\vec{n}\cdot\nabla \phi$ where $\vec{n}$ is some random 3-vector). ...

7

First I want to remind you what is going on behind the scenes. You know where the particle is at some initial time $t_1$, and you know where the particle is at some final time $t_2$, and the question you are asking is, which path will get me from the initial position at the initial time to the final position at the final time in a way that minimizes the ...

6

I) At least three different quantities in physics are customary called an action and denoted with the letter $S$. The (off-shell) action $$\tag{1}S[q]~:=~ \int_{t_i}^{t_f}\! dt \ L(q(t),\dot{q}(t),t)$$ is a functional of the full position curve/path $q^i:[t_i,t_f] \to \mathbb{R}$ for all times $t$ in the interval $[t_i,t_f]$. See also this question. ...

6

I) Here we will assume that we ultimately want to consider the full quantum theory, usually written in terms of a gauge-fixed path integral $$Z~=~\int \!{\cal D}\phi~ \exp\left(\frac{i}{\hbar}S_{\rm gf}[\phi]\right)$$ rather than just the classical action and the corresponding classical equations of motion (with or without gauge-fixing terms). If the ...

6

The symmetry is required to leave the full action $A= \int L dt$ invariant. As can be seen the action is invariant because you get the missing $(1+\epsilon)^2$ factor from the measure $dt$. Now, you can apply the Noether theorem to find the conserved charge which is in this case: $Q = 2 E t - Px$ with $E = T+V = \frac{1}{2}m\dot{x}^2 + ... 6 Here's what I perceive to be a mathematically and logically precise presentation of the theorem, let me know if this helps. Mathematical Preliminaries First let me introduce some precise notation so that we don't encounter any issues with "infinitesimals" etc. Given a field$\phi$, let$\hat\phi(\alpha, x)$denote a smooth one-parameter family of fields ... 6 When you get a nonsensical equation of motion like this, it means that the action has no local extrema over the space of all possible paths. (Like how the curve$y = x$has no local minimum or maximum.) There's no path$x(t)$such that varying the path produces only a second-order (or higher) variation in the action. In fact, you can see this explicitly for ... 6 Let's compare classical mechanics and GR to attempt to get at the intuition you're looking for. Classical mechanics. Recall that in the classical mechanics of a system of$N$particles, the configuration of the system at every point in time is represented by a point$x\in\mathbb R^{3N}$. The configuration manifold$\mathcal Q$, namely the set of possible ... 5 1) Firstly, the Lagrangian$L(q(t),v(t),t)$at some time$t$is a function of: the instantaneous position$q(t)$at the time$t$; the instantaneous velocity$v(t)$at the time$t$; and the time$t$(also known as explicit time-dependence). 2) Secondly, the (off-shell) action $$\tag{1} S[q]~:=~ \left. \int_{t_i}^{t_f}\! dt \ ... 5 Dear amc, first, write your Lagrangian density as$$ L = -\frac{1}{4} F_{\mu\nu}F^{\mu\nu} = -\frac{1}{2} (\partial_\mu A_\nu) F^{\mu\nu}$$Is that fine so far? The$F_{\mu\nu}$contains two terms that make it antisymmetric in the two indices. However, it's multiplied by another$F^{\mu\nu}$that is already antisymmetric, so I don't need to antisymmetrize ... 5 Answering the third question, in any mature branch of mathematics or physics, there are always equivalent or near-equivalent formulations of the same structure (it's pretty much a definition of maturity, that some of the relations of a new topic to older topics have been worked out). The Lagrangian can be regarded as a type of functional anti-derivative of a ... 5 The dimensions of the Planck constant$\hbar$, the action$S$, and the angular momentum, are constrained by the following important facts: A conjugated pair of two observables is quantum mechanically related to the Planck constant$\hbar$via a Heisenberg uncertainty relation. A conjugated pair of two variables is classically related to the action ... 5 Short answer: Lagrangian mechanics only applies to a subset of classical mechanics problems, but when it does, it is mathematically equivalent to Newtonian mechanics (which I take to mean direct application of$\vec{F} = m\vec{a}\$). More detailed rambling: Your class and whatever text you're using ought to cover the equivalence of the Newtonian and ...

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