# Tag Info

44

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. ...

32

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 ...

29

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 ...

24

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 ...

23

The action $S$ is not a well-known object for the laymen; however, when one seriously works as a physicist, it becomes as important and natural as the energy $H$. So the action is probably unintuitive for the inexperienced users - and there's no reason to hide it - but it is important for professional physicists, especially in particle and theoretical ...

19

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 ...

18

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 ...

18

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 $S_V[\phi]=\int_V \! d^nx~{\cal L}$. The Euler-Lagrange equations = the equations of motion (EOM). A solution $\phi$ of EOM. If any of the items 1-3 are ...

15

We vary the action $$\delta \int {L\;\mathrm{d}t} = \delta \int {\int {\Lambda \left( {A_\nu ,\partial _\mu A_\nu } \right)\mathrm{d}^3 x\;\mathrm{d}t = 0} }$$ ${\Lambda \left( {A_\nu ,\partial _\mu A_\nu } \right)}$ is the density of lagrangian of the system. So, $$\int {\int {\left( {\frac{{\partial \Lambda }}{{\partial A_\nu }}\delta A_\nu + ... 15 Dear amc, first, write your Lagrangian density as$$ L = -\frac{1}{4} F_{\mu\nu}F^{\mu\nu} = -\frac{1}{2} (\partial_\mu A_\nu) F^{\mu\nu} $$Is that fine so far? The F_{\mu\nu} contains two terms that make it antisymmetric in the two indices. However, it's multiplied by another F^{\mu\nu} that is already antisymmetric, so I don't need to antisymmetrize ... 15 General approach First recall that Euler-Lagrange equations are conditions for the vanishing of the variation of action S. For a scalar field \Phi with Lagrangian density \mathcal L on some open subset U we have$$S[\Phi] = \int_U {\mathcal L}(\Phi(x), \partial^{\mu}\Phi(x)) {\rm d}^4 x$$Consider a variation of the field in direction \chi and ... 14 There is also Feynman's approach, i.e. least action is true classically just because it is true quantum mechanically, and classical physics is best considered as an approximation to the underlying quantum approach. See http://www.worldscibooks.com/physics/5852.html or http://www.eftaylor.com/pub/call_action.html . Basically, the whole thing is ... 13 As that lovely article linked by dfan says the virial theorem comes from varying the action S[x] by x\rightarrow(1+\epsilon)x$$\frac{1}{T}\delta S = \frac{1}{T}\epsilon\int_{0}^{T} dt\{m\dot{x}^2 -x\frac{\partial V}{\partial x}\}$$This is a variation of the action and therefore must vanish up to some boundary terms if x is a solution of the ... 12 Well, you are almost there. Use the fact that$$ {\partial (\partial_{\mu} A_{\nu}) \over \partial(\partial_{\rho} A_{\sigma})} = \delta_{\mu}^{\rho} \delta_{\nu}^{\sigma}$$which is valid because \partial_{\mu} A_{\nu} are d^2 independent components. 12 The action functional and Hamilton's principal function are two different mathematical objects related to the same physical quantity. The action along a trajectory \gamma:[t_1,t_2]\rightarrow Q is given by$$ S[\gamma] = \int_{t_1}^{t_2}L(\gamma(t'),\dot\gamma(t'),t')dt' $$whereas the principal function is the solution of the Hamilton-Jacobi equation$$ ...

12

I) Initial value problems and boundary value problems are two different classes of questions that we can ask about Nature. Example: To be concrete: an initial value problem could be to ask about the classical trajectory of a particle if the initial position $q_i$ and the initial velocity $v_i$ are given, while a boundary value problem could be to ask ...

11

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 ...

11

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 ...

11

Setting the einbein to $1$ corresponds to a diffeomorphism of the metric, as the einbein is given by $e_{\tau\tau}=\sqrt{g_{\tau\tau}}$, which can be easily deduced from the fact that a the vielbein is given as the transformation coefficients from the coordinate basis to a non-coordinate basis. Hence, the dimensionality of the einbein depends on that of the ...

10

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. ...

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 ...

10

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 ...

9

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 ...

9

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). ...

8

As a layman, I can't offer too much, but I can offer how I think about action. $$S=\int L \;\mathrm{d}t$$ The key distinction in classical mechanics between the Hamiltonian and the Lagrangian is that Hamiltonian (H) is the sum of kinetic energy (T) and potential energy (V), where as the Lagrangian (L) is the difference: $$H=T+V$$ $$L=T-V$$ In the case ...

8

I) Not all equations of motion (eom) are variational. A famous example is the self-dual five-form in type IIB superstring theory. In classical point mechanics, frictional forces typically lead to non-variational problems. II) Consider for instance $n$ variable $q^i$ and $n$ eoms, $$\tag{1} E_i~\approx~ 0, \qquad i~\in~\{1, \ldots, n\}.$$ A simplified ...

8

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 ...

8

Some time after Newton described the laws of nature in terms of an instantaneous relationship, others noticed that the history, rather than the instantaneous state, of a system could, in at least one case, be described by saying that it obeyed a certain relationship: a particular function describing the history must always be the one that (a) starts and ends ...

8

The term vanished because we can translate this term to one making a statement about the fields at the boundary and assume that the fields themselves vanish in spatial and temporal infinity. By Stokes' Theorem, we can translate volume integrals into surface integrals. More specifically Gauss' Theorem states that the integral of a divergence of a field over ...

Only top voted, non community-wiki answers of a minimum length are eligible