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1

In particular, I want to know if the fact that accelerated charges radiate (Larmor's formula) can be derived from the Hamilton's equation of the system. If the Hamilton's equation include the electric and magnetic fields as dynamical, then yes, it should be do-able... However, if you are just including the electrostatic interaction between the ...


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Kane's Method is another accepted formalism (Thomas R. Kane) which is a method for formulating equations of motion.


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I'm not familiar with "Modern Analytical Mechanics" by Pellegrini & Cooper so I can't comment on that one but I'm very familiar with the other two books you mentioned. Landau's books are generally excellent but tend to be shorter in length and sometimes very dense. Nearly every paragraph has some profound insight that you'll miss if you don't ponder ...


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If all you are looking for is a basic introduction without the calculus of variations, then the following article (which, however, assumes knowledge of elementary calculus as a prerequisite) may be of help: Hanc, Jozef, Edwin F. Taylor, and Slavomir Tuleja. "Deriving Lagrange’s equations using elementary calculus." American Journal of Physics 72.4 (2004): ...


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Maybe you can give a rough idea about what the subject is about. You can introduce first for example the Fermat's principle of least time and maybe kind of make an analogy like. "There is a similar principle of minimization in mechanics where you minimize another quantity called action". Maybe if an student is interested you can give him more information. ...


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The most fundamental parts of Lagrangian mechanics involve calculus. The action principle involves an integral and the Euler-Lagrange equation is a partial differential equation. Unless the students are pretty good with calculus it will be quite hard to teach.


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I) Disclaimer: In this answer we will use the (traditional) physicist's definition of tensors using indices and their transformation properties under coordinate transformations. Moreover, let us suppress time dependence $t$ for simplicity. II) Let the manifold $Q$ be the configuration space. The Lagrangian $L:TQ\to \mathbb{R}$ transforms as a scalar ...


2

Gauge theories become constrained Hamiltonian systems when passing from the Lagrangian $L(q,\dot{q},t)$ to the Hamiltonian $H(q,p,t)$ where $p = \frac{\partial L}{\partial \dot{q}}$. Generically, you get a constrained Hamiltonian system whenever the matrix/operator with components $$ \frac{\partial^2 L}{\partial \dot{q}^i\partial\dot{q}^j}$$ is singular, ...


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In a rather general approach you can consider the Poisson bracket ${g,f}$ as expressing the rate of change of $g$ as a consequence of a flow induced by $f$. As mentioned by AngusTheMan in the comments, you get the time variation of $g$ if $f=H$ (assuming that quantities are not explicitly time-dependent). Here $g$ and $f$ are any (smooth) functions on the ...


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Regarding question 2, in Weinberg's The Quantum Theory of Fields page 295, the author defines the functional derivative of an arbitrary bosonic functional $F[q,p]$: $$\frac{\delta F[q,p]}{\delta q}=i[p,F[q,p]]$$ $$\frac{\delta F[q,p]}{\delta p}=i[F[q,p],q]$$ where $q,p$ are a pair of operators that satisfy canonical conmutation relations. If that is indeed ...


2

If e.g. we consider a 1D non-relativistic free particle with kinetic energy $$\tag{1} T(q,v,t)~=~\frac{1}{2}mv^2$$ the information that the velocity is a constant$^1$ $$\tag{2} v~\approx~\text{constant},$$ only came afterwards from solving the Lagrange eqs. $$\tag{3} \left(\frac{d}{dt}\frac{\partial T}{\partial v}-\frac{\partial T}{\partial ...


4

Comment to the question (v4): Classically, the Lagrangian for a fermion system reads $$ L ~=~ \int\! d^3x~ i\psi^{\dagger}\dot{\psi}-H.\tag{A}$$ The Legendre transformation from the Lagrangian to the Hamiltonian formalism is tricky for at least three reasons: The traditional Dirac-Bergmann analysis leads to constraints. See e.g. my Phys.SE answers here ...


1

Yes. The action functional $S[q]=\int \mathcal{L} dt$ is something that can be applied to ANY [differentiable] path. So $\mathcal{L}$ is to be understood as an object which can really take n any value of $q$ and $\dot{q}$ independently. This makes it so that $\mathcal{L}$ really is a function of two variables, and so that it really does have to be defined ...


3

If $Q$ is configuration space, then the Lagrangian is a function $L: TQ\times \mathbb{R}\to \mathbb{R}$. Let the cotangent bundle $M:=T^{\ast}Q$ be the corresponding phase space. The Hamiltonian/phase space Lagrangian is a function $L_H: TM\times \mathbb{R}\to \mathbb{R}$.


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Let $X$ be the phase space. Then $L_\text{ph}(q,p,\dot{q},\dot{p},t)$ is a function on $TX\times \mathbb{R}$1, since the coordinates of $TX\times\mathbb{R}$ are precisely the coordinates of $X$, i.e. $(q,p)$ and their derivatives $(\dot{q},\dot{p})$ (and time $t$). If Hamilton's equations are fulfilled, there are relations among $q,\dot{q},p,\dot{p}$ (the ...


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OP wrote (v3): Is there anything in particular I should be careful of? Yes. Watch out for secondary constraints, cf. e.g. this Phys.SE post. Below follows a brief partial derivation. Let Greek letters $\mu,\nu,\ldots$ denote spacetime indices, while Roman letters $i,j,\ldots$ denote only spatial indices. The Lagrangian density $$ {\cal L}~=~ ...


1

I did not get my copy of Srednicki out but from what you have written... Srednicki is referencing the method of steepest descent. Although these notes look to be better than wikipedia. Another page that is directly applicable to the quantum field theory case is here. In short, exponential integrals may be estimated by the saddle points of the integrand. ...


1

Comments to the question (v1): Let there be given an $n$-dimensional manifold $M$ with a smooth vector field $X\in \Gamma(TM)$. If $(x^1, \ldots, x^n)$ is some local coordinates on $M$, then the vector field takes the form $$\tag{A} X~=~X^i(x)\frac{\partial}{\partial x^i},$$ and one may study the autonomous first-order ODE $$\tag{B} ...


0

To obtain the result $\frac{\text d \rho }{\text d t}=0$ you need two facts: the first is that the hamiltonian flow preserves the volume of phase space. The second fact is the conservation of probability, that is, the probability that the system is found in a volume $U$ at time $t=0$ equals the probability of finding it within $\Phi _t U$ at time $t$, where ...


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You can think about it as in the case of a simple fluid. You need to enforce conservation of probability (in the same way you could enforce mass conservation in fluid dynamics for instance or charge conservation in electrodynamics). As far as I understand it, probability conservation is an additional premiss to the conservation of phase space volume under ...


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The phase-space density $\varrho(p,q)$ tells us how many dynamical possible trajectories (DPT) pass through a given unit volume of phase space. Therefore it is a measure for the probability of finding a system in state $(p,q)$ since if there are more DPTs, it is more likely to find a system in this state (if we assume that every DPT is equally probable, that ...


2

Comments to the question (v2): On one hand, let there be given a configuration space $(Q,g)$ endowed with a metric $g$. (As ACuriousMind points out in a comment, there is a 1-1 correspondence between a metric $g$ and the kinetic term in a Lagrangian.) On the other hand, note that the canonical symplectic 2-form $\omega$ on the cotangent bundle $T^{\ast}Q$ ...


1

No, but you are most likely to get one from the kinetic term of the Lagrangian itself. In most cases one requires it to be a convex function in the $\dot q$ variables. You then get a metric if such kinetic term is quadratic in $\dot q$ (and of course sensible kinetic energy is positive-definite). The metric and symplectic structures on a manifold are ...


1

The Hamilton's eqs. $$\dot{z}^I(t)~\approx~ \{z^I(t), \mathbb{H}\}_{PB},\tag{1}$$ for non-local Hamiltonian theories have the same form as for local theories. Here $\mathbb{H}$ is a (possibly non-local) Hamiltonian functional, cf. my Phys.SE answer here. It follows that the symplectic integrator program carries over essentially un-modified, if the ...


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In this answer we apply the general non-local theory developed in my Phys.SE answer here to OP's non-local example. Let us for simplicity assume that time belongs to the unit interval $[t_i,t_f]=[0,1]$. OP's non-local Lagrangian action functional reads (modulo some sign conventions$^1$) $$ \left. S[q,v]\right|_{v=\dot{q}}, \tag{A} $$ where $$ ...


0

I've found the inconsistency. What follows is a derivation of the Hamiltonian for convolutional Lagrangians. Starting with the total variation of a convolutional Lagrangian, we have: $$\delta \mathbb{L}=\int^t_0 \left(\frac{\delta \mathbb{L}}{\delta \dot{q}}\delta\dot{q}(t-\tau)+\frac{\delta \mathbb{L}}{\delta q}\delta q(t-\tau)\right)\,\text{d}\tau ...


4

Let here consider point mechanics (as opposed to field theory) for simplicity, i.e. the generalization to field theory is left as an exercise. I) Bad news. If the Lagrangian action functional $S[q]$ is non-local, the usual definition of Lagrangian momenta as a partial derivative $$p_i~:=~\frac{\partial L}{\partial v^i} \tag{1}$$ does not apply. II) Good ...



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