# Tag Info

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Lagranges equations are: ${\partial L \over \partial q_i} = {d \over dt }{ \partial L \over \partial q_i'}$ where $q_i'={dq_i \over dt}$ You can find constants of motion using lagranges equations and Hamiltons equations. You already know that the Hamiltonian is conserved-time is not explicit. (The energy is not always equal to the Hamiltonian) You can ...

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I think you made an mistake in the algebra. Let's focus on the PDE $$p \frac{\partial \rho }{\partial q} - q \frac{\partial \rho }{\partial p} = 0.$$ You claim that $\rho(p,q) = \exp^{\, f(p+q)}$ is a solution to this equation for any function $f$, but that's simply not true. If you plug that Ansatz in the PDE, you get $$(p-q)\; f'(p+q) = 0$$ which is ...

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This is a partial answer which I will hopefully come back to and expand. The property of being its own Legendre transform is unique to the pure quadratic kinetic energy $T(v)=\frac12 mv^2$. As a simple example, consider $T(v)=\frac14Av^4$. Here the Legendre momentum is $$p=\frac{\partial L}{\partial v}=\frac{\partial T}{\partial v}=Av^3,$$ so the velocity ...

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It isn't "obtained". It's a definition of the function $g$ and this definition is useful because it leads to the nice symmetric relationships on the rest of the page 3. They exchange the two Legendre-dual variables. So the definition of $g$ wasn't really "derived" in any straightforward way. It was a clever guess that Legendre made at some point of his life....

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Hamiltonian gauge symmetries usually come up in the context of lattice gauge theory, in which the system is defined on a discrete lattice. Such a Hamiltonian is defined to have a gauge symmetry if it commutes with some extensively-scaling set of local (i.e. finitely-supported) unitary operators $\{ U_i = e^{-i Q_i} \}$. Operators that commute with the ...

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There is no mistake. The Hamiltonian satisfies the relation $$\frac{\text{d}H}{\text{d}t} = \frac{\partial H}{\partial t}.$$ This follows immediately from Hamilton's equations: $$\frac{\text{d}H}{\text{d}t} = \dot{H} = \frac{\partial H}{\partial q}\dot{q} + \frac{\partial H}{\partial p}\dot{p} + \frac{\partial H}{\partial t} = \frac{\partial H}{\partial q}... 2 The naive thing to do would be to just replace x by \hat{x} and p by -i\hbar \nabla. Note however that (\hat{x}\hat{p})^{\dagger} = \hat{p}\hat{x} \neq \hat{x}\hat{p}, i.e. the operator is not Hermitian due to the noncommutativity of \hat{x} and \hat{p}, which is a problem of course. So what can we do to fix this? One possible way is to ... 2 I believe the term "stochastic web" was first employed in this article by Zaslavsky, where he studied not the quasi-integrable stochasticity you're talking about, but stochastic processes in 1-dimensional systems. The reason is that most people can't see dimensions higher than three, and he was particularly interested in the fractal structure of chaos (... 6 There is also the routhian formalism of mechanics which is described as being a hybrid of lagrangian and hamiltonian mechanics. The routhian is defined as$$R = \sum_{i=1}^n p_i\dot{q}_i - L$$You can learn more about it by clicking this link for wikipedia's description of it. Reading more in regards to the routhian because I was bored, I realized it is ... 3 It's worth pointing out that the Hamiltonian and Lagrangian formalisms are independent, even though they're usually taught as if the former were a filtering of the latter (here enter Legendre transforms). Both formalisms are as independent as the notions of tangent and cotangent bundles in differential geometry: independent, but intrinsically connected. ... 2 The first part of OP's construction is directly related to the covariant Hamiltonian formalism for a real scalar field with Lagrangian density$$ {\cal L} ~=~ \frac{1}{2}\partial_{\alpha} \phi ~\partial^{\alpha} \phi -{\cal V}(\phi), \tag{CW4} $$see e.g. Ref. [CW] and this Phys.SE post. See also the Wronskian method in this Phys.SE post. [In this answer we ... 2 Often X is a coadjoint orbit of a Lie group. These have a natural symplectic structure; see https://en.wikipedia.org/wiki/Symplectic_reduction 2 That the Hamiltonian is zero is completely correct. The system is time-reparametrization invariant - changing \tau to \xi(\tau) transforms$$ n(q(\tau))\mapsto n(q(\xi)),\quad \dot{q}\mapsto \frac{\mathrm{d}\xi}{\mathrm{d}\tau}q', \quad \mathrm{d}\tau\mapsto \frac{\mathrm{d}\tau}{\mathrm{d}\xi}\mathrm{d}\xi and the action is invariant under this ...

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