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From here, how do we say that probability distribution function is constant as we flow in the phase-space? More accurately, value of probability distribution function $f_t(p,q)$ at representative point $p^*(t),q^*(t)$ moving along any Hamiltonian trajectory in phase space is constant in time. The function itself generally changes in time. This value is ...


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Comments to the question (v2): First of all, let us stress that OP is correct, that a given set of equations of motion does not necessarily have a variational/action principle, cf. e.g. this Phys.SE post and links therein. On one hand, if there exists a Lagrangian formulation, then one may in principle obtain a Hamiltonian formulation via a (possible ...


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The field equations must be conservative in a fairly precise sense in order that this can be done in a physically appropriate sense. Then there are several Hamiltonian approaches to field theory: the De Donder-Weyl formalism and the multisymplectic formalism. Although both formalisms can accommodate Lagrangians, the can also be understood without any ...


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The Hamiltonians $H_S$ and $H_R$ both implicitly depend on their respective volumes (or confining potential strength). To allow volume exchange between the two systems, you simply impose the constraint $V_R = V_{tot}-V_S$. The joint Hamiltonian is always given by $H_S+H_R$. You can check that in mechanical equilibrium, $\partial_{V_S} ...


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It implies that total energy is conserved along the paths specified by the trajectories, except for whatever explicit changes are written in the Hamiltonian. It is not sufficient to get the rest of the physics because there may be many energy-conserving paths. For example, for the free-particle Hamiltonian, the energy is the kinetic energy and the predicted ...


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$$\frac{d\mathcal{H}}{dt}= \dot{q_i}\frac{\partial\mathcal{H}}{\partial q_i}+\dot{p_i}\frac{\partial\mathcal{H}}{\partial p_i}+\frac{\partial\mathcal{H}}{\partial t}$$ And physically this is related to how the Hamiltonian experienced, $\mathcal{H}(t)=\mathcal{H}(q_i(t),p_i(t),t),$ changes as you move around in phase space, compared to how the Hamiltonian ...


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For $\mathcal{H}(q,p,t)$ to have in any meaningful sense a partial derivative $\partial_t$ that is different from the total time derivative, you have to consider it applied to a curve $(q(t),p(t))$ in phase space. The equation $$ \frac{\mathrm{d}\mathcal{H}}{\mathrm{d}t} = \partial_t \mathcal{H}\tag{1}$$ then says that $\mathcal{H}$ is "constant" along the ...


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Assuming the existence of a smooth function $\mathcal{H}(q_i,p_i)$ in $(q_i(t), \,p_i(t))$ phase space, such that it obeys the following (taken as a postulate): \begin{equation} \frac{d\mathcal{H}}{dt}=0 \end{equation} Therefore: \begin{equation} \dot{q_i}\frac{\partial\mathcal{H}}{\partial q_i}+\dot{p_i}\frac{\partial\mathcal{H}}{\partial ...


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Thanks guys, i followed your seggestion. Using $\dot{z} = const. =\alpha ' $ and $\dot{\phi}² = (\frac{\beta - a\rho²}{m\rho²})²$. Now applying on $H$ we have $\frac{d\rho}{dt} = \sqrt{\frac{H}{m} - \alpha'² - \rho²(\frac{\beta -a\rho²}{m\rho²})²}$ then $t = \int\frac{d\rho}{\sqrt{\alpha - \frac{(\beta - a\rho²)²}{m²\rho²}}}$ where $\alpha = \frac{H}{m} - ...


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I think you have a logical contradiction in your question. When H is independent of time, then H=E. Please refer to this previous answer. When is the Hamiltonian of a system not equal to its total energy?


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In my opinion, the Mikael Fremling's comment is precisely it. The trick is not just to "relabel" lagrangian variables in terms of $p$. The meaning is "to change the formalism", not just "substitute variables". I agree that the quoted book formulation might be a bit confusing, but seen from this point of view it does make sense.


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When you say that $L_x$ and $L_y$ vanish for a point confined to move in the plane $z=0$, you mean that the the solution $\vec{x}=\vec{x}(t)$, $\vec{p}=\vec{p}(t)$ describes a curve in the given plane with tangent vector parallel to that plane. So that, exactly along that curve, $$L_x(\vec{x}(t),\vec{p}(t))= L_y(\vec{x}(t),\vec{p}(t))=0\quad \forall t \in ...


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Refs. 1 and 2 only discuss a generic/bulk notion of a constant of motion in phase space; not any refinement thereof restricted to a subsurface of phase space. Let us here carefully explain the generic/bulk notion of a constant of motion used in Refs. 1 and 2. Definition 1: An on-shell constant of motion $F(q,p,t)$ satisfies $$\tag{1} ...


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The lapse function is not defined by the metric alone, but instead depends on both the metric $g_{ab}$ and its slicing into timelike hypersurfaces. One way to "slice" a spacetime $\mathcal{M}$ into timelike hypersurfaces is to define a timelike coordinate $f$, which is just a function $f: \mathcal{M} \to \mathbb{R}$ such that $\nabla_a f$ is a timelike ...


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For an Hamiltonian $ H $, given by \begin{equation} H(q,p) = T(q,p) + U(q), \end{equation} the Hamilton eqs. read \begin{align} \dot{q} &= \frac{\partial{H}}{\partial{p}}=\frac{\partial{T}}{\partial{p}},\\ \dot{p} &= -\frac{\partial{H}}{\partial{q}}=-\frac{\partial{U}}{\partial{p}}. \end{align} and remembering that the kinetic term written with ...


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As for your difficulty, take the kinetic energy term to be $$ K = \dfrac{p^2}{2m}$$ and you will get the correct answer. However there are a few more issues with your derivation like you wrote $$F= p/t $$ etc. Please resolve them.


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Two reasons A generating function by definition generates canonical transformations. If you take terms like $F_{5}(p,q,t)$ then it doesn't relate coordinates involving two separate canonical transformations. The area preserving property of canonical transformations imply the existence of generating function. Then there is no way you can construct ...


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Let us suppress $t$-dependence in this answer for simplicity. A canonical transformation$^1$ $$(q,p)\quad\longrightarrow\quad (Q(q,p),P(q,p))$$ can be viewed as a graph, thereby yielding a $2n$-dimensional submanifold $M$ embedded inside a $4n$-dimensional total manifold $N$. The total manifold $N$ has $4n$ local coordinates $(q,p,Q,P)$. The submanifold $M$ ...


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I think it's due to the fact that your transformation equations between $x_{\alpha ,i}$ and $q_{j}$ do explicitly contain the time. $$ x = Rsin(\phi )cos(\theta )=Rsin(\phi )cos(\omega t) $$ $$ y = Rsin(\phi )sin(\theta )=Rsin(\phi )sin(\omega t) $$ $$ z = Rcos(\phi ) $$ And, according to the theorem mentioned by Oiale, your kinetic energy does not present ...



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