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How do I figure out if the energy in a Hamiltonian is conserved or not? I have found the conditions for $H=E$ in Goldstein's Analytical Mechanics that the equations defining the generalized coordinates mustn't depend on t explicitly and that the forces have to be derivable from a conservative potential $V$. And further that H is conserved if the time-derivative is 0. However, I'm working a problem where I only know the Hamiltonian (and not the Lagrangian):

$$H(p,q) = \frac{p^2}{2m}*q^4+\frac{1}{2}*k*\frac{1}{q^2}.$$

I know that $p$ and $q$ are canonically conjugated and that $m$ is mass and $k$ is a constant. However, I don't know how I should verify whether or not this is the total energy?

Qmechanic
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basilikum
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    . I am asking how to determine wheter the energy is conserved, but per Goldstein et al (2014) p. 339 the Hamiltonian is only the total energy when the specified conditions are met. :-) – basilikum Oct 29 '21 at 15:47
  • Interesting! Then I wonder how Goldstein et al are defining "total energy." The question When is the Hamiltonian of a system not equal to its total energy? is related to this, but I didn't find a definition of "total energy" there, either, unless it's defined as "T+V" where T is kinetic and V is potential energy, but that requires having definitions of kinetic and potential energy... Anyway, I retract my request-for-clarification comment (deleted). – Chiral Anomaly Oct 29 '21 at 15:56

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For any physical observable $A(q,p,t)$ you can determine its total time derivate by calculus: $$\dot{A} =\sum_i\frac{\partial A}{\partial q_i}\dot{q_i} +\sum_i\frac{\partial A}{\partial p_i}\dot{p_i} +\frac{\partial A}{\partial t}$$

You can do this also for the Hamiltonian itself: $$\dot{H} =\sum_i\frac{\partial H}{\partial q_i}\dot{q_i} +\sum_i\frac{\partial H}{\partial p_i}\dot{p_i} +\frac{\partial H}{\partial t}$$

By using Hamilton's equations of motion ($\dot{q}_i=\frac{\partial H}{\partial p_i}$ and $\dot{p}_i=-\frac{\partial H}{\partial q_i}$) here you get $$\dot{H} =\sum_i\frac{\partial H}{\partial q_i}\frac{\partial H}{\partial p_i} -\sum_i\frac{\partial H}{\partial p_i}\frac{\partial H}{\partial q_i} +\frac{\partial H}{\partial t}$$

The first and the second sum cancel each other. So you are left with: $$\dot{H}=\frac{\partial H}{\partial t}$$

Because your special Hamiltonian only depends on $q$ and $p$, but not on $t$ explicitly you have $$\dot{H}=0,$$ meaning that $H$ is conserved.

Thomas Fritsch
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  • Thank you, but I should maybe have formulated myself clearer ;-) I am looking for a way to determine whether or not the energy is conserved/in this case if H=E. – basilikum Oct 29 '21 at 16:31
  • @basilikum In case of $\frac{\partial H}{\partial t}=0$ Hamiltonian and energy are the same thing. – Thomas Fritsch Oct 29 '21 at 16:34
  • ahhh thanks -> how is that? – basilikum Oct 29 '21 at 16:35
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    @basilikum Simply said, it is by definition. Or saying more sophisticated with Noether's theorem: Energy is the observable which is conserved because of the time-invariance of a system. – Thomas Fritsch Oct 29 '21 at 16:46