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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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1 Answers1

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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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