# Hamiltonian = total energy? [duplicate]

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?

• . 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. :-) Oct 29, 2021 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). Oct 29, 2021 at 15:56

## 1 Answer

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.

• 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. Oct 29, 2021 at 16:31
• @basilikum In case of $\frac{\partial H}{\partial t}=0$ Hamiltonian and energy are the same thing. Oct 29, 2021 at 16:34
• ahhh thanks -> how is that? Oct 29, 2021 at 16:35
• @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. Oct 29, 2021 at 16:46