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An answer to the question When is the Hamiltonian of a system not equal to its total energy? is:

In an ideal, holonomic and monogenic system (the usual one in classical mechanics), Hamiltonian equals total energy when and only when both the constraint and Lagrangian are time-independent and generalized potential is absent.

(from Siyuan Ren (https://physics.stackexchange.com/users/3887/siyuan-ren), URL (version: 2011-07-06): https://physics.stackexchange.com/a/11918/ )

that seems the most exact one I've found in several similar posted questions.

I'm trying to translate these conditions to equations.

  1. Lagrangian is time-independent could be: $L(q,\dot{q},t)=L(q,\dot{q})$.
  2. constraint is time-independent: ?
  3. generalized potential is absent: ?
Qmechanic
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pasaba por aqui
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1 Answers1

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The function $h$ defined by the equation $$h=\sum_j\left( \frac{\partial \mathcal{L}}{\partial \dot{q}_j}\dot{q}_j\right)-\mathcal{L}$$ is called the energy function of the system $\mathcal{S}$.

If $\mathcal{L}=\mathcal{L}(q,\dot{q},t)$, then $\partial \mathcal{L}/\partial t\not=0$ and $h$ is not conserved.

If $\mathcal{L}=\mathcal{L}(q,\dot{q})$ then $\partial \mathcal{L}/\partial t =0$ so that $h$ is a constant. The conservation formula $$\sum_j\left( \frac{\partial \mathcal{L}}{\partial \dot{q}_j}\dot{q}_j\right)-\mathcal{L}=\mathrm{constant}$$ is called the energy integral of the system $\mathcal{S}$.

If $\mathcal{S}$ is a conservative standard system, then $\mathcal{S}$ is autonomous and so $h$ is conserved. In addition, energy integral can be written in a more familiar form, if

$$T=\sum_{j,k}a_{jk}(q)\dot{q}_j\dot{q}_k$$

and $V=V(q)$. The energy integral becomes $$h=T+V=\mathrm{constant}$$ In this case the total energy $E$ of the system.

Now All you need to do is to take Hamiltonian in place of the Energy function. $$\mathcal{H}=\sum_j\left( \frac{\partial \mathcal{L}}{\partial \dot{q}_j}\dot{q}_j\right)-\mathcal{L}=\sum_j p_j\dot{q}_j-\mathcal{L}$$

Young Kindaichi
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