5
$\begingroup$

If Lagrangian of the motion is

$$\mathcal{L}=\frac{1}{2}m\left(a^2\dot\phi^2+a^2\dot\theta^2\sin^2\phi\right)+mga\cos\phi,$$

how can I show that total mechanical energy is conserved? I've read this:

If the time $t$, does not appear [explicitly] in Lagrangian $\mathcal{L}$, then the Hamiltonian $\mathcal{H}$ is conserved. This is the energy conservation unless the potential energy depends on velocity.

Potential energy of this motion doesn't depend on velocity. Also, $t$ does not appear explicitly in Lagrangian. Is this enough to say that total mechanical energy is conserved?

$\endgroup$
3
  • $\begingroup$ Comment to the question (v2): Please provide reference of quote. The reference is probably talking about the Lagrangian energy function $h(q,\dot{q},t):=\dot{q}^i\frac{\partial L}{\partial \dot{q}^i}-L$ rather than the Hamiltonian $H(q,p,t)$. $\endgroup$
    – Qmechanic
    Jul 31, 2013 at 19:26
  • $\begingroup$ In my case, generalized coordinates are $q_1=\theta$ and $q_2=\phi$. Kinetic energy is $T=\frac{1}{2}m(a^2\dot\phi^2+a^2\dot\theta^2\sin^2\phi)$ Potential energy is $V=-mga\cos\phi=V(q_2)$ Lagrangian is $L=\frac{1}{2}m(a^2\dot\phi^2+a^2\dot\theta^2\sin^2\phi)+mga\cos\phi$ $p_1=\frac{\partial L}{\partial \dot q_1}=\frac{\partial L}{\partial \dot\theta}=ma^2\dot\theta\sin^2\phi$ $p_2=\frac{\partial L}{\partial \dot q_2}=\frac{\partial L}{\partial \dot\phi}=ma^2\dot\phi$ $T=\frac{1}{2}\dot\phi p_2+\frac{1}{2}\dot\theta p_1=T(p_1,p_2)$ That's it? $\endgroup$
    – gov
    Jul 31, 2013 at 19:34
  • $\begingroup$ Seems fine to me! Although, you can always put $\dot{q}_i$ in terms of $p_i$, or viceversa, but also $T$ can explicitly depend on $q_i$, I messed it there. Try to prove explicitly that the hamiltonian is the energy and that it is conserved, so that you get convinced of it. $\endgroup$
    – user24999
    Jul 31, 2013 at 20:32

1 Answer 1

12
$\begingroup$

The reason for the asseveration

If time $t$, does not appear in Lagrangian $\mathcal{L}$, then the Hamiltonian $\mathcal{H}$ is conserved. This is the energy conservation unless the potential energy depends on velocity.

is that, from the definition of the hamiltonian as the Legendre transformation, $$\mathcal{H}\equiv\sum_i\dot{q}_i\frac{\partial\mathcal{L}}{\partial\dot{q}_i}-\mathcal{L}\hspace{1in}(\dagger)$$ and knowing that for any function in phase space, $F=F(q_i,p_i,t)$, $$\frac{dF}{dt}=\frac{\partial{F}}{\partial{q}_1}\frac{d{q_1}}{d{t}}+\ldots+\frac{\partial{F}}{\partial{q}_n}\frac{dq_n}{dt}+\frac{\partial{F}}{\partial{p}_1}\frac{d{p_1}}{d{t}}+\ldots+\frac{\partial{F}}{\partial{p}_n}\frac{dp_n}{dt}+\frac{\partial{F}}{\partial{t}}\\=\sum_{j=1}^n\left(\frac{\partial{F}}{\partial{q}_j}\dot{q}_j+\frac{\partial{F}}{\partial{p}_j}\dot{p}_j\right)+\frac{\partial{F}}{\partial{t}}\\=\left\{F,\mathcal{H}\right\}+\frac{\partial{F}}{\partial{t}}$$ where $\left\{F,\mathcal{H}\right\}$ is the Poisson bracket of $F$ and $\mathcal{H}$, defined as $$\left\{F,\mathcal{H}\right\}\equiv\sum_{j=1}^n\left(\frac{\partial{F}}{\partial{q}_j}\dot{q}_j+\frac{\partial{F}}{\partial{p}_j}\dot{p}_j\right)=\sum_{j=1}^n\left(\frac{\partial{F}}{\partial{q}_j}\frac{\partial{\mathcal{H}}}{\partial{p}_j}-\frac{\partial{F}}{\partial{p}_j}\frac{\partial{\mathcal{H}}}{\partial{q}_j}\right)$$ if $\mathcal{L}=\mathcal{L}(q_i,p_i)$, i.e. the Lagrangian does not depend explicitly on time, which in turn means, from the definition $\mathcal{L}\equiv{T}-V$, that kinetic energy $T$ and potential $V$ does not depend explicitly on time, then $\frac{d\mathcal{H}}{dt}=\left\{\mathcal{H},\mathcal{H}\right\}=0$. Now, a constant of motion is precisely some function $F$ of phase space that is independent of time, i.e. such that $\frac{dF}{dt}=0$, so in this case the hamiltonian would be conserved. Now, from the definition $(\dagger)$, you may verify that the hamiltonian equals the energy, $$\mathcal{H}\equiv{T}+V$$ only if $V=V(q_i)$ alone. So if that is the case, then energy would be conserved.

So identify that in your Lagrangian and get your conclusions, anyway you can always verify it this way for your particular case.

$\endgroup$
1
  • $\begingroup$ +1 for the word asseveration (also the answer is genuinely helpful). $\endgroup$
    – alessandro
    Apr 22, 2021 at 21:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.