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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?

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  • $\begingroup$ 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 '13 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 '13 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 '13 at 20:32
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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.

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