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So I know that in Noether's theorem, it is stated that

$$\frac{d}{dt}\left(\frac{\partial L}{\partial \dot x}\dot x - L\right)=0$$

where the Hamiltonian $H$ is defined as

$$H = \frac{\partial L}{\partial \dot x}\dot x - L.$$

Which would mean that the time derivative of the Hamiltonian is $0$. However, I do wonder how to just calculate $\frac{d}{dt}\left(\frac{\partial L}{\partial \dot x}\dot x - L\right)$ and show that it is equal to $0$. Could it be possible for someone to show how to fully calculate $\frac{d}{dt}\left(\frac{\partial L}{\partial \dot x}\dot x - L\right)$ and prove that it equals to zero?

Frobenius
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Jungwoon Song
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2 Answers2

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The variation of $L$ with infinitesimal change $\delta t$ in $t$ is \begin{align} L(t+\delta t,x(t+\delta t),\dot{x}(t+\delta t))-L(t,x(t),\dot{x}(t)) & = {\partial L \over \partial t}\delta t+{\partial L \over \partial x}\dot{x}\delta t+{\partial L \over \partial \dot{x}}\ddot{x}\delta t \\ & = \bigg({\partial L \over \partial x}\dot{x}+{\partial L \over \partial \dot{x}}\ddot{x}\bigg)\delta t \\ & = \bigg({d \over dt}\bigg({\partial L \over \partial \dot{x}}\bigg)\dot{x}+{\partial L \over \partial \dot{x}}\ddot{x}\bigg)\delta t \end{align} where in the second equality we use ${\partial L \over \partial t}=0$. This is what guarantees the applicability of Noether's theorem, and some conserved quantity correspondingly exists. Therefore, $${dL \over dt}={d \over dt}\bigg({\partial L \over \partial \dot{x}}\dot{x}\bigg)$$ So you have ${dH \over dt}=0$.

Andy Chen
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You made a mistake in your definition of the Hamiltonian but I guess that's simply a copy-paste error. Regardless, using the Hamilton equations:

$$\frac{dH}{dt}=\frac{\partial H}{\partial t}+\frac{\partial H}{\partial p_i}\dot{p_i}+\frac{\partial H}{\partial q_i}\dot{q_i}= \frac{\partial H}{\partial t}-\frac{\partial H}{\partial p_i}\frac{\partial H}{\partial q_i}+\frac{\partial H}{\partial q_i}\frac{\partial H}{\partial p_i}=\frac{\partial H}{\partial t}$$

On the other hand, you have:

$$dH=\frac{\partial H}{\partial p_i}dp_i + \frac{\partial H}{\partial q_i}dq_i + \frac{\partial H}{\partial t}dt $$

But also, form the definition of the hamiltonian:

$$dH=\dot{q_i}dp_i + p_id\dot{q_i}-dL= \dot{q_i}dp_i + p_id\dot{q_i}-\frac{\partial L}{\partial q_i}dq_i-\frac{\partial L}{\partial \dot{q_i}}d\dot{q_i}-\frac{\partial L}{\partial t}dt$$

Identifying terms, we can find a third Hamilton equation (as well as the other two, but we already used them above):

$$\frac{\partial H}{\partial t}=-\frac{\partial L}{\partial t}$$

And from our first result:

$$\frac{dH}{dt}=-\frac{\partial L}{\partial t} $$

Which means that the time derivative of the Hamiltonian is only $0$ when the time partial derivative of the Lagrangian is $0$. Fortunately, this is often the case, as kinetic energy isn't explicitly time dependent, and most potentials you'll consider aren't either.

agaminon
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