The partial time derivative of Hamiltonian vs Lagrangian

[erore]

TL;DR

Why does partial derivative of hamiltonian wrt time equal to minus partial derivative of Lagrangian wrt time?

I have been reading a book on classical theoretical physics and it claims:
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If a Lagrange function depends on a continuous parameter $\lambda$, then also the generalized momentum $p_i = \frac{\partial L}{\partial\dot{q}_i}$ depends on $\lambda$, also the velocity $\dot{q}_i=f_i(q_j,p_j,\lambda)$ and so the Hamiltonian depends on $\lambda$. If we calculate the derivative of Hamiltoniat wrt $\lambda$:

$$ \frac{\partial H}{\partial \lambda}=\frac{\partial}{\partial{\lambda}}\left(\sum_j p_j f_j - L\right)=\sum_j p_j\frac{\partial f_j}{\partial\lambda}-\sum_j \frac{\partial L}{\partial\dot{q}_j}\frac{\partial f_j}{\partial\lambda}-\frac{\partial L}{\partial\lambda}= -\frac{\partial L}{\partial\lambda}$$

which if we put $\lambda=t$ gives:

$$\frac{\partial H}{\partial t}=-\frac{\partial L}{\partial t}$$
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I wonder, why there are no $\frac{\partial p_j}{\partial\lambda} f_j$ terms?

 

[stevendaryl]

Partial derivatives are a little tricky to talk about, because you have to be absolutely clear about what are the independent variables. We are considering the following variables: $q, \dot{q}, p, \lambda$. (I'm going to leave off the indices on the momenta, velocities and coordinates).

You can consider $q, \dot{q}, \lambda$ to be the independent variables, and then $p$ is a function of those, or you can consider $q, p, \lambda$ to be the independent variables, and then $\dot{q}$ is a function of those.

In thermodynamics, you have the same sort of options. You can consider volume and entropy to be independent, in which case temperature and pressure are functions of those, or you can consider temperature and pressure as independent, in which case volume and entropy are functions of those.

Borrowing notation from thermodynamics, we can write things this way:

$\frac{\partial H}{\partial \lambda}|_{q, p}$

This means the derivative of $H$ with respect to $\lambda$ holding $q$ and $p$ constant. Now, if we express the hamiltonian in terms of the Lagrangian:

$H = p \dot{q} - L$

If we now take a derivative with respect to $\lambda$, we have:

$\frac{\partial H}{\partial \lambda}|_{q, p} = p \frac{\partial \dot{q}}{\partial \lambda}|_{q, p} - \frac{\partial L}{\partial \lambda}|_{q,p}$

There is no term involving $\frac{\partial p}{\partial \lambda}$ because we're holding $p$ constant.

Now, $L$ is an explicit function of $q, \dot{q}$ and $\lambda$. So we can write:

$\frac{\partial L}{\partial \lambda}|_{q,p} = \frac{\partial L}{\partial q}|_{\dot{q}, \lambda} \frac{\partial q}{\partial \lambda}|_{q,p} + \frac{\partial L}{\partial \dot{q}}|_{q, \lambda} \frac{\partial \dot{q}}{\partial \lambda}|_{q,p} + \frac{\partial L}{\partial \lambda}|_{q, \dot{q}}$

The first term on the right is 0, because if you hold $q$ constant, then $\frac{\partial q}{\partial \lambda} = 0$

So we have:

$\frac{\partial H}{\partial \lambda}|_{q, p} = p \frac{\partial \dot{q}}{\partial \lambda}|_{q, p} - \frac{\partial L}{\partial \dot{q}}|_{q,\lambda} \frac{\partial \dot{q}}{\partial \lambda}|_{q,p} - \frac{\partial L}{\partial \lambda}_{q,\dot{q}}$

By definition: $p = \frac{\partial L}{\partial \dot{q}}|_{q, \lambda}$, so the first two terms cancel, leaving:

$\frac{\partial H}{\partial \lambda}|_{q, p} = - \frac{\partial L}{\partial \lambda}_{q,\dot{q}}$

 

[Steve123]

I'm not quite sure how you're applying the derivatives but we begin by taking the total derivative H using its definition:

\begin{equation*}
H=\sum_i\dot{q_i}\frac{\partial L}{\partial \dot{q}} - L
\end{equation*}

When applying derivatives on the sum, product rule is used. On the L term, the multivariable chain rule applies so you get the following:

\begin{equation*}
\frac{dH}{dt}=\sum_i \left(\ddot{q_i}\frac{\partial L}{\partial \dot{q}}+\dot{q_i}\frac{d}{dt}\frac{\partial L}{\partial \dot{q}}\right) - \left(\sum_i\frac{\partial L}{\partial \dot{q}} \ddot{q_i} + \frac{\partial L}{\partial q} \dot{q_i}\right) -\frac{\partial L}{\partial t}
\end{equation*}

The first from each sum are the same so they cancel. If the Euler-Lagarange equations hold then the second terms in each sum are also equal so they cancel are you are left with

\begin{equation*}
\frac{dH}{dt}= -\frac{\partial L}{\partial t}
\end{equation*}