- #1
erore
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- TL;DR Summary
- 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?
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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?