Quick Question- Hamiltonian constant proof

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The discussion centers on proving that if the Lagrangian does not explicitly depend on time, the Hamiltonian remains a constant of motion. Participants clarify the transition from the differential form dH to its time derivative dH/dt, emphasizing that this involves division by dt rather than differentiation. The conversation highlights the importance of understanding the product rule in calculus when analyzing the relationship between Lagrangian and Hamiltonian mechanics.

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Homework Statement



Show that if the Lagrangian does not explicitly depend on time that the Hamiltonian is a constant of motion.

Homework Equations



see below

The Attempt at a Solution


method attached here:
hcom.png


Apologies this is probably a bad question, but just on going from the line ##dH## to ##dH/dt## I see the##d/dt## has hit the ##dq_0## terms only, I don’t understand why a product rule hasn’t been used, so I would get:##\frac{dH}{dt}=\sum_{u} \dot{q_u} (\frac{dp_{u}}{dt}-\frac{\partial L}{\partial q_u} )+ \ddot{q_u}dp_{u}-\frac{d}{dt}(\frac{\partial L}{\partial q_u}) dq_u ##Many thanks
 

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You are not taking the derivative of ##dH##. ##dH## in itself is a differential
 
In going from dH to dH/dt, you are not differentiating, but dividing by dt. Differentiation was in the previous step.
Thus e.g. if f = uv
df = udv + vdu
df/dx = udv/dx + vdu/dx (not uddv/dx + dv du/dx +vddu/dx + du dv/dx)

Edit: Beat me to it!
 
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