How to Find ∂L/∂q: A Simple Guide for Calculating Lagrangian Equations

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The discussion focuses on deriving the relationship ∂L/∂q = dot{p}, where L is the Lagrangian, p is momentum, and q is the general coordinate. It is established that ∂L/∂q = dot{p} because the Lagrangian depends on q primarily through the potential energy, U, leading to the expression -∂U/∂q = dot{p}, which aligns with Newton's second law. Additionally, the conversation touches on identifying translational symmetry, where momentum must be constant (p-dot = 0) for the corresponding coordinate. The canonical momentum is defined as p = ∂L/∂dot{q}, highlighting the distinction between canonical and mechanical momentum, particularly in contexts like motion in a magnetic field. Understanding these relationships is crucial for deriving Hamiltonian equations and analyzing systems with symmetries.
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hi, silly question but would someone please show me how \frac{∂L}{∂q}=\dot{p}?

L being the lagrangian, p being the momentum, and q being the general coordinate.
 
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(i hope this doesn't qualify as a homework question this actually has nothing to do with my homework.. I'm just trying to derive the hamiltonian equation and this was just part of the steps)
 
iScience said:
please show me how \frac{∂L}{∂q}=\dot{p}?

\frac{∂L}{∂q}=\dot{p} because the Lagrangian, L=T-U, depends on q only in the potential energy, U.

-\frac{∂U}{∂q}=\dot{p} is Newton's 2nd law of motion expressed in terms of the potential.
 
thanks!

also, i had another question i hope i can just ask it in the same thread; if not let me know (moderators/admins) and i'll just make a new thread.



how do i know when there is translational symmetry? in other words what quantity has to be zero?
 
You want p=constant (p-dot = 0) for the momentum conjugate to that coordinate.
 
"momentum that is conjugate to that coordinate" meaning just the momentum corresponding to the particular coordinate at hand right? (just checking)
 
The canonical momentum of the generalized coordinate q is by definition given by
p=\frac{\partial L}{\partial \dot{q}}.
It's important to keep in mind that also for a Cartesian coordinate it is the canonical and not necessarily the mechanical momentum. An interesting example for that both need not be the same is the motion of a particle in a magnetic field.
 
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