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Laplace equation derivation, where does the potential go

 
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Jan22-12, 09:51 PM   #1
Uku
 

Laplace equation derivation, where does the potential go


1. The problem statement, all variables and given/known data
Since the potential field is only a function of position, not velocity, Lagrange's equations are as follows:


(Wikipedia, image 1)

2. Relevant equations


(Wikipedia, image 2)

3. The attempt at a solution

Now, [itex]-\frac{\partial{V}}{\partial{q_{j}}}[/itex]

How is speed involved in this derivative of potential by generalized coordinate (upper left, image 1)? Potential only depends on the position, we are assuming a conservative field, and that is what we are doing, how does this term disappear?
Feels like saying [itex]F=-\nabla V=0[/itex].
I mean, I can understand eg. generalized impulse depending on speed: [itex]p_{j}=\frac{\partial{L}}{\partial{\dot q_{j}}}[/itex], but not the potential-evaporating transition:
[itex]\frac{d}{dt}\left(\frac{\partial{L}}{\partial{\dot q_{j}}}\right)-\frac{\partial{L}}{\partial{q_{j}}}=0[/itex]
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Jan24-12, 12:57 AM   #2
 
So what is your question exactly? Cant quite follow.
Jan24-12, 03:53 AM   #3
Uku
 
My problem is in the sentence:
Since the potential field is only a function of position, not velocity, Lagrange's equations are as follows:
And in:
[itex]\frac{\partial{V}}{\partial{g_{j}}}[/itex] disappearing from the very first statement.
So - where does it go, [itex]g_{j}[/itex] is a generalized coordinate, not generalized velocity..?
Jan24-12, 05:26 PM   #4
 
Recognitions:
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Laplace equation derivation, where does the potential go


For a conservative field, which is only dependent on position, the lagrangian is defined as: ##\mathcal{L} = \mathcal{L}(\dot q_j, q_j, t) = T(\dot q_j, q_j, t) - V(q_j)##.

In particular, this means that:
$${\partial \mathcal{L}\over \partial \dot q_i} = {\partial \over \partial \dot q_i}(T(\dot q_j, q_j, t) - V(q_j))
= {\partial \over \partial \dot q_i}T(\dot q_j, q_j, t) - {\partial \over \partial \dot q_i}V(q_j)
= {\partial \over \partial \dot q_i}T(\dot q_j, q_j, t) - 0$$
Jan24-12, 05:43 PM   #5
Uku
 
Thanks! That makes perfect sense, but it means that there is an error in:


since here, very literally, [itex]\frac{\partial V}{\partial q_{j}}=0[/itex]
Jan24-12, 05:45 PM   #6
 
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Well, what is the following?
$${\partial \mathcal{L}\over \partial q_i} = {\partial \over \partial q_i}(T(\dot q_j, q_j, t) - V(q_j)) $$


Quote by Uku View Post
since here, very literally, [itex]\frac{\partial V}{\partial q_{j}}=0[/itex]
Uhh
How do you get that?
Jan26-12, 12:25 AM   #7
Uku
 
I think I mixed up, the potentials subtract. Thanks!
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