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Laplace equation derivation, where does the potential go |
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| Jan22-12, 09:51 PM | #1 |
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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] |
| Jan24-12, 12:57 AM | #2 |
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So what is your question exactly? Cant quite follow.
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| Jan24-12, 03:53 AM | #3 |
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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 |
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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 |
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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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Recognitions:
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Well, what is the following?
$${\partial \mathcal{L}\over \partial q_i} = {\partial \over \partial q_i}(T(\dot q_j, q_j, t) - V(q_j)) $$ ![]() How do you get that? |
| Jan26-12, 12:25 AM | #7 |
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I think I mixed up, the potentials subtract. Thanks!
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