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Laplace equation derivation, where does the potential go

  1. Jan 22, 2012 #1

    Uku

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    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:
    592afce1200ba232f3ce737cb32f2e25.png

    (Wikipedia, image 1)

    2. Relevant equations
    6aa558a2b076e81bd4c97d6d5f168838.png

    (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]
     
    Last edited: Jan 23, 2012
  2. jcsd
  3. Jan 24, 2012 #2
    So what is your question exactly? Cant quite follow.
     
  4. Jan 24, 2012 #3

    Uku

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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..?
     
  5. Jan 24, 2012 #4

    I like Serena

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    Homework Helper

    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:
    $${\partial \mathcal{L}\over \partial \dot q_i} = {\partial \over \partial \dot q_i}(T(\dot q_j, q_j, t) - V(q_j))
    = {\partial \over \partial \dot q_i}T(\dot q_j, q_j, t) - {\partial \over \partial \dot q_i}V(q_j)
    = {\partial \over \partial \dot q_i}T(\dot q_j, q_j, t) - 0$$
     
  6. Jan 24, 2012 #5

    Uku

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    Thanks! That makes perfect sense, but it means that there is an error in:
    592afce1200ba232f3ce737cb32f2e25.png

    since here, very literally, [itex]\frac{\partial V}{\partial q_{j}}=0[/itex]
     
  7. Jan 24, 2012 #6

    I like Serena

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    Homework Helper

    Well, what is the following?
    $${\partial \mathcal{L}\over \partial q_i} = {\partial \over \partial q_i}(T(\dot q_j, q_j, t) - V(q_j)) $$


    Uhh :confused:
    How do you get that?
     
  8. Jan 26, 2012 #7

    Uku

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    I think I mixed up, the potentials subtract. Thanks!
     
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