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Lagrangian mechanics - derivation doubt.

  1. Mar 11, 2013 #1
    In the attached snip, the last few steps of the lagrangian equation is shown. I don't understand how the [itex]\frac{\delta V}{\delta\dot{q_j}}= 0[/itex]. As an example let me take gravitational force. With change in velocity ( along the downwards direction obviously), there sure is a change in gravitational potential energy. When [itex]q_j[/itex] changes then [itex]\dot{q_j} [/itex]definitely changes. So what am I missing here?

    nothumb.jpg
     
  2. jcsd
  3. Mar 11, 2013 #2
    The fact is that V is not *explicit* function of the [itex]{q_j} [/itex]s. Once you have found a set of (usually) independent variables [itex]{q_j} [/itex] and [itex]\dot{q_j} [/itex] you have to find how the quantities you are considering depends on these variables; you find, in this case, that V doesn't depend on the [itex]\dot{q_j} [/itex] but on the [itex]{q_j} [/itex] only.

    Another consideration: why writing L as function of [itex]{q_j} [/itex], [itex]\dot{q_j} [/itex] and t? If it's function of the [itex]{q_j} [/itex], you could think it's superfluous to write even their time derivative; and if all of the [itex]{q_j} [/itex] at the end are only function of t, why not simply saying that L is function of t only?
    There are precise reasons too technical to write here (at night, for me :smile:) you will have to make a little effort in your study, make many excercises and go on...
     
    Last edited: Mar 11, 2013
  4. Mar 11, 2013 #3

    Fredrik

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    Not if you're slowly lowering a massive object attached to a rope, with a fixed velocity. The potential function is supposed to be useful regardless of what else is going on.

    Note that if ##F=-1/x^2## and ##F=-dV/dx##, we have
    $$\frac{dV}{dx}=-F=\frac{1}{x^2} =\frac{d}{dx}\left(-\frac{1}{x}\right).$$ So
    $$V=-\frac 1 x.$$ Note that there's no ##\dot x## in this formula.
     
  5. Mar 11, 2013 #4
    But you are taking a specific case and some assumptions. They haven't done that. So...
     
  6. Mar 11, 2013 #5

    Fredrik

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    Maybe they didn't. (There's no link and no attachment in your OP). But you did. You said that the velocity "definitely changes", so you must have assumed that the object is falling or something. I showed that there is a situation in which the velocity doesn't change.

    My point is that the gravitational potential is the same (proportional to 1/r2) regardless of what non-gravitational forces are acting on the object.
     
  7. Mar 11, 2013 #6
    Oh sorry, the attachment was too big. Click on that thumbnail. It takes you to a image hosting site.
     
  8. Mar 12, 2013 #7

    Fredrik

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    There is no thumbnail either.
     
  9. Mar 12, 2013 #8
    I've read it copying the link which appears answering his post. It simply says that if V doesn't depend on the generalized velocities, it can be brought into the partial derivative with respect to the velocities as a subtractive term to kinetic energy T, becoming the lagrangian L so the lagrange equations results in the standard form:
    (d/dt)@L/@q_j' = @L/@q_j
     
  10. Mar 12, 2013 #9
    Glory!...thank you for seeing the link somehow. :D
    Yeah that is my doubt. I understand why they are doing it but is it right? Doesn't potential energy change when velocity changes?
     
  11. Mar 12, 2013 #10

    AlephZero

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    The point is that you form the Lagrangian at each particular instant in time. For any fixed value of t, the PE doesn't depend on the velocity, it only depends on position.

    As an example, think about throwing a ball vertically. The (gravitational) PE is just mgh, the same as it was when you learned about "energy methods" in a first dynamics course without using the Lagrangian formulation. if the ball is at a particular height above the ground, the PE is the same whether it is moving up, down, or not moving at all.
     
  12. Mar 12, 2013 #11
    The ball not moving is a trivial case of velocity being zero. But consider any other possibility of velocity. Then h depends on the velocity. So the potential energy does depend on the velocity, right?
     
  13. Mar 12, 2013 #12

    Fredrik

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    No, it does not. Even if you define h as a function, and write something like V(h(v)), then V is still just a function from the set of possible positions into ℝ. ##V\circ h## on the other hand is a function from the set of possible velocities into ℝ.

    You should try to think in terms of functions instead of in terms of variables that depend on each other.
     
  14. Mar 12, 2013 #13

    AlephZero

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    Go back to the basic definition of a potential function in mechanics. At any point in space, it is equivalent to a force on an object equal to the gradient of the potential.

    If a potential function exists at all, then the work done by the force when you move between two points A and B is always the same. It doesn't depend on the path you take between the points, or on the speed you travel. So the potential function can only depend on the position, not on the velocity, or anything else.

    Of course there are other types of force in physics which DO depend on the velocity, but they can't be described by a potential function, by definition. You may learn how to handle them in the Lagrangian formulation later, but don't get connfused by trying to deal with "everything all at once". The best way to learn any subject is one step at a time!
     
  15. Mar 14, 2013 #14
    That did it. Now I get it.
    Thanks a lot!
     
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