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How can I know when the Lagrange multiplier is a constant?

  1. Mar 20, 2015 #1
    Consider a holonomic system where I have ##n## not independent variables and one constraint ##f(q1,q2,...,qN,t)=0##. One can rewrite the minimal action principle as:

    ##\frac{\partial L}{\partial q_i} - \frac{d}{dt} \frac{\partial L}{\partial q'_i} - \lambda \frac{\partial f}{\partial q_i} = 0 ##

    My question: how can one check if lambda is a constant or a function? Because in general it does not have to be a constant in this reasoning according to my understanding.

    Example:

    The harmonic oscillator given by the constaint that the distance of the mass to the origin is constant. The small angle approximation is allowed. So here I write out the equations and I indeed do find a harmonic oscillation under the assumption that ##\lambda## is constant. How to show that this is indeed constant? If I try doing this and do some small angle approximations using the tension I eventually find that the ##y##-coordinate is constant so I'm probably doing something wrong.

    Is the lambda not constant when not using the small angle approximation?
     
  2. jcsd
  3. Mar 21, 2015 #2

    Orodruin

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    This depends on your problem. For a pendulum, lambda is related to the tension in the pendulum, for a mass on a circle lambda is related to the constraining force. Now obviously, if you move with constant speed on a circle, you need a constant (magnitude) centripetal force and your lambda is constant. For the pendulum, the tension changes and so does lambda.
     
  4. Mar 21, 2015 #3
    Assuming a non constant lambda won't give me a harmonic motion in the x coordinate though .
     
  5. Mar 21, 2015 #4

    Orodruin

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    Why don't you show what you get? In that way it will be easier to judge where you meet a problem.
     
  6. Mar 21, 2015 #5
    http://imgur.com/kLeIwfN
     
  7. Mar 21, 2015 #6

    Orodruin

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    So, small oscillations imply that you can neglect higher-order terms in the deviation from the equilibrium solution. What does this imply for ##\lambda##?
     
  8. Mar 21, 2015 #7
    Well ##\lambda## is equal to the tension up to a sign. It seems that one implicitly assumes the tension to be ##mg## in the standard derivation and thus assuming ##\lambda## to be constant?
     
  9. Mar 21, 2015 #8

    Orodruin

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    Yes, but you can argue it in a more formal way in the small angle approximation.
     
  10. Mar 22, 2015 #9

    stevendaryl

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    In general, [itex]\lambda[/itex] is not constant. And in the particular case of a rock on a string being twirled around in a vertical plane in a circle of radius [itex]R[/itex], the constraint force is the tension, which is NOT constant; the tension is highest at the bottom of the circle, and is lowest at the top of the circle.

    The use of Lagrangian multipliers to work with a constraint [itex]f(\vec{x}) = 0[/itex] is equivalent, at least in the simplest cases, to doing Newtonian physics (F = ma) with an unknown constraint force [itex]\vec{F}_c[/itex] that is in the direction of [itex]\nabla f[/itex]. The magnitude of [itex]\vec{F}_c[/itex] is determined by the criterion that the particle never leaves the surface [itex]f(\vec{x}) = 0[/itex]. If the other forces are sufficient to keep the particle within the surface, then the constraint force is zero. Otherwise, it's whatever magnitude is necessary.
     
  11. Mar 22, 2015 #10

    Orodruin

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    It is however constant in the small angle approximation, which is what was currently under discussion.
     
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