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In general, when dealing with mechanics problems using a function ##f(q1,q2,...)=0## that represent constraints one is minimizing the action ##S## while adding a term to the Lagrangian of the not-independent coordinates ##L + \lambda f ##. One can show that this addition doesn't change the minimal action and that the desired equations follow under a correct choice of ##\lambda##. Anyway this is the known theoretic approach.
Now consider the following treatment of the catenary problem on page 5: http://physics.ucsd.edu/students/courses/fall2009/managed/physics110a/documents/Lecture11-9.pdf
Don't waste the time on the details because my question is about the method:
Why does one add the ##\lambda f## term to the function ##V## one wants to minimize? As you see from my introduction above, when minimizing the action ##S##, that term is not added to the action but to the Lagrangian. Here ##V## is supposed to be the analogue of the action since this is what we want to minimize. So it seems to me that in this problem Lagrange multipliers are used differently than we had used in previous problems and different from the description of the theory.
Please help me lift the confusion. To show that I have thought about it, this is my idea but I'm not sure:
Since we are talking about the potential energy function ##V## and the kinetic energy is 0, we are basically talking about the Lagrangian and not the analog of the action. As we know from the theory we add this ##\lambda f## term to the Lagrangian, or the potential energy here. We know that the physics haven't changed from this addition. And now we minimize the new expression with the added term and forget that we even ever used lagrange multipliers?
Now consider the following treatment of the catenary problem on page 5: http://physics.ucsd.edu/students/courses/fall2009/managed/physics110a/documents/Lecture11-9.pdf
Don't waste the time on the details because my question is about the method:
Why does one add the ##\lambda f## term to the function ##V## one wants to minimize? As you see from my introduction above, when minimizing the action ##S##, that term is not added to the action but to the Lagrangian. Here ##V## is supposed to be the analogue of the action since this is what we want to minimize. So it seems to me that in this problem Lagrange multipliers are used differently than we had used in previous problems and different from the description of the theory.
Please help me lift the confusion. To show that I have thought about it, this is my idea but I'm not sure:
Since we are talking about the potential energy function ##V## and the kinetic energy is 0, we are basically talking about the Lagrangian and not the analog of the action. As we know from the theory we add this ##\lambda f## term to the Lagrangian, or the potential energy here. We know that the physics haven't changed from this addition. And now we minimize the new expression with the added term and forget that we even ever used lagrange multipliers?
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