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Lagrangian with one constraint

  1. Jun 27, 2013 #1

    I have the functional

    J = ∫ L(ψ, r, r') dψ, where r'=dr/dψ. L is written in polar coordinates (r,ψ).

    Now I want to constrain the motion to take place on the polar curve r = r(ψ). Can I write the constrained lagrangian as

    Lc=L(ψ, r, r') - λ(r - r(ψ)) and then solve the Euler-Lagrange equation

    Does this make sense?

  2. jcsd
  3. Jun 27, 2013 #2


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    You should probably call the curve something else, like ##\rho(\psi)##. But yes, you should be able to do that, the extra ##\lambda(r - \rho)## term will give you a generalized force constraining the object to the curve.
  4. Jun 27, 2013 #3
    Thanks CompuChip...
    Along these lines - say, the Lagrangian is, as agreed, Lc = L(a, ψ, r, r') - λ(r - f(ψ))
    where r=f(ψ) is the constraint and "a" is a parameter.

    The question is - I know that when imposing the constraint, it so happens that a = a(r). I mean, "a" becomes a function of r under the imposed constraint.

    This is a challenge... Do I write

    Lc = L(a(r), ψ, r, r') - λ(r - f(ψ)) and then solve the Euler-Largange eq.; or do I solve the E-L eq. and not worry about a(r)??? It's a puzzle. Would you or someone know?
    Last edited: Jun 27, 2013
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