How to Handle Constraints in Lagrangian Mechanics?

AI Thread Summary
The discussion focuses on formulating a constrained Lagrangian in polar coordinates for a system constrained to a curve defined by r = r(ψ). The proposed constrained Lagrangian is Lc = L(ψ, r, r') - λ(r - r(ψ)), which is deemed valid as it introduces a generalized force to enforce the constraint. A further complication arises when a parameter "a" is introduced, which becomes a function of r under the constraint. The key question is whether to incorporate this dependency in the constrained Lagrangian or to solve the Euler-Lagrange equation without considering a(r). The participants agree that the formulation makes sense and provide insights into handling the dependency of "a" in the equations.
Sunfire
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Hello,

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?

Thanks!
 
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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.
 
CompuChip said:
... yes, you should be able to do that, ...

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?
 
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