# Lagrangian with one constraint

1. Jun 27, 2013

### Sunfire

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!

2. Jun 27, 2013

### CompuChip

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.

3. Jun 27, 2013

### Sunfire

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