# Understanding Lagrange multipliers in the Lagrangian

1. Sep 30, 2012

### mjordan2nd

In Goldstein, the action is defined by $I=\int L dt$. However, when dealing with constraints that haven't been implicitly accounted for by the generalized coordinates, the action integral is redefined to

$$I = \int \left( L + \sum\limits_{\alpha=1}^m \lambda_{\alpha} f_a \right) dt.$$

f is supposed to be an equation of constraint. I do not understand the significance of the new term. It kind of seems to take the form of a generalized force, but I have not quite been able to convince myself of this. Why is this new term being added? Why is it that solving for lambda gives you the constraint force. And how exactly does one take a derivative of a constraint equation? For instance, say your constraint equation is $r \theta = x$. This seems like a holonomic constraint if x and theta are you generalized coordinates. Now if I take the derivative with respect to x I should get 1 on one side and 0 on the other. I'm a bit confused by this. Any help would be appreciated. Thanks.

2. Sep 30, 2012

### Muphrid

A typical case would be, say, let $r, \theta$ be your generalized coordinates, and constrain $r = a$, a constant. The equation of constraint is then $f = r-a = 0$, and the equation is enforced by saying that $\partial L'/\partial \lambda = 0$ must be true (which follows directly from the Euler-Lagrange equations when applied to the modified Lagrangian $L'$).

3. Oct 1, 2012

### medthehatta

I believe that the sum $\lambda_a f_a$ is the work that would be done on the system due to a displacement away from the constraints (except that those motions don't actually happen).

That is, the $f_a$ terms are generalized displacements which are disallowed by the constraints (which is why you arrange the form of $f_a$ so that the RHS is zero: the displacement "away" from the allowed displacements is zero).

Then the $\lambda_a$ are constraint forces: they multiply the displacements from allowable configurations to yield the work done. (Of course, no work is actually done by the constraint forces because the displacements $f_a$ are 0.)

It's still not totally clear to me why -- formally -- the constraints wind up being zero (except that you're minimizing the augmented Lagrangian, and there isn't any real reason for the constraint term to not vanish).

4. Oct 6, 2012

### mjordan2nd

Thank you folks. After some searching, I believe I have worked things out.

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