Understanding Lagrange multipliers in the Lagrangian

[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.

 

[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 $L'$).

 

[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).

 

[mjordan2nd]

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

 

[PJC01]

Thank you for your question. The addition of the term \sum\limits_{\alpha=1}^m \lambda_{\alpha} f_a in the action integral is known as the method of Lagrange multipliers. It is used to incorporate constraints into the Lagrangian formulation of mechanics.

The Lagrangian, L, represents the kinetic and potential energies of a system, and it is typically a function of the generalized coordinates and their time derivatives. However, in some cases, there may be additional constraints on the system that cannot be expressed in terms of the generalized coordinates. These constraints can be in the form of equations, as in the example you provided, or they can be inequalities.

The purpose of the Lagrange multipliers is to introduce these constraints into the Lagrangian in a way that preserves the equations of motion. The \lambda_{\alpha} terms are known as the Lagrange multipliers, and they act as undetermined multipliers that are used to enforce the constraints.

Solving for the Lagrange multipliers allows us to determine the values of the constraint forces or torques that are needed to maintain the constraints. This is because the Lagrange multipliers are related to the constraint forces through the constraint equations. By solving for them, we can obtain the values of the constraint forces that are necessary for the system to satisfy the constraints.

Taking the derivative of a constraint equation is not always straightforward, as you have observed in your example. This is because the constraint equation may involve multiple variables and their derivatives. In general, the derivative of a constraint equation with respect to one of the variables will involve the partial derivatives of the other variables. In the example you provided, taking the derivative with respect to x would involve the partial derivative of r with respect to x and the partial derivative of \theta with respect to x.

I hope this helps to clarify the significance of the Lagrange multipliers in the Lagrangian and how they are used to incorporate constraints into the formulation. If you have any further questions, please do not hesitate to ask. Thank you for your interest in this topic.