# Lagrangian Mechanics: When to Use Lagrange Multipliers?

I've seen a few other threads on here inquiring about what is the point of Lagrange Multipliers, or the like. My main question though is, how can I tell by looking at a system in a problem that Lagrange Multipliers would be preferred compared to generalized coordinates. I'm in a theoretical mechanics course, and we are just doing very basic systems (pendulums, points constrained to some shape).

The book I have just outlines Lagrange Multipliers incorporated into the Lagrangian Equation.

$$\frac{\partial L}{\partial q_j} -\frac{d}{dt}\frac{\partial L}{\partial \dot{q_j}} + \sum_k \lambda_k(t) \frac{\partial f_k}{\partial q_j}=0.$$

The book gives about 2 examples of using these, but I wouldn't know whether or not to use them over just using the regular generalized coordinate example.

References:

1. Thornton & Marion, Classical Dynamics of Particles and Systems, Fifth Ed.; p.221.

In the context of Lagrange equations

\begin{align}\frac{d}{dt}\frac{\partial (T-U)}{\partial \dot{q}^j}-\frac{\partial (T-U)}{\partial q^j}~=~& Q_j-\frac{\partial{\cal F}}{\partial\dot{q}^j}+\sum_{\ell=1}^m\lambda^{\ell} a_{\ell j}, \cr j~\in~&\{1,\ldots, n\}, \end{align}\tag{L}

in classical mechanics, the Lagrange multipliers are used to impose semi-holonomic constraints

\begin{align}\sum_{j=1}^n a_{\ell j}(q,t)\dot{q}^j+a_{\ell t}(q,t)~=~0, \cr \ell~\in \{1,\ldots, m\}. \end{align}\tag{SHC}

See my Phys.SE answer here for notation.

Even if a semi-holonomic constraint is holonomic there might be various reasons to keep the constraint and corresponding Lagrange multiplier in the model:

1. One might not be able to solve the constraint.

2. One might want to keep a symmetry/locality of the model.

3. One may be interested in calculating the corresponding constraint force.

References:

1. H. Goldstein, Classical Mechanics; Chapter 1 & 2.