Can the Lagrange Multipliers depend on the coordinates?

Question

When dealing with Lagrange multipliers to solve systems with constraints we usually have two ways if the constraints are holonomic:

• Differentiate the constraint and add the appropiate term to the Euler-Lagrange EOM's: $$ g_j(q_i,t) \rightarrow d g_j= \frac{\partial g_j}{\partial q_i}dq_i + \frac{\partial g_j}{\partial t} $$ then the EOM read: $$ \frac{d}{dt}\left(\frac{\partial L}{\partial \dot{q}_i}\right) - \frac{\partial L}{\partial q_i} = \sum_j\lambda_j\frac{\partial g_j}{\partial q_i},\quad \forall i $$

• Or add the constraint to the Lagrangian and treat the multipliers $\lambda_j$ as new coordinates: $$ L'(q_i,\dot{q}_i,t;\lambda_j) = L(q_i,\dot{q}_i,t) + \sum_j \lambda_j g_j(q_i,t) $$ and we would get the same equations. (See for example David Tong's Chapter 2 on Classical Dynamics)

But by the second way, it seems to imply that the multipliers $\lambda_j$ don't depend in the generalized coordinates, but for example in the simple pendulum: $$ L = \frac{1}{2}m (\dot{r}^2 + r^2\dot{\phi}^2) + mgr(1-\cos(\phi)) $$ with the constraint $g= r-l$, if we solve for $\lambda$ we can arrive to the following expresion: $$ \lambda = -ml\dot{\phi} -mg(1-\cos(\phi)) $$ and that it is also the reaction force, as the reaction force is written $Q_i = \sum_j \lambda_j\frac{\partial g_j}{\partial q_i}$ (See Classical Mechanics: Systems of Particles and Hamiltonian Dynamics Walter Greiner, Ch. 16)

Then, can we say that the multipliers $\lambda_j$ depend on the coordinates and their velocities and may be their accelerations? Wouldn't this contradict the second way to do the derivations?

Answer 1

1. Off-shell, meaning without assuming the Lagrange equations and the constraints, the Lagrange multipliers $\lambda^a(t)$ does by definition not depend$^1$ on the dynamical variables $q^j(t)$.

   (Hence, in the context of point mechanics, which we assume here, the Lagrange multipliers $\lambda^a(t)$ depend on time $t$. Correspondingly, in the context of field theory, the Lagrange multipliers depend on the spacetime coordinates.)

2. On-shell, meaning using the Lagrange equations and the constraints, the Lagrange multipliers $\lambda^a(t)$ may, as a consequence, depend on the dynamical variables $q^j(t)$ and derivatives thereof.