information:classification, linguistic:translation, structural:expansion, information:filtering
Lagrange multipliers and constraints

The Lagrangian L can be varied in the Cartesian r coordinates, for N particles, ∫ t 1 t 2 ∑ k = 1 N (∂ L ∂ r k − d d t ∂ L ∂ r ˙ k) ⋅ δ r k d t = 0. {\displaystyle \int _{t_{1}}^{t_{2}}\sum _{k=1}^{N}\left({\frac {\partial L}{\partial \mathbf {r} _{k}}}-{\frac {\mathrm {d} }{\mathrm {d} t}}{\frac {\partial L}{\partial {\dot {\mathbf {r} }}_{k}}}\right)\cdot \delta \mathbf {r} _{k}\,\mathrm {d} t=0.}

Hamilton's principle is still valid even if the coordinates L is expressed in are not independent, here r, but the constraints are still assumed to be holonomic. As always the end points are fixed δ r (t) = δ r (t) = 0 for all k. What cannot be done is to simply equate the coefficients of δ r to zero because the δ r are not independent. Instead, the method of Lagrange multipliers can be used to include the constraints. Multiplying each constraint equation f (r, t) = 0 by a Lagrange multiplier λ for i = 1, 2,..., C, and adding the results to the original Lagrangian, gives the new Lagrangian L ′ = L (r 1, r 2, …, r ˙ 1, r ˙ 2, …, t) + ∑ i = 1 C λ i (t) f i (r k, t). {\displaystyle L'=L(\mathbf {r} _{1},\mathbf {r} _{2},\ldots,{\dot {\mathbf {r} }}_{1},{\dot {\mathbf {r} }}_{2},\ldots,t)+\sum _{i=1}^{C}\lambda _{i}(t)f_{i}(\mathbf {r} _{k},t).}

The Lagrange multipliers are arbitrary functions of time t, but not functions of the coordinates r, so the multipliers are on equal footing with the position coordinates. Varying this new Lagrangian and integrating with respect to time gives ∫ t 1 t 2 δ L ′ d t = ∫ t 1 t 2 ∑ k = 1 N (∂ L ∂ r k − d d t ∂ L ∂ r ˙ k + ∑ i = 1 C λ i ∂ f i ∂ r k) ⋅ δ r k d t = 0. {\displaystyle \int _{t_{1}}^{t_{2}}\delta L'\mathrm {d} t=\int _{t_{1}}^{t_{2}}\sum _{k=1}^{N}\left({\frac {\partial L}{\partial \mathbf {r} _{k}}}-{\frac {\mathrm {d} }{\mathrm {d} t}}{\frac {\partial L}{\partial {\dot {\mathbf {r} }}_{k}}}+\sum _{i=1}^{C}\lambda _{i}{\frac {\partial f_{i}}{\partial \mathbf {r} _{k}}}\right)\cdot \delta \mathbf {r} _{k}\,\mathrm {d} t=0.}

The introduced multipliers can be found so that the coefficients of δ r are zero, even though the r are not independent. The equations of motion follow. From the preceding analysis, obtaining the solution to this integral is equivalent to the statement ∂ L ′ ∂ r k − d d t ∂ L ′ ∂ r ˙ k = 0 ⇒ ∂ L ∂ r k − d d t ∂ L ∂ r ˙ k + ∑ i = 1 C λ i ∂ f i ∂ r k = 0, {\displaystyle {\frac {\partial L'}{\partial \mathbf {r} _{k}}}-{\frac {\mathrm {d} }{\mathrm {d} t}}{\frac {\partial L'}{\partial {\dot {\mathbf {r} }}_{k}}}=0\quad \Rightarrow \quad {\frac {\partial L}{\partial \mathbf {r} _{k}}}-{\frac {\mathrm {d} }{\mathrm {d} t}}{\frac {\partial L}{\partial {\dot {\mathbf {r} }}_{k}}}+\sum _{i=1}^{C}\lambda _{i}{\frac {\partial f_{i}}{\partial \mathbf {r} _{k}}}=0,} which are Lagrange's equations of the first kind. Also, the λ Euler-Lagrange equations for the new Lagrangian return the constraint equations ∂ L ′ ∂ λ i − d d t ∂ L ′ ∂ λ ˙ i = 0 ⇒ f i (r k, t) = 0. {\displaystyle {\frac {\partial L'}{\partial \lambda _{i}}}-{\frac {\mathrm {d} }{\mathrm {d} t}}{\frac {\partial L'}{\partial {\dot {\lambda }}_{i}}}=0\quad \Rightarrow \quad f_{i}(\mathbf {r} _{k},t)=0.}

For the case of a conservative force given by the gradient of some potential energy V, a function of the r coordinates only, substituting the Lagrangian L = T − V gives ∂ T ∂ r k − d d t ∂ T ∂ r ˙ k ⏟ − F k + − ∂ V ∂ r k ⏟ N k + ∑ i = 1 C λ i ∂ f i ∂ r k = 0, {\displaystyle \underbrace {{\frac {\partial T}{\partial \mathbf {r} _{k}}}-{\frac {\mathrm {d} }{\mathrm {d} t}}{\frac {\partial T}{\partial {\dot {\mathbf {r} }}_{k}}}} _{-\mathbf {F} _{k}}+\underbrace {-{\frac {\partial V}{\partial \mathbf {r} _{k}}}} _{\mathbf {N} _{k}}+\sum _{i=1}^{C}\lambda _{i}{\frac {\partial f_{i}}{\partial \mathbf {r} _{k}}}=0,} and identifying the derivatives of kinetic energy as the (negative of the) resultant force, and the derivatives of the potential equaling the non-constraint force, it follows the constraint forces are C k = ∑ i = 1 C λ i ∂ f i ∂ r k, {\displaystyle \mathbf {C} _{k}=\sum _{i=1}^{C}\lambda _{i}{\frac {\partial f_{i}}{\partial \mathbf {r} _{k}}},} thus giving the constraint forces explicitly in terms of the constraint equations and the Lagrange multipliers.