Question about constraint in lagrangian

Question

We know that the lagrangian can be written using constraints.

Suppose I have constraint functions $$f_1=\cos(x) - (x+t)/R=0,\qquad f_2=\sin(x)-y/R=0 .$$

But I know that $1=\cos^2+\sin^2$, so do we use the the $f_1$ and $f_2$ constraints with this one, or only $1=\cos^2+\sin^2$?

In other words, can we combine two constraints into one constraint for the lagrangian formalism?

Answer 1

Calculating the wedge product $df_1\wedge df_2$ gives $$ df_1\wedge df_2=\left(\sin x+\frac{1}{R}\right)\frac{1}{R}dx\wedge dy +\frac{1}{R^2}dt\wedge dy-\frac{\cos x}{R}dt\wedge dx, $$ which is nonzero, hence the constraints $f_1$ and $f_2$ are independent. But then they cannot be replaced by a single constraint function.

Said differently, in the $(t,x,y)$-space the equations $f_1=f_2=0$ determine a curve, while a single constraint function would give a two-dimensional surface.

In fact, unless there is another coordinate (eg. $z$) that does not appear in the constraints, the constraints determine the motion completely and no further degrees of freedom are left.

Answer 2

you can combine the two holonomic constraints into one non- holonomic constraint equation.

from

$$ \mathbf F_c=\left[ \begin {array}{c} \cos \left( x \right) -{\frac {x+t}{R}} \\ \sin \left( x \right) -{\frac {y}{R}}\end {array} \right] =\mathbf 0\tag 1$$ thus

$$\mathbf{\dot{F}}_c= \underbrace{ \left[ \begin {array}{cc} -{\frac {\sin \left( x \right) R+1}{R}}&0 \\ \cos \left( x \right) &-\frac 1R\end {array} \right]}_{\mathbf C} \,\begin{bmatrix} \dot{x} \\ \dot{y} \\ \end{bmatrix}-\begin{bmatrix} \frac 1R \\ 0 \\ \end{bmatrix}=\mathbf 0\tag 2$$

to solve this equation for $~\dot x~,\dot y~$ the determinate of the matrix $~\mathbf C~$ must be unequal zero this mean also that the constraint equations (Eq. (1) are independent .

the solution of eq. (2) is:

$$\dot x=-\frac{1}{\sin(x)\,R+1}\quad, \dot y=-\frac{R\,\cos(x)}{\sin(x)\,R+1}\quad\Rightarrow$$

$$ \dot y-\cos(x)\,R\dot x=0\tag 3$$

Eq. (3) is non holonomic constraint equation that equivalent to the two holonomic constraint equations Eq. (1) .

from here you can also see that you obtain from two degrees of freedom $~(x~,y)~$ one generalized coordinate .