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?


2 Answers 2


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.

  • $\begingroup$ First time I've seen this description of it, which book you got it from? $\endgroup$ May 25 at 13:48
  • $\begingroup$ Hello, reposting in case you didnt' see my message. Could you share what you got the differential form approach to this problemo? $\endgroup$ May 25 at 21:27
  • $\begingroup$ @Aplateofmomos I didn't respond because I don't know, and did not have the adequate time and resources to look one up right now. There are many textbooks that treat constraints via differential forms and Frobenius' theorem but this is really just elementary differential geometry. $\endgroup$ May 25 at 21:30

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


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

  • $\begingroup$ but you know that in a right triangle, we can extract 3 constraints (cos(angle), sin(angle), phythagoras theorem) the first two leads to the thirds, we add them all or the first two? $\endgroup$
    – Hady
    May 28 at 10:25

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