I'm interested in books going more over the approach described in this post. Here is a quote of the total answer in case anything happens to it:

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.

The author said it was elementary differential geometry, but his steps seemed to be totally non-obvious to me. Hopefully someone can help me out here :)

  • $\begingroup$ is this what you need reed.edu/math/wieting/essays/Frobenius.pdf ? $\endgroup$
    – hyportnex
    May 26 at 0:14
  • 3
    $\begingroup$ this really is just elementary differential geometry, and I would even say multivariable calculus. It is the inverse/implicit function theorem at play here. Showing $df_1\wedge df_2$ is non-zero is equivalent to showing that $df_1$ and $df_2$ are linearly independent, or equivalently that $F= (f_1,f_2)$ has derivative of full rank. If this is the case, the IFT allows you to show that the level set $\{F=0\}=\{f_1=f_2=0\}$ is a codimension 2 embedded-submanifold. This is just the non-linear analogue of basic linear algebra: for a surjective linear map $T:V\to W$, $\dim \ker T=\dim V-\dim W$. $\endgroup$
    – peek-a-boo
    May 26 at 4:47


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