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Paraphrasing Goldstein

If we've $N$ number of particles and there exist holonomic constraints, expressed in $k$ equations, then we may use these equations to eliminate $k$ of the $3 N$ coordinates, and we are left with $3 N-k$ independent coordinates. We introduce $3 N-k$ generalized coordinates $q_{1}, q_{2}, \ldots, q_{3 N-k}$ in terms of which the old coordinates $\mathbf{r}_{1}, \mathbf{r}_{2}, \ldots, \mathbf{r}_{N}$ are expressed by equations of the form $$ \begin{aligned} \mathbf{r} &=\mathbf{r}_{1}\left(q_{1}, q_{2}, \ldots q_{3 N-k}, t\right) \\ & \vdots \\ \mathbf{r}_{N} &=\mathbf{r}_{N}\left(q_{1}, q_{2}, \ldots, q_{3 N-k}, t\right) \end{aligned} $$ It is always assumed that we can also transform back from the $\left(q_{l}\right)$ to the $\left(\mathbf{r}_{l}\right)$ set, i.e., that above equations combined with the $k$ equations of constraint can be inverted to obtain any $q_{i}$ as a function of the $\left(\mathbf{r}_{l}\right)$ variable and time.

My doubt: The above set of $N$ vector equations are $3N$ scalar equations. If we want to find all the $q_i$ we can solve $3N-k$ equations from the above set of equations and be done. In other words it's given that there are already $3N$ equations in $3N−k$ variables, which we can solve without the k constraints.

Why then does the author say that he also needs the $k$ equations of constraint?

Kashmiri
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1 Answers1

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Referring to the third edition of Goldstein's Classical Mechanics, he said that the $3N$ scalar equations which are the parametrization of the $n=3N-k$ generalized coordinates, $ q' s$, contains the $k$ constraints implicitly. Adopting the notation, $$x_{\nu}=\xi_{3\nu-2},\;\;\;y_{\nu}=\xi_{3\nu-1},\;\;\;z_{\nu}=\xi_{3\nu}\tag{1}\label{1},$$the holonomic constraint equations are written $$f_i(\xi_1,\dots,\xi_{3N};t)=0\qquad (i=1,2,\dots,k).\tag{2}\label{2} $$ After introducing the $n$ generalized coordinates, the only requirement imposed on $q's$ is that they allow determination of independent Cartesian coordinates (I have arranged all the independent Cartesian components from the index $k+1$ to $3N$, so $\xi_{k+l}$ need not be $x_{k+l}$ or $y_{k+l}$ or $z_{k+l})$, $$\xi_{k+l}=\xi_{k+l}(q_1,\dots,q_n;t)\qquad (l=1,\dots,3N-k),\tag{3}\label{3}$$ where the first $r$ $\xi's$ denote the dependent Cartesian components, $$\xi_r=\xi_r(\xi_{k+1},\dots,\xi_{3N};t)\qquad (r=1,\dots,k).\tag{4}\label{4}$$ Upon substituting Eq.\eqref{3} into Eq.\eqref{4}, we have, returning to Goldstein's notation, the above $3N$ scalar equations, which is Eq.$(1.38)$ in his aforesaid book. But how did you determine the set of independent Cartesian coordinates in the first place? You use $k$ constraints to set up a choice of this set from the $3N$ Cartesian coordinates, but the choice of this set is not unique. Let say a certain choice of this set is identified, then according to Eq.\eqref{3}, we have the invertible transformation that can solve $q's$ in terms of $\xi_{k+1},\dots,\xi_{3N}$ without resorting to the $k$ constraints. However, if such a choice of the set has been relaxed and you now look back at the $3N$ scalar equations, you need to combine the scalar equations with $k$ constraints to make up a certain choice of the independent Cartesian set. This is why Goldstein didn't refer to the Cartesian set explicitly when he wrote that statement as there could be many different choices of the set, depending on the $k$ constraints.

Kksen
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