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I have a question about the Lagrangian-formalism; they state that the holonomic constraints are expressed as:

$$f_\alpha(x_1,y_1,z_1,...,x_n,y_n,z_n,t) = 0 $$ where $\alpha = 1,...,L$

They state now that these constraints may not contain irrelevant information. They express this by saying: $$rank(\frac{\partial{f^{\alpha}}}{\partial{x_k}}\frac{\partial{f^{\alpha}}}{\partial{y_k}}\frac{\partial{f^{\alpha}}}{\partial{z_k}}) = L$$ With $\alpha = 1,..,L $ and $k = 1,...,N $. This looks like a Jacobi-matrix. I don't understand however how this expresses that the constraints may not contain irrelevant information. (I see that they may not be linear dependent, is there a relation between linear dependence and the Jacobi-matrix?)

Further they state that under these premises, we can solve the constraints and we can say: $z_k = z_k(x_1,y_1,z_1,...,t) $ (where $z_k$ is a symbolical notation for a parameter (so it can also be for instance $x_5$)), and $z_k$ depends on $3n-L$ parameters and the time $t$. With a regular transformation these parameters can be transformed to other generalized parameters. This should follow out of the theorem of implicit functions, but these theorem only states that we can solve $f_\alpha$ local: so I don't understand why they assume that these generalized parameters should completely determine the configuration of the system. It seems that these coordinates only determine the configuration of the system in the surrounding of a point that satisfies the constraints.

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  • $\begingroup$ Who are they? Which reference are you using? $\endgroup$
    – Qmechanic ♦
    Commented Jun 17, 2013 at 21:55
  • $\begingroup$ Just my university syllabus about Lagrangian and Hamilton mechanics. $\endgroup$
    – yarnamc
    Commented Jun 18, 2013 at 14:10
  • $\begingroup$ $\uparrow$ Is it on-line? $\endgroup$
    – Qmechanic ♦
    Commented Jun 18, 2013 at 14:18
  • $\begingroup$ No, I'm sorry ;(. It's also in Dutch by the way. $\endgroup$
    – yarnamc
    Commented Jun 18, 2013 at 21:59

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Implicit function theorem states that a relation can be transformed in a function, that is from the relation:

$$f(x_1,y_1,z_1,........x_N,y_N,z_N, t) = 0$$ You could express $x_d$ (for instance) as a function:

$x_d = x_d ((x_1,y_1,z_1,...,x_{d -1},y_{d -1},z_{d -1}, y_d,z_d,x_{d +1},y_{d +1},z_{d +1}.....x_N,y_N,z_N, t))$

The rank of the Jacobi Matrix indicates the number of linearly independent columns ( or linearly independent rows), This means that the holonomic constraints are independent equations, that is : there are really $L$ different constraints.

Furthermore, considering space variables,you have $3N$ variables $x_1,y_1,z_1,........x_N,y_N,z_N$, with $L$ true constraints, so in fact, you have only $m = 3N - L$ freedom degrees for space variables.

Adding time, we see that you have $m = 3N - L$ freedom degrees for space variables and one freedom degree for time.

Setting $q_1, q_2, .........q_m$ for the space freedom degrees, this means that every $x_d$ or $y_d$ or $z_d$ could be written as a function of $q_1, q_2, .........q_m$ and $t$

$x_d = x_i(q_1, q_2, .........q_m, t)$

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  • $\begingroup$ Thank you, but my problem was that implicit function theorem doesn't state that we can express $x_d$ as a function of the other variables. It states that we can do this in a surrounding of a point that satisfies your first equation. So I wonder, are we sure that if we have L constraints, that we can find 3N-L independent parameters that describe the motion? $\endgroup$
    – yarnamc
    Commented Jun 12, 2013 at 19:04
  • $\begingroup$ Well, OK, I would have been more precise in saying that one relation corresponds to several discrete functions. Take the implicit function $x^2 + y^2 = 1$. So, obviously, you have $y = \pm \sqrt{1 - x^2}$. But this will not infirm my caveat (because the possible space of solutions is discrete $(1,2...n)$ ,if you want, but not a continuous dimension), and the counting of $m = L - 3n$ is correct. $\endgroup$
    – Trimok
    Commented Jun 12, 2013 at 19:16
  • $\begingroup$ Aha thanks, but this implies that the motion can't be described using $L-3n$ generalized coordinates, because for every set of these $L-3n$ coordinates, multiple functions are possible to determine $x,y,...$. I feel like, that this undermines the idea of a generalized coordinate. $\endgroup$
    – yarnamc
    Commented Jun 12, 2013 at 19:36
  • $\begingroup$ There is a discrete possible set of functions, OK, but these functions are functions of $L - 3N$ generalized coordinates plus time. So the idea of generalized coordinates is correct, but there is some subtelty... $\endgroup$
    – Trimok
    Commented Jun 12, 2013 at 19:40

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