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I am self-studying Goldstein's book "Classical Mechanics", and I need some help understanding the part where Goldstein discusses using Hamilton's principle to solve systems with holonomic constraints (Section 2.4). He writes on pg 46 (International Edition):

First consider holonomic constraints. When we derive Lagrange's equation from either Hamilton's or D'Alembert's principle, the holonomic constraint appear in the last step when the variations in the $q_i$ were considered independent of each other. However, the virtual displacements in the $\delta q_{I}$'s may not be consistent with constraints. If there are $n$ variables and $m$ constraint equations $f_\alpha$ of the form Eq. (1.37), the extra virtual displacements are eliminated by the method of Lagrange undetermined multipliers.

I do not understand the parts that the virtual displacements may not be inconsistent with constraints because earlier on in the book he defines virtual displacement as the infinitesimal change of the coordinates consistent with the forces and constraints imposed on the system at the given instant $t$ (pg 16).

What am I missing?

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    $\begingroup$ Comments to the post (v1): What edition are you reading? Are you translating the quotes from another language? E.g. Goldstein uses the word instant rather than time in the last sentence (pg. 16). More seriously, your first quote (pg. 46) seems like an incomplete truncated version of the original paragraph. $\endgroup$
    – Qmechanic
    Jul 26, 2016 at 19:11
  • $\begingroup$ @Qmechanic (1) I am reading the 3rd edition (International Edition-India Version ISBN: 978 81 317 58915) (2) It is my mistake that I wrote the word "time" instead of "instant". My copy of the book does say the word "instant" on pg 16. I just edited it. (3) I truncated the last sentence of that paragraph. I have just added that sentence, but I still don't understand. Why, or how it is possible the virtual displacements may not be consistent with holonomic constraints? $\endgroup$
    – mononono
    Jul 26, 2016 at 19:33
  • $\begingroup$ Your first quote (v4) is still quite far from the original paragraph. $\endgroup$
    – Qmechanic
    Jul 26, 2016 at 19:44
  • $\begingroup$ @Qmechanic I will grab the original book and read the original paragraph $\endgroup$
    – mononono
    Jul 26, 2016 at 19:50
  • $\begingroup$ @Qmechanic I had the change to read the original paragraph. Am I correct that in case of non holonomic constraints, the virtual displacements of the generalized coordinates could be inconsistent with the constraints since the coordinates may not contain the constraints in them implicitly? $\endgroup$
    – mononono
    Jul 27, 2016 at 19:59

2 Answers 2

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Consider a bead sliding on a thin, rigid rod in the $x$-direction. An example of virtual infinitesimal displacement that is inconsistent with the constraints is $$ \delta \mathbf r = (0, \delta y, 0), \qquad \delta y \neq 0 $$ because this displacement represents the bead moving away from the rod.

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    $\begingroup$ But, isn't a virtual displacement "defined" as an infinitesimal displacement that is consistent with constraints? $\endgroup$
    – mononono
    Jul 26, 2016 at 19:49
  • $\begingroup$ @hl0202 In the bead example, if we consider the configuration manifold of the bead to $\mathbb R^3$, then there are virtual displacements (tangent vectors in $\mathbb R^3$) that are inconsistent with the constraints. On the other hand, if we consider the configuration manifold to be $\mathbb R$ (coinciding with the rod), then there are no virtual displacements (tangent vectors in $\mathbb R$) that are inconsistent with the constraints because the constraints have already been used to specify a smaller cofiguration manifold that automatically encodes the constraints. $\endgroup$ Jul 26, 2016 at 19:55
  • $\begingroup$ So basically there are two ways of modeling the motion of the bead (1) the configuration manifold is $\mathbb R^3$ and there are forces from the rod on the bead that always keep it there -- in this case it makes sense to talk of virtual displacements that are inconsistent with constraint forces. (2) The configuration manifold of the bead is $\mathbb R$ and we can completely ignore the constraint forces of the rod on the bead -- in this case it doesn't make sense to talk of virtual displacemets that are inconsistent with constraints. $\endgroup$ Jul 26, 2016 at 20:01
  • $\begingroup$ I would define a virtual displacement as a tangent vector to the configuration manifold. In the case of the bead, by modeling the system by a sufficiently large configuration manifold, we were able to produce virtual displacements inconsistent with the constraints. $\endgroup$ Jul 26, 2016 at 20:04
  • $\begingroup$ Goldstein doesn't even mention any manifold. Is it necessary to bring it in? This book is very confusing $\endgroup$
    – Kashmiri
    Sep 12, 2021 at 14:29
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I understand this comment by Goldstein as a motivation for Lagrangian multipliers.

In fact virtual displacements have to be, by definition, consistent with the constraints. Now let us consider a system with $N$ generalized coordinates $q_k$ and let us apply the Hamilton's principle. In order to get the Euler-Lagrange equations we need to consider that all the variations $\delta q_k$ are arbitrary and independent of each other. Suppose we have one constraint, $f(q_1,\ldots,q_N)=0$. This means that one of the coordinates, say $q_N$, is no longer independent and its variation $\delta q_N$ cannot be arbitrary, it has to be consistent with constraint. In order to obtain the Euler-Lagrange equation for the $q_N$ coordinate we must use the method of Lagrangian multipliers. Of course we could avoid this if we have started with $N-1$ independent generalized coordinates.

Moreover, the expression the virtual displacements in the $\delta q_k$ may not be consistent with constraints may not be the best. I suppose that by virtual displacement he really means virtual change (in the sense of Cornelius Lanczos' book, i.e. an infinitesimal change in $q_k(t)$ that does not come from an infinitesimal change in its argument but comes from an arbitrary change of the type $\epsilon\phi(t)$).

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  • $\begingroup$ So you think it's a poor choice of words from Goldstein? $\endgroup$
    – Kashmiri
    Sep 16, 2021 at 5:39
  • $\begingroup$ Quarter of my savings for your answer sir. :) $\endgroup$
    – Kashmiri
    Sep 19, 2021 at 3:14

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