Lagrange Multipliers and Virtual Work: Are Joos & Freeman wrong?

Question

I have come to suspect that the treatment of virtual work in configuration space using Lagrange multipliers given here "Theoretical Physics, by Georg Joos & Ira M. Freeman, pg 114" is not correct. He begins with the expression

$$\sum_i F_i\delta x_i=0 \tag{17'}$$

for a system of $N$ particles in static equilibrium. Introduces $l$ equations of condition

$$f_k(x_1,x_2,...,x_{3N})=0\tag{17''}$$

which reduces the number of independent $\delta x_i$ by $l$.

"Differentiation" of the equations of condition produces

$$\frac{\partial f_k}{\partial x_1}\delta x_1 + \frac{\partial f_k}{\partial x_2}\delta x_2 + ... + \frac{\partial f_k}{\partial x_{3N}}\delta x_{3N} = 0.\tag{18}$$

These $l$ null expressions are then multiplied by corresponding $\lambda_k$ which are provisionally undetermined constants, then added to the original equation of static equilibrium. The resulting summation is then rearranged to produce

$$\sum_i \left( F_i + \lambda_1 \frac{\partial f_1}{\partial x_i} + \lambda_2 \frac{\partial f_2}{\partial x_i} + ... +\lambda_l \frac{\partial f_l}{\partial x_i} \right)\delta x_i= 0.\tag{19}$$

I can make sense of all of the above. But then he adjusts the $\lambda_k$ so that the last $l$ terms in the last summation vanish. He then asserts that the $\delta x_i$ multiplying the remaining $3N-l$ terms are independent.

A cursory reading might lead one to accept that claim, but it appears to me that some or all of the $f_k$ might be constant for $x_i|i=3N-l,...,3N$. The values of the corresponding $\lambda_k$'s would thereby be completely arbitrary. Furthermore, I don't see how removing the last $l$ of the $\delta x_i$ from consideration renders the remaining $\delta x_i$ mutually independent.

Is Joos & Freeman's presentation valid?

Answer 1

I) When Ref. 1 writes

let there be $\ell$ equations (17'') of constraints,

it is implicitly assumed that they are independent, as also noted in Peter Diehr's answer. Obviously this implies that $\ell\leq 3N$. Moreover, the rectangular $\ell \times 3N$ matrix $$\tag{A} \left( \frac{\partial f_k}{\partial x_i}\right)_{1\leq k\leq \ell, ~1\leq i \leq 3N} $$ must have rank $\ell$.

II) However, OP has a point. Ref. 1 forgets to inform the reader that it is assumed that the square $\ell \times \ell$ submatrix $$\tag{B} \left( \frac{\partial f_k}{\partial x_i}\right)_{1\leq k\leq \ell, ~3N-\ell+1\leq i \leq 3N}$$ is invertible. This can be achieved by possibly relabelling the $3N$ coordinates $x_i$ appropriately. Then one may uniquely find Lagrange multipliers $(\lambda_1,\ldots ,\lambda_{\ell})$ such that the $\ell$ last terms in eq. (19) vanish. (The rest of the proof in Ref. 1 is correct.)

References:

1. Georg Joos & Ira M. Freeman, Theoretical Physics, p. 114.

Answer 2

The equations of constraint must be independent of each other - otherwise they would represent the same constraint; this independence is what guarantees that Lagrange's theorem is satisfied. Joos' method is correct if the conditions are met.

An Introduction to Lagrange Multipliers provides a geometric analysis which is often helpful in understanding mathematical concepts. The constraints become gradients, and as they are independent, each points in a different direction; their removal allows the subspace represented by the constraints to be removed from the solution space without all of the work required to transform the equations via a new set of independent coordinates.