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

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?

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.
• While I agree that relabeling the coordinates as you suggest will work, I note that Joos turns around in the next section and attacks the first $l$ terms, rather than the last ones. As if to say, "if you think I made a mistake, think again". – Steven Thomas Hatton May 11 '16 at 20:00
• @StevenHatton : Which next Section? Which page? – Qmechanic May 11 '16 at 21:11
• I stand corrected. It's in the same section at the bottom of page 116. "Exactly as in the statical case, we can dispose of the $l$ quantities $\lambda_k$ so that the first $l$ parentheses vanish." – Steven Thomas Hatton May 13 '16 at 0:04
• Obviously the argument can be fitted to any subset of $\ell$ columns of the rectangular matrix, e.g. the $\ell$ last columns, or the $\ell$ first columns, but yeah, that's confusing writing. – Qmechanic May 13 '16 at 3:59
• The presentation may have lost something in translation. The salient point I take from it is that the $\delta x_i$ can all be treated as arbitrary if the values of the $\lambda_k$ are appropriately chosen. To me, this seems important because the principle of virtual work does not necessitate displacement consistent with the given system of constraint. There are times when varying a constraint produces that desired result. A simple example is the imagined variation in the length of a member in a truss, and determining the change in potential energy that would result. – Steven Thomas Hatton May 13 '16 at 17:20

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.

• I understand the general idea of reducing the number of dimensions in the configuration space by introducing the equations of condition. I'm just not following the details of reasoning in the approach given. Take the simple case of a mass in a frictionless cylinder of unit radius with the y-axis as its axis of symmetry. $x^2 + z^2 = 1$ is the equation of condition. Its gradient at a point of static equilibrium is a normal vector parallel to the $z$-axis. If we choose to remove the term associated with the $y$ component using the stated method, $\delta x$ still depends on $\delta z$. – Steven Thomas Hatton May 11 '16 at 18:28
• @StevenHatton: But you are not simplifying the system of equations by elimination of unneeded variables; Lagrange's method allows you to remove the degrees of freedom associated with the constraints while retaining all of the variables! If you follow the process for your example (which I recommend as an exercise), you will find that it all works out -- and the geometric analysis will show you why. – Peter Diehr May 11 '16 at 22:30
• That's my point. He simply attacks the last $l$ terms without regard for their relationship to the system of constraints. I intentionally chose the $y$ term in order to illustrate what happens if the coordinate associated with the term is unconstrained. That could very well happen if the coordinates are not reindexed as Qmechanic suggested doing. – Steven Thomas Hatton May 12 '16 at 23:46
• Yes, qmechanic has given the requirements correctly. Mathematics texts tend to be more precise than those in physics, because it obscures the physical content. – Peter Diehr May 12 '16 at 23:52