# Joos's Treatment of Virtual Displacement

1. Oct 28, 2012

### Hetware

I having a bit of trouble understanding Joos's treatment of virtual displacements. I am referring to pages 114 and 115 of Theoretical Physics, By Georg Joos, Ira M. Freeman

If I consider, for example, an iron mass suspended by a rigid non-magnetic rod, length $a$ of negligible mass, and universally pivoted at the origin of my coordinate system, I can write the equation of constraint as:

$f(x,y,z)=x^2+y^2+z^2-a^2=0$

Now $\partial _xf=2x=0$, $\partial _yf=2y=0$, $\partial _zf=2z=0$. So forming the expression corresponding to equation VI-18 leads to the astounding conclusion that:

$\partial _xf\delta x+\partial _yf\delta y+\partial _zf\delta z=2(x\delta x+y\delta y+z\delta z)=0(\delta x+\delta y+\delta z)=0$

According to VI-19, I form:

$(\vec{F}+\lambda \vec{\nabla f})\cdot \vec{\delta r}=0$

Which stands to reason since, by the assumption of static equilibrium $\vec{F}\cdot \vec{\delta r}$, and we also have $\lambda \vec{\nabla f}\cdot \vec{\delta r}=\vec{0}\cdot \vec{\delta r}=0$.

The text following equation VI-19 states that I can choose λ such that $F_{z}+\lambda \partial _zf=0$.

Either I'm missing something stated in the text, or there are unstated assumptions being made.

From my limited understanding of Lagrange multipliers, I might form $g(x,y,z)=x^2+y^2+z^2$ and insist that $\vec{F}+\lambda \vec{\nabla g} =0$. I could then choose $\lambda$ such that $F_{z}+\lambda \partial _zg=0$, so long as $\partial _zg≠0$.

Am I correct in understanding that the $\partial _xf,\partial _yf,\partial _zf$ are to be interpreted as behaving as my function $g(x,y,z)=x^2+y^2+z^2$ would have them behave?

If my understanding is correct, then what proof do I have that "In this sum of 3N terms we can select the multipliers in such a way that the last $l$ terms vanish."?

2. Oct 28, 2012

### AlephZero

Yes the partial derivatives are $\partial _xf=2x$ etc, but why do you say $x = y = z = 0$? Clearly they are not all 0 if the particle is at distance $a$ from the origin.

3. Oct 28, 2012

### Hetware

Well, if $\delta x, \delta y, \delta z$ are arbitrary, and $\partial_{x}f,\partial_{y}f, \partial_{z}f$ are non-zero, then $\partial_{x}f \delta x+\partial_{y}f \delta y + \partial_{z}f \delta x \neq 0$. Also, $\partial_{x}f \Longleftrightarrow \frac{d}{dx}x^2=-\frac{d}{dx}c$ where $c=y^2+z^2-a^2$ which is a constant by the definition of partial differentiation. I really don't know how to interpret the result. I agree that it appears contradictory.

4. Oct 28, 2012

### AlephZero

$\delta x, \delta y, \delta z$ are not all arbitrary. There is a constraint between them, because the particle can only move on the surface of a sphere.

The point of doing it this way is that it's easier to keep all 3 of $\delta x, \delta y, \delta z$ and impose the constraint, rather than pick any two as arbitrary quantities and eliminate the other one.

The constraint that the particle DOES stay on the surface of the sphere is the equation
$\partial_{x}f \delta x+\partial_{y}f \delta y + \partial_{z}f \delta z = 0$ or $2x\delta x+ 2y\delta y + 2z\delta z = 0$.

For example if the particle is at (a,0,0), the constraint is then $2a\delta x = 0$, in other words $\delta x = 0$ and $\delta y$ and $\delta z$ are arbitrary. But if the particle is at some "random" point like (a/3, 2a/3, -2a/3) the constraint is a linear relationship between $\delta x$, $\delta y$ and $\delta z$.

Last edited: Oct 28, 2012
5. Oct 28, 2012

### Hetware

Lanczos treats this somewhat differently, beginning on Page 43 of The Variational Principles of Mechanics.

He explicitly states that the $\delta {u}_i$ are not all independent of each other. Indeed, he argues that if they were mutually independent then all $\partial_{u_{i}}f=0$. He also insists that these partials do not all vanish, which leads back to my problem with how to interpret their meaning. I'm almost certain that I should understand them to mean $\partial_{u_{i}}f=0$ evaluated in the neighborhood of $f=0$, and not to treat f as a constant value 0.