1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Joos's Treatment of Virtual Displacement

  1. Oct 28, 2012 #1
    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

    http://books.google.com/books?id=vI...ce=gbs_ge_summary_r&cad=0#v=onepage&q&f=false

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

    [itex]f(x,y,z)=x^2+y^2+z^2-a^2=0[/itex]

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

    [itex]\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[/itex]

    According to VI-19, I form:

    [itex](\vec{F}+\lambda \vec{\nabla f})\cdot \vec{\delta r}=0[/itex]

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

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

    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 [itex]g(x,y,z)=x^2+y^2+z^2[/itex] and insist that [itex]\vec{F}+\lambda \vec{\nabla g} =0[/itex]. I could then choose [itex]\lambda[/itex] such that [itex]F_{z}+\lambda \partial _zg=0[/itex], so long as [itex]\partial _zg≠0[/itex].

    Am I correct in understanding that the [itex]\partial _xf,\partial _yf,\partial _zf[/itex] are to be interpreted as behaving as my function [itex]g(x,y,z)=x^2+y^2+z^2[/itex] 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 [itex]l[/itex] terms vanish."?
     
  2. jcsd
  3. Oct 28, 2012 #2

    AlephZero

    User Avatar
    Science Advisor
    Homework Helper

    Yes the partial derivatives are [itex]\partial _xf=2x[/itex] 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.
     
  4. Oct 28, 2012 #3
    Well, if [itex]\delta x, \delta y, \delta z[/itex] are arbitrary, and [itex]\partial_{x}f,\partial_{y}f, \partial_{z}f[/itex] are non-zero, then [itex]\partial_{x}f \delta x+\partial_{y}f \delta y + \partial_{z}f \delta x \neq 0[/itex]. Also, [itex]\partial_{x}f \Longleftrightarrow \frac{d}{dx}x^2=-\frac{d}{dx}c[/itex] where [itex]c=y^2+z^2-a^2[/itex] 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.
     
  5. Oct 28, 2012 #4

    AlephZero

    User Avatar
    Science Advisor
    Homework Helper

    [itex]\delta x, \delta y, \delta z[/itex] 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 [itex]\delta x, \delta y, \delta z[/itex] 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
    [itex]\partial_{x}f \delta x+\partial_{y}f \delta y + \partial_{z}f \delta z = 0[/itex] or [itex]2x\delta x+ 2y\delta y + 2z\delta z = 0[/itex].

    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
  6. Oct 28, 2012 #5
    Lanczos treats this somewhat differently, beginning on Page 43 of The Variational Principles of Mechanics.

    http://books.google.com/books?id=ZW...ce=gbs_ge_summary_r&cad=0#v=onepage&q&f=false

    He explicitly states that the [itex]\delta {u}_i[/itex] are not all independent of each other. Indeed, he argues that if they were mutually independent then all [itex]\partial_{u_{i}}f=0[/itex]. 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 [itex]\partial_{u_{i}}f=0[/itex] evaluated in the neighborhood of [itex]f=0[/itex], and not to treat f as a constant value 0.

    That is how I have previously understood this subject, but when I tried to write out an example, I realized I'm not completely clear on what the assumptions are. I know that the final result is that each constraint equation eliminates a degree of freedom. It also says that the constraint forces are normal to allowable trajectories, and therefore do no work.
     
    Last edited: Oct 28, 2012
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Joos's Treatment of Virtual Displacement
  1. Virtual displacement (Replies: 1)

  2. Virtual displacement (Replies: 3)

  3. Virtual displacement (Replies: 1)

Loading...