- #1
Hetware
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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."?
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."?