lecture 19. the method of zs - university of rochester · when problems get complicated numerical...
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Lecture 19. The Method of Zs
When problems get complicated numerical complexity makes computa7on SLOW
The method of Zs speeds the computa7on up
We will eventually look at the third problem of problem set 5 which is computa7onally imprac7cal (at least in Mathema7ca) without Zs
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We set this in the context of Hamilton’s equa7ons with nonholonomic constraints
I generally replace connec7vity constraints with pseudononholonomic constraints
When I do this we recall that we can write
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˙ q i = S ji u j
The momentum can be wriIen in terms of uj.
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pi =∂L∂ ˙ q i
/.{˙ q i → S ji u j}
where I am borrowing Mathema7ca’s subs7tu7on operator
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The game is to isolate the us, wri7ng the deriva7ves as coefficients 7mes us
I will denote the coefficients in the momentum (Hamilton and reduced Hamilton) equa7ons by Zs
This is going to seem a wee bit weird Bear with me. It’ll make sense eventually
We note that pi is linear in uj, which we can write as
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pi = Zijuj
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We can obtain the Zs from the usual momentum expression
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pi = Mij ˙ q j = Mik ˙ q k = MikS jku j ⇒ Zij = MikS j
k
or
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Zij =∂pi∂u j
The laIer is oQen easier
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We already have the equa7ons for the evolu7on of q:
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˙ q i = S ji u j
Hamilton’s equa7ons become
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˙ p i = Zij ˙ u j + ˙ Z ijuj =
∂L∂qi + Qi
where any explicit
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˙ q i = S ji u j
We need to replace
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˙ Z ij =∂Zij
∂qk ˙ q k =∂Zij
∂qk Smk um
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So that we have Hamilton’s equa7ons in the form
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Zij ˙ u j +∂Zij
∂qk Smk umu j =
∂L∂qi + Qi
Remember that these are not actually correct because I have not wri@en the Lagrange mulBpliers
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Zij ˙ u j +∂Zij
∂qk Smk umu j =
∂L∂qi + Qi + λkCi
k
is the correct form
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We have learned to mul7ply by S to obtain the correct form the reduced Hamilton’s equa7ons
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Zij ˙ u jSni +
∂Zij
∂qk Smk umu jSn
i =∂L∂qi Sn
i + QiSni
We need to solve these simultaneously with the q equa7ons
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˙ q i = S ji u j
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∂L∂qi
is, formally, a terribly complicated expression
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∂L∂qi =
12
˙ q p∂M pq
∂qi ˙ q q − ∂V∂qi
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∂L∂qi
=12Smpum
∂Mpq
∂qiSnqun − ∂V
∂qi
However, if we relegate the connec7vity constraints to the pseudononholonomic world most of the deriva7ves of M are zero, and it is not at all bad We will see this in context
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Zij ˙ u jSni +
∂Zij
∂qk Smk umu jSn
i =12
Smp um ∂M pq
∂qi Snqun −
∂V∂qi
Sn
i + QiSni
The full formal expression for the reduced Hamilton’s equa7ons
This is a set of nonlinear first order ordinary differen7al equa7ons in u the coefficients of which are func7ons of q, oQen quite complicated func7ons
This would be a snap to integrate numerically were the coefficients to be constant
The method of Zs pretends that they are
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I have adapted the method of Zs from Kane and Levinson’s 1983 paper, cited in the text
Their work used Kane’s method, which I explore in the text, but not in class
My method of Zs is not as complete as it might be, because I allow some of the coefficients to remain as explicit func7ons of q
This relies on
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∂L∂qi
and Sij actually being preIy simple func7ons of q
(and also the generalized forces)
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ZijSni ˙ u j +
∂Zij
∂qk Smk Sn
i umu j =∂L∂qi Sn
i + QiSni
Write Hamilton’s equa7ons as follows
We replace the Z variables by constants
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Zij → Tij , ∂Zij
∂qkSmk → Tijm
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TijSni ˙ u j + TijmSn
i umu j =∂L∂qi Sn
i + QiSni
This choice of subs7tu7ons aligns with Hamilton’s equa7ons, not the reduced equa7ons
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Of course, they are not constants, so we have to add algebraic equa7ons to our system to allow us to calculate them as we go forward
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Tij = Zij qk( ), Tijm =
∂Zij qp( )
∂qkSmk qq( )
That has been a lot to digest, and we’ll go through an example shortly
Let me try to summarize this first
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the simple holonomic constraints
generalized coordinates
Lagrangian
nonholonomic and pseudononholonomic constraints
Hamilton’s equa7ons
Start from square one
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˙ q i = S ji u j
introduce the Zs
reduced Hamilton’s equa7ons with Zs
generalized forces
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How do we set this up?
Set the K axis of the big sphere horizontal as before for rolling
I can choose the iner7al axis such that the small sphere is on the i axis without loss of generality
If all I care about is the ball rolling down star7ng from rest then the simplest thing to do is but its K axis horizontal as well
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It might be fun to start the ball rolling on a line of la7tude
The easiest way to do that is to put the K axis on a line of longitude which we can do by the proper choice of φ and θ.
We want K to be
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K 2 =
−cosζcosη−cosζsinηsinζ
Pick
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φ2 =η+π2
, θ2 = ζ −π2