Re: [math-fun] Minsky Harmonic Motion
I should have pointed out that the Trinsky analysis of Minsky essentially produces Pythagorean triples/Gaussian primes for certain rational numbers; e.g., if d=1, delta*eps = 4/(1+4*d^2) = 4/(1+4*1^2)=4/5, and w = 2*asin(sqrt(4/5)/2) = 2*asin(1/sqrt(5)) = asin(4/5) -- i.e., we get a 3,4,5 triangle. When we compute arbitrarily high powers ((3+4*i)/5)^n, we conserve the length (i.e., energy) of the system, but we don't actually get an "orbit", since Q(i) is a field of characteristic 0. I now need to analyze the much harder case where we execute the _floor(x)_ and/or _round(x)_ function at each Minsky step. This system is the one actually discovered by Minsky, and produces a slightly blurry circle/ellipse, because it touches on a number of close by integer points and may require several "cycles" before it literally repeats.
From: Henry Baker <hbaker1@pipeline.com>
Here's one way to do "Minsky Motion", aka Minsky Physics. I haven't tried it with inverse square orbits, but here's a version for harmonic orbits.
The Minsky method of drawing a "circle", a la Minskys & Trinskys:
http://nbickford.wordpress.com/2011/04/03/the-minsky-circle-algorithm/
http://www.blurb.com/bookstore/detail/2172660?alt=Minskys+%26+Trinskys+3rd+E...
x := x - delta * y ; y := y + eps * x /* _sequential_, not parallel, updates */
or equivalently,
[x] [ 1 -delta ][x] [y] := [eps 1-eps*delta][y]
or
[x] [x] [y] := M [y], where
[ 1 -delta ] M = [eps 1-eps*delta].
But we can diagonalize M with P, s.t. P^-1 M P = diag(lambda_1,lambda_2),
where lambda_1, lambda_2 are the eigenvalues of M.
Since det(M) = 1 = lambda_1*lambda_2, lambda_2 = 1/lambda_1. Also,
trace(M) = 2-eps*delta = lambda_1+lambda_2 = lambda_1+1/lambda_1
If 0 < eps*delta < 4, then -4 < -eps*delta < 0, and -2 < 2-eps*delta < 2, and -1 < 1-eps*delta/2 < 1.
So we can choose real w=acos(1-eps*delta/2), and the eigenvalues will be complex conjugates lambda_1 = cos(w)-i*sin(w) = exp(-iw), lambda_2 = cos(w)+i*sin(w) = exp(iw).
However, it will be slightly more convenient to choose angle phi=w/2.
Since cos(w) = 1-eps*delta/2,
sin(phi) = sin(w/2) = sqrt(1-cos(w))/sqrt(2) = sqrt(1-(1-eps*delta/2))/sqrt(2) = sqrt(eps*delta/2)/sqrt(2) = sqrt(eps*delta)/2
So, phi = w/2 = asin(sqrt(eps*delta)/2), and w = 2*asin(sqrt(eps*delta)/2).
Using Maxima's eigenvectors command and a lot of manipulation,
[ delta delta ] P = [1-exp(-iw) 1-exp(iw)]
We can now construct a real matrix
[cos(-w) sin(-w)] [-sin(-w) cos(-w)]
with the same eigenvalues exp(-iw), exp(iw), and
[ cos(-w) sin(-w)] [exp(-iw) 0 ] Q^-1 [-sin(-w) cos(-w)] Q = [ 0 exp(iw)]
[1 1] with Q = [i -i].
Combining these two efforts, we get
[ 1 -delta ] [ cos(-w) sin(-w)] Q P^-1 [eps 1-delta*eps] P Q^-1 = [-sin(-w) cos(-w)]
So if with compute (Q P^-1 M P Q^-1)^n, we can get the effect of n steps, i.e.,
[ cos(-n*w) sin(-n*w)] [-sin(-n*w) cos(-n*w)]
Thus, the effect of the n-th Minsky step is
M^n = P Q^-1 (Q P^-1 M P Q^-1))^n Q P^-1
In effect, we slightly skew the space, perform n iterations, and then reskew back.
Slightly more perspicuously,
x_n = x_0*cos(-n*w)+(x_0/2-y_0/eps)*sin(-n*w)/d and y_n = y_0*cos(-n*w)+(x_0/delta-y_0/2)*sin(-n*w)/d,
where d=sqrt(1/(delta*eps)-1/4) is a real number.
---------
Now instead of considering two spatial coordinates x,y, we consider instead a spatial coordinate x and its velocity -- i.e., we consider the circle in _phase space_:
[x] [1 -delta][x] [v] := [eps 1-eps*delta][v]
This phase plot can be interpreted as a simple harmonic oscillator consisting of a weight moving forward & back on a spring.
Let k be the spring constant, and m be the mass on the end of the spring.
Then m*x'' = m*v' = m*a = F = -kx, where x'' is d^2 x/dt^2.
Also, conservation of energy requires that
E = (1/2)*k*x^2 + (1/2)*m*v^2 = (1/2)*(k*x^2+m*v^2)
If k=m in suitable units, then the phase plot of such an harmonic oscillator will be a perfect circle.
Following Minsky/Trinsky, as above, we slightly skew our (digital) phase space, compute n iterations, and then reskew back to our digital phase space.
Note that since the the harmonic oscillator in two dimensions is separable, we can do exactly the same thing for y, v_y, and simply include these 2 additional dimensions in our expanded phase space.
For a differential equation which _isn't_ separable, we could (for n=2 objects), utilize quaternions (2 position coordinates, 2 velocity coordinates), and try to keep the _length_ of the quaternion (~ total energy) constant.
On Tue, Jul 5, 2011 at 3:14 PM, Henry Baker <hbaker1@pipeline.com> wrote:
I should have pointed out that the Trinsky analysis of Minsky essentially produces Pythagorean triples/Gaussian primes for certain rational numbers;
e.g., if d=1, delta*eps = 4/(1+4*d^2) = 4/(1+4*1^2)=4/5,
and w = 2*asin(sqrt(4/5)/2) = 2*asin(1/sqrt(5)) = asin(4/5)
-- i.e., we get a 3,4,5 triangle.
When we compute arbitrarily high powers ((3+4*i)/5)^n, we conserve the length (i.e., energy) of the system, but we don't actually get an "orbit", since Q(i) is a field of characteristic 0.
While I agree we don't get an orbit, because tan(3/4)/tau is not rational, I don't see the connection with the characteristic of the field. Q((sqrt(3)) is a field of characteristic 0, but it includes (sqrt(3) + i)/2, and the twelfth power of this is 1. Andy
How does a nonreal number manage to muscle its way into Q(sqrt(3)), a real field? -- Gene
________________________________ From: Andy Latto <andy.latto@pobox.com> To: math-fun <math-fun@mailman.xmission.com> Sent: Tuesday, July 5, 2011 3:22 PM Subject: Re: [math-fun] Minsky Harmonic Motion
Q((sqrt(3)) is a field of characteristic 0, but it includes (sqrt(3) + i)/2, and the twelfth power of this is 1.
Andy
On Tue, Jul 5, 2011 at 6:40 PM, Eugene Salamin <gene_salamin@yahoo.com> wrote:
How does a nonreal number manage to muscle its way into Q(sqrt(3)), a real field?
Sorry; the rule that any post correcting an error must itself contain an error applies here. I meant Q(sqrt(3), i). Andy
-- Gene
________________________________ From: Andy Latto <andy.latto@pobox.com> To: math-fun <math-fun@mailman.xmission.com> Sent: Tuesday, July 5, 2011 3:22 PM Subject: Re: [math-fun] Minsky Harmonic Motion
Q((sqrt(3)) is a field of characteristic 0, but it includes
(sqrt(3) + i)/2, and the twelfth power of this is 1.
Andy
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-- Andy.Latto@pobox.com
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Henry Baker