[math-fun] Galileo and the cycloid
I find unanimity on the web that Galileo thought the semicircle was tautochronous, i.e., that the pendulum period was amplitude-invariant. I find this incredible. He was a great experimenter. One site excuses his error by quoting some small % difference between low and high amplitudes. But I calculate that for a string length of 1.00544 ft = g/32 sec^2, the small amplitude period is 2 π/√32 = 1.107 sec, whereas the theoretical period for a full semicircular swing is ¼!^2 √(8/π) = 1.311 sec. A related mystery is why Galileo was fascinated by the cycloid. He *named* it. He couldn't compute its area, so he made a metal one and weighed it. Could he have at least suspected it was the tautochrone? Neil points out that this image http://www.scitechantiques.com/cycloidhtml/images/chrisy_cycloid_SML.jpg from a slightly dodgy website, is probably photoshopped. The page (with a broken video) claims that this beautiful piece of cabinetry admirably demonstrates tautochronicity. How could it, with rolling spheres? Their centers of mass do not follow a cycloid, and they have angular momentum. An interesting problem would be to manufacture heavy balls with very low moment of inertia. Neil suggests instead a lightweight cylinder with an osmium spindle. To reduce friction, maybe taper the ends slightly and run it on a pair of thin rails. A challenge to Veits' Mechanics geeks might be to fudge the museum cycloid to make a true tautochrone curve for a sphere of specific mass and diameter. Would the the mass actually matter? Hard to see how, given the Leaning Tower results. This exhibit would convince all the average kids, and confuse the really smart ones. How would George Hart handle this moral dilemma? --rwg
Bill, For the ball-rolling exhibit at MoMath, I specified custom balls with a metal core and foamy "mantle" to reduce the moment of inertia. But I'm no longer working with MoMath, so I'm not sure about the status of them. There are plenty of sources of error and the timing differences between different tracks are not large enough to make detailed comparisons, but the main lesson which does come through is that the straight line track, though shortest in space, is not shortest in time. Tangentially related: An excellent recent book on Galileo is "Galileo's Muse" by Mark Peterson. It doesn't say much about the cycloid, but explains the importance of mathematical ideas from art that aided his development of science. George http://georgehart.com/ P.S. Check out some of my cool math videos: http://www.youtube.com/channel/UCTl0dASnxto6j2wlVs5Bs2Q On 11/21/2012 12:33 AM, Bill Gosper wrote:
I find unanimity on the web that Galileo thought the semicircle was tautochronous, i.e., that the pendulum period was amplitude-invariant. I find this incredible. He was a great experimenter. One site excuses his error by quoting some small % difference between low and high amplitudes. But I calculate that for a string length of 1.00544 ft = g/32 sec^2, the small amplitude period is 2 π/√32 = 1.107 sec, whereas the theoretical period for a full semicircular swing is ¼!^2 √(8/π) = 1.311 sec.
A related mystery is why Galileo was fascinated by the cycloid. He *named* it. He couldn't compute its area, so he made a metal one and weighed it. Could he have at least suspected it was the tautochrone?
Neil points out that this image http://www.scitechantiques.com/cycloidhtml/images/chrisy_cycloid_SML.jpg from a slightly dodgy website, is probably photoshopped. The page (with a broken video) claims that this beautiful piece of cabinetry admirably demonstrates tautochronicity. How could it, with rolling spheres? Their centers of mass do not follow a cycloid, and they have angular momentum.
An interesting problem would be to manufacture heavy balls with very low moment of inertia. Neil suggests instead a lightweight cylinder with an osmium spindle. To reduce friction, maybe taper the ends slightly and run it on a pair of thin rails.
A challenge to Veits' Mechanics geeks might be to fudge the museum cycloid to make a true tautochrone curve for a sphere of specific mass and diameter. Would the the mass actually matter? Hard to see how, given the Leaning Tower results. This exhibit would convince all the average kids, and confuse the really smart ones. How would George Hart handle this moral dilemma? --rwg _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
On Nov 21, 2012, at 12:33 AM, Bill Gosper <billgosper@gmail.com> wrote:
A challenge to Veits' Mechanics geeks might be to fudge the museum cycloid to make a true tautochrone curve for a sphere of specific mass and diameter. Would the the mass actually matter?
It'll have to wait till next semester -- right now I'm teaching statistical mechanics. Homework: Find the probability a random NxN complex Hermitian matrix has only positive eigenvalues.
Homework question: Is there a more-or-less obvious unique natural probability measure on the space of NxN complex Hermitian matrices? --Dan On 2012-11-21, at 5:46 PM, Veit Elser wrote:
Homework: Find the probability a random NxN complex Hermitian matrix has only positive eigenvalues.
On Wed, Nov 21, 2012 at 9:14 PM, Dan Asimov <dasimov@earthlink.net> wrote:
Homework question: Is there a more-or-less obvious unique natural probability measure on the space of NxN complex Hermitian matrices?
There's a canonical left-translation-invariant measure, up to a constant factor, on any topological group, called Haar measure. For a compact Lie group, it's also right translation invariant, and the measure of the whole group is finite, so you can normalize to make the measure of the whole group 1. Andy
Sure, but . . . the space of NxN Hermitian matrices is noncompact, so using Haar measure seems likely to result in a space of infinite measure. A method I think would work might be to use the standard normal distribution on the real vector space of dimension 2N^2 (total measure = 1), and then use the usual restriction-of-a-normal-distribution-to-a-subspace to define the measure on the (real dim N^2) subspace of Hermitian matrices. But maybe there's something simpler. --Dan On 2012-11-21, at 8:32 PM, Andy Latto wrote:
On Wed, Nov 21, 2012 at 9:14 PM, Dan Asimov <dasimov@earthlink.net> wrote:
Homework question: Is there a more-or-less obvious unique natural probability measure on the space of NxN complex Hermitian matrices?
There's a canonical left-translation-invariant measure, up to a constant factor, on any topological group, called Haar measure. For a compact Lie group, it's also right translation invariant, and the measure of the whole group is finite, so you can normalize to make the measure of the whole group 1.
Andy
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
On 11/21/2012 5:46 PM, Veit Elser wrote:
On Nov 21, 2012, at 12:33 AM, Bill Gosper<billgosper@gmail.com> wrote:
A challenge to Veits' Mechanics geeks might be to fudge the museum cycloid to make a true tautochrone curve for a sphere of specific mass and diameter. Would the the mass actually matter? It'll have to wait till next semester -- right now I'm teaching statistical mechanics.
Homework: Find the probability a random NxN complex Hermitian matrix has only positive eigenvalues.
2^(-N). Brent
I have to agree with Brent, since any reasonable measure should respect the symmetry M -> -M (and matrices with at least one 0 eigenvalue should constitute a set of measure 0 among the Hermitians). Nice! --Dan Brent wrote: Veit wrote:
Homework: Find the probability a random NxN complex Hermitian matrix has only positive eigenvalues.
2^(-N).
On Nov 22, 2012, at 2:26 AM, Dan Asimov <dasimov@earthlink.net> wrote:
I have to agree with Brent, since any reasonable measure should respect the symmetry M -> -M (and matrices with at least one 0 eigenvalue should constitute a set of measure 0 among the Hermitians).
Nice!
--Dan
That only shows the probability of 5 negative and 9 positive eigenvalues is the same as 9 negative and 5 positive. The claim 2^(-N) assumes much more, that the eigenvalues are independently distributed. In fact, they are very far from independent, making the actual probability much smaller: c^(-N^2). This result is asymptotic, for large N (the limit of interest in statistical mechanics). The number c is well known, but it is not 2. As Andy pointed out, the probability measure should be invariant under arbitrary unitary transformations, i.e. M -> U M U^(-1). But the Hermitian matrices live in an N^2 dimensional space while the space of unitary matrices has only N(N-1) dimensions. The extra N dimensions correspond to the eigenvalues of M. Wigner had the idea of using the maximum entropy probability distribution, constrained by just two properties: the expectation values of Tr M and Tr M^2. If we want the expectation value of Tr M to be zero, then our probability distribution is simply the Gaussian e^(-Tr M^2) times the unitary-invariant measure. If you marginalize this distribution on just the eigenvalues (i.e. integrate out the unitary transformations) you get, say in the case of N=3 (unnormalized) dP = e^(-E1^2-E2^2-E3^2) (E1-E2)^2 (E2-E3)^2 (E3-E1)^2 dE1 dE2 dE3. It's the product over all eigenvalue pairs -- their differences squared -- that ruins the independence of the eigenvalue distribution. BTW, this very same distribution seems to perfectly model the distribution of Riemann zeta function zeros, but nobody understands why! -Veit
Brent wrote:
Veit wrote:
Homework: Find the probability a random NxN complex Hermitian matrix has only positive eigenvalues.
2^(-N).
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
I believe that Veit is referring to the *Gaussian Unitary Ensemble" in which the choose the strictly upper triangular entries independently with a N(0,1)_C distribution, and choose the diagonal entries (which must be real for a Hermitian matrix) is a N(0,1)_R distribution (each of the distributions is independent), where by N(a,sigma)_F we mean a normal distribution with mean a, and variance sigma (in the C case the covariance is sigma). Victor PS. A slight spoiler. Here's Terry Tao's take on this: http://terrytao.wordpress.com/2010/02/23/254a-notes-6-gaussian-ensembles/ On Thu, Nov 22, 2012 at 9:38 AM, Veit Elser <ve10@cornell.edu> wrote:
On Nov 22, 2012, at 2:26 AM, Dan Asimov <dasimov@earthlink.net> wrote:
I have to agree with Brent, since any reasonable measure should respect the symmetry M -> -M (and matrices with at least one 0 eigenvalue should constitute a set of measure 0 among the Hermitians).
Nice!
--Dan
That only shows the probability of 5 negative and 9 positive eigenvalues is the same as 9 negative and 5 positive. The claim 2^(-N) assumes much more, that the eigenvalues are independently distributed. In fact, they are very far from independent, making the actual probability much smaller: c^(-N^2). This result is asymptotic, for large N (the limit of interest in statistical mechanics). The number c is well known, but it is not 2.
As Andy pointed out, the probability measure should be invariant under arbitrary unitary transformations, i.e. M -> U M U^(-1). But the Hermitian matrices live in an N^2 dimensional space while the space of unitary matrices has only N(N-1) dimensions. The extra N dimensions correspond to the eigenvalues of M.
Wigner had the idea of using the maximum entropy probability distribution, constrained by just two properties: the expectation values of Tr M and Tr M^2. If we want the expectation value of Tr M to be zero, then our probability distribution is simply the Gaussian e^(-Tr M^2) times the unitary-invariant measure.
If you marginalize this distribution on just the eigenvalues (i.e. integrate out the unitary transformations) you get, say in the case of N=3 (unnormalized)
dP = e^(-E1^2-E2^2-E3^2) (E1-E2)^2 (E2-E3)^2 (E3-E1)^2 dE1 dE2 dE3.
It's the product over all eigenvalue pairs -- their differences squared -- that ruins the independence of the eigenvalue distribution. BTW, this very same distribution seems to perfectly model the distribution of Riemann zeta function zeros, but nobody understands why!
-Veit
Brent wrote:
Veit wrote:
Homework: Find the probability a random NxN complex Hermitian matrix
has only positive eigenvalues.
2^(-N).
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
Yes. Sorry about that -- as I was falling asleep I realized I'd completely overlooked the question of independence, but didn't have the energy to get out of bed and post that . . . --Dan On 2012-11-22, at 6:38 AM, Veit Elser wrote:
I have to agree with Brent, since any reasonable measure should respect the symmetry M -> -M (and matrices with at least one 0 eigenvalue should constitute a set of measure 0 among the Hermitians).
Nice!
--Dan
That only shows the probability of 5 negative and 9 positive eigenvalues is the same as 9 negative and 5 positive. The claim 2^(-N) assumes much more, that the eigenvalues are independently distributed. In fact, they are very far from independent, making the actual probability much smaller: c^(-N^2). This result is asymptotic, for large N (the limit of interest in statistical mechanics). The number c is well known, but it is not 2.
On 11/22/2012 6:38 AM, Veit Elser wrote:
On Nov 22, 2012, at 2:26 AM, Dan Asimov<dasimov@earthlink.net> wrote:
I have to agree with Brent, since any reasonable measure should respect the symmetry M -> -M (and matrices with at least one 0 eigenvalue should constitute a set of measure 0 among the Hermitians).
Nice!
--Dan That only shows the probability of 5 negative and 9 positive eigenvalues is the same as 9 negative and 5 positive. The claim 2^(-N) assumes much more, that the eigenvalues are independently distributed.
No it doesn't assume that. It only assumes that for every Hermitian matrix there is another one that has the same eigenvalues except for a reversal of all signs and that these two matrices have equal probability. Brent
In fact, they are very far from independent, making the actual probability much smaller: c^(-N^2). This result is asymptotic, for large N (the limit of interest in statistical mechanics). The number c is well known, but it is not 2.
As Andy pointed out, the probability measure should be invariant under arbitrary unitary transformations, i.e. M -> U M U^(-1). But the Hermitian matrices live in an N^2 dimensional space while the space of unitary matrices has only N(N-1) dimensions. The extra N dimensions correspond to the eigenvalues of M.
Wigner had the idea of using the maximum entropy probability distribution, constrained by just two properties: the expectation values of Tr M and Tr M^2. If we want the expectation value of Tr M to be zero, then our probability distribution is simply the Gaussian e^(-Tr M^2) times the unitary-invariant measure.
If you marginalize this distribution on just the eigenvalues (i.e. integrate out the unitary transformations) you get, say in the case of N=3 (unnormalized)
dP = e^(-E1^2-E2^2-E3^2) (E1-E2)^2 (E2-E3)^2 (E3-E1)^2 dE1 dE2 dE3.
It's the product over all eigenvalue pairs -- their differences squared -- that ruins the independence of the eigenvalue distribution. BTW, this very same distribution seems to perfectly model the distribution of Riemann zeta function zeros, but nobody understands why!
-Veit
Brent wrote:
Veit wrote:
Homework: Find the probability a random NxN complex Hermitian matrix has only positive eigenvalues. 2^(-N).
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
participants (7)
-
Andy Latto -
Bill Gosper -
Dan Asimov -
George Hart -
meekerdb -
Veit Elser -
Victor Miller