June 2015 - mathfun@mailman.xmission.com

[math-fun] Salamin polarization puzzle
by Warren D Smith 03 Jun '15

03 Jun '15

ES is correct that mirror+sugar does not solve his problem (sorry). More generally mirror plus anything will not solve the problem because Maxwell eqns are time-reversible and the mirror converts a beam into its time reverse. Hence, any incoming beam that gets frobulated on the way to the mirror, will get unfrobulated on the way back. So how about, (1) sugar, (2) mirror, (3) now beam is both helicity reversed and 90degree plane of polarization rotated, (4) beam now re-enters sugar, which ruins it. But, regarding (3) as the "output" it seems like succeeded in some sense.

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[math-fun] Fwd: harder 7x7x7?
by Bill Gosper 03 Jun '15

03 Jun '15

I sent this to the kids last week. Neil 3D printed two 2x3x1s and searched manually, with Burrtools, and with a too-slow Mathematica program, and was unable to find a solution in which the new pieces relaxed the constraints on the other key pieces. (Not firm) conclusion: Replacing the three 2x2x1s by two 2x3x1s makes the puzzle harder (for anyone who doesn't already know how to solve it). --rwg But replacing the the 2x2x2 and 2x2x1 by two 2x3x1s makes Conway's 5x5x5 *much* easier.http://gosper.org/7%5E3leg.pdf ---------- Forwarded message ---------- From: Bill Gosper <billgosper(a)gmail.com> Date: Wed, Oct 24, 2012 at 12:00 PM To: Neil Bickford <techie314(a)gmail.com> Cc: Julian Ziegler Hunts <julianj.zh(a)gmail.com>, Jon Ziegler < jonathan.zh(a)gmail.com>, Corey Ziegler Hunts <corwin.zh(a)gmail.com>, Michael Beeler <mikebeeler(a)verizon.net>, rcs(a)xmission.com, stan(a)isaacs.com Working Stan [Isaac]'s copy at the Gardner thing, I was reminded that it always seems easy to permute a solution so that the three 2x2x1s form a 2x2x3. The puzzle might be harder if we force this configuration by replacing the 2x2x1s with two 2x3x1s (e.g. by truncating two spare 2x4x1s), assuming that the temptation to use a 2x3x1 singly in the solution proves both frequent and fatal. If nonfatal, it would probably make the puzzle way too easy. --Bill

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[math-fun] Are magnetars baloney? What is the maximum allowed magnetic field?
by Warren D Smith 02 Jun '15

02 Jun '15

OK, looks like this has been a comedy of errors by physicists. 1.Naohiro Kanda http://arxiv.org/pdf/1106.0592v1.pdf claims the QED calculation of light by light scattering cross section was done wrong by everybody until now (year 2011) when Kanda finally did it right getting the right answer for the first time, apparently nobody ever checked it during 70 years, no matter how many times it appears in no matter how many textbooks, except once when Karplus+Kroll did it wrong out of a too-strong desire to duplicate the previous wrong result. Unless Kanda is wrong. 2.Various papers gave mutually-contradicting formulas for pair creation rate in constant electric field E: (A) (m/(hbar*c))^4 * c / (4*pi^3) * (E/Ecrit)^2 * SUM(n=1,2,...) n^(-2) * exp(-n*pi*Ecrit/E). where Ecrit = m^2*c^3/(e*hbar)=1.3*10^18 volt/meter was stated as EQ1 by Hagen Kleinert: http://users.physik.fu-berlin.de/~kleinert/kleiner_re396/396.pdf and he says it was derived by Euler+Heisenberg and later by Schwinger. This formula was also stated as EQ13 of Dunne+Schubert: http://arxiv.org/pdf/hep-th/9907190v1.pdf And it looks plausible to me, EXCEPT that I suspect there ought to be an extra factor of alpha because this effect should depend on the 4th power of e, i.e. 2nd power of alpha. (B) A different, and obviously-wrong, formula for this is EQ 15 p786 of N.D. Hari Dass: Strong magnetic fields, in "Quantum Electrodynamics of Strong Fields" edited by W.Greiner 1981. https://books.google.com/books?id=MnQECAAAQBAJ&pg=PA786#v=onepage&q&f=false (C)A third different, also obviously wrong, formula is EQ 6.41 p677 in Schwinger's paper [Phys Rev 82 (1951) 664] where he instead says the pair creation rate per time & volume, is (alpha/pi)^2 * E^2 * SUM(n=1,2,...) n^(-2) * exp(-n*pi*m^2/(eE)) which is clearly dimensionally wrong. But it and the Hari Dass formula both involve the extra factor alpha that I wanted. WHAT DO I THINK? I think intuitively that the correct answer is probably of the form K * (m/(hbar*c))^4 * c^2 * alpha^2 * (E/Ecrit)^2 * SUM(n=1,2,...) n^(-2) * exp(-n*pi*Ecrit/E) where Ecrit = m^2*c^3/(e*hbar)=1.3*10^18 volt/meter and K is some dimensionless numerical constant, but none of the above lit cites agree with my intuition. I have not attempted to redo the calculation, though. But we know for sure that at least 2 of those three physicist-camps were wrong. 3.Pair creation in constant magnetic field. On p785 Hari Dass says "a pure magnetic field is stable against pair creation." Chung Ngoc Leung, Shang-Yung Wang: Phys.Lett.B674 (2009) 344-347 say "there does not exist a maximum magnetic field in QED and the magnetized vacuum is stable for all values of the magnetic field." Bulanov et al: Phys.Rev.Lett.105 (2010) 220407 http://arxiv.org/abs/1007.4306 say "Pair creation is determined by the Poincare invariants (E^2 - B^2) and (E.B) and requires the first invariant be positive." WHAT I THINK: Two possible resolutions: (A) every single one of those bastards is wrong. A sustained magnetic field above m_e * c^2 / mu_e = 8.8*10^9 tesla should be impossible due to pair creation. The Schwinger-esque pair-creation mechanism indeed as Bulanov says really does depend on sqrt(E^2-B^2) not E [it was derived for constant E-field and zero B-field, but the resulting formula, if anybody ever managed to state it correctly, then should be correct using sqrt(E^2-B^2) in place of E, more generally] and indeed should only operate when the argument of the sqrt() is positive; BUT what Bulanov forgot to say is, there is a DIFFERENT pair creation mechanism that also operates, and it is not based on the charge*potential interaction energy, but rather on the moment*field interaction energy. No previous author, apparently, has examined this mechanism. (B) The ground state energy for an electron moving slowly in the z-direction in a huge z-directed magnetic field, is m_e*c^2 PROVIDED its moment is aligned with the field, according to http://www.springer.com/cda/content/document/cda_downloaddocument/978364236… EQ 2.24 with n=0 and beta=|e*B| and section 2.2. Note this does not depend on B. You only get energy dependence on B if your moment is anti-aligned with the field. If so, then it is NOT energetically favored to create a pair in a B-field, no matter how strong that B-field is. Which hypothesis is right -- A or B? I'm favoring B, because: Imagine this thought-experiment. 1. Start with a superconducting circular loop generating some big magnetic dipole. 2. Create N pairs near the centerpoint of the loop, making N large enough that the total magnetic dipole moment of all the pair members, cancels out the dipole moment of the loop. 3. Pull on the loop and the pairs until the loop is moved far away from all the pairs. 4. Annihilate all the pairs. If we believe hypothesis B then we believe step 2 cost energy 2*N*me*c^2, which is exactly regained in step 4. But we know step 3 cost energy. Oops. Therefore, it seems to me, we are forced to believe instead in hypothesis A. And therefore to believe a strong magnetic field will cause pair creation, and all those bastards were wrong. A big cause of their wrongness was their starting point that the B-field extended infinitely far at constant value. This confused matters and screwed up energy bookkeeping. If use realistic model with a finite dipole field, then saved from that confusion, hopefully. So I seem to have arrived at a state of: big dispute with the physicists, with me on one side, but all of them on the other for the last 85 years. -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)

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[math-fun] Salamin polarization puzzle
by Warren D Smith 01 Jun '15

01 Jun '15

Here is a puzzle concerning the optics of polarized light. Every state of polarization has its opposite. For linear polarization, it's linear but rotated 90 degrees. For ere is a puzzle concerning the optics of polarized light. Every state of polarization has its opposite. For linear polarization, it's linear but rotated 90 degrees. For circular polarization, it's circular with opposite helicity. For general elliptic polarization, it's elliptic with the ellipse rotated 90 degrees, and the helicity reversed. On the Poincaré sphere, opposite states of polarization are represented by diametrically opposite points. The puzzle is to construct an optical device that reverses the polarization state. For any input, the output is the opposite polarization. Or, prove that it can't be done. -- Gene. For general elliptic polarization, it's elliptic with the ellipse rotated 90 degrees, and the helicity reversed. On the Poincaré sphere, opposite states of polarization are represented by diametrically opposite points. The puzzle is to construct an optical device that reverses the polarization state. For any input, the output is the opposite polarization. Or, prove that it can't be done. -- Gene --WDS: There is a sense in which it cannot be done for a flow-thru device. For linear polarization, it can be done, you just rotate the plane of polarization by 90 degrees by passing it thru a sugar solution; also faraday effect works. But when you said "for circular polarization, it's circular with opposite helicity" that seems to be the death blow. Because the input is photons with clockwise spin while the output is photons with anticlockwise spin. This means, in view of conservation of angular momentum, your device will be forced into rotation, faster the longer you run it. I.e. it is applying torque to your light, and the light is applying torque to it. So this is not, and cannot be, a "passive device" of the usual 100% efficiency optical variety. Having said that, maybe this does not bother you. And anyhow here is a "back at you" device: An ordinary mirror in combination with sugar solution layer of the correct thickness will input a light beam (normal to mirror) and output one (reflected) that is both circularly helicity-reversed and also linear polarization rotated 90. Does that satisfy you?

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Re: [math-fun] A polarization puzzle
by Henry Baker 01 Jun '15

01 Jun '15

Are mirrors allowed (in addition to other types of polarizers) ? Don't mirrors invert most everything about light? Rotations in geometry are a combination of 2 reflections; perhaps mirrors can help here. At 01:19 PM 6/1/2015, Adam P. Goucher wrote: >I don't know enough about optics to know whether this is correct or not, but I can believe that a negative solution might be of the form `polarising filters can only apply orientation-preserving transformations to the PoincarÃ© sphere, whereas the antipodal map is orientation-reversing'. > >Sincerely, Adam P. Goucher >> Sent: Monday, June 01, 2015 at 9:10 PM >> From: "Eugene Salamin via math-fun" <math-fun(a)mailman.xmission.com> >> To: math-fun <math-fun(a)mailman.xmission.com> >> Subject: [math-fun] A polarization puzzle >> >> Here is a puzzle concerning the optics of polarized light. >> Every state of polarization has its opposite. For linear polarization, it's linear but rotated 90 degrees. For circular polarization, it's circular with opposite helicity. For general elliptic polarization, it's elliptic with the ellipse rotated 90 degrees, and the helicity reversed. On the PoincarÃ© sphere, opposite states of polarization are represented by diametrically opposite points. >> The puzzle is to construct an optical device that reverses the polarization state. For any input, the output is the opposite polarization. Or, prove that it can't be done. >> -- Gene

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[math-fun] Wilson's grid problem (for scaled, not rotated, grids)
by Warren D Smith 01 Jun '15

01 Jun '15

The ultimate generalization of the result would be if, in any dimension, for any nondegenerate point lattice, for any translation and volume-preserving-affine transformation of grid G1 (result: G2), there exists a matching of the points of G1 to the points of G2, such that each distance is bounded by a constant (but this constant may vary depending on the lattice, etc -- but for any particular choice of all that stuff, a finite constant exists). So far, I, following Wilson's 2D idea, proved this provided the "affine" was a "rotation." We now want to remove that restriction. The theory of the SVD (singular value decomposition) from linear algebra shows (I claim) that it will suffice to solve this problem in the (other) restricted case when the affine is a pure-scaling, i.e. a diagonal matrix. So: PROBLEM: Given a lattice G1, apply a diagonal matrix with determinant=1 to it to get lattice G2. To prove: always exists a matching of G1's to G2's points, with distance bounded by a finite constant depending on G1 and the diagonal matrix. This in turn would follow if we could show the LEMMA: there must exist some nonsingular affine transformation A, which causes both G1 and G2 to become the same (i.e. isometric) lattice, up to a rotation. I believe I can prove this in 2 dimensions, but my proof does not work in any higher dimension. This would confirm what (I think) Wilson was claiming, i.e. that he'd proved the maximally generalized claim at top, but in 2 dimensions only. -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)

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[math-fun] Negatively-curved toast
by James Propp 01 Jun '15

01 Jun '15

My daughter Eliana was holding her cream-cheese-coated banana bread by two opposite corners this morning, creating a saddle shape that I realized could be imposed on the bread more durably if it were toasted. So, I put a fresh piece of bread in the toaster oven with two support-pillars made up of bread: one short pillar for the middle of the intact piece of bread, and two tall pillars for two diagonally-opposite corners. Press the button and what do you get? (drumroll) CURVED TOAST! (Cf. https://www.youtube.com/watch?v=lRlcpKu9aqI ) Anyway, the result looks like http://jamespropp.org/NegativelyCurvedToast.jpg . Note the ironic juxtaposition of a positively-curved globe in the background. Jim Propp

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[math-fun] Pairings in bipartite graphs
by Allan Wechsler 01 Jun '15

01 Jun '15

I thought of this problem when musing about Wilson's grid problem. Suppose we have a bipartite graph, with N vertices in each of two sets A and B. Further suppose that the graph is regular of degree k. For what values of N and k can we prove that there exists a bijection f between A and B such that a vertex and its partner are always connected by an edge of the original graph? The application to Wilson's problem is that there exists some radius r for which the points of one grid are always within r of at least k points of the other grid. This creates an infinite bipartite graph, and if such graphs always support pairings, that would establish r as an upper bound on the distance in Wilson's problem.

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