November 2011 - mathfun@mailman.xmission.com

Re: [math-fun] Tiling puzzle
by Dan Asimov 13 Nov '11

13 Nov '11

I wrote to Gene (inadvertently not sending it to all math-fun) about his solution to the tiling problem: << Yes, that's essentially what I had in mind -- nice solving. and then I added: << Now for Part B: Is there a way to solve this puzzle (for 2 sets) so that one of the sets is a subgroup of R ? >> That is: Can R be partitioned into two congruent subsets that are each dense, one of which is a subgroup of R ? --Dan Even though kleptomaniacs can't help themselves, they do.

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Re: [math-fun] To Gene Re: Tiling puzzle
by Eugene Salamin 13 Nov '11

13 Nov '11

Dan, you are correct. My construction failed to consider the case of x = 1/4. -- Gene >________________________________ >From: Dan Asimov <dasimov(a)earthlink.net> >To: Eugene Salamin <gene_salamin(a)yahoo.com> >Sent: Saturday, November 12, 2011 7:30 PM >Subject: To Gene Re: [math-fun] Tiling puzzle > >Gene, > >I'm not sure if or where you may have erred, >but I thought I had an argument making it >impossible that one of the two sets is a >subgroup of R: > >----- >Let x |-> x + t be the translation taking >subgroup H bijectively onto H'. > >If H is a subgroup, then it >contains 0 and x in H implies -x in H. > >And so the subgroup has a self isometry >via x |-> -x. > >But this means there is also a map of the >form x |-> t - x taking H bijectively >onto H'. But this is impossible since >x -> t-x always has a fixed point at x = t/2. >----- > >Or have I made a mistake? > >--Dan > ><< >It suffices to find an additive subgroup H of the reals of index 2, as then the two cosets H and H' are translates of each other. Start with H[0] = {0}, H'[0] empty, and let R be well ordered. H and H' will be built up by transfinite induction. > >For α a successor ordinal, let x[α] be the smallest real in R - H - H'. Make H[α] by appending to H[α-1] all reals (y + n x[α]), with y in H[α-1] and n an integer. Make H'[α] by appending to H'[α-1] all reals (y' + (n+1/2) x[α]), with y' in H'[α-1]. Then H[α] and H[α] + H'[α] are subgroups of R and the former has index 2 in the latter. > >For β a limit ordinal, let H[β] be the union of all H[α] for α < β, and let H'[β] be the union of all H'[α] for α < β. Continuing until the reals are exhausted yields the desired subgroup H. >>> > >Even though kleptomaniacs can't help themselves, they do. > > >

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Re: [math-fun] Science fiction (some of which is permanently gonna be fiction)
by Henry Baker 13 Nov '11

13 Nov '11

"THERE IS NO PRACTICAL WAY TO STORE A CHUNK OF ANTIMATTER." Somebody better tell NASA... http://www.bbc.co.uk/news/science-environment-14405122 Antimatter belt around Earth discovered by Pamela craft "Dr Bruno said that, aside from confirming theoretical work that had long predicted the existence of these antimatter bands, the particles could also prove to be a novel fuel source for future spacecraft - an idea explored in a report for Nasa's Institute for Advanced Concepts." http://arxiv.org/abs/1107.4882 ---- I've asked the following sort of question before on math-fun: I realize that matter annihilates antimatter when they come into 'contact', but what is 'contact' in this context, and what are the time sequence details -- e.g., when a molecule of glucose 'contacts' a molecule of anti-glucose ? At 07:05 AM 11/13/2011, Warren Smith wrote: >There's a lot of garbage science fiction out there that has nothing to do with science, >especially in hollywood... but in this post I want to demolish two >sci-fi ideas that occur in works of Larry Niven (et alia) which maybe >did sound achievable to the naive. >But a third such idea (super strong materials) will also be considered >and argued to maybe have hope. > >1. Sci-fi antimatter. >You just keep a chunk of antimatter magnetically suspended in vacuum or something, >where it's an excellent energy source for your super-rocket, ray-gun, >also an excellent mega-bomb when you turn off the magnetic levitation. > Stores the maximum possible amount of energy for that amount of mass. > >Actually: >let's say a single normal-matter gas atom contaminates the vacuum. Hits >the antimatter. Boom. Way more energy is released than you need to >eject some antimatter atoms at high speed. They in turn hit >the normal matter container. Boom boom boom. The resulting normal >atom ejecta fly out to hit more antimatter. Etc. This is an exponentially >amplifying process. > >THERE IS NO PRACTICAL WAY TO STORE A CHUNK OF ANTIMATTER. > >Get over it.

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Re: [math-fun] Science fiction (some of which is permanently gonna be fiction)
by Henry Baker 13 Nov '11

13 Nov '11

I thought that fermions couldn't occupy the same point in spacetime. I can see the problems with electron/anti-electron and proton/anti-proton, since they have opposite charges, but what about neutron/anti-neutron ? How do they annihilate? Do they temporarily "decay", so that the products then annihilate each other?? Also, what is a wavelength in this context? The de Broglie wavelength? But of what? The molecule as a whole? Some electron within the molecule? Etc., etc. At 08:38 AM 11/13/2011, Mike Stay wrote: >On Sun, Nov 13, 2011 at 8:23 AM, Henry Baker <hbaker1(a)pipeline.com> wrote: >> I've asked the following sort of question before on math-fun: >> >> I realize that matter annihilates antimatter when they come into 'contact', but what is 'contact' in this context, and what are the time sequence details -- e.g., when a molecule of glucose 'contacts' a molecule of anti-glucose ? > >Contact is occupying the same point in spacetime (to within a wavelength). >-- >Mike Stay - metaweta(a)gmail.com >http://www.cs.auckland.ac.nz/~mike >http://reperiendi.wordpress.com

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Re: [math-fun] Science fiction (some of which is permanently gonna be fiction)
by Henry Baker 13 Nov '11

13 Nov '11

Agreed, but "photon rockets" are pretty cool, because their specific impulse is better than any matter-propelling rocket. The bit problem with "photon rockets" is how to produce enough of them aimed at a specific direction, without producing things like neutrinos, which can also carry away momentum, but are much harder to direct. At 07:05 AM 11/13/2011, Warren Smith wrote: >There's a lot of garbage science fiction out there that has nothing to >do with science, >especially in hollywood... but in this post I want to demolish two >sci-fi ideas that occur in works of Larry Niven (et alia) which maybe >did sound achievable to the naive. >But a third such idea (super strong materials) will also be considered >and argued to maybe have hope. > >2. "Bussard Ramjet." >A magic interstellar travel ship which somehow attracts interstellar >hydrogen (with a "magnetic field"?) into its inscoop, pipes it thru a >fusion reactor, and spits the hot helium out the back. "Keeps >accelerating forever" (preferably at 1G) need not carry fuel, and if >you wait long enough you'll achieve speeds arbitrarily near lightspeed >enabling you to visit the whole universe in your lifetime thanks to >time dilation. > >Actually: >Even if this could be made to work, the impinging hydrogen would slow the ship down, >it's a jet not a rocket, and the ultimate speed would be upper-bounded by the ejection >velocity of the helium, no matter how long you waited. That is way below lightspeed >since the mass loss (hydrogen-->helium) is very small fractionally. >Meanwhile, if you just carried a tank of hydrogen fuel massing e times >your unfueled mass (genuine rocket), you could achieve slightly >greater speed without all the technical difficulty... and if it were >deuterium you'd get a good deal faster speed. > >THIS WHOLE "BUSSARD RAMJET" IDEA IS JUST IRREDEEMABLY STUPID.

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[math-fun] Science fiction (some of which is permanently gonna be fiction)
by Warren Smith 13 Nov '11

13 Nov '11

There's a lot of garbage science fiction out there that has nothing to do with science, especially in hollywood... but in this post I want to demolish two sci-fi ideas that occur in works of Larry Niven (et alia) which maybe did sound achievable to the naive. But a third such idea (super strong materials) will also be considered and argued to maybe have hope. 1. Sci-fi antimatter. You just keep a chunk of antimatter magnetically suspended in vacuum or something, where it's an excellent energy source for your super-rocket, ray-gun, also an excellent mega-bomb when you turn off the magnetic levitation. Stores the maximum possible amount of energy for that amount of mass. Actually: let's say a single normal-matter gas atom contaminates the vacuum. Hits the antimatter. Boom. Way more energy is released than you need to eject some antimatter atoms at high speed. They in turn hit the normal matter container. Boom boom boom. The resulting normal atom ejecta fly out to hit more antimatter. Etc. This is an exponentially amplifying process. THERE IS NO PRACTICAL WAY TO STORE A CHUNK OF ANTIMATTER. Get over it. 2. "Bussard Ramjet." A magic interstellar travel ship which somehow attracts interstellar hydrogen (with a "magnetic field"?) into its inscoop, pipes it thru a fusion reactor, and spits the hot helium out the back. "Keeps accelerating forever" (preferably at 1G) need not carry fuel, and if you wait long enough you'll achieve speeds arbitrarily near lightspeed enabling you to visit the whole universe in your lifetime thanks to time dilation. Actually: Even if this could be made to work, the impinging hydrogen would slow the ship down, it's a jet not a rocket, and the ultimate speed would be upper-bounded by the ejection velocity of the helium, no matter how long you waited. That is way below lightspeed since the mass loss (hydrogen-->helium) is very small fractionally. Meanwhile, if you just carried a tank of hydrogen fuel massing e times your unfueled mass (genuine rocket), you could achieve slightly greater speed without all the technical difficulty... and if it were deuterium you'd get a good deal faster speed. THIS WHOLE "BUSSARD RAMJET" IDEA IS JUST IRREDEEMABLY STUPID. 3. Super-strong materials. There's lots of them in sci-fi novels. So now let me actually do a little math (order of magnitude estimates); I'm going to attempt to derive upper bounds on the yield strength of any conceivable material. First try: relativity. If you pull on your material with force F enough to stretch it by a factor of (say) 2, then that stores elastic energy or order F*StretchDistance. This energy is bounded by m*c^2. Hence F<m*c^2/length is a rough upper bound on tensile strength of any mass=m object. Expressed in the usual units (pascals) of force per unit cross-sectional area, Strength < MassDensity * c^2. Now most sci-fi applications for superstrong spaceship hulls (armor), superstrong cables, etc envision materials with densities similar to normal materials, let's say 7.85 grams/cc which is the density of steel. Sci-fi super-steel would have a strength upper limit of Strength < 7*10^20 pascals as opposed to the actual strength of available ultra-high-strength steels 5*10^9 Pascals and maybe super-fibers could exceed that by factor 10-100 in our wildest dreams. So... not bad: sci-fi might hope if it could meet the relativity bound, to get strengths exceeding today's strongest stuff by a factor of 10^10. That's pretty good... (but never will be strong enough to protect spacecraft from collisions at 0.01c)... Second try: quantum mechanics. Material is made out of atomic (or subatomic, whatever) particles. Particles are trapped in potential-energy wells. Say the energy of your mass=m particle at position x is E(x) which has a minimum at x=0 locally like (m/2)*w^2*x^2 where w is the angular frequency of the harmonic oscillator. Considering the ground state of harmonic oscillator http://en.wikipedia.org/wiki/Quantum_harmonic_oscillator the particle then is localized with std.deviation=DeltaL=(hbar/(m*w))^(1/2) and energy-gap to the next quantum level DeltaE=hbar*w [I may ignore small constant factors like 2]. If you pull on it with force of order F=DeltaE/DeltaL=(hbar*m)^(1/2)*w^(3/2) you are going to be able to excite the thing into the next state. This corresponds to a tensile strength (force per unit area) of order at most Strength < hbar^(-1/2) * w^(5/2) * m^(3/2) achieved by a material with Density of order Density = m^(5/2) * hbar^(-3/2) * w^(3/2). So with relativity, Strength < c^2 * Density. With quantum mechanics, Strength < (hbar * w / m) * Density. Roughly. The QM bound actually might be able to reach the same order as the relativity bound if we were dealing with particles bound by energies comparable to their mass-energy -- which is the case for quarks (the mass of the proton derives mostly from binding energies, not quark masses). For atoms, though, the fact electrons do the binding and they have way smaller masses than the nuclei, immediately kills you by a factor of at least 2000 versus the relativity bound. Actually only the OUTER electrons do the binding and the nuclei of the useful atoms are around 20000 times heavier than an electron, killing us by a factor of around 20000. This if we had a material whose bonds could only be broken by gamma rays in the 100 KeV range (electron positron annihilation radiation) would be tight, but all the materials we know of have bonds breakable by w in the UV light range of frequencies (<10eV), killing you by a further factor of 10000. This is not merely an empirical observation but actually theoretically inherent (actually "10000" is about 137^2 where the fine structure constant is 1/137) because electron-binding is via photons and the electron-photon nondimensional coupling constant in QED is 1/137. So our QM bound in view of that predicts the highest strength materials one could make with chemistry ought to have strengths 2*10^8 smaller than the relativity bound... which is only at most about 50X stronger than today's strongest stuff. I.e. the fundamental order of magnitude arguments get fairly close to the truth. However, a hypothetical material made of quarks bound into fibers (material would have extremely high density within the fibers, but low density if fibers surrounded by a lot of empty space) could, far as I can see from this crude analysis, genuinely achieve vastly greater strengths than today's materials with comparable densities. To rule that out you'd have to argue every such construct would be inherently unstable or something, which is an argument probably out of reach of today's physicists. Nuclear matter has a lot of surface tension which is why nuclei all are ball-shaped (isoperimetric theorem at work). Cylindrical fibers have too much surface. However, if you made one, then trying to cut it would cause even more surface -- so you might be able to hope for a quark fiber stable against all small-enough perturbations if you could somehow create it in the first place. I'd be worried though, that if anybody DID cut one of your fibers, then that'd cause big energy release, whacking nearby fiber hard enough to destabilize it... a chain reaction... and your chunk of material would turn into a nuclear bomb. CONCLUSION: this 3rd sci-fi fantasy of super-strong materials conceivably is achievable -- anyhow cannot be ruled out using simple arguments that I can see -- but don't try this at home!

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Re: [math-fun] Tiling puzzle
by Warren Smith 13 Nov '11

13 Nov '11

while impressed by Salamin's solutions for some abstract notion of "impressed"... (I.e. how much it boggles my mind) they seem to me to be too "nonconstructive." I tried to read http://en.wikipedia.org/wiki/Transfinite_induction but my mind remained boggled. Can we find a "constructive" way to partition the reals into two congruent dense subsets -- i.e. so it is somehow "algorithmic" to tell which set a given real is in? Well, ok, maybe that seems hopeless, but somehow Salamin's solution seems really really nonconstructive to me. Would a Brouwerian "intuitionist" accept it? I think not. (Agree?) Can we reduce the number of "really"s? (I have a lot of sympathy with the Brouwerian point of view at times. Such as now.) So: can we find a solution which happens also to partition the rationals, or quadratic algebraics, or algebraics... into two congruent dense subsets... and then asking for an algorithm is a perfectly sensible thing to do. I guess the biggest subfield of the reals for which we could have this hope, would be the "computable reals." All these subfields are countable. OK, so Salamin's solution can be rewritten to pertain to these countable subfields in a way that I think would not make Brouwer turn over in his grave: H = {0} H' = emptyset for all rational (algebraic, whatever) numbers X in order do if(X is not in H and X is not in H') H = {y + n X where y in old H and n integer} H' = {y + (n+1/2) X where y in old H and n integer} fi od After running that "algorithm" we get that S = H U H' is the full set of rational (algebraic, whatever) numbers, and H is a subgroup of index 2 inside it (additive groups) and H and H+1/2 form the desired partition. Now note that for any particular rational (algebraic, whatever) R this "algorithm" will find out which set R is in (red or blue?), in finite time. However... this won't work for the computable reals since it is uncomputable whether two are equal (right?). I can't help suspecting there is something a lot cleaner waiting to be found. We've just been discussing a super-nice way to order the rationals with Calkin-Wilf tree. Well, maybe there is a comparable super-nice thing just for the red rationals. You need to find it, or Brouwer's going to kick your ass.

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Re: [math-fun] Tiling puzzle
by Dan Asimov 13 Nov '11

13 Nov '11

Even Gene's solution (as does mine) to the original puzzle uses the Axiom of Choice, in the step where a special set is chosen to include exactly one member of each coset of R/Q. I doubt that it can be avoided. --Dan << while impressed by Salamin's solutions for some abstract notion of "impressed"... (I.e. how much it boggles my mind) they seem to me to be too "nonconstructive." >> Even though kleptomaniacs can't help themselves, they do.

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Re: [math-fun] Tiling puzzle
by Eugene Salamin 13 Nov '11

13 Nov '11

>From: Dan Asimov <dasimov(a)earthlink.net> >To: Eugene Salamin <gene_salamin(a)yahoo.com> >Sent: Saturday, November 12, 2011 2:24 PM >Subject: Re: [math-fun] Tiling puzzle > >I wrote: ><< >PUZZLE: Can the real line be >partitioned into two congruent >sets that are each dense? >>> >_______________________________________________ > >Yes, and not just 2, but any finite or countable number. Define an equivalence relation on the real numbers, x == y if x-y is rational, and let S(0) be a set containing one number from each equivalence class. S(0) can be chosen to be dense. Let S(r) be the translate of S(0) by r. For rational r, the countably many S(r) are disjoint and their union is all of the real line. The S(r) give a countable dense partition of the real line. For a finite partition, first let T[k] for 0<=k<n be a partition of the real line into n congruent sets, e.g. the union of k+mn <= x < k+mn+1 for all integers m. Let S[k] for 0<=k<n be the union of S(r) for r in T[k]. Then these S[k] give a finite dense partition. > >To show that the uncountable set S(0) can be chosen to be dense, let r[n] be the n-th rational, and choose real number x[n] in the interval I[n]: r[n] - 1/2^n < x[n] < r[n] + 1/2^n, but such that x[n] =/= x[k] for k<n. Let S(0) contain the x[n]. Now let x be a real number, let ε > 0, and let I be the open interval from x-ε to x+ε. Since there are infinitely many rationals in I, some interval I[n] is contained within I. Then x[n] is in I, so |x - x[n]| < ε, and thus S(0) is dense in the reals. -- Gene >>> >_______________________________________________> >Yes, that's essentially what I had in mind -- nice solving. > >Now for Part B: >Is there a way to solve this puzzle (for 2 sets) >so that one of the sets is a subgroup of R ? > >--Dan >_______________________________________________ It suffices to find an additive subgroup H of the reals of index 2, as then the two cosets H and H' are translates of each other. Start with H[0] = {0}, H'[0] empty, and let R be well ordered. H and H' will be built up by transfinite induction. For α a successor ordinal, let x[α] be the smallest real in R - H - H'. Make H[α] by appending to H[α-1] all reals (y + n x[α]), with y in H[α-1] and n an integer. Make H'[α] by appending to H'[α-1] all reals (y' + (n+1/2) x[α]), with y' in H'[α-1]. Then H[α] and H[α] + H'[α] are subgroups of R and the former has index 2 in the latter. For β a limit ordinal, let H[β] be the union of all H[α] for α < β, and let H'[β] be the union of all H'[α] for α < β. Continuing until the reals are exhausted yields the desired subgroup H. -- Gene

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[math-fun] bicoloring the plane (or any manifold) so all curves bicolored
by Warren Smith 12 Nov '11

12 Nov '11

Somebody a few months ago wanted a way to 2-color the plane so that no continuous curve could be monochromatic. My solution, which works in any-dimensional space: Pick two dense subsets A and B of the reals R, with A subset B subset R. [Example B=terminating decimal expansion, A=and the last digit is 1.] If X has either all coordinates in complement(B) or at least one coordinate in A, then color red. Else blue. This actually works on any Riemannian manifold of any dimensionality by using the Nash embedding theorem http://en.wikipedia.org/wiki/Nash_embedding_theorem

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