October 2013 - mathfun@mailman.xmission.com

[math-fun] Pick formulas generalized to noncubic lattices: trivial
by Warren D Smith 21 Oct '13

21 Oct '13

> From: Joerg Arndt <arndt(a)jjj.de> > To: math-fun <math-fun(a)mailman.xmission.com> > Subject: [math-fun] Pick theorems > Pick's theorem (cf. http://en.wikipedia.org/wiki/Pick's_theorem): > > The area A of a simple polygon with all corners on a square grid is > A = i + b/2 - 1 > where i is the number of lattice points in the interior > and b is the number of lattice points on the boundary. > > > For the following A is the area as number of unit cells covered. > > Triangular lattice: > A = 2*i + b - 2 > > Hexagonal lattice: > A = i/2 + b/4 - 1/2 > > Square grid, every second column shifted by a half unit: > A = i/2 + b/2 - 1 > where b counts only the boundary points whose local neighborhood > is not concave (i.e., 90 degrees of outside, 270 degrees of inside). > > I obtained these scribbling on a train ride yesterday. > > Best, jj --it seems to me that any "Pick theorem" (in any space-dimension D) can be trivially generalized to work for a lattice L which differs from the plain integer lattice Z^D by just linearly transforming. Whatever lattice L you have, it is just an invertible linear transformation of D-space of the plain Z^D lattice. Wlog L is scales so its voronoi cells have unit D-volume in which case the linear transform is volume-preserving. Now note, the polygon (or polytope) counts of lattice points within, on vertices, most geenrally on k-faces (0<=k<=D), all are preserved by the linear transformation. Also, the volume is preserved (as we just postulated wlog). Hence, any Pick formula should generalize to work for any lattice. QED

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[math-fun] Pick formula
by Warren D Smith 21 Oct '13

21 Oct '13

I was just saying, if there were a Pick formula in higher dimensions than 2, (or in 1D ior 2D) it'd be trivial to generalize it to other lattices. I did not claim any such Pick formula existed. :) But to answer RCS: Wikipedia: "The Reeve tetrahedron shows that there is no analogue of Pick's theorem in three dimensions that expresses the volume of a polytope by counting its interior and boundary points. However, there is a generalization in higher dimensions via Ehrhart polynomials."

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Re: [math-fun] GR question
by Keith F. Lynch 21 Oct '13

21 Oct '13

Henry Baker <hbaker1(a)pipeline.com> wrote: > A bigger black hole has a bigger Schwarzschild radius, and space in > the vicinity of the Schwarzschild radius of a very large black hole > is relatively flat. Right. You could survive a fall through the event horizon of a quiescent galactic-mass black hole. (Of course you wouldn't survive for long after that, as you approached the singularity. Nor could you avoid the singularity.) > I.e., suppose the Sun were a black hole, whose Schwarzschild radius > is quite small, so the Earth is very far from this radius. Now > consider a Sun' whose mass is, e.g., twice as big as the Sun. Its > Schwarzschild radius is bigger than before, but if we are still at > 1 AU, would we be able to tell _just from the local curvature_ how > much mass is in the center of the solar system? Certainly, given that you know the distance to the sun. Just measure the local tidal force, i.e. how much less a test mass on Earth's surface weighs when the sun is overhead or underfoot than when it's on the horizon, after correcting for the moon's tidal force. This could perhaps be most sensitively measured by timing a pendulum clock. Getting back to black holes, I wonder if anyone has tried to measure the tidal force on Earth from the black hole in the center of our galaxy. I think it ought to be measurable. As an aside, since the tidal force is differential gravity, it falls off with the inverse cube of distance. Conversely, the volume of an object of given angular size increases with the cube of its distance. So the two effects cancel out. Hence the relative tidal forces from two objects of the same angular size is in proportion to their relative densitites. If the moon, which is about the same angular size as the sun, has about three times the tidal force as the sun, it's "because" the moon has about three times the density as the sun. This scaling law must be generally known, but I've never seen it. (I worked it out for myself years ago.) Another fun scaling law is that the ratio of the sun's absolute surface temperature to the Earth's ought to be the fourth root of the proportion of Earth's sky the sun's disk takes up. (This assumes the sun and Earth are both black bodies or both fall short of being black bodies by the same proportion, that the sun is Earth's only source of heat, that the sun-Earth distance is constant, and that everything is in equilibrium. By "Earth's sky" I include the half that's hidden by the ground.) > ... curvature of space alone? Nitpick: Curvature of spacetime, not of space alone. Fred Lunnon <fred.lunnon(a)gmail.com> wrote: > What about its charge? And (quantum-theoretically) temperature? For a small black hole, any charge would rapidly dissipate by Hawking radiation. For any size black hole, its charge would disippate by preferential absorption of the opposite charge from surrounding material. But I suppose a large black hole in a very good vacuum could retain a charge for quite some time. The temperature depends entirely on the mass. Smaller is hotter. As a rough rule of thumb, the peak wavelength of the radiation is about equal to the apparent diameter of the event horizon.

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[math-fun] The Guardian quiz
by Eric Angelini 21 Oct '13

21 Oct '13

How good is your mathematical sense of humour? Best, É. http://www.theguardian.com/science/quiz/2013/oct/21/mathematical-humour-qui…

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[math-fun] Stupid GR question
by Henry Baker 20 Oct '13

20 Oct '13

I've been watching Leonard Susskind's lectures on GR (available on the Internet) & had a question. A non-rotating black hole is completely characterized by its mass & therefore its Schwarzschild radius. A bigger black hole has a bigger Schwarzschild radius, and space in the vicinity of the Schwarzschild radius of a very large black hole is relatively flat. Q: Can a point observer outside the Schwarzschild radius of a black hole tell how big the black hole is by examining the curvature of space very near the observer? I.e., suppose the Sun were a black hole, whose Schwarzschild radius is quite small, so the Earth is very far from this radius. Now consider a Sun' whose mass is, e.g., twice as big as the Sun. Its Schwarzschild radius is bigger than before, but if we are still at 1 AU, would we be able to tell _just from the local curvature_ how much mass is in the center of the solar system? Of course, in order to stay in a "stable" orbit, the Earth would have to speed up to match the new mass in the center, but other than the period of the orbit, is there any way for Earth-bound scientists to measure the mass of the Sun' via curvature of space alone?

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Re: [math-fun] GR question
by Keith F. Lynch 20 Oct '13

20 Oct '13

Whitfield Diffie <whitfield.diffie(a)gmail.com> wrote: > "Keith F. Lynch" <kfl(a)KeithLynch.net> wrote: >> You could survive a fall through the event horizon of a quiescent >> galactic-mass black hole. (Of course you wouldn't survive for long >> after that, as you approached the singularity. > I don't see why. If a black hole is massive enough, (my crude > calculation, which those with better ALUs are welcome to correct, is > 3.65x10^18 solar masses or about four million times the mass of the > Miky Way galaxy) it has about the density of our galaxy as a whole > and might be habitable. Time and space are swapped inside the event horizon. Once you're inside, the singularity is not some distance ahead of you, it's some time in your future. (As such, the more powerful your engine, the *sooner* you'd hit the singularity. To live as long as possible, remain in free fall.) Of course if the black hole is big enough, you would die of old age before you reached the singularity. But it would have to be very large indeed. Even a galactic mass black hole only gives you a few minutes. > What this leads to is the question can the universe be distinguished > from a black hole? The fact that it's expanding rather than contracting. Charles Greathouse's post reminded me of the firewall conjecture. If that's true, you can't fall through any event horizon without being incinerated. Henry Baker <hbaker1(a)pipeline.com> wrote: > According to a recent talk given at UCSB, the largest known black > holes have a Schwarzschild radius of the same order of magnitude as > the radius of the Earth's orbit -- i.e., 1 AU. > Light would orbit this black hole in a few minutes; I would presume > that an Earth-sized object would have to orbit at nearly the speed > of light even at 5 AU (Jupiter's distance from the Sun). That orbit wouldn't be stable. The planet would lose energy by emitting gravitational radiation, and within a few orbits would spiral into the black hole. Of course everything about black holes is conjectural. Some of it -- the division by zero part -- pretty much has to be wrong. GR has never been tested except in the weak field limit. What is known for certain is that there are non-luminous objects so dense that no known force could prevent their total collapse, and that one of them is in the center of our galaxy. There are several non-GR theories about these objects, but all such theories are, in my opinion, uglier than GR, and there's no evidence for them.

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Re: [math-fun] GR question
by Henry Baker 20 Oct '13

20 Oct '13

According to a recent talk given at UCSB, the largest known black holes have a Schwarzschild radius of the same order of magnitude as the radius of the Earth's orbit -- i.e., 1 AU. Light would orbit this black hole in a few minutes; I would presume that an Earth-sized object would have to orbit at nearly the speed of light even at 5 AU (Jupiter's distance from the Sun). At 12:17 PM 10/20/2013, Whitfield Diffie wrote: >> You could survive a fall through the event horizon of a quiescent >> galactic-mass black hole. (Of course you wouldn't survive for >> long after that, as you approached the singularity. > > I don't see why. If a black hole is massive enough, (my crude >calculation, which those with better ALUs are welcome to correct, is >3.65x10^18 solar masses or about four million times the mass of the >Miky Way galaxy) it has about the density of our galaxy as a whole and >might be habitable. > >> Nor could you avoid the singularity.) > > Why couldn't you be in orbit inside the event horizon. > > What this leads to is the question can the universe be >distinguished from a black hole? > > Whit

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[math-fun] Egan's absurd conjecture
by Keith F. Lynch 20 Oct '13

20 Oct '13

Eric Angelini <Eric.Angelini(a)kntv.be> wrote: > Thank you, Keith -- I've just bought > on Amazon.fr the Greg Egan books > you've indicated. Which books? I mentioned two short stories, "Luminous" and "Dark Integers." The website isfdb.org will tell you what books and magazines they appear in. If you're read them in the week since you posted that, what do you think of them? I've been thinking about how Egan's conjecture could be tested. In its weakest form, that no conjecture is true or false until it somehow makes a difference to some physical system, it can't be. But what about in a slightly less weak form? My modest proposal is that someone try testing a large number of conjectures that have almost certainly never been tested, or made any difference to any physical system, ever before. For instance the conjecture that the thousandth binary digit of the fractional part of some random hundred-digit number is 1. Test many such hundred-digit numbers using several very different algorithms. See if there's any correlation between which algorithm is used and the odds of the conjecture being true. If there isn't, that doesn't prove or disprove anything, but if there is, then something is going on. Probably the something is an uninteresting bug in your program. But suppose that is ruled out. First, there's the "most intuitive" possibility that whichever algorithm determines its value first, any other algorithm run later will find the same value, even if the value is biased by the first algorithm. If so, this could be used for communications. The sender and recipient agree ahead of time on a bank of numbers for each bit of the message. How many numbers are needed per bit depends on how strong the bias is. The recipient has to agree not to test the numbers until after the scheduled transmission time. If, after that, he finds the bias using an algorithm with the opposite bias, then a 1-bit has been sent, otherwise a 0-bit has been sent. If that's how it works, what if the two algorithms are run simultaneously, in the sense that no light-speed signal could get from either computer after it starts the computation to the other computer until after it had finished the computation? Would you somehow cause a contradiction? Would the conjecture be both true and false? One possible solution to that paradox is that no bias appears unless the computation will *never* be done with the other algorithm. But this could give rise to classic time-travel paradoxes instead. By deciding to later test or not test the banks of numbers with the other algorithm depending on, say, what next week's winning lottery numbers will be, you could send messages to yourself back through time. (If the numbers are large enough, and if you are careful to leave no record of them, it's all but certain that if you don't calculate on them using the other algorithm, nobody else ever will.) Self-reference: Since Egan's conjecture is itself a conjecture, then by its own rule it's neither true nor false until tested, and maybe how we test it will change the odds of its being true or false. Which do we want it to be? The conjecture is of course absurd, and in the unlikely case that there's any bias in the digits of square roots it's shared by all algorithms. But I could be wrong, so it might be worth testing. Maybe I'll test it myself if I get a TUIT with fewer corners. If you hear that I've won the lottery.... > Propuls? d'un aPhone Was the question mark supposed to be an e with an acute accent? Nobody seems to have mentioned, in the recent discussion about UTF-8, that in the digest option all high-bit characters are replaced by question marks. So in the digest option, only plain old 7-bit ASCII works.

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Re: [math-fun] Stupid GR question
by Henry Baker 20 Oct '13

20 Oct '13

(Pure) GR don't know nuthin' 'bout charge, temperature, etc. I suspect that the answer to my question is yes -- the local curvature is greater at a distance of 1 AU if m(Sun')=2*m(Sun), so it should be possible to detect this even without calculating the period of revolution. What is true (according to Susskind), is that an Earth observer can't tell if the Sun is a black hole or not by simple local measurements of curvature. I.e., Newton's theorem holds even in GR: a spherical shell of matter still affects outside observers exactly the same as if all of its mass were located at the center of the sphere. At 09:23 AM 10/20/2013, Fred Lunnon wrote: ><< A non-rotating black hole is completely characterized by its mass & >therefore its Schwarzschild radius. >> > >What about its charge? And (quantum-theoretically) temperature? See > http://en.wikipedia.org/wiki/No-hair_theorem > >WFL > >On 10/20/13, Henry Baker <hbaker1(a)pipeline.com> wrote: >> I've been watching Leonard Susskind's lectures on GR (available on the >> Internet) & had a question. >> >> A non-rotating black hole is completely characterized by its mass & >> therefore its Schwarzschild radius. >> >> A bigger black hole has a bigger Schwarzschild radius, and space in the >> vicinity of the Schwarzschild radius of a very large black hole is >> relatively flat. >> >> Q: Can a point observer outside the Schwarzschild radius of a black hole >> tell how big the black hole is by examining the curvature of space very near >> the observer? >> >> I.e., suppose the Sun were a black hole, whose Schwarzschild radius is quite >> small, so the Earth is very far from this radius. Now consider a Sun' whose >> mass is, e.g., twice as big as the Sun. Its Schwarzschild radius is bigger >> than before, but if we are still at 1 AU, would we be able to tell _just >> from the local curvature_ how much mass is in the center of the solar >> system? >> >> Of course, in order to stay in a "stable" orbit, the Earth would have to >> speed up to match the new mass in the center, but other than the period of >> the orbit, is there any way for Earth-bound scientists to measure the mass >> of the Sun' via curvature of space alone?

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Re: [math-fun] math-fun Digest, Vol 128, Issue 17
by Warren D Smith 20 Oct '13

20 Oct '13

> Message: 6 > Date: Sun, 20 Oct 2013 06:03:58 -0700 > From: Henry Baker <hbaker1(a)pipeline.com> > To: math-fun(a)mailman.xmission.com > Subject: [math-fun] Stupid GR question > Message-ID: <E1VXsgM-0003Il-25(a)elasmtp-banded.atl.sa.earthlink.net> > Content-Type: text/plain; charset="us-ascii" > > I've been watching Leonard Susskind's lectures on GR (available on the > Internet) & had a question. > > A non-rotating black hole is completely characterized by its mass & > therefore its Schwarzschild radius. > > A bigger black hole has a bigger Schwarzschild radius, and space in the > vicinity of the Schwarzschild radius of a very large black hole is > relatively flat. > > Q: Can a point observer outside the Schwarzschild radius of a black hole > tell how big the black hole is by examining the curvature of space very near > the observer? --time the period of your circular orbit about the hole and measure the length of your orbital path, thus deducing, e.g. your acceleration (which is a "local curvature" measure) and the mass of the hole. If that was not local enough, then measure your acceleration directly by moving your arm in a straight line (following light beam) while rest of you moves in orbit; difference is acceleration to good approximation anyhow. Also, several copies of you slightly displaced will move apart (differing orbits) at a rate of acceleration (tidal) which again allows inference both of local curvature & the hole mass, at least if you know your orbit circular.

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