View Full Version : Beautiful Math

darkdev

2004-Apr-16, 10:10 PM

I refer to the Golden Ratio: (1 ± 5^.5) / 2

(1 + 5^.5) / 2 = 1.618034

(1.618034)^2 = 2.618034

1/(1.618034) = 0.618034

0.618034

1.618034

2.618034

Mathematically delicious! I even read somewhere that e, as in ln (log sub e) can be derived from this number, but don't trust the source.

Since so much in nature seems to fit so well into this ratio, I am wondering why I don't see this relationship in cosmology? When I read that some particles could have 1/3 or 2/3 spin I was thinking... hmm... closer to .382 and .618 perhaps?

Is this just a local math anomoly (i.e. "center" of nature for us), or is this really a worthy universal constant, like c?

ToSeek

2004-Apr-16, 11:16 PM

Mario Livio's book on the subject shows that phi is not as ubiquitous as urban legend would have it. It's definitely not as prevalent in works of art as people make out that it is.

mike alexander

2004-Apr-16, 11:19 PM

Phi is one of those terrifically fun things to play around with. It pops out of the Fibbonaci series, you can make spirals with it, and so on. But its not a physical constant like c. And I agree with ToSeek, I've seen some examples of post hoc phi bashing-to-fit.

JohnOwens

2004-Apr-17, 12:49 AM

Isn't it in the spiral arms of typical spiral galaxies? If so, I wouldn't be surprised to see it in accretion or planetary disks as well.

The spin numbers are definitely integral ratios, though. There are even some who'd prefer to see 1/3 made the new unit, so multi-quark particles would have total spins of -3, 0, 3, etc. (if I recall correctly).

Grey

2004-Apr-17, 07:12 AM

When I read that some particles could have 1/3 or 2/3 spin

Just to nitpick, it's charge that comes in thirds. Spin is always an integer (for bosons) or half integer (for fermions).

milli360

2004-Apr-17, 05:30 PM

I refer to the Golden Ratio: (1 ± 5^.5) / 2

(1 + 5^.5) / 2 = 1.618034

(1.618034)^2 = 2.618034

1/(1.618034) = 0.618034

0.618034

1.618034

2.618034

Mathematically delicious!

Check this out (notice the progression of both sets of coefficients is just the Fibonnacci series:

phi^1 = 0 + 1*phi

phi^2 = 1 + 1*phi

phi^3 = 1 + 2*phi

phi^4 = 2 + 3*phi

phi^5 = 3 + 5*phi

phi^6 = 5 + 8*phi

...etc.

These sort of games with phi are so numerous that they created a journal just for discussion of the Fibonnacci series.

Since so much in nature seems to fit so well into this ratio, I am wondering why I don't see this relationship in cosmology? When I read that some particles could have 1/3 or 2/3 spin I was thinking... hmm... closer to .382 and .618 perhaps?

I see grey has already addressed your main point, but just out of curiousity, where'd the .382 come from?

darkdev

2004-Apr-17, 09:30 PM

I see grey has already addressed your main point, but just out of curiousity, where'd the .382 come from?

Rounded 1-.618, it's "an" a in (a/b)==(b/(a+b))

(0.381966/0.618034) == 0.618034

(0.618034/1) == 0.618034

(1/1.618034) == 0.618034

(0.618034/1.618034) == 0.381966

AstroRockHunter

2004-Apr-19, 07:28 PM

Grey wrote:

Just to nitpick, it's charge that comes in thirds. Spin is always an integer (for bosons) or half integer (for fermions).

Isn't charge quantizied???

Kaptain K

2004-Apr-19, 08:40 PM

Yep! 0, +/-1/3, +/-2/3, +/-1, etc.

countrywideoptionone

2004-Apr-19, 10:47 PM

It also has the property that it equals it's reciprical + 1. If that's all you know, you can set up a simple quadratic equation to solve it.

Let's use g for "golden."

Assume: g = 1/g + 1

Multiply by g: g^2 = 1 + g

Solve for 0: g^2 - g - 1 = 0

Use the Quadratic Equation to solve for g: g = [- (-1)+-sqr{(-1)^2 - 4(1)(-1)}]/2(1)

Simplify: g = [1+-sqr{5}]/2

We take only the postive root because it usually represents a ratio of lengths or other things that only make sense when they are positive.

g = [1+sqr{5}]/2

Mathematically delicious! I even read somewhere that e, as in ln (log sub e) can be derived from this number, but don't trust the source.

e is a transcendental number, which means it can't be the solution of an algebraic equation like phi is.

Disinfo Agent

2004-Apr-20, 11:52 AM

So?

HenrikOlsen

2004-Apr-20, 01:16 PM

The one I always enjoy the most for its simple beauty is

e^(i * pi) = -1

mike alexander

2004-Apr-20, 11:54 PM

One of those things that just knocked my socks off was learning that a straight line of constant slope in Cartesian coordinates is a logarithmic spiral in polar coordinates. And suddenly the growth patterns of things like nautilus shells make perfect sense.

countrywideoptionone

2004-Apr-21, 06:02 PM

One of those things that just knocked my socks off was learning that a straight line of constant slope in Cartesian coordinates is a logarithmic spiral in polar coordinates. And suddenly the growth patterns of things like nautilus shells make perfect sense.

That's not exactly true.

A straight line of constant slope in Cartesian coordinates can be put in the general form Ax + By + C = 0, where A, B, and C and real number constants. To convert it to polar coordinates we need to find r and θ where θ is the angle between the polar axis and the ray from the origin that goes through our point. r is the distance our point is along the ray. To convert from Cartesian coordinates (x,y) to polar coordinates (r,θ) we can't just assume x=r and y=θ. We use the Pythagorean Theorem to find that r = sqr(x^2 + y^2), and some simple trig to find that θ = arctan(y/x) if x>0, θ = arctan(y/x) + Pi if x<0, θ = Pi/2 if x = 0 and y>0, and θ = 3Pi/2 if x = 0 and y<0. If x = 0 and y = 0, then r = 0 and θ has no meaning. Also x = r cosθ and y = r sinθ

So from our line Ax + By + C = 0 we can write it as A(r cosθ) + B(r sinθ) + C = 0. Or simplify to A cosθ + B sinθ +C/r = 0. It should not appear as a spiral.

Here’s a nice little applet (http://www.ies.co.jp/math/java/calc/sg_kyok/sg_kyok.html) where you can try some polar graphing. I tried y = 2x + 1, enter as r(t) = -1/(2*cos(t)-sin(t)) and you get a nice line that appears exactly as y = 2x + 1 appears on a Cartesian graph.

A logarithmic spiral can be expressed in polar form as r = ae^bθ where a and b are constants, or parametrically as x = r cosθ = a cosθe^bθ and y = r sinθ = a sinθe^bθ

Hope that helps.

mike alexander

2004-Apr-21, 09:22 PM

Aw, gee. Does that mean I have to put my socks back on? :(

What I meant to say was that if you make a polar plot where the radius is directly proportional to the angle you will get a spiral.

countrywideoptionone

2004-Apr-22, 03:51 PM

Aw, gee. Does that mean I have to put my socks back on? :(

What I meant to say was that if you make a polar plot where the radius is directly proportional to the angle you will get a spiral.

Right, but that wouldn't be a straight line, would it? :o

milli360

2004-Apr-22, 04:15 PM

Right, but that wouldn't be a straight line, would it?

He was generalizing the concept of constant slope.

mike alexander

2004-Apr-22, 07:23 PM

Y'know, milli, if you keep explaining things to me I may one day actually get smart! :D (I much appreciated your patience when you were explaining reference frames a while back. Thanks again.).

What I was trying to say (is this my third iteration?) is that in nature growth forms like shells follow spiral patterns. If you look at some of the fossil ammonites you will see they closely approximate Archimedian spirals (constant growth). If you look at, say, a nautilus, they exhibit a constant rate of growth, resulting in a logarithmic spiral (although I've always heard it called the equiangular spiral). My phrasing was wrong originally; I meant to imply the derivative and botched it.

Sorry about my confusion, countrywide.

Kevinito

2004-Apr-23, 01:35 AM

Since so much in nature seems to fit so well into this ratio, I am wondering why I don't see this relationship in cosmology?

Well, if you look at not only the golden ratio and Fibonacci's sequence related to the golden spiral, you will see the logarithmic spiral form over typical spiral galaxies.

-Kevin

darkdev

2004-Apr-23, 01:49 AM

Well, if you look at not only the golden ratio and Fibonacci's sequence related to the golden spiral, you will see the logarithmic spiral form over typical spiral galaxies.

-Kevin

Exactly, if this happens on such a large scale, all the way down to trees and snail-shells, why isn't it part of atomic and subatomic relationships...

It would seem to me that the spiral galaxy is a macro-effect, but of what?

milli360

2004-Apr-23, 03:24 PM

Exactly, if this happens on such a large scale, all the way down to trees and snail-shells, why isn't it part of atomic and subatomic relationships...

The Fibonacci sequence (and their ratio phi) are related to growth--that's why they pop up in that relationship I posted earlier (http://www.badastronomy.com/phpBB/viewtopic.php?p=243486&sid=dde7692f2ad91acbf969d63 264ab1a08#243486). It's not just shape, but shape influenced by growth tends to that ratio. In nature, it's not perfect, but it can be found.

Disinfo Agent

2004-Apr-23, 04:31 PM

Here (http://www.lhup.edu/~dsimanek/pseudo/fibonacc.htm) is a rather skeptical opinion about the prevalence of Fibonacci numbers and the like in nature...

milli360

2004-Apr-23, 04:40 PM

Here (http://www.lhup.edu/~dsimanek/pseudo/fibonacc.htm) is a rather skeptical opinion about the prevalence of Fibonacci numbers and the like in nature...

Heh. :)

From that website:

A favorite obsession of numerologically-inclined folk is to look for pi in man-made structures such as the Pyramids of Egypt. Look and ye shall find, if you are willing to select data and fuge a bit.

I know he means fudge.

countrywideoptionone

2004-Apr-23, 06:44 PM

Right, but that wouldn't be a straight line, would it?

He was generalizing the concept of constant slope. A spiral does not have a constant slope.

JohnOwens

2004-Apr-23, 07:31 PM

Right, but that wouldn't be a straight line, would it? He was generalizing the concept of constant slope. A spiral does not have a constant slope.

Depends on what kind of spiral you're talking about. There are different kinds, you know.

mike alexander

2004-Apr-23, 08:38 PM

I'm just trying to enjoy beautiful math here :D

Wouldn't a spiral with constant dr/d(angle) be an Archimedian spiral (constant spacing between turns)?

Any how, just pretty math for the perplexed. If you take a coordinate plane and draw a series of spokes through the origin, pick an arbitrary point on a spoke and draw a line normal to the spoke until it intersects the next spoke, then draw a normal line from that point, etc. you also get a spiral. This is a great time-waster at long meetings. Drawing nested golden triangles and inscribing a spiral is also fun.

(Edit)

Rectangles! Golden RECTANGLES! Not Triangles!

milli360

2004-Apr-23, 09:01 PM

Right, but that wouldn't be a straight line, would it?

He was generalizing the concept of constant slope. A spiral does not have a constant slope.

I didn't say it did. :)

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