Arcadian Functor

occasional meanderings in physics' brave new world

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Marni D. Sheppeard

Sunday, January 06, 2008

Riemann Rekindled III

I haven't had a chance to watch GRT Lecture 19 yet, but the puzzle is to find a groupoid with cardinality $\pi$ and one with cardinality $e^e$. What fun! As it happens, I was thinking about $e^e$ last night, because

$e^{e^{i \pi}} = \frac{1}{e} = \textrm{exp} (1 + 2 + 4 + 8 + \cdots) = e \cdot e^2 \cdot e^4 \cdots$

and that made me wonder about rescalings of the Riemann zeta function, such as

$F(s) = \sqrt{\frac{1}{2} (s+1) (s+2) \zeta (s) }$

in terms of which

$i = \frac{\textrm{log} (1 + 2 + 4 + 8 + \cdots)}{F(2)}$

with $F(2) = \pi$, and this looks something like a count of binary trees, with an increasing number of branches at each step. What are the higher dimensional analogues of $i$? What if we took the $s$-th root, so that $F(2n)$ was some multiple of $\pi$ for all $n \in \mathbb{N}$, just like the volumes of spheres?

12 Comments:

Blogger Kea said...

P.S. At some point we want to consider the zeta function (or a scaled version) as a map from the Riemann sphere to itself...

January 06, 2008 9:49 AM  
Blogger nige said...

Normally I read your blog using Explorer, but using Mozilla Firefox now makes the equations come alive! Maybe you should consider making little jpeg pictures of equations and posting them in the post so that readers using any browser can see them (i.e., control+alt+print screen, then paste into Paint using control+v, edit image size and save as jpeg; it's fairly quick).

But now I'm wondering if there is something wrong with my eyes, browser or the way the equations are being displayed. Your first one says 1/e = e^{1 + 2 + 4 + 8 + ...) = e(e^2)(e^4)...

Clearly 1/e ~ 1/2.7 ~ 0.3-0.4. This obviously can't equal e^{infinity} which is what you get from the sum e^{1 + 2 + 4 ...} or e(e^2)(e^4)...

I've no problem with e^(e^{i*Pi}) = 1/e, seeing that it's well known that i^i = e^{-Pi/2}.

I only have a problem with the first equation seeming to say that 0.3 = infinity. Please let me know what my problem is, when or if you have time. Thanks!

January 07, 2008 1:04 PM  
Blogger nige said...

(I'm not claiming that e^(e^{i*Pi}) = 1/e, is a consequence of i^i = e^{-Pi/2}, just that I'm not surprised that any such weird results can come out of maths. This is not in exactly the same league as claiming that the number 0.3 equals infinity.)

January 07, 2008 1:11 PM  
Blogger Kea said...

Nigel, such sums are only infinity if we blindly add them up using ordinary counting, but as I often discuss on this blog, categorified sums can behave a little differently.

January 08, 2008 7:46 AM  
Blogger nige said...

I'll have to concentrate on this a lot more, I guess. At present categorical theory is still way over my head. I think in school we did a bit of very basic set/group maths, like Venn diagrams and the just the abstract symbols for union (U) and intersection (upside down U), but they the whole area was dropped. From there on it was algebra, trig and calculus (particularly the nightmare of integrating complex trig functions like cot or cosec theta, without having a good memory for trivia like definitions of abstract jargon). There was no set or group theory in the pure maths A-level, and at university the quantum mechanics and cosmology (aka elementary general relativity) courses didn't use anything more advanced than calculus with a bit of symbolic compression (operators).

The kind of maths where you get logical arguments with lots of abstract symbolism from set theory and group theory is therefore completely alien. I can see the point in categorizing large numbers of simple items, if that is as actually a major objective of categorical theory. It would be nice if it were possible to build up solutions to complex problems like quantum gravitation by categorizing large numbers of very simple operations, i.e. if individual graviton exchanges between masses could be treated as simple vectors and categorized according to direction or resultant to simplify the effect. Smolin had a Perimeter lecture on quantum gravity where he showed how he was getting the Einstein field equation of general relativity by summing all of the interaction graphs in an assumed spin foam vacuum. I'm not sure that a spin foam vacuum is physically the correct, but the general idea of building up from a summing of lots of resultants for individual graviton interaction graphs is certainly appealing from my point of view.

"with $F(2) = \pi$, and this looks something like a count of binary trees, with an increasing number of branches at each step. What are the higher dimensional analogues of $i$? What if we took the $s$-th root, so that $F(2n)$ was some multiple of $\pi$ for all $n \in \mathbb{N}$, just like the volumes of spheres?"

I may be way off topic in my physical interpretation here, but if you are considering how graviton exchanges occur between individual masses (particles, including particles of energy since these interact with gravity and thus have associated with them a gravitational charge field), then you could well have a tree structure to help work out the overall flow of energy in a gravitational field from a theory of quantum gravity.

I.e., each mass (or particle with energy) radiates gravitons to several other masses, which radiate to still more, in an geometric progression. This loss of energy is balanced by the reception of gravitons. Presumably this kind of idea just sounds too naive and simplistic to people in the mainstream, who assume (without it ever having been correctly proved) that such simplistic ideas must be wrong because nobody respectable is working on them.

I'm studying the maths of the SU(2) lagrangian as time allows. It's nice that the lagrangian is simplest for the case of massless spinor fields (massless gauge bosons). The most clear matrix representations of U(1) and SU(2) to particle physics I've come across are equations 8.59 and 8.65 (which are surprisingly similar) in Ryder's "Quantum Field Theory". The Dirac lagrangian for a massless field just summed for the particles: e.g., right handed electron, left handed electron, and also the neutrino which only occurs in the left-handed form. Given some time, it should be possible to understand the massless SU(2) lagrangian since it is relatively simple maths (pages 298-301 of Ryder's 2nd edition, also the first 3 chapters of Ryder were excellent lucid introductions to gauge fields in general and the Yang-Mills field in particular).

But one problem I do have with the whole gauge theory approach is that it is built on calculus to represent fields; ideal for a vacuum that is a continuum, but inappropriate for quantized fields. There's an absurdity in treating the acceleration of an electron by quantized, individual discrete virtual photons or by gravitons as a smooth curvature of spacetime! It's obviously going to a bumpy (stepwise) acceleration, with a large number of individual impulses causing an overall (statistical) effect that is just approximated by differential geometry. I think it's manifestly absurd for anyone to be seeking a unification of general relativity and quantum field theory that builds on differential geometry. Air pressure, like gravity, appears to be a continuous variable on large scales where the number of air molecule impacts per unit area per second is a very large number. But it breaks down for small numbers of impacts, for example in Brownian motion, where small particles receive chaotic impulses not a smooth averaged out pressure. Differential equations are usually good approximations for classical physics (large scales), but they are not going to properly model the fndamental physical processes going on in quantum gravity. You can do quite a lot with the calculus of air pressure (such as fnding that it falls off nearly exponentially with increasing altitude, and finding the relationship between wind speed and pressure gradients in hurricanes), but you can't deduce anything about air molecules from this non-discrete (continuum) differential model. It breaks down on small scales. So does differential geometry when applied to small numbers of quantum interactions in a force field. This is why the classical physics breaks down on small scales, and chaos appears.

It would be nice if it were possible to replace differential geometry in QFT and GR with some kind of quantized geometry and show how the approximations of QFT and GR are valid, emerging for the limiting case whereby very large numbers of field quanta interact with the particle of interest, so that the averaging of many chaotic impulses produces a deterministic average effect every time on large scales.

January 08, 2008 2:01 PM  
Blogger kneemo said...

Clearly 1/e ~ 1/2.7 ~ 0.3-0.4. This obviously can't equal e^{infinity} which is what you get from the sum e^{1 + 2 + 4 ...} or e(e^2)(e^4)...

The sum 1 + 2 + 4 + 8 + 16 + ... actually converges to -1. This is one of those non-intuitive results one finds when learning about geometric sequences.

January 09, 2008 4:54 PM  
Blogger nige said...

Hi Kneemo,

"The sum 1 + 2 + 4 + 8 + 16 + ... actually converges to -1."

A geometric series doesn't converge, and this one tends toward infinity.

It's quite easy to show that if you keep adding positive real numbers to positive real numbers, you can only end up with a positive real number.

Do you know if the proof to the contrary exists anywhere on the internet? The kind of mathematical groupthink where people make claims without proving them, and then on closer inspection the proof is of something else entirely, is the claptrap which puts some logical people off maths.

In this case, I'd guess that the series you're referring to is not what you claim (the sum of positive real numbers), but must include negative numbers, e.g

... -16 - 8 - 4 - 2 - 1 + 0 + 1 + 2 + 4 + 8 + 16 + ...

Is clearly not infinity (it is a sum that is zero).

This kind of thing gives a non-infinite result because negative numbers are included to cancel out positive numbers.

January 10, 2008 9:40 AM  
Blogger nige said...

BTW, it's great to get non-intuitive results if they're right. Hopefully you're right and you can prove it. I think that 11 dimensional supergravity would also be a great non-intuitive result if it could be proved, but it can't. I look forward to seeing physically how adding up an infinite series of positive numbers can result in a negative number. Cheers

January 10, 2008 9:47 AM  
Blogger nige said...

Kneemo, taking your series of numbers

1 + 2 + 4 + 8 + 16 + ...

this series is just

2^0 + 2^1 + 2^2 + 2^3 + 2^4 + ...

In a normal geometric progression,

r^0 + r^1 + r^3 + r^3 + r^4 + ...

= r^n

(see any school textbook on algebra or http://en.wikipedia.org/wiki/Geometric_progression#Geometric_series )

Hence, in your case r = 2 and n = infinity terms.

Thus, you are claiming that

2^{infinity} = -1.

It will be interesting to see what you can provide to back this claim up with.

A lot of claptrap in physics is falsely defended as being non-intuitive genius, when in fact it's just not even wrong.

January 10, 2008 10:05 AM  
Blogger nige said...

Sorry a typing error exists above:

1 + 2 + 4 + 8 + 16 + ...

this series is just

2^0 + 2^1 + 2^2 + 2^3 + 2^4 + ...

In a normal geometric progression,

r^0 + r^1 + + r^2 + r^3 + r^4 + ...

= r^n

January 10, 2008 10:43 AM  
Blogger kneemo said...

In this case, I'd guess that the series you're referring to is not what you claim (the sum of positive real numbers), but must include negative numbers, e.g

The sum converges to -1 over the 2-adics, not the reals. I guess 2-adics seem fancy, but 2-adic series is taught in calculus these days, along with this non-intuitive result. Kea's "categorified sums" might be related to this 2-adic result.

January 10, 2008 6:56 PM  
Blogger kneemo said...

Thus, you are claiming that

2^{infinity} = -1.

It will be interesting to see what you can provide to back this claim up with.


Take a look at this calculus webpage to see how the -1 comes about using the 2-adic norm. The argument rests on the fact that the partial sum after n terms is the 2-adic expansion of 2^n - 1. You then take the 2-adic norm of 2^n and find that it is of zero length for large n. When this happens, all you're left with is the -1.

January 10, 2008 9:47 PM  

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