Arcadian Functor

occasional meanderings in physics' brave new world

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Marni D. Sheppeard

Saturday, February 24, 2007

M Theory Lesson 17

In the recent comments, Mahndisa wondered what I meant by mentioning Euler's relation

$e^{i \pi} = -1$

in M Theory Lesson 16. To be honest, it isn't entirely clear to me. But rather than going ahead and expanding the exponential in the usual way, I was asking why this function has the properties it does. We need to view Euler's relation anew, to see complex analysis in the light of diagrammatic reasoning. Non-standard analysis may take us a long way into topos methods, but it doesn't dissect the good old complex plane in the simple geometric way we require. Remember that those associahedra tilings were about real moduli. Now we need to understand the complex case, using 2-operads.

2-operads involve two-level trees. If we replace the upper level by the ordinal which counts the number of leaves, we obtain one-level trees with labels. It is known that such trees are appropriate for complex moduli. Now we would like to take the 2-operad polytopes, such as the hexagon, and tile spaces with them.


Blogger L. Riofrio said...

The Euler relation deserves attention too. The concept of "i" and imaginary time was used by Einstein, but forgotten in today's relativity textbooks. That has prevented generations of physicists from improving upon Einstein, or predicting a changing speed of light. Good posts!

February 24, 2007 4:57 PM  
Blogger Kea said...

Thanks, Louise. I'm happy for the Nene bird, too.

February 24, 2007 5:16 PM  
Blogger Mahndisa S. Rigmaiden said...

02 23 07

Interesting comment Louise. I have seen time describe in the padic context as well and I am not sure if such possibilities are excluded if one considers extending to irrationals! Thanks for putting this up Kea. I will think more about what you have asked. And I appreciate your response!

February 24, 2007 5:27 PM  
Anonymous Doug said...

gHi Kea

1 - With due respect, I think Euler's Identity is:
e^(i * PI) = -1.

[On my screen, your post reads:
e^(i / PI) = -1]'s_identity

2 - Imaginary time is discussed by Terry Gannon in 'Moonshine Beyond the Monster'. Although Einstein may have known about this concept, Steven Hawking made this popular in 'A Brief History of Time'.

Imaginary is a misnomer since Caspar Wessel demonstrated that this was an existent number in about 1797 [or 1799]. It is rotated one unit counterclockwise from a reference line and is therefore more invisible than imaginary.

David Hestenes is among those who recognize this history [See chart within ‘On the Evolution of Geometric Algebra and Geometric Calculus‘]. Hestenes tends to use “imaginary unit“ in ‘Grassmann's Vision’ as he discusses the improvements of both Grassmann and Clifford.

February 25, 2007 12:08 AM  
Blogger Kea said...

Hi Doug. Only Firefox with mathML fonts gives correctly formatted equations. Sorry.

February 25, 2007 11:57 AM  
Blogger nige said...

From Euler's relationship,

i^i = e^{-Pi/2) = 0.20787957635...

It's weird that you get real numbers come out of i^a

(1) when a = 2n where n is a positive integer (i.e., i^2 = -1, i^4 = +1, i^6 = -1, etc.), and

(2) when a = i (i.e., i^i = 0.20787957635...).

Presumably these are the only possible values of a which give a real number from i^a?

We had to do complex numbers and the Argand diagram in school because they were on the A-level maths syllabus, but I never understood them. This sort of abstract maths, that is totally weird compared to previous maths topics you have learned, needs to be taught with the historical background plus a few real applications (phase relationships in AC electricity and electronics, quantum mechanics, etc.). Otherwise it seems as crazy as M-theory!

February 26, 2007 3:30 AM  
Blogger nige said...

(I did pass the A-level maths later, however I just skipped the questions on complex numbers, and only studied them years later when doing a quantum mechanics course.)

February 26, 2007 3:33 AM  
Blogger nige said...

Sorry, above in i^a where a = 2n, n can be an either positive or negative integer since i^{-2} = 1/[(-1)^0.5]^2 = 1/-1 = -1.

February 26, 2007 5:02 AM  

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