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Author Topic: Is Euler's identity all about spin?  (Read 493 times)

Online jeffreyH

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Is Euler's identity all about spin?
« on: 12/06/2016 18:23:32 »
Euler's identity is

60f46a1ea13a8254cf88154ce48ec491.gif

This can be written as

089de068666bb4f30f6ea23be63868de.gif

then

45fc84d7336cd5cf1656b9973cc3a394.gif

We then end up with

26547a68cd71677e1c0a41e1c5088c80.gif

So we now have two equations that relate to one half and one full rotation of the unit circle respectively.


 

Offline chiralSPO

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Re: Is Euler's identity all about spin?
« Reply #1 on: 12/06/2016 19:27:26 »
I think it has more to do with trigonometry than spin, but I could be mistaken.

e = 1 and ei2π = 1 are both just special cases of eix = cos(x) + i*sin(x)

Is there more to it? Maybe, but I don't see it...
 

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Re: Is Euler's identity all about spin?
« Reply #2 on: 12/06/2016 19:56:10 »
I think it has more to do with trigonometry than spin, but I could be mistaken.

e = 1 and ei2π = 1 are both just special cases of eix = cos(x) + i*sin(x)

Is there more to it? Maybe, but I don't see it...

I will be following up on this later.
 

Online jeffreyH

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Re: Is Euler's identity all about spin?
« Reply #3 on: 12/06/2016 20:59:39 »
I think it has more to do with trigonometry than spin, but I could be mistaken.

e = 1 and ei2π = 1 are both just special cases of eix = cos(x) + i*sin(x)

Is there more to it? Maybe, but I don't see it...

The special case applies to the expression einπ where n is in the set of integer values. So that multiples of half and integer rotation round the unit circle will assume the identity format. We then have +1 for odd integers and -1 for even integers. The fluctuation in values is binary as is spin up/spin down or right handed/left handed. So I do believe there can be a deeper meaning to the identity. Not sure exactly what it is though. It may be one of those thoughts like the beta function and string theory. I am not a fan of string theory.
« Last Edit: 12/06/2016 21:01:47 by jeffreyH »
 

Offline chiralSPO

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Re: Is Euler's identity all about spin?
« Reply #4 on: 12/06/2016 21:08:06 »
I think it has more to do with trigonometry than spin, but I could be mistaken.

e = 1 and ei2π = 1 are both just special cases of eix = cos(x) + i*sin(x)

Is there more to it? Maybe, but I don't see it...

The special case applies to the expression einπ where n is in the set of integer values. So that multiples of half and integer rotation round the unit circle will assume the identity format. We then have +1 for odd integers and -1 for even integers. The fluctuation in values is binary as is spin up/spin down or right handed/left handed. So I do believe there can be a deeper meaning to the identity. Not sure exactly what it is though. It may be one of those thoughts like the beta function and string theory. I am not a fan of string theory.

But particles aren't just spin up or spin down. There are some particles that are spin 0 (and there is no x that satisfies eix = 0), and there are particles that are or 3/2... (also not producible from eix)
 

Online jeffreyH

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Re: Is Euler's identity all about spin?
« Reply #5 on: 13/06/2016 22:15:10 »
I am only considering the subset where x is integer multiples of pi. Where x can be positive or negative. We can then take n/2 to indicate spin. With plus and minus one as handedness/polarity. So that fractional spin continually switches handedness/polarity and integer spin maintains the same polarity. An offset of pi will then swap the polarity of bosons. It is not an ideal model by any means.
 

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Re: Is Euler's identity all about spin?
« Reply #6 on: 14/08/2016 18:39:43 »
If we look at particles of spin 1/2, 1 and 2 we can use values of pi to represent them. With 4*pi for spin 1/2 (fermions), 2*pi for spin 1 (photons) and pi for spin 2 (gravitons).

So that

spin 1/2 is e^(i*4*pi) = 1
spin 1 is e^(i*2*pi) = 1
spin 2 is e^(i*pi) = -1

spin 2 is then distinguished by having the opposite sign. Since spin 1 photons have only polarity and no charge we could amend its definition as

spin 1 is e^(i*2*pi) - 1 = 0

This is now a modified Euler identity.

The field of the electron can be given by

e^(i*4*pi) -2 = -1

The proton is then given by

e^(i*4*pi) = 1

As with the masses of the proton and neutron there is an imbalance between the derivations of positive and negative involving the value of -2. Elsewhere I investigated this mass/charge discrepancy in a slightly different way but the results were similar in nature.

Does this indicate a dual nature of the gravitational field? It may have no bearing at all. I will leave that for others to judge.
« Last Edit: 14/08/2016 18:41:49 by jeffreyH »
 

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Re: Is Euler's identity all about spin?
« Reply #7 on: 14/08/2016 19:57:13 »
Also the graviton can be represented by Euler's identity of e^(i*pi) + 1 = 0.
 

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Re: Is Euler's identity all about spin?
« Reply #8 on: 14/08/2016 21:02:30 »
One last thing to note. Why no 3/2 spin particles? Well if we use this method

Spin 3/2 is e^(i3pi/2) = -i

So that this super symmetry partner is complex.
« Last Edit: 14/08/2016 21:04:54 by jeffreyH »
 

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Re: Is Euler's identity all about spin?
« Reply #8 on: 14/08/2016 21:02:30 »

 

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