Is Euler's identity all about spin?

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jeffreyH

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

$$e^{i\pi}+1\,=\,0$$

This can be written as

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

then

$$\left[\left(e^\pi\right)^i\right]^2\,=\,1$$

We then end up with

$$e^{i2\pi}-1\,=\,0$$

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

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...

jeffreyH

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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.

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 »

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 ±½ or ±3/2... (also not producible from eix)

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.

jeffreyH

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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 »

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.

jeffreyH

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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 »