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Unique Lorentz Boost?
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mark_horn@sbcglobal.net
science forum beginner


Joined: 18 Dec 2005
Posts: 6

PostPosted: Mon Jun 19, 2006 8:51 pm    Post subject: Unique Lorentz Boost? Reply with quote

18-JUN-2006

In the context of a Lornetz boost in the x,x' dimension, and in
dimensionless terms where the velocity of light is given as c = 1, I've
found for parameter 0 < v < 1, a solution that uniquely satisfies the
following condition:

(v^-1 - v) = p / (E + m), m = 1, p = mv, E = v^2,

where p is momentum, E is total energy and m is rest mass.

Then, also uniquely satisfied we have,

v = (1 / v) / (1 / v^2)

and

(v^2 + c) = (c / v^2).

I have it on good authority that there is absolutely no physics in this
solution... Why that is, was left as an exercise (the royal road of
knowledge is paved with self-help)...

Naively I've arrived at two interpretations:

A. Special relativity specifies a unique, non-trivial linear velocity
for the observable universe, treated as a unitary mass;

B. I am continually falling victim to a reasonable ineffectiveness of
mathematics, owing to my own ignorance.

I'm leaning towards B, in deference to the authorities, despite the
beauty, simplicity and jaw-dropping symmetry I find in my mathematical
results. I'd like to get unlost.

Any suggestions?

Cheers,

mark jonathan horn
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mark_horn@sbcglobal.net
science forum beginner


Joined: 18 Dec 2005
Posts: 6

PostPosted: Mon Jun 19, 2006 10:53 pm    Post subject: Re: Unique Lorentz Boost? Reply with quote

mark_horn@sbcglobal.net wrote:
Quote:
18-JUN-2006

In the context of a Lornetz boost in the x,x' dimension, and in
dimensionless terms where the velocity of light is given as c = 1, I've
found for parameter 0 < v < 1, a solution that uniquely satisfies the
following condition:

(v^-1 - v) = p / (E + m), m = 1, p = mv, E = v^2,

where p is momentum, E is total energy and m is rest mass.

Then, also uniquely satisfied we have,

v = (1 / v) / (1 / v^2)

as written, this is equivalent to the first equation; the correct
derivation of unique v is

v = (1 / v) / gamma,

gamma = (1 - v^2 / c^2)^-1/2.

For the unique solution we find

1 / v^2 = gamma. A little confusing, I guess.

[snip rest of original post]

cheers, m
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mark_horn@sbcglobal.net
science forum beginner


Joined: 18 Dec 2005
Posts: 6

PostPosted: Thu Jun 22, 2006 6:42 am    Post subject: Re: Unique Lorentz Boost? Reply with quote

21-JUN-2006

The really interesting thing here is that, we can derive the unique
velocity directly from a proof in differential geometry (Thomas
Friedrich, "Solutions of the Einstein-Dirac Equation on Riemannian
3-Manifolds with Constant Scalar Curvature," arXiv: math.DG/0002182),
wherein we find the non-Einstein Sasakian metrics on S^3 admitting weak
Killing spinors can be given as

+C = v^2
- C = - (1 / v^2)

+lambda = (v^-4 - 1 / 2)
- lambda = (-v^2 + 1 / 2)

where +/- C are the eigenvalues of the Ricci tensor, and +/- lambda are
the real eigenvalues of the Einstein-Dirac equation. In this way we
have for the unique velocity in the Lorentz transform,

v = (+C)^1/2.

So, for the sake of argument:

If the observable universe, idealized as a Riemann sphere with unit
mass, has a linear velocity on the order of 79% light velocity, what
measurable signature of such relativistic energy might be imprinted on
the intrinsic observables? Or perhaps more appropriately, what
artifacts of its acceleration to v would be observable? How would we
compare this to a trivial velocity?

With respect to the possible meaninglessness of the idea of cosmic
velocity, how would acceptance of such meaninglessness compare with
assuming v = 0?

Can a unique solution of the Lorentz transform really be written off as
a numerical coincidence possessing no physical meaning?

Best,
mark jonathan horn
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