Science > Physics > If space is non-Euclidean, then are the inverse-square laws really inverse-square?
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Science > Physics |
| User: |
"Casey Hawthorne" |
| Date: |
25 Aug 2006 07:43:16 AM |
| Object: |
If space is non-Euclidean, then are the inverse-square laws really inverse-square? |
If space is non-Euclidean, then are the inverse-square laws really
inverse-square?
e.g. F<sub G> = (G * m * M) / (r^2)
and Coulomb's Law?
--
Regards,
Casey
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| User: "Dirk Van de moortel" |
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| Title: Re: If space is non-Euclidean, then are the inverse-square laws really inverse-square? |
25 Aug 2006 08:07:13 AM |
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"Casey Hawthorne" <caseyhHAMMER_TIME@istar.ca> wrote in message news:asrte2tipu2t6ga85k9101e9f4g8fo1bn2@4ax.com...
If space is non-Euclidean, then are the inverse-square laws really
inverse-square?
e.g. F<sub G> = (G * m * M) / (r^2)
and Coulomb's Law?
Even if space is made of molten furlongs per chocolate fortnight,
inverse-square laws are really inverse-square. Remarkable, isn't it?
Dirk Vdm
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| User: "Timothy Golden BandTechnology.com" |
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| Title: Re: If space is non-Euclidean, then are the inverse-square laws really inverse-square? |
25 Aug 2006 08:39:18 AM |
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Casey Hawthorne wrote:
If space is non-Euclidean, then are the inverse-square laws really
inverse-square?
e.g. F<sub G> = (G * m * M) / (r^2)
and Coulomb's Law?
--
Regards,
Casey
I have focused on getting a product relationship close to the
simplistic force laws.
This requires a transformation of the usual distance by
p = 1 / ( d + 1 ) .
I don't know if this new space is non-Euclidean; it is transformable
back to the usual.
Our notion of distance in general is so fundamental a concept that to
deny the Euclidean sense seems paralyzing. Yet space is what physics
concerns itself with. Some fundamental changes down here could resolve
a lot of problems.
When things are far away they get smaller, not larger. Yet according to
the Cartesian system their distances grow larger to us. Informationally
there is a conflict here. Informationally the influence of or on a body
far away is nill. The only exception to this that seems possible is
electromagnetic effects, but even those dissipate. So the usage of
Cartesian distance causes the reciprocal relationship. Whether this is
really a problem of Euclidean versus non-Euclidean may be an
appropriate context, but this little transformation step above allows
force to be expressed as
F = q1 q2
where these q's hold distance in p and their signs are their charges
and are inherently quantized.
These distances are unity when adjacent and zero when very far away.
This eliminates infinities at either close distances or far away
distances. Strangely enough the same tape measure can be used; just cut
off the first inch and take the reciprocal. That is all that the
transform does. The notion of unity as local as opposed to zero is
pleasing.
This (p) is a space which corresponds to the second derivative in the
usual space.
The two spaces are necessary. The product space cannot perform vector
addition easily whereas the ordinary space does this naturally. This is
the next operation that has to be done; add up all of the quantized
forces. So these two spaces appear naturally along with the arithmetic
product. This suggests that we exist in a product space. The space
beneath may be topologically different. I call this the substrate. This
is part of my own agenda tied into:
http://bandtechnology.com/PolySigned/PolySigned.html
So I try to look at things on the topology
0D 1D 2D ...
and seek geometric product correspondences. Above is a semi-classical
approach.
-Tim
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| User: "Mike" |
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| Title: Re: If space is non-Euclidean, then are the inverse-square laws really inverse-square? |
25 Aug 2006 10:23:30 AM |
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Casey Hawthorne wrote:
If space is non-Euclidean, then are the inverse-square laws really
inverse-square?
e.g. F<sub G> = (G * m * M) / (r^2)
and Coulomb's Law?
--
Regards,
Casey
Whether space has actually a geometry or the geometry of space is a
matter of convention, is a matter of philosophical debate given the
name Conventionalism.
However, Newton's inverse square law has been confirmed experimentally
to about 1 part in a trillion accuracy:
http://prola.aps.org/abstract/PRL/v69/i12/p1722_1
If there is a deviation from the inverse square law it must take place
either in very small distances in the order of Planck length (10^-35 m
or so) or at very long ranges or both. Even is that is the case, it
will be hard to tie that to the "geometry" of space because it is very
hard to prove space has such property.
Mike
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| User: "Richard Tobin" |
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| Title: Re: If space is non-Euclidean, then are the inverse-square laws really inverse-square? |
25 Aug 2006 10:07:44 AM |
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In article <asrte2tipu2t6ga85k9101e9f4g8fo1bn2@4ax.com>,
Casey Hawthorne <caseyhHAMMER_TIME@istar.ca> wrote:
If space is non-Euclidean, then are the inverse-square laws really
inverse-square?
If space were the surface of a sphere, you would feel no force from an
object at the antipodes.
-- Richard
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