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Loren Booda
Aug30-04, 10:11 PM
Closed spacetime with its dispersed masses and open spacetime under dark energy interfere dynamically at the Euclidean boundary. This zero curvature surface distinguishes where the cosmic geometry contracts from where it expands, and represents a significant unexplored structure within familiar cosmology. What general pattern do flat geodesics manifest throughout the magnitudes of the observable universe?

chroot
Aug30-04, 10:23 PM
I'm afraid this is gibberish, to me anyway.

- Warren

Loren Booda
Aug31-04, 12:17 AM
chroot,

What patterns do null geodesics follow in spacetime generally, and does this structure follow a geometry of physical significance cosmologically? For instance, lightlike worldlines describe a network of rays that juxtaposes matter and dark energy. Looking at light trajectories in toto may not only reveal a novel perspective on local collapse and expansion of spacetime, but also discover an overlooked configuration of utility in the pursuit of attractive vs repulsive gravity.

Chronos
Aug31-04, 02:05 AM
That did not aid in my understanding the question, comment, or point.

Loren Booda
Aug31-04, 11:00 AM
Take all zero-curvature manifolds in spacetime. As an aggregate, do these surfaces indicate a physical entity (like large-scale structure, galaxies or stars), but one whose form actively distinguishes between positive ("massive") and negative ("dark energy") curvature?

Loren Booda
Aug31-04, 07:09 PM
Once more: do lightlike geodesics themselves trace out an overall path that not only defines spacetime but manifests a physical system correspondent to, e. g., that of orbiting masses or vacuum energy creation?

chroot
Aug31-04, 08:30 PM
This is also gibberish. There is no such thing as a "lightlike geodesic." Geodesics are geodesics. Geodesics do not define spacetime directly, but one may use the "geodetic effect," the failure of initially parallel geodesics to remain parallel in curved space, to quantitatively describe the curvature. I have no idea what "manifests a physical system" means.

- Warren

Loren Booda
Aug31-04, 11:36 PM
Consider only light in our universe, not other particles massive and massless. What general form do all of these rays take, and can one predict their evolution give only this form as an initial condition?

Chronos
Sep1-04, 12:08 AM
I assume you have a mathematic or observational model to support your assertion. On the other hand, I doubt it. The 'initial condition' is poorly defined.

Loren Booda
Sep1-04, 10:29 PM
Is it possible for two photons to co-orbit stably solely by attraction between their mutual energies (photonium)?

jcsd
Sep2-04, 06:24 AM
This is also gibberish. There is no such thing as a "lightlike geodesic." Geodesics are geodesics. Geodesics do not define spacetime directly, but one may use the "geodetic effect," the failure of initially parallel geodesics to remain parallel in curved space, to quantitatively describe the curvature. I have no idea what "manifests a physical system" means.

- Warren

lightlike geodesic = null geodesic I imagine.

da_willem
Sep2-04, 11:34 AM
Is it possible for two photons to co-orbit stably solely by attraction between their mutual energies (photonium)?

As this is the only thing in all of your postst that makes any sense to me: I guess it is possible for two photons to orbit each other, at least in theory. But I imagine they would have to have a very large energy (frequency).

Loren Booda
Sep3-04, 01:26 AM
Does a null geodesic oscillate in spacetime correlative to the E-M frequency of the photon which defines it? This indicates to me that null geodesics rely not only on the constant speed of light to define them, but also on the given photon's wave character.

On small scales, the test for geodesics would need to be modified. Can multiple yet equal extremal lightlike intervals simultaneously satisfy Feynman's path integral, giving geodesics a probabilistic character? Equivalently, if photons of differing wavelengths thus follow slightly different sinusoidal paths, would not the requirement of extremal length itself, a la quantum mechanics, compel multifold geodesics? Would the number of these geodesics satisfying such a condition be finite?

Please give the above a fair reading.

chroot
Sep3-04, 01:42 AM
Loren,

Please be careful with your posts. While you seem to be mostly asking questions, you are also bordering on pseudoscientific personal theories, which are no longer allowed on this forum.

Geodesics do not oscillate, and are not defined by photons.

- Warren

Chronos
Sep3-04, 01:45 AM
They do not satisfy the 'Feynman' path integral. 'probabilistic' or otherwise. That is why that model has been abandoned. Try a new paradigm. You apparently know enough to rethink that position.

Loren Booda
Sep3-04, 10:57 AM
Thank you for keeping me in-bounds, chroot and Chronos. At least you gave me this opportunity to thresh out my ideas.

Would most of this material be more appropriate to Theory Development? Even there Physics Forums seems to have "locked down" upon genuine speculation. Where can I best read about this new policy?

chroot
Sep3-04, 02:59 PM
There's a sticky at the top of the Theory Development subforum outlining the new policies. The same text also appears in the posting guidelines, viewable in the Feedback forum.

- Warren