Requiem for Relativity

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17 years 5 months ago #19461 by Joe Keller
Replied by Joe Keller on topic Reply from
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote"><i>Originally posted by cosmicsurfer</i>
<br />Hi Joe,

I agree with your statement from above post:

"...This is consistent with a new, gravitational theory of the CMB, which would revolutionize our understanding of the relationship between gravity and electricity."

Wondering if you have any source documents or links pertaining to this research area?

My own views are a modification of MM theory in that the Graviton/Antigraviton is a charged particle of higher frequency above the speed of light spectrum and is in itself the driving force behind all mass fluctuations as shown in collider experiments revealing three trillion times per second flipping between states of matter and antimatter in sub B Mesons.

Thanks, John

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Dear Cosmicsurfer (John),

Thanks for the timely comment! See next post!

- Joe Keller

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17 years 5 months ago #19060 by Joe Keller
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Throughout almost all of the solar system out to about 52.6 AU, the direction of the gravitational field is approximately constant over nuclear distances. The gravitational field in Australia is opposite that in England, but the gravitational field is practically the same, in adjacent nucleons.

At 52.6 AU, the sun's gravity becomes equal to the strongest gravity achievable within a proton (the Heisenberg uncertainty principle limits the mass density, achievable with DeBroglie waves). Outside 52.6 AU, the direction of gravity varies up to 180 degrees, within an atomic nucleus; inside 52.6 AU, the direction of gravity is never more than 90 degrees from sunward. This is a quantized change. Almost anywhere inside 52.6 AU, a space-crystal can exist such that all gravity dipoles within the crystal are aligned parallel to the crystal axis, and also within 90 degrees of parallel to the microscopic gravitational field. Outside 52.6 AU, some gravity dipoles (in crystals near protons) must align antiparallel with the crystal axis in order to be within 90 degrees of parallel to the microscopic gravitational field. The energy penalty for this might correspond to the CMB temperature.

A probe could be sent to the region of space between Earth and Sun, 160,000 mi from Earth, where Earth's gravity cancels the Sun's, to within one part in 2500, thereby simulating the 52.6 AU barrier. This region would be about 65 mi thick and 130 mi wide. At sunrise/set this "agravitational region" would be 1.4 deg from the sun, on-center, and 2.6 arcminutes diam. (Never look at or near the sun with or through anything at any time, not even sunset! Let experts make photographs, then look at the photographs; or, project the sun on grey cardboard and look at the cardboard.)

Wikipedia has accurate information on the discovery of Vulcan by Lescarbault & LeVerrier. More complete information is found in the book, "In Search of Planet Vulcan" by Baum & Sheehan, 1997. The original English source is Rev. Prof. Challis, Proc. Camb. Phil. Soc. 1:219-222, March 12, 1860 (cites Comptes Rendus January 2, 1860, p. 40). The transit time observed for "Vulcan" differs only 2.67% from that expected for projections (see next par.) from the center of the above "agravitational region". This would seem to be within observer error; Vulcan's chord gave only a 37 degree sector. So the transit time was about right; though at 48N latitude, the phenomenon should have missed the sun by 1.0 degree, on-center. Instead it apparently was Liais' serendipitous simultaneous negative observation, from Rio de Janeiro, which missed.

In the preceding paragraph, I assume that the "agravitational region" is a sidereally nonrotating transparent optical structure. For some reason, it blocks wavefronts propagating through it in a certain narrow cone of directions. Lescarbault estimated Vulcan's apparent diameter as 1/4 Mercury's. Thus the angular diameter of this black cone ("Vulcan") happens to be in about the same ratio to the distance from the "region" to Earth, as the angular diameter of the sun is to the distance from the "region" to the sun. Relative to the sun, the region's optical structure rotates retrograde. This causes the black cone to appear as a dot moving prograde as if the region were moving around the sun at the same speed as Earth, not one part in 500 slower. This apparent motion, which is in the plane of the ecliptic, causes about 1/3 of the dot's apparent motion.

The other 2/3 of the dot's apparent motion is due to Earth's rotation, and lies in Earth's equatorial plane. This results in a 15 degree inclination to the ecliptic, tolerably close to Lescarbault's estimate, 6.3 +/- 1 degree (both ascending).

LeVerrier, too, calculated an orbital inclination and ascending node for Vulcan. His results differ from Lescarbault's because, at least according to one anonymous internet source, LeVerrier combined Lescarbault's data with data from five other similar sightings gleaned by searching the literature and eliminating sightings that likely were sunspots, or which were Mercury or Venus. LeVerrier found:

Vulcan (LeVerrier) inclination to ecliptic: 12# 10'
(Barbarossa/Frey (Keller) inclination: 12# 9.7')

Vulcan (LeVerrier) ascending node: 12# 59'
(Barbarossa/Frey (Keller) asc. node: 283# 41' = 13# 41' - 90#)

Maybe both Vulcan and Barbarossa are manifestations of the same real phenomenon which is not a planet. Of five sky surveys I have seached for Barbarossa/Frey, the two surveys on which I have found both of them, were the two surveys on which they were closest to opposition: the survey dates were February 25, 1954, and March 15, 1986. The only prospective photo showing both Barbarossa & Frey, also is the first good-quality photo taken, and the closest one to opposition: its date is March 25, 2007.

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17 years 5 months ago #19587 by Bill_Smith
Replied by Bill_Smith on topic Reply from William Smith
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote"><i>Originally posted by Joe Keller</i>
<br />Barbarossa and Cassini's third law.

Barbarossa manifests the Tifft period.

"The problem in working with short [redshift] periods...is not measurement, it is physical intuition. ...it seems inconceivable that fluctuations within systems would not mask periodic order on a very small scale. Despite the near universality of this feeling, the data appear to speak otherwise."

- WG Tifft, Astrophysical Journal 468:491+, 1996, p. 506
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Hi Joe,

I don't pretend to know jack about theory - I just observe and apply logic. So my logic says that the above is a bit of a stretch. 1. it applies to galactic redshift, that is the acceleration away from us, not their actual velocity in space, and 2. He's talking "multiples of" rather than "a value of".

As a side note, what did you actually ask those amateurs to do? I'm looking at your open letter to the MOD(N) and it doesn't appear to be a very practical request. You might get a better response by asking to observe a region of sky within a certain timeframe to a given detection limit!

Also, talking from a experienced observers perspective, looking at current images and comparing them to historic images is only useful to a certain extent, particularly if there was something on the old image that is not on the new image. It is near impossible to rule out asteroids or even space junk that may have been captured on the old image (and they can mimic both fast moving NEO's and very slow moving objects at all brightness levels requiring observations over a number of days to rule them out).

Cheers

Bill

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17 years 5 months ago #19588 by Bill_Smith
Replied by Bill_Smith on topic Reply from William Smith
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote"><i>Originally posted by Joe Keller</i>
A probe could be sent to the region of space between Earth and Sun, 160,000 mi from Earth, where Earth's gravity cancels the Sun's, to within one part in 2500, thereby simulating the 52.6 AU barrier. This region would be about 65 mi thick and 130 mi wide. At sunrise/set this "agravitational region" would be 1.4 deg from the sun, on-center, and 2.6 arcminutes diam. (Never look at or near the sun with or through anything at any time, not even sunset! Let experts make photographs, then look at the photographs; or, project the sun on grey cardboard and look at the cardboard.)
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Hi Joe,

Isn't this the L1 point of the Lagrange points? According to the reference we already have 4 probes/satellites at the L1.

Cheers

Bill

en.wikipedia.org/wiki/Lagrangian_point

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17 years 5 months ago #19589 by Joe Keller
Replied by Joe Keller on topic Reply from
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote"><i>Originally posted by Bill_Smith</i>
<br /><blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote"><i>Originally posted by Joe Keller</i>
A probe could be sent to the region of space between Earth and Sun, 160,000 mi from Earth, where Earth's gravity cancels the Sun's, to within one part in 2500, thereby simulating the 52.6 AU barrier. This region would be about 65 mi thick and 130 mi wide. At sunrise/set this "agravitational region" would be 1.4 deg from the sun, on-center, and 2.6 arcminutes diam. (Never look at or near the sun with or through anything at any time, not even sunset! Let experts make photographs, then look at the photographs; or, project the sun on grey cardboard and look at the cardboard.)
<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">

Hi Joe,

Isn't this the L1 point of the Lagrange points? According to the reference we already have 4 probes/satellites at the L1.

Cheers

Bill

en.wikipedia.org/wiki/Lagrangian_point
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Dear Bill,

Thanks for your input! One of the Lagrangian points, is where the sun has enough gravity left over, even after subtracting Earth's, to swing a satellite in an orbit slightly smaller than Earth's but with period one year. What I discussed, is a different point at which Earth cancels approximately all the sun's gravity. That's much closer to Earth, than the Lagrangian point is.

Regarding magnitudes: Uranus is +5.7 at opposition and almost 20 AU out. Barbarossa is almost 200 AU out. The brightness goes as the fourth power of distance because the inverse square law holds both ways (it's farther away, and also the sun is dimmer there). Four powers of 10, is 10 magnitudes. So, if Barbarossa were just like Uranus, it would be about +15.7 (more precisely, +15.9).

Uranus has a rather high albedo, 66% according to a 1983 college textbook. Albedos of 4 to 8% often have been used in recent years as "canonical" albedos in academic journal articles about comet nuclei and Kuiper Belt Objects. Also, a recent article theoretically estimated the albedo of one type of borderline brown dwarf, as 1%! So, it would be fair to use 6.6%, i.e., about 1/10 the albedo of Uranus: +15.9+2.5=+18.4. Using 1% albedo would give +20.45.

The objects I've found and suspected of being Barbarossa, have comparison albedos (these are necessarily somewhat inaccurate) ranging from +17.3 to +20, but mainly +18 to +19. Theoretically, Barbarossa (assuming 1 to 10 Jupiter masses) should be the size of Jupiter only if Barbarossa has a H/He composition like the sun or Jupiter. The same article predicts that if a medium-weight brown dwarf isn't made of hydrogen (even if it's helium or oxygen), it has 1/2 to 1/3 the diameter of the hydrogen version. Of course, this is only theory. That's why observation is so important.

- Joe Keller

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17 years 5 months ago #19462 by Bill_Smith
Replied by Bill_Smith on topic Reply from William Smith
Looks like the forum has gone a little haywire with duplication.

About the magnitudes, so this is why the object is at or near the limit of Bobs images. I think I quoted mag 18.5 to Bob when I measured them so it leaves a large gap between 18.5 and 20.

If the object is a brown dwarf would you be relying on albedo only?

Cheers

Bill

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