Paradoxes Resolved, Origins Illuminated - Requiem for Relativity
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Larry Burford

USA
2220 Posts

Posted - 25 Feb 2010 :  18:26:58  Show Profile  Reply with Quote
Joe,

Sorry for the deletions, but we had to have our database restored from a four or five day old back-up. We still have some messes to clean up, but so far things seem to be stable and no new attacks have been detected. (We are pretty sure the BEMs and the LGM are not involved, but who knows for sure?)

(... Sigh)

Regards,
LB
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Joe Keller

USA
957 Posts

Posted - 25 Feb 2010 :  19:58:10  Show Profile  Reply with Quote
quote:
Originally posted by Jim

Dr Joe, How do you determine the gravity of a proton? ...I think the solar field becomes equal to the gravity of the universe at some distance near 50 or 100AU.



Thanks, Jim, for the post. This "gravity of the universe" figure, is interesting! Do you recall the source?

Regarding the gravity of a proton, I sketched the calculation in other posts on this messageboard years ago, and also in my 2002 article in "Aircraft Engineering and Aerospace Technology", but here's a review:

1. Let's try to construct a proton of the smallest size possible, using college quantum mechanics to make the most compact possible radial Gaussian mass distribution allowed by the proton's mass, according to the Heisenberg uncertainty principle. Let's start with the formula E = sqrt(m^2*c^4 + p^2*c^2). I'm considering the proton's constituent deBroglie waves; in this view, there is no rest mass because all the energy is in the internal "p". Because of equipartition of energy, one part of the proton's mass is available for the deBroglie waves in each of the three directions, and a fourth part is available for "spin". Along, say, the x-axis, I have momentum p such that the root-mean-square value of p satisfies p^2*c^2 = 0.25 * m^2*c^4, i.e. p = 0.5 * m*c. A deBroglie wave of wavenumber k, has momentum p = hbar*k. So, the rms value of k (along the x-axis) is 0.5*m*c/hbar.

2. Heisenberg's uncertainty principle is really only a theorem in the first year graduate mathematical course called "real analysis". The theorem says, that in Gaussian wave packets, the bigger the rms k value (i.e. wavenumber) along, say, the x-axis, the smaller the packet's x dimension. The formula is k*x = 0.5, where the k & x are the rms values for the wave packet. (There's another theorem that says that Gaussian wave packets are the best you can do; for other kinds of wave packets, k*x > 0.5.)

3. Now I have a Gaussian wave packet, the smallest possible ( x = 0.5/k, where k is given in #1, so x = hbar/(m*c), the Compton radius for the proton), and it's radially symmetric. I used a statistical table of the normal distribution, but with modern computers, if one knows how to program BASIC or some other language, it's fairly easy, to have the computer add up the effects of the "shells" and find the gravitational force inside the proton at various radii, bearing in mind that the proton's mass is distributed in this Gaussian form. There is a radius at which the gravitational force is maximum.

4. Note that I call hbar/(m*c), the Compton radius. Usually, h/(m*c) is called the Compton wavelength. So my "Compton radius" is the Compton wavelength divided by 2*pi, recalling that hbar is defined as h/(2*pi). The gravitational force for the above Gaussian packet is maximum at 1.3688 Compton radii, and the mass inside this radius, is 0.40096 times the total proton mass. So, for this compressed proton, the maximum interior gravitational acceleration is G*m/(Compton radius)^2 * 0.40096/1.3688^2 = (using physical constants from my old Handbook of Chemistry & Physics) 5.398/10^5 cm/sec^2.

5. The gravitational acceleration due to the sun, at a distance of 1 AU, is omega^2 * r = 0.5930 cm/sec^2. This would seem to imply that the gravitational acceleration due to the sun at 104.8 AU, equals the maximum gravitational self-acceleration inside a proton.

6. Now let's recall the Copenhagen interpretation of quantum mechanics. If only one dimension, say, x, can be measured, then the energy available for localization along the x-axis isn't m*c^2/4, as I assumed above; it's simply m*c^2. The proton effectively can be sqrt(4) = 2x smaller, and its internal gravitational acceleration 4x greater. This corresponds to 104.8/2 = 52.4 AU.

Edited by - Joe Keller on 26 Feb 2010 15:17:35
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Joe Keller

USA
957 Posts

Posted - 25 Feb 2010 :  20:04:32  Show Profile  Reply with Quote
quote:
Originally posted by Larry Burford

Joe,

Sorry for the deletions, but we had to have our database restored from a four or five day old back-up. ...


Thanks!
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Jim

1848 Posts

Posted - 25 Feb 2010 :  21:04:06  Show Profile  Reply with Quote
Dr Joe, I misstated the distance where the solar gravity field merges with the background gravity of the universe. Its much more distant than 100AU so would not be relevant to the Pioneer anomaly at 53AU. Sorry for the interuption
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Stoat

United Kingdom
964 Posts

Posted - 26 Feb 2010 :  06:50:08  Show Profile  Reply with Quote
Hi Joe, that makes the idea of an f.m particle all the more interesting. If we write the Lorentzian as sqrt(1 - 1/rho^2) and allow for the possibility of a phase change at the speed of light, then the sign changes to a plus rather than a minus. A negative refractive index in other words. Then we can have a frequency modulated wave particle. The cosine of the natural log of the Lorentzian. it will have two anti parallel phase velocities, one very fast the other slower.

Now I did look at this in a grapher program but didn't try the same thing for a sine. That was Because I wanted a peak at x = 0 and a cosine is symmetrical about zero, a sine isn't. I didn't bother graphing the sine wave.

Such a particle will become denser on accepting a photon but the anti parallel movement, I assume to be a graviton being emitted. This runs both ways, and I think a particle accepts gravitons from huge distances, then emits a photon which is taken up by a particle nearby. That particle has to be graviton rich in order to take on precisely that photon energy. No violations of causality then.
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Joe Keller

USA
957 Posts

Posted - 26 Feb 2010 :  15:37:42  Show Profile  Reply with Quote
quote:
Originally posted by Jim

...the distance where the solar gravity field merges with the background gravity of the universe. ...much more distant than 100AU


Thanks for letting me know. Most people know only about the Pioneer anomalous acceleration, but there was another anomaly shown on a graph in one of JD Anderson's articles, and discussed by Anderson in the text. This "other" Pioneer anomaly, which I discussed years ago in the same places I discussed the proton gravity, amounted to large erratic frequency shifts near 52 or 53 AU. Anderson speculated that these shifts were caused by gravitational acceleration from multiple close encounters with Kuiper belt objects.

However, my estimate showed that the magnitude, duration and number of these anomalies would be extremely unlikely unless the number and mass of Kuiper belt objects were much more than believed. Also, the anomalies occurred near the Kuiper belt dropoff, not near the densest part of the belt.
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Joe Keller

USA
957 Posts

Posted - 26 Feb 2010 :  15:52:31  Show Profile  Reply with Quote
quote:
Originally posted by Stoat

Hi Joe, ...cosine of the natural log of the Lorentzian. ...



I agree with your strategy of using these important, "elementary" mathematical functions, because they give "dimensionless" quantities that are likelier to be valid. Reference books like the CRC Math Tables or the compact Cambridge math tables have "infinite series" approximations of these functions. Maybe the lowest-order term or two of the infinite series, or series of a series, would give another expression that could be recognized in experimental or observational data.
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Larry Burford

USA
2220 Posts

Posted - 27 Feb 2010 :  18:03:00  Show Profile  Reply with Quote
[Joe Keller] "However, my estimate showed that the magnitude, duration and number of these anomalies would be extremely unlikely unless the number and mass of Kuiper belt objects were much more than believed. Also, the anomalies occurred near the Kuiper belt dropoff, not near the densest part of the belt."

DRP expects the local elysium to be entrained by local massive objects such as stars planets and moons. But as you move farther away from such entraining masses there will have to be a "transition zone" between the entrained mass of elysium and the larger background mass of elysium through which the average mass and its entrained elysium are moving.

For purposes of visualizating this transition zone, imagine something like the magnetopause around Earth, where the solar wind (analogous to the flow of background elysium) interacts with the Earth's magnetic field (analogous to the locally entrained elysium). This observationally verified transition zone has pockets of turbulence in it, as will any transition zone between two media, or between two regions of one medium that have velocity and/or other differences in properties.

Anomalies such as you describe might be evidence of this predicted transition zone. It ought to be spherical to a first approximation, but have a tail like a rain drop or that magentopause I mentioned above. We expect to see turbulence and/or chaotic flow patterns that would show up as, for example, odd twinkle patterns of the background stars, intermittent distortions of larger background objects, unusual and intermittent delay or advance patterns in EM waves sent to and from spacecraft within or beyond the zone. And so on.


Regards,
LB
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Joe Keller

USA
957 Posts

Posted - 28 Feb 2010 :  01:37:34  Show Profile  Reply with Quote
quote:
Originally posted by Larry Burford
...DRP expects the local elysium to be entrained by local massive objects...there will have to be a "transition zone"...odd twinkle patterns of the background stars, intermittent distortions of larger background objects, unusual and intermittent delay or advance patterns in EM waves sent to and from spacecraft within or beyond the zone. ...



This hits the nail on the head. I'll try to think along these lines.
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Joe Keller

USA
957 Posts

Posted - 28 Feb 2010 :  02:10:40  Show Profile  Reply with Quote
Sled dogs

If another "Younger Dryas" is here, it might be good to have sled dog(s). Here in rural central Iowa, despite a large and well-manned fleet of county snowplows, plus privately owned tractors and four-wheel drive pickup trucks, with attachable scoops and blades, roads and driveways often have been impassable this winter due to snowfall and drifting. If we lose our petrol supply, then the foregoing will be useless, as will snowmobiles.

From my research, I gather that almost any medium-sized to large, athletic dog with warm fur, can pull a sled over the snow, though some breeds are better adapted than others. Nansen, Scott and Amundsen apparently preferred Samoyeds. Modern competitive sledding seems to use mainly Siberian Huskies and Alaskan Malamutes, but I gather that Samoyeds are more docile and safer as pets.

I gather that in N. America, all our Samoyeds are descended from about 20 animals, mainly survivors of Arctic expeditions, and though they were originally good stock, the lack of genetic diversity has led to four serious genetic diseases: hips, retina, kidney, and severe diabetes. Breeders screen for the first two of these, but apparently it's difficult to screen for the latter two diseases, because they don't appear until after reproductive maturity. It might be best to cross Samoyeds with compatible similar breeds in the "Spitz" dog family, to defeat these genetic diseases, which seem to be mainly recessives.

A neighboring farmer had a "Spitz" type dog (he got it at a shelter and thought it might be a Norwegian elkhound) which was extremely well-behaved around people. Apparently these dogs are not really of "wolf" (dangerous) ancestry. Unfortunately Spitz dogs tend to be hard to train to stay out of the road, he didn't believe in tying the dog (at least not for very long), couldn't afford to fence the dog, and eventually another farmer's tractor couldn't stop in time and ran over it.

Edited by - Joe Keller on 28 Feb 2010 02:22:10
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Jim

1848 Posts

Posted - 01 Mar 2010 :  13:50:11  Show Profile  Reply with Quote
Dr Joe, The gravity of the proton must somehow depend on its radius. So how can that be nailed when dense matter like gold has many more protons(and neutrons)than hydrogen for example? It seems to me your approch is mostly smoke and mirrors so far(bad QM at best). The dog post is a good Don Quixote type interlude.
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Stoat

United Kingdom
964 Posts

Posted - 03 Mar 2010 :  04:25:18  Show Profile  Reply with Quote
Hi Joe, I was thinking about what Robert Carroll said, after I read your post on the proton. He says that we should have half of the energy as the particle's "space." There's a fourth power fall off of this to infinity. We can write E = 0.5mc^2 + 0.5mc^2

But if that's e.m. energy, and there's a phase change at the speed of light, then we have to consider that the e.m. energy has to be added to the gravitational energy, and that gravitational energy is half and half as well. So E = 0.25mc^2 + 0.25mc^2 + 0.25mb^2 + 0.25mb^2 where b is the speed of gravity.

This has to raise some questions about momentum. A radius of one gravitational second is much much larger than a radius of one light second but it's informationally a smaller "size."
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Joe Keller

USA
957 Posts

Posted - 04 Mar 2010 :  20:24:11  Show Profile  Reply with Quote
quote:
Originally posted by Jim

Dr Joe, ...gold has many more protons(and neutrons)than hydrogen for example? ...


Hi Jim, Thanks for mentioning this. The volume of a nucleus, is roughly proportional to the number of nucleons. The binding energy is usually 0.8 - 0.9% of the total mass-energy, so in a nucleus, the smallest possible gaussian protons, as calculated above, could be slightly lighter, larger and have slightly less internal gravity. On the other hand, the tidal gravitational force due to the other nucleons cancels about a third of the difference between free protons' and bound protons' internal gravity. (This tidal force is about the same within all nuclei because they all have about the same density.)

My hypothesis, in my 2002 paper on the Pioneer probes, published in "Aircraft Engineering and Aerospace Technology", was that when macroscopic gravity = microscopic gravity, the radio frequency anomaly charted by Anderson at 53 AU, occurs. Indeed the two regions where the anomaly occurred, correspond to the distance at which the Sun's gravity equals either the maximal internal gravity of a proton in a hydrogen-1 or lithium nucleus, or of a proton in other nuclei.

The only other significant source of macroscopic gravity, was the probe itself. Using the mass and average radius of the probe (including antennas but not booms) I find that the probe's self-gravity at its surface, is about half that needed to explain the range of distance over which the anomaly occurred (when it occurred the first time, at the distance corresponding to the internal gravity of a lone proton). The internal arrangement of the probe is rather like what's under the hood of my car. So it's believable that in some places, near large dense components, the internal gravity would be twice that on the surface. If so, then the duration of the anomaly perfectly matches the range of gravitational forces existing within the probe.

The second time the anomaly occurred, at the slightly greater distance corresponding to protons within nuclei, it occurred over a longer distance interval. The extra interval, corresponded to the range of proton masses due to different binding energy in common nuclei such as oxygen and iron.

Edited by - Joe Keller on 05 Mar 2010 14:38:21
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Joe Keller

USA
957 Posts

Posted - 04 Mar 2010 :  20:31:41  Show Profile  Reply with Quote
quote:
Originally posted by Stoat

...we should have half of the energy as the particle's "space." There's a fourth power fall off of this to infinity. ...


Hi Bob! Thanks for the post. Could this fourth power falloff govern the proton's size in a nucleus?

Edited by - Joe Keller on 04 Mar 2010 20:32:29
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Joe Keller

USA
957 Posts

Posted - 04 Mar 2010 :  21:45:15  Show Profile  Reply with Quote
Sled dogs (cont.)

I gather that "Greenland dogs", "Canadian Eskimo dogs", "Alaskan Malamutes", and "Northeasterly [Siberian] Hauling Laikas" etc., are all roughly the same thing and "not for first time dog owners", nor are Siberian Huskies. To me this isn't just a statistic: I once spoke with two teenagers who had just seen "pit bulldogs" owned by a man rather experienced with dogs, escape and kill a housewife in her own front yard.

The best registry I could find, lists zero serious bites by Samoyeds. I did find a post entitled "Owww! My Samoyed bites!" but the author was complaining about being bitten daily, so surely these were like puppy discipline nips. I've found much anecdotal evidence not only from Samoyed fans but also from polar explorers, that Samoyeds are tamer than the other important sled dog breeds. Not that all dogs aren't dangerous.

Beef cattlemen routinely cross purebreds like Angus and Hereford to get an F1 generation for market that is predictable, yet with hybrid vigor and not homozygous for any harmful recessives. The benefit in the F2 and subsequent generations is much less, because the results become unpredictable, and little hybrid vigor occurs. This is much like hybrid corn.

Veterinarian Bonnie Wilcox's encyclopedia of dog breeds, indicates that of the spitz breeds west of the Samoyed (i.e. tamer, not wilder, spitz breeds) the Finnish or Swedish "Lapphund" is closest to the Samoyed, but this isn't an American Kennel Club breed. Of the AKC registered breeds (dogs, owners, and breeders easier to find) the two large European spitz breeds are the Norwegian Elkhound and the Keeshond.

The Elkhound has been recognized by the Norwegian government as a sled dog that can be commandeered in wartime. From the pictures, I wonder if they can get their tails in front of their faces. The AKC Elkhound breed standard says the tail should be "not brush"; on the other hand, the AKC Keeshond standard says the tail should be "feathered", which sounds better to me.

The AKC Samoyed standard says the tailbone should reach to the hocks. A polar explorer tried some dogs whose tails had been mistakenly docked, and they all died of pneumonia. Shackleton's men built individual igloos for their dogs (they seem to have been Canadian Eskimo dogs; see Lansing, "Endurance", pp. 23, 31-32) but sometimes this isn't going to be possible. Even in Iowa we now get several nights per year of -20F and if another Younger Dryas strikes, make that -30F.

All three breeds - Samoyed, Elkhound, Keeshond - can get the recessive hereditary retinal blindness, but I don't know if it's the same locus. If it's not the same locus, outcrossing will prevent it.

Above, I mentioned nephropathy in Samoyeds. This quickly kills young males, but causes only rather mild disease in the female carriers.

Edited by - Joe Keller on 04 Mar 2010 21:54:15
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Jim

1848 Posts

Posted - 04 Mar 2010 :  22:58:58  Show Profile  Reply with Quote
Dr Joe, I hope this won't evolve into an Abbot&Costello skit but you can estimate the sun is 10E57 protons and if you know the sun's gravity you can calculate the proton's gravity-right? I never been able to determine what the sun's gravity amounts to in any terms. What would your estimate render for the gravity of the sun?
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Stoat

United Kingdom
964 Posts

Posted - 06 Mar 2010 :  07:22:40  Show Profile  Reply with Quote
Hi Joe, up to now I've just been looking at the electron. lets say that it's a ball of a certain radius,, then it has its own "space" rather like an atmosphere. We've got the Compton wavelength of an electron at about 2.4E-12 that would give the electron an angular momentum of h and an angular velocity of c h = mcr

We've also got the Schwarzschild radius, where I think the gravitational energy of the particle is hidden behind a phase change. That gives us a W shaped energy profile, rather than a V shaped profile. It also means that we can say that gravitational energy is "imaginary" or that e.m energy is "imaginary."

Rotate that W out of the page about the central point, and we get crater shape with a extremely thin spike at it centre. No, let's make that a whirlpool with a thin water spout at its centre. Fall in and you have to go round and round it. You cannot zoom up the water spout and escape that way. (You might be ale to jump the gap onto the top of the spike and out though)

From that we can have a perfectly sensible toroidal electron. Imagine the charge corkscrewing round the toroid, left or right handed, to give a positron or electron. The charge has a magnetic field to it with a cube fall off. The flux density at the centre of a electron toroid would be huge huge. (The Kanarov electron has six "turns" per cycle, rather than 2pi turns. So on the next cycle it's in a different position.)

With that set up we have an electromagnetic field zooming round at the speed of light. At the centre of the toroid we'll assume that the magnetic field lines cannot cross. If we look down on an electron, the charge will trace out a rough flower shape. Its magnetic field we can imagine as a disk coming out of the paper plain. Sometimes, when we see it as a line, it points to the centre of the toroid. If we want to have an exchange of energy between grav and e.m energy, it simply has to be via the magnetic field. it also means that the magnetic disk has to be bent, otherwise the outer edge would have to travel faster than light.

Back again to the "space" of a particle. It moves translationally with the particle but at it's outer "edge" its angular momentum is sensibly zero. That would have to be an average though, as the charge is sometimes closer to the toroid centre and sometimes further out. Like wise for the gravitational "space." there has to be both e.m and grave space have equal angular momentum. No Aether drift at all.

What can we make of that? We know its angular momentum is zero, so we cannot know its position? I would argue that we simply don't know that radius, rather than we "cannot" know that radius. It's going to be the radius at which grav energy and e.m energy can be exchanged, it's going to be close but larger than the schwarzschild radius. If we animate that cosine of the natural log of the Lorentzian we get anti parallel group velocities, with two values where we always have a peak, x = 0 and x = 1


So; sorry about this being such a long post; on to the proton. First though, I'm saying that the vacuum is complex, we have to think in terms of "i" or j notation. Now I know a lot of people on this board are unhappy about the idea of "imaginary" space but it just means that the aether is a viscoelastic. We could use Buckminster Fuller's tetrahedral coordinate system and make 3d models of four dimensional models. Let's put a proton at the mass centre of a tetrahedron. But we're supposed to have only three quarks. Look down on this structure, we have a star delta. Could it be that there's a null quark? This quark would be at zero potential most of the time but it could have a slight magnetic field at other times i.e. a neutron. We use star delta starters to control inertia, do protons do the same I wonder?




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Pluto

Australia
53 Posts

Posted - 10 Mar 2010 :  02:31:18  Show Profile  Reply with Quote
G'day

I'm looking for the post that has the ways redshift can be formed.

I read it before, but! I cannot find it.

Must be blind.

Smile and live another day
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Joe Keller

USA
957 Posts

Posted - 11 Mar 2010 :  23:19:51  Show Profile  Reply with Quote
Galaxy image affected by Barbarossa

Previously, I mentioned on this thread that two of the nearby stars (Theta Crateris & 61 Leonis) showing the very strongest "interstellar" absorption lines, happen to be very near Barbarossa. Tonight I noticed that the galaxy on Wil Tirion's star map, NGC 3672, which is nearest Barbarossa, is one of three spiral galaxies known (as of 1996 or maybe 2007) with severely abnormal apparent internal radial velocities (RVs).

"Kinematic subsystems in disk galaxies (found so far only in NGC 3672, NGC 4826, and NGC 253) appear to be relatively rare compared to ellipticals in which more than a dozen or so cases are known."

- Anantharamaiah & Goss, ApJ 466:L13-L16, 1996, p. L16

"[kinematic subsystems in disk galaxies] are yet rare..."

- Anantharamaiah, ApJ, 2007 (full article unavailable online)


NGC 4826 (a.k.a. M64, aka the "Black Eye Galaxy") and NGC 253 (aka "The Sculptor Galaxy") are bright nearby galaxies (mag 9.4 & 7.1 resp., 24 & 10 Mltyr distant resp.). Not only are they among the nearest galaxies, they both are almost precisely aligned with the pole of the Milky Way:

North pole of Milky Way (J2000.0 coords): RA 12:51, Decl +27:08
Black Eye Galaxy: 12:57, +21:41
Sculptor Galaxy: 00:48, -25:17

Thus the Sculptor Galaxy is only about two degrees from our S galactic pole, and the Black Eye Galaxy only about six degrees from our N galactic pole. It would seem that proximity to the galactic pole, distorts apparent RVs observed within a galaxy.

If proximity to the galactic pole can distort a galaxy's image, then maybe so can proximity to Barbarossa's direction. Indeed the third galaxy of this type known in 1996, has been within a few degrees of Barbarossa in recent decades. (The bulk of the Virgo Cluster is between Leo and northern Virgo, much farther from Barbarossa).

NGC 3672 is at RA 11:25, Decl -9:48. This is about a degree from Barbarossa's 2012 position, and roughly a degree per decade farther than that, from Barbarossa in other years. NGC 3672 is magnitude 12.2, 90 Mltyr distant.

Edited by - Joe Keller on 11 Mar 2010 23:34:49
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Larry Burford

USA
2220 Posts

Posted - 12 Mar 2010 :  11:04:56  Show Profile  Reply with Quote
[Joe Keller] "Thus the Sculptor Galaxy is only about two degrees from our S galactic pole, and the Black Eye Galaxy only about six degrees from our N galactic pole. It would seem that proximity to the galactic pole, distorts apparent RVs observed within a galaxy.

If proximity to the galactic pole can distort a galaxy's image, then maybe so can proximity to Barbarossa's direction. Indeed the third galaxy of this type known in 1996, has been within a few degrees of Barbarossa in recent decades."


As mentioned before, DRP expects there to be a transition zone between the entrained elysium associated with our solar system and the much larger mass of background elysium. EM waves reaching us from within or outside of that zone should exhibit some sort of anomalous distortion. It is expected to be very weak, and transient.

And, using Earth's megnetopause analogy as an aid to visualizing things, the anomalies ought to be different, perhaps stronger, in the polar areas.

Proximity to the Sol system's transition zone (Barbarosa? maybe, but we don't really know where the zone is at this point) might be another reason to expect "anomalous anomalies" ;-) Or, perhaps Barbarosa is well outside of the Sol system's transition zone, in which case it ought to have its own (much smaller) entrained mass of elysium and its own (much smaller) transition zone.

LB
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