Requiem for Relativity

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14 years 8 months ago #23925 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 Jim</i>
<br />Dr Joe, ...gold has many more protons(and neutrons)than hydrogen for example? ...<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">

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.

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14 years 8 months ago #23837 by Joe Keller
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<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote"><i>Originally posted by Stoat</i>
<br />...we should have half of the energy as the particle's "space." There's a fourth power fall off of this to infinity. ... <hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">

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

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14 years 8 months ago #23838 by Joe Keller
Replied by Joe Keller on topic Reply from
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.

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14 years 8 months ago #23839 by Jim
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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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14 years 8 months ago #23840 by Stoat
Replied by Stoat on topic Reply from Robert Turner
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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14 years 8 months ago #23844 by Pluto
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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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