Requiem for Relativity

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13 years 10 months ago #20995 by Jim
Replied by Jim on topic Reply from
Sloat, The charge is a unit of energy and the number you want(~1.6-E19) is just a number. How can you make energy from a number? You never explain or account for units-what happens to all the units used in your calculations? If you just use numbers that look good and dump the units the numbers belong to don't you have nothing at all? It seems to me all you have is numbers with no meaning. 1.602x10E-19Joules is the basic charge found everywhere but without knowing what a Joule is or dumping the Joule from the charge unit the number has no meaning at all.

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13 years 10 months ago #20996 by Joe Keller
Replied by Joe Keller on topic Reply from
Ephemeris for Barbarossa: 2007, 2008, 2009, 2011 observing seasons

This ephemeris, is so those who have taken photos, can check them more efficiently than with my earlier, probably less accurate ephemerides. Linear interpolation suffices for Feb. & Mar. The intervals in each year, are 30 days, always at 12h GMT (12h UT). The ephemeris is accurate for Jan. and Apr., with quadratic extrapolation.

This is the most exact ephemeris I now can make for Barbarossa. It's based on my orbit (found in spring 2009) through the four sky survey points. Earth parallax was corrected using a first-order correction for Earth's eccentricity.

Coordinates are J2000.0 celestial, geocentric.

date / Right Ascension / Declination

Feb. 1, 2007 12h GMT 11h 26m 06s -9deg 05.4'
Mar. 3, 2007 12h GMT 11h 25m 38s -9deg 06.2'
Apr. 2, 2007 12h GMT 11h 25m 07s -9deg 06.4'

Feb. 1, 2008 12h GMT 11h 26m 31s -9deg 08.8'
Mar. 2, 2008 12h GMT 11h 26m 02s -9deg 09.6'
Apr. 1, 2008 12h GMT 11h 25m 31s -9deg 09.7'

Feb. 1, 2009 12h GMT 11h 26m 54s -9deg 12.1'
Mar. 3, 2009 12h GMT 11h 26m 26s -9deg 12.9'
Apr. 2, 2009 12h GMT 11h 25m 55s -9deg 13.0'

Here is the ephemeris for those who want to look this season:

Jan. 2, 2011 12h GMT 11h 28m 00s -9deg 17.3'
Feb. 1, 2011 12h GMT 11h 27m 43s -9deg 18.6'
Mar. 3, 2001 12h GMT 11h 27m 14s -9deg 19.5'
Apr. 2, 2011 12h GMT 11h 26m 43s -9deg 19.6'

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13 years 10 months ago #24168 by Stoat
Replied by Stoat on topic Reply from Robert Turner
Hi Jim, I did give a link to an article on Hawking radiation and I mention power several times throughout the posts on this. The Hawking equation for power, where we look at the Planck mass, gives us a power in Joules per Planck second. When we alter the equation and put in the speed of gravity, then we get 2.898E-19 Watts.

What does this mean? Well, a Planck mass micro black hole evaporates in about 1E-41 seconds! yet if we allow for a ftl speed of gravity then it radiates very slowly, it would take about 70 billion years to disappear.

It looks rather like a mini supernova. Interesting questions would be, this thing is a source, does it have a twin that's a sink? Or is it its own antiparticle? Could protons/antiprotons and electrons/positrons be slightly different types of mini nova?

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13 years 10 months ago #20997 by Jim
Replied by Jim on topic Reply from
Hi Sloat, I try to follow and have yet to get more than a few thoughts into a post before I'm lost. This latest post has me wondering what a Planck second is. Can you define that term? I am going to study the post while avoiding that term until you define what it is. I'll try to figure out the part about the universal charge and the watt unit you have uncovered.

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13 years 10 months ago #20998 by Joe Keller
Replied by Joe Keller on topic Reply from
Why a binary's eccentricity should be ~0.6

Otto Struve, Aitken and others long have noted that there might be poorly understood systematic errors in measuring binary orbits. Barbarossa's eccentricity (0.6106, which I calculated from the sky surveys) is, in any case, near the measured mean for wide (>100 yr) binaries (details in my previous posts).

Suppose the early Sun were surrounded by a disk of distant particles. For simplicity suppose these particles are of equal, small mass. Suppose that the mean energy (potential plus kinetic) of the particles is zero, but that every particle has the same angular momentum.

Suppose interactions occur between the particles, involving only forces radial to the Sun, so the particles keep the same angular momentum but their energies acquire a Boltzmann (i.e. exponential) distribution with mean zero.

Let a0 be the radius of a circular orbit which has the given angular momentum. Those particles having negative energy (i.e. bound, i.e. elliptical orbits) will have semimajor axes between a0 and infinity, corresponding to eccentricities between 0 and 1. The Boltzmann (i.e. exponential) distribution of energy levels, contains a number of particles, between energies E and E + deltaE, that is proportional to exp(-k*E) where k is a constant determined so that the mean energy will be zero. Elementary integral calculus shows that the mean energy of the bound particles (not of all the particles, only of the bound particles) corresponds to a semimajor axis 2.71828... - 1 = 1.71828... times as large as a0.

If all the particles then agglomerate into one body, while conserving energy and angular momentum, that body's eccentricity must satisfy the equation

sqrt(1-eccentricity^2) = 1/sqrt(2.71828... - 1)

i.e., eccentricity = 0.64655... .

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13 years 10 months ago #20999 by Jim
Replied by Jim on topic Reply from
Dr Joe, Binary orbits are determined by assuming the mass center is also the location of mass whereas the mass is located far from the point called the barycenter of a binary. If all of the system's mass was located at the barycenter the Kepler Law exactly predicts the orbits of all other bodies is the system. But, as the mass of the system can never by exactly located on any one point the Kepler Law cannot be expected to explain everything. I would rather like to see the 2/3 relationship pop up(~.6 or so in your terms) but mostly for my own sense of neatness. In the real universe how would the particles you want to exist work-they seem to violate natural laws in some ways in order to merge into an independent body bound to it's partner because it cannot gain enough angular momentum to go off on its own. The rules of of the force we call gravity have not yet been totally revealed or discovered.

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