Paradoxes Resolved, Origins Illuminated - Requiem for Relativity
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Stoat

United Kingdom
1065 Posts

Posted - 28 Oct 2010 :  05:10:23  Show Profile  Reply with Quote
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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Jim

1880 Posts

Posted - 28 Oct 2010 :  13:49:02  Show Profile  Reply with Quote
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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Joe Keller

USA
1061 Posts

Posted - 28 Oct 2010 :  22:15:12  Show Profile  Reply with Quote
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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Jim

1880 Posts

Posted - 29 Oct 2010 :  14:01:54  Show Profile  Reply with Quote
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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Joe Keller

USA
1061 Posts

Posted - 29 Oct 2010 :  20:52:15  Show Profile  Reply with Quote
quote:
Originally posted by Jim

Dr Joe, ...how would the particles you want to exist work - they seem to violate natural laws in some ways in order to merge...



Hi Jim,

Thanks for the post. You make several relevant points, but this one hits the nail on the head. I'm hoping that though the behavior of real particles would differ from mine in several important ways, those differences would not affect the final eccentricity much.

- Joe Keller
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Jim

1880 Posts

Posted - 29 Oct 2010 :  21:19:26  Show Profile  Reply with Quote
Dr Joe, The real final structure would be a galactic disk and lesser mass particles would be captured by greater mass particles-right? The lesser structures within the galactic disk such as our solar system or any other star system would be bound to the disk structure which must have some unknown effect on our Kepler model. Our star has a tiny effect on the galactic structure which might be covered by Newton's 3rd law. If you find action you can assume it will be observed in some other form in some other place.
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Joe Keller

USA
1061 Posts

Posted - 29 Oct 2010 :  23:23:25  Show Profile  Reply with Quote
The two relations between Barbarossa's orbital parameters, and fundamental astrophysical constants

I mention both these relations in recent posts but will restate them here:

1. speed / acceleration = A * time

Here the speed is sqrt(k*T/m) = 14996.77 cm/s, where k is Boltzmann's constant, 1.38053/10^16 erg/deg, T is the cosmic microwave background temperature, estimated as 2.72553deg Kelvin; and m is the mass of the hydrogen atom, 1.00797 atomic mass units = 1.67302/10^24 gm. This is the isothermal speed of sound in atomic hydrogen, however rarified, at the CMB temperature. The acceleration is the Hubble acceleration, i.e. the Hubble parameter times the speed of light: H0 * c; in this post, I'll find the value of the Hubble parameter that fits both equations. The time is the Barbarossa period; my best estimate is from the Egyptian & Mayan calendars, 6339.5 tropical yr, adjusted to Newcomb's mean tropical year during the interval 4328BC - 2012AD: 6339.5yr*365.2424d/yr = 2.00055*10^11 sec (the 1.8 day correction for Earth's orbital eccentricity's effect on this solstice-to-solstice time, is insignificant at this precision). "A" is an adjustable parameter (see below).

2. speed * A * time = distance

The speed and time are as above. The distance is Barbarossa's latus rectum, a*(1-e^2) = 343.84AU * (1-0.610596^2) = 3.22619*10^15 cm. The parameters of Barbarossa's orbit are from my own work; the value of the CMB constant is as I recall it from a recently published journal article; and other physical constants are from the CRC Handbook, 44th ed., 1962.

To exactly satisfy both equations simultaneously, requires two adjustable parameters. The first adjustable parameter is the Hubble parameter, which is known observationally only to within ~5%. The second adjustable parameter is the constant, A, by which Barbarossa's period is multiplied.

Eqn. #2 implies that A = 1.075331 = approx. 1 + 1/13.

With this value of A, eqn. #1 implies that the acceleration is 6.97118/10^8 cm/s^2 = 71.7556 km/s/Mpc. This is the 72 km/s/Mpc value for H0 which long has been the most accepted, and still is within the error bar of the latest best Hubble telescope effort.
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Joe Keller

USA
1061 Posts

Posted - 31 Oct 2010 :  15:28:11  Show Profile  Reply with Quote
More reasons a binary's eccentricity should be ~0.6

In my Oct. 28, 2010, post, I explain that if all the original particles have the same angular momentum, and if their mean orbital energy is zero, and if the orbital energies follow a Boltzmann distribution, then, if all the bound (i.e., negative orbital energy) particles agglomerate without loss of orbital energy, the eccentricity will be 0.64655. Particles are likelier to agglomerate when they intersect with similar velocities; at similar velocities, rather little orbital energy is lost. Some of the orbital energy lost when the particles agglomerate, might be recoverable (e.g. orbital energy of one particle about the other). Yet some orbital energy will be lost, and this will cause the final eccentricity to be < 0.64655.

If the apses of the original particles are distributed symmetrically in the plane, then the final eccentricity (the eccentricity after all particles are agglomerated into one) must be zero, because there is no preferred direction. On the other hand, Barbarossa's aphelion aligns with the galactic center, so, plausibly the original particles all aligned with the galactic center for the same reason; let's suppose that the original particles did have the same aphelion direction (and the same angular momentum vector).

The simplest particle interaction, is that of point masses, as discussed in the dynamics chapter of most freshman college physics texts. Particles with the same aphelion direction and angular momentum vector, have the same latus rectum segment, and for different eccentricities, intersect only at the ends of the latus rectum. At the latus rectum, the potential energy is the same for all particles because the position is the same; the energy of transverse velocity is the same because the angular momenta are the same; only the energies of radial velocity differ, proportional to eccentricity^2. The eccentricity is proportional to the radial velocity; so the mean eccentricity, will be the eccentricity of the agglomeration. This mean eccentricity, e, can be found as the ratio of two integrals:

Numerator: integral of d(e^2)*exp(-e^2)*e, from 0 to 1

Denominator: integral of d(e^2)*exp(-e^2), from 0 to 1

(The exponential function gives a Boltzmann distribution, with mean corresponding to eccentricity = 1.) Integrating the numerator numerically (use the substitution u = e^2), I find

eccentricity = 0.59948... . The eccentricity might be higher than this if some particles in hyperbolic orbits agglomerate also, on their one-time pass through the system; this describes Barbarossa (e = 0.6106 according to the orbit I fit to the sky surveys last year).

Edited by - Joe Keller on 03 Nov 2010 20:37:01
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Stoat

United Kingdom
1065 Posts

Posted - 01 Nov 2010 :  07:16:55  Show Profile  Reply with Quote
Hi Joe, with Jim I'd like to see that 2/3 but I think we have to allow for the fact that our orbit round the galactic centre is not going to be a perfect circle. In which case the apses will sometimes lead or lag by a tiny amount. i can't help thinking that this has something to do wuth the neutrino, that temperature of about 2.752 Kelvin looks to e in the mass range of the neutrino. We have basically hydrogen atoms orbiting in a sea of neutrinos with little interaction but won't there be slightly more neutrinos coming from galactic centre? A particle diving in an elliptical orbit through this soup wouldn't see the mass "above" it, it would want to describe a roseate (it would look like the petals of a flower) orbi, all things being equal but that slight excess of neutrinos could tend to lock it.
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Jim

1880 Posts

Posted - 01 Nov 2010 :  23:34:25  Show Profile  Reply with Quote
Well, there must be a way to determine the angular momentum of the sun relative to the galatic center and using that to determine the orbit of the sun around the galaxy. You guys have unused observational details of the Milky Way that can be useful to prove your hypothesis
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Jim

1880 Posts

Posted - 04 Nov 2010 :  14:03:20  Show Profile  Reply with Quote
Dr Joe, I don't understand what the "isothermal speed of sound" means. How can sound exist in the vacuum of space? I suppose the existence of sound needs matter unlike light that exists even without matter.
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Joe Keller

USA
1061 Posts

Posted - 04 Nov 2010 :  17:29:00  Show Profile  Reply with Quote
quote:
Originally posted by Jim

Dr. Joe, ...what the "isothermal speed of sound" means. ...vacuum of space?...



Hi Jim,

My source for the formula, was Hausmann & Slack's old college physics text. There isn't any definite cutoff at which sound disappears. For example, a rocket can exceed the speed of sound even though the atmosphere is very thin at very high altitude.

There are a few hydrogen atoms and/or protons per cubic centimeter in interplanetary space. It would be difficult to impart much sound energy to such a medium using, say, stereo speakers, but something huge like a nova could impart much sound energy to it. Also, the continuum approximation used for studying sound, would apply (in interplanetary space) only to long wavelengths.

Isothermal means that the temperature of the medium is constant over time. The speed of sound isn't the same, if the sound isn't isothermal (i.e. if the medium is significantly heated and cooled by the compression waves each cycle).

- Joe Keller
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Jim

1880 Posts

Posted - 05 Nov 2010 :  13:38:32  Show Profile  Reply with Quote
Dr Joe, Sound waves transmit energy but, not much even in dense matter. How much of an effect would sound waves have when the density is a few particles per cubic meter?
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Joe Keller

USA
1061 Posts

Posted - 05 Nov 2010 :  18:15:33  Show Profile  Reply with Quote
quote:
Originally posted by Joe Keller

Roland, Iowa March 16, 2007

Open letter to the Director of the Lowell Observatory

...Assuming a circular orbit and making first order approximations to correct for Earth parallax, Barbarossa has period 2640 yr. and is 191 AU from the sun. Accordingly, the resonances of the orbital periods of the outer planets have discrepancies which advance prograde with periods

Jupiter:Saturn 5:2 2780 yr
...



In March, 2007, I told the Lowell Observatory (in direct emails similar to the above "open letter" on p. 8 of this messageboard thread) that Barbarossa's orbital period is similar to the period of advance of a Jupiter-Saturn conjunction; that is, Barbarossa shepherds the Great Inequality.

In 2009 I fitted an elliptical orbit to the sky surveys. At the latus rectum of an elliptical orbit, the relationship between distance and angular speed, is the same as for a circular orbit. So, near the latus rectum, as Barbarossa has been in recent decades, a circular orbit will fit the (geocentric) observational data with only a small error. In its actual elliptical orbit, Barbarossa's angular speed at its latus rectum, matches the average speed of advance of the Jupiter-Saturn conjunction.

Using the 6339.36 Julian yr orbital period and 0.610596 eccentricity, I find at Barbarossa's latus rectum (reached at 2003.94 AD) an angular speed consistent with a period of 3148.7 Julian yr. At Dec. 21, 2012 (true anomaly = 91.022deg) I find (heliocentric) angular speed consistent with a period of 3218.4 Julian yr.

The mean period of advance of a given Jupiter-Saturn conjunction point (i.e. a given corner of the "trigon") is exactly 3x the "Great Inequality", i.e. 3 * 1/(5/S - 2/J) where J & S are the periods of Jupiter & Saturn, resp. The present-day Great Inequality (as in the time of Laplace) is about +900 yr, but

"...the GI's average period is...1092.9 years;..."

- Ferenc Varadi et al, "Jupiter, Saturn, and the Edge of Chaos", Icarus 139:286-294, 1999, p. 287, col. 2

Three times the average GI given by Varadi, is 3278.7 yr, only 1.9% longer than period corresponding to the heliocentric angular speed of Barbarossa at the end of the Mayan Long Count.

Most of the discrepancy disappears when the motion of the trigon point, is projected onto Barbarossa's orbital plane. Barbarossa is inclined ~14.5deg to the angular momentum-weighted mean orbital plane of Jupiter + Saturn (about the same as the principal plane of the solar system) and at 2013.0AD is ~68deg past its descending node on that plane. So the angular speed of the trigon point, projected onto Barbarossa's orbital plane, corresponds to a period of roughly

3278.7 * cos(14.5*sin(90-2*(90-68))) = 3224.5 Julian yr

only 0.19% longer than the period, P = 3218.4 yr, corresponding to Barbarossa's angular speed at the end of the Mayan Long Count. A more precise calculation of the projection, using a modern value of the Jupiter + Saturn combined angular momentum vector, and spherical trigonometry, gives

3278.7 / 1.02287 = 3205.4 Julian yr

which is 0.40% shorter than P. A 0.4% error in the speed of the trigon point, corresponds to only 1 part in 30,000 error in Jupiter's average orbital period.

According to NASA's Fact Sheets, for sidereal years of the planets, J = 4332.589 days, S = 10,759.22 d, U = 30,685.4 d, N = 60,189 d, P = 90,465 d, M=686.980 d, and V = 224.701 d (I'll use Wikipedia's V = 224.70069 d). Also, E = 365.25636 d. The Neptune-Pluto inequality is 1/(3/P-2/N); this is -41,068 Julian yr (the same length as the Milankovitch cycle), and also equals the period of regression of the Neptune-Pluto conjunction point. The Jupiter-Neptune inequality is 1/(14/N-1/J) = +1528.0 Julian yr. To get the period of progression of the conjunction point, this must be multiplied by 14-1=13, giving +19,864 yr, (negative) half the Neptune-Pluto regression period. Likewise 1/(11/N-2/S) = -874.9 yr; *(11-2) = -7873.8 yr, a fifth of the Pluto-Neptune regression period. We have

S / J = 2.48 and
19864 / 7907.9 = 2.52

that is, the mean conjunction points of Jupiter (13 points, i.e. a 13-gon) and Saturn (9 points, a 9-gon) with Neptune, move in the same 5::2 ratio as do Jupiter and Saturn themselves, though the J-N conjunction is the slower, and the S-N conjunction moves retrograde.

Earth's conjunction points with Jupiter (an 11-gon), move prograde with period equal to the Great Inequality:

11*1/(12/J-1/E) = 944.0 Julian yr

Varadi gives the average value of the Jupiter-Saturn Great Inequality as 1092.9 yr, but the current value is

1/(5/S-2/J) = 883.19 yr

so the Jupiter-Earth mean conjunction progression is between the current and average values of the GI.

The Venus-Earth conjunction point moves retrograde, with period 5*1/(13/E-8/V) = -1192.8 Julian yr. This is also rather near the long term average value of the GI.

The mean conjunction of Saturn and Uranus moves prograde with period 2*1/(3/U-1/S) = 1135.4 yr. Suppose that originally the PI had its average value, and the Saturn-Uranus conjunction moved forward with period equal to the PI. If the semimajor axes of Jupiter and Saturn are conserved, and their eccentricities remain small, then conservation of angular momentum between Jupiter and Saturn requires that the increase in Saturn's angular speed almost exactly equals the decrease in Jupiter's angular speed. So, half the change in the PI is due to Saturn and half to Jupiter. If the Saturn-Uranus interaction is comparatively small, then the decrement in the frequency of the Saturn-Uranus conjunction progession would be due only to Saturn and would be 1/2 * (3/2)/(5/3) = 9/20 as much as the increment for the Jupiter-Saturn progression (proportionally, 9/20*1/3 = 3/20):

1092.9 / (1 - 3/20*(1092.9/883.19-1)) = 1133.3 yr

only 0.19% less than the actual value. Because the (proportional) change in frequency due to the Jupiter-Saturn interaction, is only 3/20 as much, this might afford a better estimate of the average PI:

1092.9*1133.5/1130.9 = 1094.925 yr

then times 3, and correcting for projection of the Jupiter+Saturn plane,

1094.71*3/1.02287 = 3211.33 yr

only 0.22% shorter than P above.

Let's consider Mars analogously to Uranus. Most online sources use NASA's value of 686.980 d for Mars' period, though some still quote 686.95 d (the older accepted value, e.g. Inglis, "Planets, Stars & Galaxies", 3rd ed., John Wiley, 1972). Mars' eccentricity and proximity to Earth and Jupiter make exact long-term average orbital period calculation hard. Proceeding anyway, I find that the Mars-Jupiter conjunction point progresses with period

(19-3)*1/(19/J-3/M) = 2376.955 yr = 2*1188.48 yr.

Analogously to Uranus, I get the fraction 1/2*(19/16)/(5/3)*2/3 = 19/80 and

1188.48*(1 - 19/80*(1092.9/883.19-1)) = 1121.46 yr

which is about 2/3 of the correction needed to get the average PI.

From Wikipedia's recent values (citing Chapront, 1991), for the various kinds of month lengths with their linear secular trends, I find P = period of progression of the Lunar apse and N = period of regression of the Lunar node. A mean conjunction of the node and perigee (one of the points of the trigon of three conjunctions), progresses with period

3*1/(1/P-2/N)
= 549.49 Julian yr = 1098.98/2 for epoch 2013.0AD
= 546.98 J yr = 1093.96/2 for epoch 2013.0AD-6339.36 J yr = ~4328BC

Using instead, the fourth degree polynomials for accumulated advance or regression, in Simon et al, Astronomy & Astrophysics 282:663+, 1994, p. 669, sec. 3.4.a.1, I find

549.48 Julian yr for epoch 2012.97 AD and
547.36 Julian yr for epoch 2012.97AD-6339.36 = ~ 4328BC

(Chapront's linear corrections for the month

Edited by - Joe Keller on 13 Nov 2010 17:22:33
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Stoat

United Kingdom
1065 Posts

Posted - 06 Nov 2010 :  04:54:08  Show Profile  Reply with Quote
Hi Joe, something that may be of interest. I was taking on a webpage to a woman who is writing her second book on myth and christianity. She asked a question about the planet Mercury, as she was looking at astrological charts. I explained a it about elongation and told her to get a copy of the free Celestia star program.

Then she came back on and cried for help. Wikipedia had given a list of solar eclipses in historic times and there was one for Jerusalem in 303 bce. She's looked at it and nothing! Then she'd found a really good eclipse for the year 301 bce with the aid of the program. She also said that it lasted all day!! I explained that thee date was to with the changes made in the calendar and there being no year zero but with a year zero for astronomy. Two years and ten days error. I also said that a total eclipse can't last all day.

So I looked at the event in celestia. Highly unusual, the moon does partially eclipse the sun for the whole day. Now I've said to her that the drop in light intensity would be difficult to notice, as the human eye is very good at adjustment but the temperature drop over the day would be noticed, and animals would be going ballistic all day. Then we've got Venus in the sky ahead of the sun and Jupiter and Sirius appearing at sun set.

Have you got anything on what the temperature drop would be in Jerusalem?

(Edited) I think I should stress here, that the partial eclipse seems to be a particularly good fit for this eclipse in Jerusalem. Saros cycle?

Edited by - Stoat on 06 Nov 2010 05:37:27
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Jim

1880 Posts

Posted - 06 Nov 2010 :  12:58:56  Show Profile  Reply with Quote
Sloat, You can use the NASA JPL horizons generator to determine when events like this occurred. I'm confident the NASA generator is the best available but not exact because basically its just a model.
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Stoat

United Kingdom
1065 Posts

Posted - 07 Nov 2010 :  07:21:28  Show Profile  Reply with Quote
Thanks Jim, I passed the link on to her but I think she's asking her university.
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nemesis

185 Posts

Posted - 08 Nov 2010 :  13:54:12  Show Profile  Reply with Quote
Joe, at some point it might be a good idea to post all your Barbarossa data in one place - period, distance from the Sun, location in the sky, etc.
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Joe Keller

USA
1061 Posts

Posted - 15 Nov 2010 :  23:14:44  Show Profile  Reply with Quote
Mars' medieval and ancient orbital period

Abstract. Vedic planetary and lunar parameters, mostly originating c. 3100BC, conform superbly to modern theory, except that the Vedic orbital period for Mars is short. Most of the best ancient and medieval information (Seleucid Babylonian, Arabic/Byzantine/Persian, and most moderns before LeVerrier) shows a swift linear time trend in Mars orbital period. However, Ptolemy and Kepler, and also the Seleucid Babylonian record of a Mars stellar conjunction at stationarity, conform well to modern theory. The apparent anomaly of Mars orbital period, shows suggestive relationships not only with Earths precession period, but also with the deviation of the Jupiter-Saturn Great Inequality from its average value.



Contents

I. Introduction
II. The 1336AD Trebizond & related 1353AD almanac tables
III. The Trebizond and Alfonsine almanac parameters
IV. Refining the almanac parameters
V. The 1590AD Heidelberg Venus-Mars occultation
VI. The 251BC Seleucid conjunction of Mars with Eta Geminorum at stationarity
VII. Seleucid and Ptolemaic parameters
VIII. Han China
IX. Vedic India
X. Discussion
Appendix 1. Plethon's mean motion of Mars
Appendix 2. Kepler's mean motion of Mars



I. Introduction.

The well-known 13::8 Venus::Earth conjunction resonance point, regresses with period 1192.8 yr, according to prevailing modern estimates of the planets' periods. Likewise the analogous 19::3 Mars::Jupiter conjunction resonance point, progresses with period 2*1188.48 yr, twice the period of the Venus-Earth point.

Suppose this -2::1 ratio, of the Venus::Earth regression to the Mars::Jupiter progression rates, is constant. Maybe an unknown force, acting on triads instead of pairs of bodies, causes this. The orbits of Venus and Earth are nearly circular and relatively close to the Sun. The orbital period of Mars might be much more affected by Jupiter or other perturbations. Suppose the orbital periods of Venus and Earth are practically constant, so the Venus::Earth regression rate stays the same. Then, the orbital period of Mars must change by the same proportion as that of Jupiter, if the Mars::Jupiter progression rate is to stay the same too.

If Jupiter and Saturn maintain constant total J+S orbital angular momentum, nearly constant semimajor axes, and small eccentricity, then their orbital frequencies must change by almost exactly equal and opposite amounts. Then the Great Inequality of Jupiter and Saturn, can be restored to its calculated average value, 1092.9 yr [1, p. 287], only by shortening Jupiter's period, by about 1 part in 3000. According to the speculation of the previous paragraph, the orbital period of Mars also would shorten by 1 part in 3000.

The published calculations based on the theory of gravity, say that planetary orbital periods change only very slowly; Mars' orbital period shortens only 1 part in 350,000,000 per 1000 yrs [2, p. 24][3]. At that rate, 10^8 yr might be needed, in a linear extrapolation, to restore the Great Inequality to its average value. If fluctuations toward or away from the average Great Inequality occur much faster, there might be historical evidence. The purpose of this paper is to examine medieval and ancient records for evidence of a change in the orbital period of Mars.

Sometimes upper and lower bounds will be found, often due to rounding error in the information source, restricting the result to an interval of a priori uniform probability. Sometimes several results found by similar methods, will be averaged with equal weight. Always, uncertainty will be given as root-mean-square ("rms") using, when appropriate, Standard Error of the Mean ("SEM").



II. The 1336AD Trebizond & related 1353AD almanac tables.

Trebizond was "the last refuge of Hellenistic civilization" [4, article "Trebizond, empire of"]. The 1336AD Almanac for Trebizond was authored by Manuel, a Christian priest of Trebizond, but apparently referred to parameters from contemporary Persian and Arabic almanacs. Mercier [5] also includes fragments of three 1353AD almanacs: the 1353AD Constantinople almanac by Manuel's student Chrysococces, and two 1353AD Persian almanacs on which that of Chrysococces might in part be based.

The Persian and Byzantine almanacs often were more accurate than mere extrapolations from Ptolemy's ancient observations. Allowing for various choices of equinox, errors usually were less than a degree.

The Trebizond almanac table gives noon planetary longitudes to the day and arcminute. Interpolating in the table, I find the geocentric ecliptic longitude of the Sun, at the time when Mars and Venus have equal geocentric longitudes. Mars' orbit is more eccentric and more subject to perturbation by Jupiter, so Mars' JPL ephemeris for a remote time would be less accurate than for Venus or Earth.

Next I find the time, according to the JPL ephemeris, when the geocentric longitude difference between Venus and the Sun, is the same as in the Trebizond almanac when Mars and Venus have equal geocentric longitude. I take only longitude differences from the Trebizond Almanac, so I need not know its equinox. I then get the longitudes of Venus and the Sun, from the JPL ephemeris.

The small inclination of Earth's 1336AD ecliptic to the 2000.0AD ecliptic (approx. 0.8' per century = 5') is neglected, but Mars' orbital inclination (approx. 2 degrees) is considered. Mars' orbital elements with secular terms according to [2] are used. My BASIC language computer program solves this problem by iteration on four variables: Mars' orbital radius = r, Mars' true anomaly = f, Mars' orbital radius projected onto the ecliptic = r0, and Mars' heliocentric ecliptic longitude = g.

I begin with rough graphical estimates of r and f. First, Mars' orbital elements imply, if f is known, a sinusoidal approximation for Mars' ecliptic latitude, the cosine of which gives r0 from r. Second, the Law of Sines applied to the ecliptic plane polygon Sun-Earth-Venus-Mars (the projections of Venus and Mars on the ecliptic are used; the polygon is a triangle because E, V and M are collinear) gives g from r0. Third, another sinusoidal approximation involving Mars' inclination, and the definitions of the orbital angles, gives f from g. Fourth, the orbital equation gives r from f. Nine iterations reach double precision.

The Kepler equation and another textbook formula [7, eqns. 4.57 & 4.60, pp. 84, 85] give the eccentric and mean anomalies, which can be compared to Kieffer's mean longitude [2] for 2000.0AD. Correction for actual day length is unneeded, because the JPL ephemeris uses Julian Days. The calculated average Martian sidereal year between the Venus-Mars conjunction in August (then the Islamic month of Muharram) 1336AD (JD 2209262.924) and 2000.0AD (JD 2451545.0) is thus

686.9813730 Julian d

based on the Trebizond almanac table together with the JPL ephemeris for Venus and Earth only. Kieffer's modern figure [2] for 2000.0AD is equivalent to

686.9798529 d

Newcomb [ 8 ] cites LeVerrier's sidereal figure, 686.9798027 d, epoch 1850.0AD (corrected for day lengthening, 2ms/century from 2000.0AD).

If the Trebizond longitudes are accurate except for rounding to the nearest arcminute, then the rms error for a given longitude, is about 2/7 arcminute; interpolation reduces this by ~sqrt(2), though use of three longitudes increases it again by ~sqrt(3). Both the Trebizond and JPL ephemerides give apparent position at the observer; the greatest aberration of light involved is, because of the position of Venus, less than the combined aberrations of Earth and Mars, which for Mars near aphelion, is about 20+15 = 35 arcsec.

These two biggest errors thus are much smaller than the discrepancies between at least some of these almanacs: in 1353AD, my same procedure finds in Chrysococces' almanac a 23.66de

Edited by - Joe Keller on 28 Feb 2011 13:27:01
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Stoat

United Kingdom
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Posted - 16 Nov 2010 :  03:21:35  Show Profile  Reply with Quote
Hi Joe, according to the Stellarium program, there was an occultation of mars and venus seen from Baghdad on the 22nd of October 864 at 3.50 a.m. Jupiter and mercury were very close together on the horizon as well. They must have seen that sky as a bit of an omen.
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