Requiem for Relativity

June 7, 2010 crop circle at Stony Littleton Long Barrow near Wellow, Somerset, England: conjunctions of Venus

An intelligent space probe might locate native astronomical monuments such as Stonehenge, and make crop circles there, to aid the natives in interpreting the crop circle. The June 7, 2010 Somerset crop circle signifies conjunctions of Venus, the last on March 29, 2013. (My source for basic information on this crop circle is Linda Howe's website, www.earthfiles.com ).

There was a transit of Venus June 8, 2004, and will be another June 6, 2012. These are represented by the concentric disks lying at opposite ends of the big circle: transits are inferior conjunctions very near the nodes of Venus. The other three concentric disks, with reversed colors, represent superior conjunctions. The center one, represents the occultation of Venus June 9, 2008 (maximal, from Earth's center, at 03:42 UT). June 7, 2010 (the date of the crop circle) is exactly halfway between June 9, 2008 and June 6, 2012.

There are four other inferior conjunctions of Venus, between the two transits. If the concentric disks at the outside ends, outside the big circle, represent the superior conjunctions immediately before the 2004 transit and after the 2012 transit, then each one is the seventh superior conjunction counting from the other, hence the seventh circle along the arcs drawn. The last superior conjunction is thus March 29, 2013.


(the below is posted here June 22, 2010, not separately, because of a server problem)

The Mayan Long Count, 61 & 243: not just another pretty calendar

The Roman calendar contains a millenium analogous to the Mayan Long Count. We all know it: 365.25 days (Julian style) * 10 yrs/decade * 10 decades/cent * 10 cent/millenium. The Gregorian style adopts, on average, 365.2425 days, the closer to match the actual tropical year, 365.24219 days. Presumably Clavius et al, knew the tropical year more accurately than their Gregorian calendar shows, but sacrificed some accuracy, for the sake of a simpler leapyear formula.

The Mayan Long Count is 360 days * 20 * 20 * 13, where there are Mayan names for the intervals analogous to the Roman decade, century and millenium. The 360 day year isn't unusual for ancient people; it's a round-number compromise between 365 days = 1 Earth year, and 354 days = 12 synodic months. Our European trigonometry of 360 degrees, derives from an Egyptian awareness of this same 360 day approximation. Likewise "base 20" (fingers and toes) is as natural as "base 10". But why 13?

The shocking answer, is that the Mayans, or their forerunners, juggled the factors, to find a simple, Roman-like calendar, usable by the common man, but which also would have a "millenium" (Long Count) equal to a fundamental astronomical period. Consider these from Lang's "Astrophysical Data", p. 41 (Springer, 1992):

Mayan Long Count = 5125.366 tropical year (see above)
MLC / sidereal orbital period, in tropical yr, of Jupiter = 5125.366/11.86223 = 432.07
MLC/ 29.4577 (Saturn) = 173.99
MLC/ 84.0139 (Uranus) = 61.006

Thus the Mayan Long Count, as I've noted in earlier posts here to Dr. Van Flandern's messageboard, is significantly close (p=0.003%) to a whole multiple of the orbital periods of Jupiter, Saturn, and Uranus. One must use 243 periods of Uranus, to get a closer common multiple with Jupiter & Saturn, than the MLC which uses 61 Uranus periods. I find that integers with no prime factors bigger than 13, are dense enough, that the existence of a way to construct, by multiplication of days, an MLC so near this desired best approximate common multiple, is not surprising.

This relationship, between the Long Count and the periods of three giant planets, tells us that the Mayans, or their forerunners, knew of Uranus. Furthermore, they knew the orbital periods of Jupiter, Saturn, and Uranus to about four significant figures or better. For Uranus, Newcomb's 1873 investigation was good to almost five figures, but required a century of telescopic data. The Mayan accuracy also can be achieved, however, by comparison of modern unaided observations, with Ptolemy's occultation records for Jupiter or Saturn, if Earth parallax is accurately calculated. The Mayans must at least have had a good Copernican solar system model, and many centuries of careful visual observation records.

The number 61 apparently does not arise by accident. An even closer common multiple, is 243 times Uranus' period:

243*84.0139(Uranus)/11.86223(Jupiter) = 1721.04
243*84.0139(Uranus)/29.4577(Saturn) = 693.04

The "243" relation also holds, using Lang's 1992 values, for Earth, Venus and Mercury:

243*365.256 day sidereal period (Earth) / 224.701(Venus) = 395.001
243*365.256 / 87.969(Mercury) = 1008.96

(addendum June 23, 2010: I also find that the "243" relation holds for Mars, Earth, and Venus:
243*686.98 day orbital period (Mars) / 365.256(Earth) = 457.04
243*686.98 / 224.701(Venus) = 742.93 )

Finally, recall that Venus' sidereal rotation period is (2007 World Almanac) 243.02 days (these are synodic Earth days). This is only slightly faster than the sidereal rotation period, 243.165d, which would result in Venus, on average, showing exactly the same face to Earth at every closest approach.

The accuracy of the common multiples in the Mayan Long Count, proves that unlike the Roman calendar millenium, the Mayan Long Count is not an arbitrary length of time, but rather is an important astronomical cycle, disguised, for convenience or for deception or both, as an arbitrary length of time. The MLC is based on 61 orbits of Uranus, but the common multiples based on 243 orbits of Uranus or 243 orbits of Earth (243 = 61*4 - 1) reveal that some unknown physical effect causes the "243" cycles and probably the Mayan Long Count too.

Hi Joe, it might be an idea to e mail Linda Howe, as she does have the ear of the media.

I thought I'd pick your brains again, see what you think about this idea. In the thread about why someone believes in an end to the universe, I got thinking about what it must be like to be in orbit round the galactic super massive body. For one thing you could be inside of a neutrino ball. Something with a radius of about a month. Now if you hopped into a space ship and orbited very close to the Swartzchild radius you could watch the universe go out, check your watch and say, "well that's it then. Ten forty gmt."

now let's suppose that the Scwartzchild radius for the universe is one gravitational second. Remember that I'm a bit torn between between it being h = c^2 / b^2 and barh = c^2 / b^2 where b is the speed of gravity. I am leaning more to it being the later.

That gives us a Scwartchild radius of about 2.9E 25 metres.
2.9E 25 = 2GM / c^2

Work out the mass and it's aout the current estimate.

g /r_s (r/c)^2 = 0.5

where r_s, r subscript s, is the Scwartzchild radius. We take g to be G, the universal constant. Working out r we get aout 139 billion light years for the radius of the universe.

t_r / t = sqrt(1 - r_s / r)

Where t_r is time for an observer near or inside Scwartzchild radius. The upshot is that if we could see a galaxy 139 billion years in the past, then it would e in a different time zone but it's not huge, a tenth.

Well, is this a big bang? Yes and no, I suppose. if we take from Dirac the idea that we can have an expanding universe in which either G decreases, or h increases, the we can have a static universe in which the both vary. A universe that always has the Planck lenght at the same value.

Pl = sqrt(barh G / c^3) = about 1.62E-35

Say that's a wavelenght and its mass will be 1.62E-35 = h /mc =1.36E-7 kg

Pop that into the equation for the Scwartzchild radius. r = 2Gm /c^2 = 2.0225E-34

Rough calculations but I reckon that's going to work out at twice barh

Hi Joe, I haven't done the sums yet but there's something interesting going on here I think. all park figures; from the equation g / r_s (r / c)^2 = 0.5 we can work out G when the radius was at about 2.4E-12 metres.

Now let's say that, I think it was de Sitter, that G = 1 /6pi rho t^2

Rho is going to be huge, as we're putting the whole mass of the universe into something the size of an electron. So t is going to be very small. In the region of 1E-60 seconds.

But suppose we say that in the past G was very large and h was proportionately much smaller than it is now. Then from sqrt(barh G / c^3 we can work out what barh was at 1E-60 seconds.

As we are talking about the lorentzian as a refractive index, it makes sense to say that the frequency of a particle doesn't change in a different refractive index space. Because h is much smaller we can have a situation where a subatomic particle has the same energy as the whole universe. I suspect that this primordial particle will turn out to be a pion rather than an electron. First particle on the scene at 1E-60 seconds is a pion which doesn't like what it sees and promptly decays. I suppose the question has to be, how promptly is promptly in this case.

Hi Bob,

I hope to look at this more carefully later, but I see that the famous British calm might endure even to the end of the universe:

<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 />...check your watch and say, "well that's it then. Ten forty gmt." <hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">

Hi Joe, it can be a problem on the battlefield. Brits and Anzacs fighting alongside American troops can sometimes get into "language" difficulties over usage. Though there was the famous "Houston we have a problem" which was a very British way of stating something.

On the idea of the Scwartzchild radius of the observable universe being one gravitational second. Then the idea of a primordial pion that has the mass energy of the universe ounds like fun to look at, if only because its decay should somehow look like the universe we inhabit. the pion's neutrino carries off the bulk of the energy but this neutrino, I think, has to e considered to have mass. i need to read up on this further, as the few links I've looked at are not saying the neutrino has mass, whereas newer research says it does.

Before getting into that though, I though I'd need to look at red shift. If the "universe" has a radius of about 139 billion light years, then most of it is empty of galaxies, we can still see them after all. eh most distant ones are very young, and if G and h vary proportionately to the planck length then Gand h differ from our values but they are in a different "time zone."

With Doppler shift we have (1-v/c)f or (1+v/c)lambda where v is much smaller than c, let's call that e.m doppler. What about gravitational red shift? (1-c/b)f where b is the speed of gravity. We expand it and can ignore c^2/^2 and C^3/b^3 etc.

So let's accept for the moment that there's a phase transition at the speed of light. Then a photon will have a plus and minus value.

So, let's take a look at the grav energy and the e.m energy again.

E = mc^2 and E =mb^2 we'll use the electron as our mass particle.

m = 9.11E-31 c = 2.998E 8

8.188E-14Js

grav energy will be much greater where I've taken b to be 2.91E 25

Divide 8.188E-14 by barh to get a frequency and we find that grav energy and e.m frequency have the same numerical value. This must mean that barh takes the value one in gravitational space. it's effectively borrowing energy form a time in its own far future. Or it borrows from its past, because with negative refractive indices there's nothing to stop that sort of borrowing. Though at the moment I prefer to have one direction to time but allow negatives with respect to positive. A bit like running the universe off a nine volt battery, set a line at 4.5 volts.

More about the "243" relation

Let's choose one of the nine planets, multiply its orbital period by 243, then divide by the orbital period of one of the other eight planets. Randomly, of these 9*8=72 numbers, 1/10 * 72 = 7.2 should be within 0.05 of an integer. So, it isn't surprising that 8 are, but it is surprising that 5 of the 8, are pairs where the second planet is one or two steps *below* the first (there are only 15 such pairs, so only 1.5 of those should be within 0.05 of an integer). The 8 are:

Earth:Venus
Earth:Mercury
Mars:Earth
Uranus:Saturn
Uranus:Jupiter

Pluto:Jupiter
Earth:Pluto
Mercury:Saturn

Also, 7.2 of the 72 should be within 0.05 of a half+integer (1/2 + n), but instead, 12 are. It's surprising that 12 are; and the excess is due to 6 pairs where the second planet is one or two steps *above* the first (there are only 15 such pairs, so only 1.5 of those should be within 0.05 of a half+integer). The 12 are:

Mercury:Earth
Venus:Earth
Venus:Mars
Earth:Jupiter
Mars:Jupiter
Mars:Saturn

Earth:Neptune
Jupiter:Neptune
Jupiter:Mars
Saturn:Mercury
Uranus:Mercury
Pluto:Mercury

Summary: Five of 15 possible pairs, where the inner planet, of the nine including Pluto, is one or two steps below the outer, have the ratio, 243*(outer planet sidereal orbital period)/(inner planet period), within 0.05 of an integer. Six of these same 15 possible pairs, have the ratio, 243*(inner planet period)/(outer planet period), within 0.05 of an (integer plus half). The binomial test gives p=1.27% for the former phenomenon and p=0.22% for the latter.