Requiem for Relativity

Update on Monterosa's axis of rotation

When I complained to astronomer W, that I thought astronomer Z's refusal to divulge the 2003 Monterosa lightcurve, amounted to theft of the data, W told me never to contact W again. So, I contacted W one more time, delivering to him a Freedom of Information Act request (which he's ignored so far) for any and all data on the lightcurves of Monterosa and Arlon.

The ordinate on the graph of the 2003 Monterosa lightcurve, which I downloaded from Behrend's website, is about "-13.80" so it can't be the magnitude. If that's the negative of the magnitude, then dimmer is at the top of the chart, brighter below. I still have no response from any of the three main listed investigators of that lightcurve, all but one of whom now have received emails from me (I haven't yet found an email address for one).

W's 2007 Monterosa lightcurve data sheet, has an ambiguous baseline, because it doesn't say whether, among several other things, a 0.7 magnitude correction that it lists, has to be added or subtracted, or not, from the listed numbers, to get the actual magnitudes measured. Without knowing the jargon, it's impossible to know unambiguously whether several corrections listed, need to be added or subtracted from the numbers, to get the actual magnitudes measured.

So, I really know only the relative magnitudes of each lightcurve. Therefore I modified my criterion. As before, I used the above computer program to find the rotation axis for which the coefficients of ten tesseral harmonic terms of "K inverse", namely (n,m)={ (0,0),(0,2),(2,2)(sin,cos),(3,2)(sin,cos),(3,3)(sin,cos),(4,3)(sin,cos) }, minimized the sum of squares of (coefficient divided by (0,0) coefficient), but this time I excluded the (0,2) coefficient, because I don't know the mean brightnesses unambiguously. I only insisted that the (0,2) coefficient be smaller in absolute magnitude than the (0,0) coefficient. I assumed that the distance-adjusted mean magnitudes were the same except for the small difference in the empirical Gehrels-Tedesco phase correction (see the posted BASIC program REM notes for the reference). This method gave, for Monterosa, a rotation axis near the ecliptic pole and inclined toward ecliptic longitude 180.

The 2003 lightcurve graph had so many scattered points that in measuring from the graph, I had to rely the midline someone had drawn. This gave almost zero second order harmonic, so, I suspect that someone assumed that the second order harmonic was zero, in fitting that line.

My computer program automatically considers all relative phases of the two observations. So, there is nothing more to be gained by reversing the polarity.

(Review emailed to a friend)

For years, I've been thinking of reasons why Barbarossa might be so dim in the far infrared. I've mentioned all this on the thread over the years, but here's a review:

I can't be a slave to the calculations that have been published, about what the temperature and diameter of a body of Barbarossa's mass should be. For one thing, those calculations assume pure hydrogen composition. Textbook formulas show, that in the hyperjovian mass range, gravitational collapse is a sensitive function of the atomic number of the predominant constituent atoms; the quantum mechanical calculation, of threshold gravitational collapse involving atoms of atomic number much greater than one, seems to be intractable, or at least, seems not to be in the literature. Since I don't know the composition of Barbarossa, even conventional quantum mechanics tells me I don't know Barbarossa's diameter either; rough estimates based on textbook formulas, show that Barbarossa's diameter could be about that of Earth, not that of Jupiter, which would make Barbarossa 100x dimmer.

For another thing, the published calculations of brown dwarf temperature, make assumptions about how they form and cool. If they form some other way (accretion?) or cool some other way (an unexpected form of convection?) they could be much colder. At Barbarossa's distance, equilibrium temperature with sunlight, is near the temperature of interstellar dust. If Barbarossa is the same temperature as interstellar dust, Barbarossa would be hard to find.

The foregoing arguments assume only textbook physics. We're still discovering "new physics". Forty years ago pulsars were unknown, and 400 years ago Newton's law of gravity was unknown. So it's hubristic to assume that anything other than observation will show what Barbarossa is or isn't.

Also, there's evidence of a nebula around Barbarossa. Evidently, enough red and infrared light penetrates this nebula, for Barbarossa to appear on the sky surveys, but there's no way to predict for certain how this nebula would affect Barbarossa's endogenous far infrared emission.

Hi Dr Joe, Your latest post had me wondering about the mass of your object. Do you know its mass?

Hi Joe, In the next thread down I suggested that your computer program might be of some help in looking at the orbit of the rather odd S2. The star going round the supposed black hole at the centre of our galaxy. I think we can scale it down, make the black hole's mass a solar mass, and scale S2 which is about 15 solar masses, the mass of a terrestrial planet. Then we could perhaps look at he properties of the space. Is the space a neutrino ball for instance?

<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 />Hi Dr Joe, Your latest post had me wondering about the mass of your object. Do you know its mass?
<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">

Thanks! You just threw a completed pass, and I think I can run this for a touchdown.

It's 0.0280 solar masses, but I'll review the information about Barbarossa's mass, that is in my old posts. In 2007, I fitted my sky survey detections of Barbarossa, to a circular orbit, and found that Barbarossa's distance from the sun, if its orbit were circular, would be about 198 AU. This implies, from "precession resonances" (see below) that Barbarossa has 0.0108 solar mass. In 2009 I found that an elliptic orbit with e=0.61 and a=344, fits the sky survey detections much better. Though the Sun's planets are in circular orbits, an eccentricity near 0.6 is common for binary stars, so my elliptical orbit is plausible too.

Calculus shows that the eccentricity makes Barbarossa 2.0x as effective at causing precession of the planets, so Barbarossa's mass must be 0.0108*(344/198)^3 /2.0 = 0.0280 (the last digit is off because I carried more significant figures in the original work).

This is a good time to ask someone, anyone, possibly with the help of a friend who is good at trigonometry, to confirm my calculation. There's no substitute for proving it to yourself, then meeting a professional astronomer face to face and saying, "Keller isn't simply a lone nut. I checked his calculations myself and they're right. What are you going to do about it? Are you going to be derelict in your duty as a public official? You're a full time government employee, aren't you? In charge of astronomy, aren't you?" They're high-handed bureaucrats, with almost an aristocratic mentality, and think they're too good to be addressed that way, but it breaks through their psychological "denial". If he tells you to go away or he'll call the campus police, then your next stop is your Congressman's nearest hometown office. Anyway that's how I've been doing it. Much can be done via the web, but keyboard work alone, won't be sufficient.


Explanation of "precession resonance" (review).

Suppose "Planet A" has a circular orbit of radius = 1. Let Planet B have a circular orbit of radius r, maybe r&gt;1 or maybe r&lt;1. Draw these on paper and let the right side (rightmost end) of the orbit of Planet B tilt a tiny amount, "dphi", upward out of the plane of the paper.

Now use trigonometry to find the upward or downward force on Planet A, from Planet B. If Planet A is at the righthand end of its orbit (i.e., angle theta = 0), this has to be summed over all positions of Planet B, from 0 to 180 (which I did in 2007 on a computer). It can be done with a calculator and a sheet of paper, accurately enough, with "3-step Gaussian quadrature". As applied to this situation, all you have to do is find the upward force on Planet A, when Planet B is at theta = 20.3deg, and add to that the upward force on Planet A, when Planet B is at theta = 180-20.3 = 159.7deg (Gaussian quadrature is like Simpson's rule but more accurate; usually it would be necessary to find the upward force at 90deg, but that's zero here). Remember, downward force is a negative upward force. So, you'll be adding a negative to a positive, i.e, subtracting.

This way, add up the upward forces (remember to multiply by the mass of Planet when Planet B is Jupiter, Saturn, and Uranus; and Planet A is Neptune. Now put Barbarossa at r=198/30, m=3300 Earth masses (i.e. approx. 0.01 solar mass) and find that the upward force due to Barbarossa is almost exactly 1/3 that due to Jupiter + Saturn + Uranus (with a small correction for Uranus' eccentricity, it's exact for Barbarossa = 0.0108 solar mass). That is, Barbarossa torques Neptune 1/3 as much, for a given inclination, as do the other planets combined; Neptune's rate of precession around Barbarossa's orbital plane is 1/3 its rate of precession around the common orbital plane of the other planets.

What we did for Neptune, can be done for a typical plutino. Pluto itself is complicated, because Pluto's orbit is eccentric, and crosses Neptune's with "avoidance" synchronization. So, use an idealized plutino with a circular orbit with Pluto's major axis. The result is, that Barbarossa torques this plutino almost exactly 1/2 as much as do J+S+U+N combined.

Finally, do it for a typical "classical Edgeworth-Kuiper belt object". Websites show that their median major axis is about 43 AU; assume a circular orbit. The result is, that Barbarossa torques this typical classical KBO, almost exactly as much as the planets do.

Barbarossa's torque, for a given planet, is M/R^3*f, where f is an eccentricity correction factor; f=1 if Barbarossa's orbit is circular, otherwise f&gt;1 (for eccentricity 0.6, f=2).

Here's where a negativistic person will say, "So what?" but a fairminded person will begin to suspect something. At this point in the argument, some professors might realize that this "precession resonance" provides a clue about how distant solar system objects (as far out as the Kuiper belt) remain within the ecliptic despite Barbarossa's torque. But the main point is, that only one value of Barbarossa's M/R^3, gives the 1: resonance for Neptune. Can it be mere chance, that this same value of M/R^3, gives 1::2 resonance for a plutino and 1::1 resonance for a classical KBO? The same value, would not satisfy all three equations, unless something were there. Barbarossa overrides Bode's law beyond Uranus; precession resonance determines the outermost orbital radii.

Really, Barbarossa's orbit is larger and elliptical. See the third paragraph of this post, for how I correct for that, to get 0.0280 solar mass.

Something else corroborates Barbarossa's mass. Paul Wesson discovered that when J is the angular momentum of any astronomical system (such as a rotating planet or a binary star orbit) J/M^2 tends to be a universal constant. Unfortunately this empirical constant varies by a factor of 2 or more. J/M^2 is about the same for rotating Jupiter and rotating Earth: Jupiter is 300x as massive, has 10x the radius & 3x the orbital frequency, and 300*10^2*3 / 300^2 = 1. But Earth+Luna have J/M^2 almost 6x bigger than Earth alone; alphaCentauriAB have J/M^2 almost twice as big as Earth+Luna.

However, there is evidence that the angular momentum of the Earth+Luna system has been the same since about the time of its formation by co-accretion (a fundamental 5hr Earth rotation period when Luna was at the Roche limit; see my post). If J/M^2 is the same for the Earth+Luna system as for the Sun+Barbarossa system (counting also the masses and orbital angular momenta of Jupiter and Saturn, and vectorially adding the angular momenta of Earth's spin and Luna's orbit), then I find that Barbarossa's mass must be 0.0305 solar. This differs only 9% from the figure deduced above from "precession resonance".

Relatively small bodies, much nearer to the Sun, could contribute a large share of the precession resonance attributed to Barbarossa. Also, relatively small bodies, much farther than Barbarossa is, from the Sun, could contribute much of the angular momentum attributed to Barbarossa. However, it then would be unlikely that the implied mass of Barbarossa would be the same, by either estimate. The equality of the two estimates, shows that probably almost all of the mass beyond Neptune, is at about the same distance from the Sun: Barbarossa and Barbarossa's moons or twins, if any.

I realized last night, that Barbarossa's aphelion is exactly in the direction of the galactic center. A year ago, when I fit a=343.8, e=0.61059, and i=12.93 exactly, I only estimated Omega, the ascending node, as 293, and omegatilde, the longitude of perihelion, as 85. This would give the ecliptic coords of aphelion as longitude 85+180=265, and latitude -(293-265)*sin(12.93)=-6. That's about the position of the galactic center.

Some books say that binary star orbits are random, but some articles, and my own research posted here a few years ago, say the orbit orientations are subtly nonrandom. Be that as it may, the conjunction of Barbarossa's calculated aphelion, with the galactic center, is one more coincidence that naysayers have to accept as chance.

In December a new infrared survey satellite was launched, and this month a new solar observation satellite, despite overall budget cuts including the cancellation of manned projects. Maybe the authorities know more than they tell us.