Requiem for Relativity

Frey is a Low-Mass, Black, Iron Dwarf

The term, "Iron Dwarf", usually refers to a white dwarf containing mainly elements near iron. The term "Black Dwarf" refers to a white dwarf which has lost enough heat that its surface has become cold. The partial collapse (partial electron degeneracy) that would be seen in an iron object of Frey's mass, gives Frey a size consistent with Frey's observed Visual and Red magnitude, assuming the usual albedo for outer solar system objects, and a surface resembling a red class Kuiper belt object.

Only one certain photograph of Frey (and none of Barbarossa) were taken with clear, not red, filter. This is the Dec. 22, 2008, U. of Iowa photo. Let's assume that this CCD response approximates the Visual band.

Thirty arcseconds from Frey in the U. of Iowa photo, is a USNO-B catalog star with Red magnitude (average of R1 & R2) +19.41 +/- 0.20, Blue magnitude +20.55, and Optical Infrared mag +18.45. According to Straizys, "Multicolor Stellar Photometry", Table 27, R-I for this star corresponds to spectral type K2V, and B-R corresponds to type G7V. The catalog documentation warns that Blue magnitudes > +21 are invalid, so I'll estimate this star closer to the former type, and say K0V; Straizys gives V-R=+0.52 for this type, so the Visual magnitude of this comparison star is +19.93 +/- 0.20. Frey seems slightly brighter, so I'll say Frey has Visual magnitude +19.8.

This is consistent with the Red magnitudes, +17.5 to +19.5, I've estimated for Barbarossa and Frey from Red sky surveys. Wickramasinghe & Hoyle, Ap & Space Sci 268:369, 1999, say that for typical "red class" Kuiper Belt objects, reflectance at 6500A is 2.50 times (1.0 magnitude) that at 4500A. Those wavelengths are near the "central wavelengths" of the Blue & Red bands, 4350A & 6800A, resp. ( www.elvis.rowan.edu ). So, with a small correction for the larger actual difference (6800-4350)/(6500-4500)=1.2, red class Kuiper belt objects have B-R=1.2+0.97=2.2 (the second term arises because Vega, which reaches the zenith at midnorthern latitudes, was chosen as a convenient color photometric standard; so, G2V stars like the sun, have B-R=0.97 in the Cousins VRI system according to Straizys' Table 27). For G2V stars (see Straizys) (V-R)/(B-R)=0.35/0.97=0.36, which times 2.2 is 0.79. So if Frey's surface resembles a typical red class Kuiper object, it has Red mag +19.8 - 0.8 = +19.0, consistent enough with my estimates on Red sky surveys.

Iron-nickel meteorites are common; some contain mainly nickel (specific gravity under ordinary conditions, 8.90). If Frey were nickel, its specific gravity certainly would be higher than this, because the pressure in Frey's interior, times the area of an atom, is comparable to atomic electrical force (the atom would be compressed as though Z, the atomic number, were increased but the number of electrons held the same). Suppose Frey has exactly the mass of Jupiter but Frey's density is 11, i.e., 8x Jupiter's. Suppose also that Frey's albedo is 0.04 (common, some say typical, for outer solar system objects, which usually are fairly black; Luna's average albedo is 0.07, facing us). Then Frey's Visual magnitude would be * +18.3 *. This still is slightly too bright, because the Red magnitude would be +17.5 (if Frey is as red as red class Kuiper objects) which is brighter than any of the comparison Red magnitudes I've found. The Visual magnitude would be 1.5 mag brighter than the one Visual magnitude I have (above). This is assuming 100% nickel "ordinary matter" composition with some compression but no collapse.

Following Stein & Cameron, "Stellar Evolution", 1966, pp. 72-73, I find that if composed of iron, with mass equal to Jupiter, Frey is already semi-collapsed, almost like a white dwarf. Here is the approximate calculation:

It's about right, to say that a pure hydrogen body of the mass of Jupiter, is on the verge of collapse. That is, the gravitational potential energy difference for a hydrogen atom, between Jupiter's surface and center, about equals the ionization energy of the hydrogen atom. Let's consider an iron atom near the center of this body. The gravitational potential for the iron atom is 56x greater, enough to ionize it to about Fe IX (see Lange's Handbook of Chemistry, 13th ed., or recent edition CRC handbooks, for atomic ionization potentials). For a conservative estimate, suppose it is Fe VI. In Fe VI, the 4s shell electrons detect 7.5 net positive charge units, vs. 1.5 for the 4s shell of calcium, so should be 5x smaller, i.e., 1.9/5=0.38A (this is tolerably consistent with the Fe II & III ionic crystal radii on www.webelements.com ) instead of 1.5A. So, Fe VI is about (?) 1/4 the diameter of neutral Fe. The free, degenerate electrons also fill space, but their size is estimated from the nonrelativistic version of Stein & Cameron eqn. 5.8, to be roughly 5x the Compton radius, negligible vs. the Fe ions. If Frey is all iron, it shrinks even more, because it starts smaller with bigger gravitational potential, ionizing the iron more.

Frey shrinks by a factor of better than 4, so its brightness decreases by more than 3 magnitudes, to dimmer than +18.3 + 3 = +21.3. This is consistent with the Visual +19.8 mag observed, if the albedo is higher than 0.04 * 4 = 0.16.

Barbarossa, if iron, would be so collapsed, that rings, a nearby moon, high albedo or faint self-luminosity, would be needed to explain its brightness. On the other hand, Barbarossa might be composed of lighter atoms than Frey. Even a hydrogen composition for Barbarossa is possible, if present estimates of hyperjovian diameter turn out to be inaccurate. Such inaccuracy might arise either from assuming too much retained heat, or from approximations made in solving the integral equations involved.

(Feb. 6, 2009)

Please Try This Yourself at Home


My (first) Feb. 3 post, explains how I could not find a real binary orbit satisfying Kepler's second law. Today I realized that luckily, the binary orbit can be solved graphically to good accuracy, if Kepler's second law is set aside for the short 1986-1987 interval (the rationale for this will be discussed in a later post).

You can do this yourself at home with graph paper and a ruler. A short table of trig functions is optional. I think this is the best opportunity there ever will be, to see for yourself that these objects are real, before many photos become available.

First, let's review Barbarossa's solar orbit. In 2007, I discovered that three pairs of "disappearing dots", on 1954, 1986 and 1987 Red sky surveys (online scans) have a center of mass which makes a straight, constant-speed arc across the celestial sphere, if the mass ratio is chosen right. One adjustable parameter, the mass ratio, solves two equations: for straightness and for constant speed. The mass ratio (most accurate, from my recent program) is Barbarossa::Frey = 0.89::0.11. The orbit (corrected for Earth parallax) cuts Declination lines at about 27 degrees southward to the east.

With the program above, I got accurate enough fitting of these three center of mass points, to an orbit accelerated only by the Sun, to make arcsecond-accuracy predictions. It turned out that a starlike "disappearing dot" of comparison Visual magnitude +19.8, on the Dec. 22, 2008 (call it 2009, since the others all were late winter or early spring) U. of Iowa photo, is in a position consistent with Frey, but the corresponding Barbarossa, assuming the predicted center of mass, is off the west edge of the photo.

Now, there are four observations of the Frey-Barbarossa binary orbit. I converted radians on the celestial sphere, to A.U., by multiplying by Barbarossa's distance. Frey's coordinates with the center of mass (predicted, for 2008; actual, for the others) at the origin, are:

1954 -0.701, +0.117
1986 -0.061, +0.332
1987 +0.041, +0.190
2008 +0.223, -0.063

The first number is how many A.U. Frey is east of the c.o.m., and the second number is how many A.U. north.

When you plot these, you'll notice that three of these "Frey" dots are almost perfectly collinear (draw a line through them; it will be sloped about 54 degrees south and to the east). This a clue that the dots are not random accidents. The orbit is roughly circular but very slanted, almost edge-on to our line of sight. The 1986, 1987 and 2008 Freys are all at about the same distance from us, on the nearer side.

The 1954 Frey is on the farther side of the orbit (thus the orbit is retrograde). As Barbarossa revolves around the Sun, that orbital position seems to move, by parallax, westward parallel to Barbarossa's solar orbit. If 1986-1987 is the reference time, then the precession correction for the 2008 Frey is small, mainly because it isn't much nearer us than the origin is. On the other hand, the precession correction for the 1954 Frey is almost (assuming a 1.4 AU Frey/c.o.m. orbit, which will be justified later) 2*1.4*(1986.5-1954)/(2009-1954)*0.112 radians of orbital travel = 0.185 AU, of which, multiplying by sin(54-27), 0.09 AU is perpendicular to the 1986-1987-2008 Frey line. On your graph, make a new "corrected 1954" Frey, 0.185 AU from the original Frey, moving along a ray sloped 27deg north and west (27deg = arctan 0.5). The corrected Frey is to the west, because that is where Frey would have been if the orbit had been in its apparent position of ~1986.5. The rearward limb of the orbit moves west due to parallax.

Draw two more lines parallel to the line through the "three Freys". One of these lines, draw through the origin, and the other, draw through the "corrected 1954 Frey". Note that the line through the corrected 1954 Frey, is four times as far from the line through the origin, as is the line through the "three amigos" Freys. This is because the 1954 Frey is near apogee, and the "three amigos" Freys near perigee. In this arrangement, perspective doesn't change the apogee::perigee ratio, which is 4::1, corresponding to eccentricity 0.6 ( (1+0.6)/(1-0.6) = 4 ).

How long is the semimajor axis, "a"? Call the angle between our line of sight, and the normal to the binary orbital plane, "i". From the graph (draw a line perpendicular to the "three amigos" line, and through the "corrected 1954 Frey") 2 * a * cos(i) = 0.74 AU. From 1954 to 1987, Barbarossa moved 1/15 radian in its solar orbit. To get 0.185 AU parallax, the 1954 Barbarossa must have been 0.185 * 15 AU closer to us, than the others. So, the semimajor axis must satisfy

2 * a * sin(i) = 2.775 AU

So, tan(i) = 2.775/0.74 and i = 75deg, and a = 2.775/2/sin(75) = 1.44 AU.

Now let's find the orbital period and investigate Kepler's second law. Draw a line from the "corrected 1954 Frey" through the origin. This line happens to go through the 2008 Frey. This shows that from 1954 to about 1987 (33 yr) is a period and a half, and from about 1987 to 2008 (really more like 2009, 22 yr), another period, for two and a half total. This satisfies Kepler's second law for the long time intervals, and shows that the period is about 22 yr.

Another way to find the period, is to recall that the three amigos are collinear. By conservation of angular momentum, their speed along that line must be constant. The time from the 1987 point (really, one orbit ahead of 1987) to the 12/22/2008 point, is the distance, divided by the speed from 3/15/1986 to 1/31/1987. Subtract this extra time from (12/22/2008 minus 1/31/1987); I found 20.4 yr. However, if for some reason Kepler's second law (conservation of angular momentum) fails for the 1986-1987 interval, thus the speed is not constant (for example, Frey might have a moon of its own), this would not be so accurate. In fact, the true Frey of 1987 probably is only 40% as far from the 2008 Frey position, as shown on the graph, and moving 3x as fast; so, my best estimate of the true period is 21.7 yr.

Knowing a = 1.44 and e = 0.6, I find (see chapter on ellipses, in any analytic geometry book) b = 1.15, and the apparent ellipse area pi*a*b*cos(i) = 1.35 sq AU. The area of the triangle formed by the origin and the 1986 & 1987 Freys should be 1.35*(0.87yr / 21.7 yr) = 0.054 sq AU. Finding the area of the small origin-1986-1987 triangle on my graph, I find only 0.013 sq AU. So, this short interval does not conform to Kepler's second law. This is hardly surprising, since the 1954 and 1986 sky surveys show additional fainter "disappearing dots" near Frey, more than ten times closer to Frey than to Barbarossa (i.e., within the Jacobi, or other, stability limits), which moons could displace Frey enough to give big deviations from Kepler's second law, in short time intervals. The demand that Kepler's second law be satisfied, even for the short 1986-1987 interval, might be what made it impossible for me to find a consistent binary orbit.

From hypothetical outer solar system precession resonances and other theories, I've estimated the Barbarossa+Frey system as 0.01 solar mass. The a = 1.44, is for the Frey-c.o.m. distance; the Frey-Barbarossa distance is a = 1.44/0.89 = 1.62 AU. The 21.7 yr orbit and 0.01 solar mass imply 1.68 AU, fair agreement for this rough graphical method.

(Feb. 8, 2009)

Fourier-Cruttenden, Pioneer 10&11, and the Barbarossa/Frey Binary Orbit

On Jan. 30 (see my first and third Jan. 30, 2009 posts above) I found that the Barbarossa/Frey binary orbit, satisfies Kepler's second law, if the 1954 & 2008 Freys are displaced perpendicular to the orbit, by 0.159 AU each, westward for the 2008 Frey and eastward for the 1954 Frey. Subsequently I tried to make this more accurate, by recognizing that the displacement due to Barbarossa's solar orbit, would be parallel to that orbit, i.e., at a 54-27 = 27deg angle to the 1986-1987-2008 Frey line, not perpendicular to it. Again (calculating now with the accurate cubic interpolation, and trying unequal displacements too) Kepler's second law can be satisfied, by various, not necessarily equal, displacements of about 0.2 AU for the 1954 & 2008 Freys, eastward & westward, resp.; but these displacements bring the 2008 Frey near to the origin, and correspond to a real orbit that is nearly parabolic with large major axis.

It is as if the component of the parallax displacement, parallel to the Frey/Barbarossa orbit, is removed. The binary orbit seems to exist in a synodic, rather than a sidereal, frame.

There is evidence, following the Fourier-Cruttenden concept of a rotating solar system frame, that the Earth/Sun orbit also exists in a rotating frame. The Sun's average acceleration at Earth is 0.593 cm/s^2. Presently the Sun and Barbarossa seem to be orbiting each other with period 3083 yr ( 2*pi / present angular speed). If the Sun/Earth frame were rotating with that period, the centrifugal force at Earth's radius, would be 0.593/3083^2 = 6.24/10^8 cm/s^2; this is near the Pioneer10&11/Ulysses/Galileo anomalous acceleration.

A similar rotating frame at Barbarossa, would convert the expected sidereal orbit into a synodic one, and remove that component of parallax motion, parallel to the binary orbit. The expected displacements needed to satisfy Kepler's second law, would become, for 1954 and 2008 resp., approx. 0.164 AU * sin(54-27) * 33/55 = 0.045 AU, and 0.164 AU * sin(54-27) * 22/55 = 0.030 AU (assuming that the 1954 & 2008 Freys are equally nearer us than the 1986 & 1987 Freys).

(cont. Feb. 9)
Trying this and other combinations, I haven't yet found a binary orbit for Frey/Barbarossa, which perfectly fits. This could be due to rapid precession of the orbit by another Barbarossa moon, Freya, or displacement of Frey by a moon of its own (which I might also call Freya).

I improved the accuracy of the sector angles in my program above, by using the BASIC sine instead of the cosine function. This showed that the center of mass orbit is planar to arcsecond accuracy. The third ("C", 1987 plate) point satisfies two equations (expected RA and Decl for circular orbit) to arcsecond accuracy with only one adjustable parameter (mass ratio).

*****

Uniformly, the many journal editors to whom I have submitted this information during the last month, have showed negligence, malfeasance, and the dereliction of their public, tax-supported trust. Maybe they are more interested in turf than truth. None forwarded my letter to any peer reviewer or to any other knowledgeable person, at least not that I know of. Not one word of intelligent criticism was returned. The editor of Icarus complained that I did not make my letter longer with more documentation, but when I offered to provide him, as a test, with all the documentation I had for any one sentence, of his choice, he did not respond, thereby proving his hypocrisy.

The department chairman says, send it to a journal; the journal says, go away. But try telling the taxman to go away when he comes to take from you, the professor's and the editor's salary.

Meanwhile in three months I have two photos from the U. of Iowa's 15 inch robotic telescope, but the best season is almost over. A professor there told me, basically, that I should quit if I couldn't predict the position within one arcminute on the second photo (although LeVerrier's prediction of Neptune within one degree was considered a great success, and Galle searched visually, equivalent to many photos). I'm also getting a photo a month (seems to be the limit, with their waiting list) from Bradford College's Tenerife telescope, but these are unstacked and it's questionable whether they can be adequate.

The $100 reward I posted, still stands, for anyone who is able to get anyone to look definitively (I'll have to be the judge of that)(one median-stacked photo with a 15 inch telescope in favorable geographic location, might be enough). This reward isn't for looking, it's just for getting someone to look.

Looking back over a few posts I realized that I've made a tiny little mistake, on what I thought was our true speed through the vacuum. I said about 1.8% of the speed of light, when it should have been 18% Oops, so about three times faster than the new estimate of how fast we are going, due to the discovery that our galaxy is actually a little larger than the Andromeda galaxy. We'd still be talking about the earth's "argument of pericentre" taking ten turns of the solar system's spiral through space but the solar system's bodies have further along the curve to travel.

(email sent this morning)

Hi *********!

Thanks!

The first coordinates I gave him were a few arcminutes off because I was trying to force [Steve Riley's and Joan Genebriera's] 2007 detections into the theory. So [his] first photo, a very good, painstakingly median-stacked one, had Frey but not Barbarossa in the field of view. His second photo, had many hot pixels, and Frey & Barbarossa were out of the field anyway. I then went "back to basics" to revise my prediction, using only the three red sky survey plates, and writing an accurate computer program for the spherical trigonometry. He has the same approximate (arcminute accuracy) prediction that you do, but wants me to give him an arcsecond accuracy prediction so he can use it to "debunk" me. That's what he's waiting for.

My new prediction, based only on the three red sky surveys, places the 1986, 1987, and 2008 "Freys" on an almost perfectly straight line (the break is < 0.01 radian, equivalent to about an arcsecond position error) relative to the center of mass. That hardly can be random. A segment of that length of the binary orbit viewed in moderately steep perspective easily could be that close to a straight line, though, if the orbit precesses in synchrony with the solar orbit (which is plausible physically).

In addition, for the third sky survey to conform accurately to the circular orbit predicted by the first two, with only one adjustable parameter (the mass ratio), amounts to solving two equations (for RA and Decl) accurately by adjusting only one parameter. So the reality of this is supported by two unlikely but plainly demonstrable coincidences: the constant-speed alignment of the three observed c.o.m.'s, and the alignment of three of the Freys.

Sincerely,
Joe Keller

********

(not part of the email)

Suppose the "three amigos" Freys (1986, 1987, & 2008) lie near perigee of a binary orbit with e = 0.6. (This is the rough situation with my graphical construction posted last week.) The apparent distance of this perigee, from the c.o.m., is 0.144 AU, but the real distance, based on the 75 degree line of sight inclination found in the construction, is 0.56 AU. For a circular orbit, this would be the radius of curvature of the real orbit. At the perigee of an elliptical orbit, the radius of curvature is greater. For a confocal parabola, the curvature is halved; considering the directrix & focus definition of a conic, let's multiply the radius of curvature by 1.6 for e=0.6, i.e. 0.90 AU. Then the apparent radius of curvature is 0.90/cos(75).5 AU. This is almost an apparent straight line, because the apparent length considered is < 0.5 AU, and 0.5/3.5 = 1/7 radian change in direction, or 0.5*(1/7)^2*3.5 = 0.036 AU deviation.

The apparent precession of the orbit due to parallax cancels some of this. The orbit segment is 1.44 AU * 0.4 * sin(75) = 0.56 AU in front of the c.o.m. In 0.112 * 2/5 = 0.045 radians of orbital travel from 1987 to 2008, its parallax is 0.025 AU, of which almost half, 0.011 AU, is perpendicular to the orbit. This cancels a third of the deviation, so the change in direction is only 1/10 radian = 6 degrees. The observed change was even closer to zero than this, but there is little need to assume that the orbit precesses, other than its apparent precession due to Barbarossa's solar orbit.

As noted in my "Do this yourself at home" post last week, the violation of Kepler's second law for the binary orbit between 1986 & 1987, could be due to a dimmer (as observed on 1954 & 1986 sky surveys) but more massive partner for Frey (to which we might shift Frey's name as well) within the modified Jacobi limit of obligate stability. This limit is about 1.6 AU * (0.11/81)^(1/3) = 0.18 AU. If the period is 0.87 * 2 = 1.74 yr, and this Frey/Freyprime orbit coplanar with the Barbarossa/Frey orbit (which implies that it is nearly edge on for us) Freyprime could be 0.18 AU (3 arcminutes) forward of Frey's expected position in 1986 and 0.18 backward in 1987. Correction for this is at least almost enough to repair Kepler's second law, because it would triple the area of the origin-1986-1987 triangle (which needs to be quadrupled, according to my rough graphical estimate above). The error in c.o.m. position would be as much as 180"/9 = 20". The implied mass of Frey (>> "Frey prime", the object we see) would be 0.18^3/1.74^2 = 0.0019 solar masses, which can be brought into agreement with the theoretical 0.0011 by skimping a little on the period, i.e., 150 deg separation instead of 180, is almost as effective at moving Freyprime, and gives 0.18^3/2.088^2 = 0.0013.

The 20" error in the c.o.m. position which this could cause, might be too big, but it plausibly is mostly removed by giving Barbarossa an inner moon, Freya, half as massive as Frey (1/18 as massive as Barbarossa) and close enough to orbit Barbarossa with the same period and sense that Freyprime orbits Frey. Freya would be 9^(1/3) = 2x as far from Barbarossa as Freyprime is from Frey. The distance between Barbarossa and the Freya/Barbarossa c.o.m., would be 1/18 * 2 = 1/9 the distance between Freyprime (let's assume its mass is negligible) and Frey. If Freyprime and Freya are in 0 deg phase (i.e., maximally avoiding each other) then the vector from Freyprime (the apparent mass position) to Frey (the actual mass position) is always -9x the vector from Barbarossa (the apparent mass position) to the Barbarossa/Freya c.o.m. (the actual mass position).

So, this plausible "solar system" for Barbarossa explains the apparent violation of Kepler's second law for the 1986-1987 interval of Frey's orbit. It also preserves the accuracy of the c.o.m. trajectory for the entire system.

Hi Joe, You do have to allow for the fact that all of these fits images have been reduced right down to the bone, just to allow people to download them. Not sure but I would think they would be in the 100 meg range in reality. Ask Iowa if they can crop out sections and send the smaller image uncompressed to you.

Another thought is that perhaps we could use hdri shop on the images and see if we get better results. An hdri image is a stacked f stop image. Say that you take a picture of the inside of a church. Take a bunch of f stopped shots. On some the windows will e washed out ut the darks will show detail, on others the windows will show detail but the darks will be jet black. hdri shop combines all of the images into one. The main man in regard to this stuff is called Paul Debevec, he might be worth e mailin as he does a lot of work for nasa, he even did work on the turin shroud.

Change the subject, have you looked at how the barycentre shift effects the orbit of Mercury?