Requiem for Relativity

Looking good Joe but now I think you need to contact your fraternity buddies to discuss the possible implications of that name. I'm sorry to bang on about it but I simply have to, it could sour the whole discovery. Freya (Freyja) and Frey (Freyjr) are okay and that would suggest the name of Odin for the brown dwarf. However, there is a general feeling that the world's mythologies are over represented by western gods and goddesses. Polynesian names are a possible, as are African, Inuit or Australian aboriginal names.

We also have to consider that this place might be a refuge for human beings, when our sun inflates into a giant. If we are still here in a few billion years time, then altering a brown dwarf to give off more light might well be possible. It definitely needs an optimistic name therefore.

The name Barbarossa, a psychotic, incompetent crusader, with only one admirer, Adolf Hitler. Barbarossa, the code name for the invasion of Russia. The other Barbarossa, is another psychotic, this time a particularly vicious pirate. I think that your friends from your university fraternity might not be too pleased with what any reasonably competent tabloid journalist could make from this. Making blood curdling oaths to the norse gods, whilst drinking vast amounts of strong lager is all when and good, when you are a nineteen year old student but such stories can become ammunition for your enemies in later years. An innocent frat group could be made to look like the vilest coven from hell.

I'll leave it there, it's up to you but I do think that it's only a matter of common courtesy to consult with your friends about the possible ramifications for them.

One of my fellow Harvard graduates, Berry Fleming (Berry with an "e"), wrote a humorous political novel, "Colonel Effingham's Raid", which became a 1945 Hollywood film starring Charles Coburn & Joan Bennett, directed by Irving Pichel. Fleming's book is dedicated "To the women of the town under whose wing - like slumbering Barbarossa in the Kyffhaueser caves - slumbers the valor of Fredericksville". Often people find it convenient to use the same language as the person who has the most to say.

Hi Joe, I'm at a loss as to what it is you're saying here. The king who sleeps, is king Arthur, the tradition of saying that any old king has this mythic property bestowed on them, is down to courtly toadying of the worst sort.

"Often people find it convenient to use the same language as the person who has the most to say." Perhaps, perhaps not. Much more common is the adage, where McCrimmon sits, is head of the table. What McCrimmon will say is, "Okay, we've got a new planet. We'll call it planet 9 for now. Find me a name that most people are happy with. I want this story to awaken the scientific imagination of a generation of kids. I don't want to see an unholy row about the name of the thing. If this guy that discovered it, looks like Brad Pitt and has the charm of Cary Grant, let him on t.v. as much as possible. If he wants to rock the boat and insist on his "right" to name it, then give him a medal and hustle him out of the frame as quickly as possible."

That's how it works Joe. It cannot be called unfair, it's simply realpolitic.

Frey's Orbit Around Barbarossa

Previously I calculated the Frey/Barbarossa orbit using the 1954, 1986 and 2007 points only, and I assumed a circular orbit. Today I included all five points, 1954, 1986, 1987, 1997 & 2007. Generally there will be a two-parameter family of ellipses in space, corresponding to five points on the celestial sphere. Generally only one of these ellipses will have Barbarossa at a focus.

Arbitrarily I chose the radius (something near 197 AU) from the sun to two of the points. An ellipse is a planar curve, and this coplanarity determines the radius to the other three points. Then I solved a 5x5 system of linear equations for the coefficients of the second-order Fourier sum, for radius as a function of theta (with Barbarossa at the origin). I evaluated all the possible ellipses according to three criteria:

1. The radii of the five points must be as monotonic as possible, that is, the cyclic sequence can switch only twice from increasing to decreasing.

2. The Fourier sum for the radius, must never become zero (impact into Barbarossa).

3. The relative standard deviation of the areal speeds for the four time intervals between the five points (i.e., std. deviation divided by mean) must be minimized.

The best fitting ellipse by these criteria had eccentricity 0.7162, so, retrospectively, the Fourier sum I used isn't very accurate. Pressing on anyway, the semimajor axis is 1.76 AU (vs. 1.2 for my circular estimate), the orbital period is 14.5 yr (vs. 14.7 for my circular estimate), and the implied mass of Barbarossa (very sensitive to semimajor axis) is 0.026 solar masses (vs. 0.008 for my circular estimate).

The relative standard deviation of areal speed, of the four time intervals, is 20% (there would be more than two full cycles between 1954 & 1986; the other intervals were less than one cycle each). Some of this might be from the inaccuracy of the Fourier sum for such large eccentricity. Some of the areal speed variation might be real, due to Frey having a moon of its own; or, Frey might have a quickly precessing, or even a chaotic orbit.

In this calculation, Frey's orbit is tilted only 9 degrees to Barbarossa's orbital plane around the sun (vs. 40 deg for my circular orbit estimate). The chance of such parallelism, between the two orbital planes, randomly happening over all angles on the celestial sphere, is only 1%.

If instead of a Fourier sum, an ellipse were fitted to the five points, then under ideal conditions, criteria #1-3 above always would be perfectly satisfied. (Kepler's area law holds for all tilts.) With large perturbations, no orbit satisfies criterion #3 and no orbit is an exact ellipse.

The Astrophysical Journal, 627:10011010, 2005

High Orbital Eccentricities of Extrasolar Planets Induced by the Kozai Mechanism

Genya Takeda and
Frederic A. Rasio

Abstract. "...at least 20% of [extrasolar] planets, including some with particularly high eccentricities, are orbiting a component of a wide binary star system. ...a large-amplitude eccentricity oscillation. This so-called Kozai mechanism is effective at a very long range...Our calculations show that the Kozai mechanism consistently produces an excess of planets with very high (~ >0.6) and very low (~ <0.1) eccentricities. Assuming an isotropic distribution of relative orbital inclination, we would expect that 23% of planets do not have sufficiently high inclination angles to experience the eccentricity oscillation [the oscillation between high inclination and high eccentricity - JK]. By a remarkable coincidence, only 23% of currently known extrasolar planets have eccentricities ~ < 0.1. ..."

Frey's Orbit Around Barbarossa (cont.)

Next, instead of approximating the orbit by the second-order Fourier sum for the radius (as a function of theta), today I did it by the second-order Fourier sum for the reciprocal of the radius (which, is exact at the five data points). This approximation should be better, because it is exact when Barbarossa is a focus.

This best fitting ellipse had eccentricity 0.6072. The semimajor axis is 1.1998 AU (vs. 1.2007 for my circular estimate), the orbital period is 13.36 yr (vs. 14.68 for my circular estimate), and the implied mass of Barbarossa+Frey is 0.00968 solar masses (vs. 0.00804 for my circular estimate; and 0.0103, for Barbarossa+Frey+remainder, based on the precession frequency ratio 3:1 surmised above for Neptune).

The relative standard deviation of areal speed, of the four time intervals, is 23.5%. If the ellipse is chosen whose focus is exactly at Barbarossa, the relative standard deviation of areal speed is 30%. The former ellipse is better, because angular momentum conservation is more fundamental than ellipse geometry.

In this calculation, Frey's orbit is tilted 34.5 degrees to Barbarossa's orbital plane around the sun (vs. 40.0 deg for my circular orbit estimate). My 5-pt. ellipse-based estimate of the orbital plane, is at a large angle to my earlier 3-pt. circle-based estimate. Following Jacobi, Tisserand and Kozai, sqr(1-e^2)*cos(i)= 0.655. If Frey's orbit originally were circular and inclined 60 deg to Barbarossa's (the median random angle), this constant would be 0.5.