Requiem for Relativity

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17 years 6 months ago #19752 by Joe Keller
Replied by Joe Keller on topic Reply from
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote"><i>Originally posted by Bill_Smith</i>
<br />Looks like the forum has gone a little haywire with duplication.

About the magnitudes, so this is why the object is at or near the limit of Bobs images. I think I quoted mag 18.5 to Bob when I measured them so it leaves a large gap between 18.5 and 20.

If the object is a brown dwarf would you be relying on albedo only?

Cheers

Bill

<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">

Hi Bill!

My May 20 posts address the self-illumination issue. Thanks for mentioning it!

- Joe

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17 years 6 months ago #19465 by Joe Keller
Replied by Joe Keller on topic Reply from
Because IRAS did not find it, Barbarossa must be much colder than Burrows' theoretical temperature (or else much less massive than believed). Burrows et al, Reviews of Modern Physics 65:301+, gives Table I (p. 316) and formula 2.58 (p. 312; see also p. 305 for a very brief explanation of the "Rosseland mean opacity") for theoretical brown dwarf temperature as a function of age and mass. Because Barbarossa's presumed 0.0090 solar mass is outside Table I, I used Burrows' formula to find (assuming age 4.6 Gyr) that Barbarossa's surface temperature should be 229K. By Wien's law, the peak emission would be at 12.65 microns.

The limit for deuterium fusion, is said to be 0.013 solar mass (Oppenheimer et al, in: Mannings et al, "Protostars & Planets IV", 2000), but deuterium fusion lasts only briefly, early in the life of the brown dwarf. At a Gyr or more old, Burrows' temperature-mass chart changes little across the 0.013 solar mass boundary (Burrows et al, 1997, in: Mannings, op. cit., Color Plate 22); so the above surface temperature formula remains fairly accurate. (Gravitational potential energy is about as important as deuterium fusion.) This chart shows that the absolute luminosity of Barbarossa should be 7.25 log10 units less than the sun's, thus its apparent (full spectrum) luminosity 7.25 + 4.6 = 11.85 log10 units less.

The infrared radiation measured by IRAS from a 229K Barbarossa (assuming "Burrows size" for its mass, i.e., approx. Jupiter size) would be roughly the same, in spectrum and intensity, as that measured from an asteroid 3 AU from the sun and 1000 mi in diameter. IRAS cataloged many asteroids much smaller than this.

"Asteroids and comets moving more slowly than 1' per hour would hours-confirm and thus reside in the Working Survey Database."

- NASA IRAS Asteroid & Comet Survey webpage

From a 229K Burrows-size (approx. Jupiter-size) Barbarossa, IRAS would have measured a 5900 Jansky peak at 12 microns. Within one degree of Barbarossa's expected position, the main IRAS catalog's brightest object at 12 microns, measured 6 Jansky; within 10 degrees, 82 Jy.

If a Burrows-size Barbarossa had a rather Neptune-like surface temperature of 48K, its radiation peak would be 55 Jy at 60 microns. Within one degree of Barbarossa's expected position, the brightest object at 60 microns is 1 Jy; within 10 deg, 27 Jy.

With no internal heat at all, the surface temperature would be about 18K (equilibrium at Barbarossa's distance from the sun; only slightly warmer than the cosmic far infrared background). Barbarossa either would be indistinguishable from background cold interstellar dust; or, if Barbarossa were seen against a 3K background, IRAS would record only 2.1 Jy even at 100mu, 0.5 Jy at 60mu, and 0 at 25 & 12mu. The lowest readings ever found are about 0.7 Jy @ 100mu, 0.3 Jy @ 60mu, and 0.2 Jy @ 25 & 12mu. So, I looked for readings within a factor of two, of 2.8 Jy @ 100mu, 0.8 Jy @ 60mu, and 0.2 Jy @ 25 & 12mu. There were 76 such IRAS objects within 10 degrees, but none within one degree, of Barbarossa's expected position. If Barbarossa's diameter were half the Burrows value, none of the brightnesses should be much above noise.

So, only a completely cold (equilibrium with solar radiation), smallish (~ 1/2 Jupiter diam) Barbarossa is readily consistent with IRAS' negative detection, given my coordinates. If my coordinates are wrong, then many IRAS objects (about 0.2 per sq degree) are consistent with even a Jupiter-size Barbarossa, if it is 18K. Some IRAS objects have been correlated with known objects and some haven't.

Burrows assumes that cooling is limited by the rate of radiation through the degenerate matter, and that convection is insignificant. On the contrary: most of the gravitational energy will be released at the boundary where the nondegenerate mantle is collapsing onto the degenerate core. The nondegenerate overlying mantle would melt and convection would carry the heat to the surface, as on Earth. Thus Burrows' temperatures might be accurate for dwarfs that have burned deuterium in their cores, but might drastically overestimate the temperature of sub-dwarfs like Barbarossa whose heat is only gravitational.

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17 years 6 months ago #19467 by Joe Keller
Replied by Joe Keller on topic Reply from
Pursuant to Kozai's and similar expressions, I found arcsin(rms sin(i)), the "rms sin(i) inclination to the ecliptic" of KBO's: it is 12# 13.5' (standard error of mean, 41')(n=220)(SEM questionable due to skew distribution). I used the top third (i.e., discovery years 2003-2006) of the "general ephemeris" list on ifa.hawaii.edu. The bottom quarter of the list, KBOs with special names, also contained some 2003-2006 discoveries but I omitted those. The absolute magnitudes of the KBOs I used were mostly +6 to +8 (vs. +3 for Varuna).

This agrees with the orbital inclinations of Barbarossa (12# 9.7') and Lescarbault/LeVerrier's Vulcan (12# 10'). Iowa State Univ. has LeVerrier's Compte Rendus communication (January 2, 1860; LeVerrier entered Lescarbault's letter to him on p. 40 and his own comment on p. 45). I can't read even scientific French well, but it seems to me that LeVerrier discussed no other sightings; he simply without fanfare corrected Lescarbault's orbital inclination calculation. The subsequent two years of Compte Rendus contained many communications by LeVerrier, but only about other subjects.

I found that the ascending nodes, omega, of the same KBOs, cluster near 14.00# and 194.00# (n=175). I omitted KBOs with i &lt; 3# because for these, the ascending node on the ecliptic correlates poorly with the ascending node on Jupiter's or Saturn's orbit. Otherwise all KBOs were weighted equally. The best fit, theta, was defined as minimizing the sum of abs(sin(omega - theta)); the criterion should emphasize the difference between 10# & 20# error more than the difference between 40# & 50#, or 70# & 80# error. This agrees with the ascending nodes of Lescarbault/LeVerrier's Vulcan (named by Babinet) (12.98#) and, except for a 90 degree shift, of Barbarossa (283.69#).

The eccentric (e &gt; 0.1) subset of KBOs (these also tend to be the ones with larger inclination) significantly clusters near omega = 20# and 200#. Again omitting those with i &lt; 3#, 40/117 had omega (rounded to the nearest degree) within the inclusive intervals [1,40] or [181,220] (p = 0.076, Poisson test). This clustering occurs despite the tendency, for small i, for omega to cluster near the ascending nodes of Jupiter & Saturn, which are at right angles to this. That is, the clustering would be seen to be even more significant if reference were made to the orbital plane of Jupiter/Saturn instead of to the ecliptic.

Using the Jupiter/Saturn reference plane instead of the ecliptic, would increase Barbarossa's inclination almost two degrees, and move the ascending nodes of Vulcan and the clustered KBOs, backward about seven degrees. Even that, would be close enough agreement to suggest that Barbarossa influenced the orbits of Vulcan (whatever Vulcan was) and the KBOs. All this is more reason to turn a big telescope there and look.

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17 years 6 months ago #19755 by Joe Keller
Replied by Joe Keller on topic Reply from
From the chart of KBO semimajor axes on Prof. Jewitt's Kuiper Belt website on ifa.hawaii.edu, one sees that some KBOs, like Pluto (hence called plutinos), cluster at 39.4 AU where their orbits have 3:2 resonance with Neptune's orbit. On this chart, KBOs scarcely occur with semimajor axes 47.8 AU where there would be 2:1 resonance with Neptune. Also the 5:3 resonance distance, 42.3 AU, intersects the distribution well off to one side: nothing peaks there.

On Prof. Jewitt's chart, the median and the mode of the distribution of semimajor axes, of the non-plutino majority set of KBOs (and of both the low- and high-eccentricity subsets separately) is about 43.5 AU. The eccentricity minimum of the high-eccentricity subset, is about 43 AU. (The high-eccentricity subset becomes more eccentric with distance; if sqrt(1-e^2) increases linearly with distance, then e=1 at about 50.5 AU.) If not resonance, what is at 43.5 AU, to attract the KBOs?

The distance at which the Barbarossa system torques KBO orbits, as effectively as the remainder of the solar system torques them, is 43.75 AU. The strength of the CMB dipole implies, from my theory above, that the Barbarossa+Frey system is 0.0104 solar mass. I totaled the torque due to all significant known solar system mass, including Pluto and Charon but excluding any other KBOs. The effect of Pluto is such that if plutinos equal to 100x Pluto's mass were added, the distance would change to 43.2 AU (for Pluto & plutinos I approximated them as circular orbits at their semimajor axis).

I used Gauss' idea of calculating (by Romberg trapezoidal numerical integration) simply the torque on Ring A due to Ring B, where Ring B is the orbit of a planet or Barbarossa. When Ring A is at 43.75 AU, then Barbarossa, vs. the rest of the solar system, cause equal torques, per degree of tilt.

Suppose all the KBOs originally lay in the Jupiter/Saturn plane. If Barbarossa's mass were negligible compared to Jupiter, then all the KBOs would have i=0 in the J/S plane (i.e., i=0 to 2 in the ecliptic plane) forever. If Jupiter's mass were negligible compared to Barbarossa, then the KBOs would precess about Barbarossa's plane, which is inclined 14# to the J/S plane, giving i=0 to 28. Because really the effect of J/S/N et al, at 43.75 AU, equals that of Barbarossa/Frey, the observed situation is a compromise. Roughly half the non-plutino KBOs are found near i=0 and roughly half are spread out between i=0 to 28.

On my list of 220 unnamed KBOs discovered 2003-2006 (it includes some plutinos), I counted 15 at inclinations [18,21], 11 at [22,25], 6 at [26,29], 4 at [30,33], and one apiece at 36, 37, & 48 (this latter surprisingly with eccentricity only 0.13). This hints at the 39 degree limit beyond which chaotic variation of eccentricity and inclination theoretically occur. It also shows that an exponential dropoff in population begins at about i=25.

In the above sense, Jupiter is an order of magnitude more effective than Barbarossa, at torquing Saturn. Thus Saturn's inclination to Jupiter is an order of magnitude less than its inclination to Barbarossa. Uranus should, on average, have larger inclination than it does, but this could be a chance time in the precession cycle when the inclination is rather small. Neptune is a paradox. Barbarossa should torque Neptune about as effectively, in the above sense, as do J/S/U. Yet Neptune is only 0.9# from the J/S plane. (Like a good KBO, Neptune's ascending node on the J/S plane is near right angles to Barbarossa's.)

Maybe the aberrant axial rotation of Uranus, allows Uranus to exchange orbital angular momentum with Neptune by some new physics. Uranus would be the linchpin holding Neptune in the plane of the ecliptic. Alternatively Neptune might simply be analogous to the low-inclination half of KBOs that seem to be influenced by J/S rather than by Barbarossa: dynamically there seems to be an all-or-nothing choice of influences.

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17 years 6 months ago #19470 by Stoat
Replied by Stoat on topic Reply from Robert Turner
Hi Joe, have you looked at Eris and Sedna yet? Their perihelion looks to be about 180 degrees from our brown dwarf. Huge eccentricities, so we can find them now as they come in close. That suggests that there are others, of about the same mass way way out there on highly eccentric orbits.

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17 years 6 months ago #19471 by Joe Keller
Replied by Joe Keller on topic Reply from
<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 />Hi Joe, have you looked at Eris and Sedna yet? Their perihelion looks to be about 180 degrees from our brown dwarf. Huge eccentricities, so we can find them now as they come in close. That suggests that there are others, of about the same mass way way out there on highly eccentric orbits.
<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">

Thanks again for your input! One of my posts above cites a plot of Trans-Neptunian Objects known as of 1998. Reviewing my notes, I see that these were shifted, in my best analysis, roughly 5.9 AU toward ecliptic longitude 182. That is, the average perihelion was at 002, 78 degrees from Barbarossa's ascending node, 284.

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