- Thank you received: 0
Requiem for Relativity
- Joe Keller
- Offline
- Platinum Member
Less
More
16 years 8 months ago #20529
by Joe Keller
Replied by Joe Keller on topic Reply from
The Barbarossa Cloud's Atmospheric Refraction
From my Feb. 22, 2008 post (original version):
"Using the same mass ratio [0.8771:0.1229] as above, the heliocentric paths A2/A-C/C9-JG/SR and A2/A-D3/D-JG/SR are both straight (i.e., great circles) within 1 arcsec. These c.o.m. angular speeds also are constant."
The straightness of the great circles comes with a proviso:
This afternoon, on another online sky survey plate, I found a sixth Barbarossa/Frey pair:
geocentric coordinates (preliminary measurement, possibly several arcsec measurement error), Julian epoch 1995.1408 optical infrared sky survey (the same survey, but a different plate, from that showing D3 & D)
Barbarossa E3 11 21 23.7
-8 24 36
Frey E4 11 21 46.8
-8 20 01
heliocentric coordinates of c.o.m.
170.2690
-8.2910
Their center of mass is displaced roughly +1" along the constant angular speed great circle orbital path A2/A-JG/SR.
The presumed center of mass of B3/B is displaced only 2" perpendicularly downward from the straight great-circle path from A2/A to JG/SR. E3/E4 is displaced 94" perpendicular upward from that path. The other optical infrared pair, D3/D, is displaced 190" perpendicular downward (exactly twice as much). The visible light pair, C/C9, is displaced 88" perpendicular downward. Yet no pair (employing the c.o.m. above) errs more than 25" from constant angular speed along that path. A variable quantized refraction occurs, perpendicular to Barbarossa's orbit. A2/A, B3/B, and JG/SR are refracted the same amount, hence were found by me to lie on one great circle.
On average, Barbarossa's orbital path 1954-2007 differs 27.77 - 23.45 * cos(11.4) = 4.78 degrees from the ecliptic. A -90" displacement perpendicular to the ecliptic (e.g. from a large error calculating heliocentric coordinates) would cause 90" * sin(4.78) = +7.5" displacement along Barbarossa's orbit, but +19.5" is observed for C/C9. Such displacement of E3/E4 (~ +90" perpendicular displacement) should be -7.5" but is roughly +1". Such displacement of D3/D (~ -180" perpendicular displacement) should be +16" but is +25". B3/B, displaced only -2" perpendicularly, should be displaced +0.2", but is displaced +11.5", along the path. Barbarossa images C, D3, and JG are streaks several arcseconds long, oriented roughly along the orbital path. This ambiguity might account for some residual displacement along the path. However, if this error in calculating heliocentric coordinates is assumed to cause the large perpendicular discrepancies, the discrepancies along the path become +11.3", +12", +9", and +8.5" for B3/B, C/C9, D3/D, and E3/E4, resp.
A quantized light refraction perpendicular to the ecliptic, simulating an error in calculating heliocentric coordinates, thus explains practically all the position discrepancy. For the c.o.m. to be displaced 9" forward in 1997, and 12" forward in 1987, implies accelerations consistent within 10%, and which could be caused by an orbital eccentricity of 0.02 or greater.
A2/A, B3/B, and JG/SR also are the unique trio for which Earth's distance from the point of Barbarossa's opposition, is, to within a few percent, directly proportional to time. Thus a parallax calculation error proportional to Earth's distance from opposition to Barbarossa, leaves their line straight and constant-speed. Yet this does not explain why the error perpendicular to the path should be quantized, or why the error perpendicular to the path should be largest for D3/D, whose relation to opposition is much closer to the trend of A2/A-B3/B-JG/SR, than is such relation for either C/C9 or E3/E4.
The synchrony of A2/A, B3/B, and JG/SR might instead be due to rotation of the Barbarossa system's, or the entire solar system's, gas/dust cloud. The diminution in light, due to absorption and refraction, would explain why the brightness magnitudes of Barbarossa and Frey are so low. In previous posts I found that the comparison magnitudes were consistent with Barbarossa and Frey having the lowest albedos known, and even then, somewhat smaller than the most widely accepted theoretical diameters for bodies of their mass. Images of Barbarossa and Frey, whether on sky survey plate scans or on electronic photographs, always are either unusually small, or unround and unusually large.
From my Feb. 22, 2008 post (original version):
"Using the same mass ratio [0.8771:0.1229] as above, the heliocentric paths A2/A-C/C9-JG/SR and A2/A-D3/D-JG/SR are both straight (i.e., great circles) within 1 arcsec. These c.o.m. angular speeds also are constant."
The straightness of the great circles comes with a proviso:
This afternoon, on another online sky survey plate, I found a sixth Barbarossa/Frey pair:
geocentric coordinates (preliminary measurement, possibly several arcsec measurement error), Julian epoch 1995.1408 optical infrared sky survey (the same survey, but a different plate, from that showing D3 & D)
Barbarossa E3 11 21 23.7
-8 24 36
Frey E4 11 21 46.8
-8 20 01
heliocentric coordinates of c.o.m.
170.2690
-8.2910
Their center of mass is displaced roughly +1" along the constant angular speed great circle orbital path A2/A-JG/SR.
The presumed center of mass of B3/B is displaced only 2" perpendicularly downward from the straight great-circle path from A2/A to JG/SR. E3/E4 is displaced 94" perpendicular upward from that path. The other optical infrared pair, D3/D, is displaced 190" perpendicular downward (exactly twice as much). The visible light pair, C/C9, is displaced 88" perpendicular downward. Yet no pair (employing the c.o.m. above) errs more than 25" from constant angular speed along that path. A variable quantized refraction occurs, perpendicular to Barbarossa's orbit. A2/A, B3/B, and JG/SR are refracted the same amount, hence were found by me to lie on one great circle.
On average, Barbarossa's orbital path 1954-2007 differs 27.77 - 23.45 * cos(11.4) = 4.78 degrees from the ecliptic. A -90" displacement perpendicular to the ecliptic (e.g. from a large error calculating heliocentric coordinates) would cause 90" * sin(4.78) = +7.5" displacement along Barbarossa's orbit, but +19.5" is observed for C/C9. Such displacement of E3/E4 (~ +90" perpendicular displacement) should be -7.5" but is roughly +1". Such displacement of D3/D (~ -180" perpendicular displacement) should be +16" but is +25". B3/B, displaced only -2" perpendicularly, should be displaced +0.2", but is displaced +11.5", along the path. Barbarossa images C, D3, and JG are streaks several arcseconds long, oriented roughly along the orbital path. This ambiguity might account for some residual displacement along the path. However, if this error in calculating heliocentric coordinates is assumed to cause the large perpendicular discrepancies, the discrepancies along the path become +11.3", +12", +9", and +8.5" for B3/B, C/C9, D3/D, and E3/E4, resp.
A quantized light refraction perpendicular to the ecliptic, simulating an error in calculating heliocentric coordinates, thus explains practically all the position discrepancy. For the c.o.m. to be displaced 9" forward in 1997, and 12" forward in 1987, implies accelerations consistent within 10%, and which could be caused by an orbital eccentricity of 0.02 or greater.
A2/A, B3/B, and JG/SR also are the unique trio for which Earth's distance from the point of Barbarossa's opposition, is, to within a few percent, directly proportional to time. Thus a parallax calculation error proportional to Earth's distance from opposition to Barbarossa, leaves their line straight and constant-speed. Yet this does not explain why the error perpendicular to the path should be quantized, or why the error perpendicular to the path should be largest for D3/D, whose relation to opposition is much closer to the trend of A2/A-B3/B-JG/SR, than is such relation for either C/C9 or E3/E4.
The synchrony of A2/A, B3/B, and JG/SR might instead be due to rotation of the Barbarossa system's, or the entire solar system's, gas/dust cloud. The diminution in light, due to absorption and refraction, would explain why the brightness magnitudes of Barbarossa and Frey are so low. In previous posts I found that the comparison magnitudes were consistent with Barbarossa and Frey having the lowest albedos known, and even then, somewhat smaller than the most widely accepted theoretical diameters for bodies of their mass. Images of Barbarossa and Frey, whether on sky survey plate scans or on electronic photographs, always are either unusually small, or unround and unusually large.
Please Log in or Create an account to join the conversation.
- Joe Keller
- Offline
- Platinum Member
Less
More
- Thank you received: 0
16 years 8 months ago #20878
by Joe Keller
Replied by Joe Keller on topic Reply from
Accurate Frey/Barbarossa Orbit Confirms Refraction Phenomenon
I tried replacing D3/D with E3/E4, then finding the best Frey/Barbarossa orbit by the method I last described above. The best fit gives a standard deviation of only 4.56% for the areal speeds during the four time intervals. The eccentricity is a plausible 0.7401, the semimajor axis is 1.223 AU, and the period is 14.28 yr. The orbit is inclined 49.9 deg to Barbarossa's orbital plane.
This orbit implies that Barbarossa+Frey = 0.01024 solar masses. This agrees perfectly with the 0.0103 solar masses predicted for the entire Barbarossa system, according to the 3:1 Neptune precession resonance theory I discussed above.
From this best orbit, I removed E3/E4 and, maintaining the same orbital plane, added D3/D, leaving the other points on this orbit as they are. Then I searched for the shift in the RA & Decl. of D (Frey) which would minimize the standard deviation of the areal speed for the new orbit. The minimum standard deviation was 18.0%, achieved when D was moved 28.25" forward parallel to Barbarossa's path on the celestial sphere, but only 0.3" closer to the path. This suggests that D is not merely a random mistake, but rather that its apparent position is affected by a physical phenomenon. Also, 28.25" * pi = 89", approximately the quantum of deviation observed perpendicular to the path. This movement of D would bring the center of gravity D3/D forward +3.5".
Refractions through a slanted window pane cancel except from objects within the pane. I've found no extremely large proper motions of stars near Barbarossa/Frey in the photos, but according to the USNO-B catalog, Barbarossa's orbital path is enriched in catalog objects with unusually high (automatically calculated) proper motions. This is the phenomenon which caused me to investigate plates of this region.
I tried replacing D3/D with E3/E4, then finding the best Frey/Barbarossa orbit by the method I last described above. The best fit gives a standard deviation of only 4.56% for the areal speeds during the four time intervals. The eccentricity is a plausible 0.7401, the semimajor axis is 1.223 AU, and the period is 14.28 yr. The orbit is inclined 49.9 deg to Barbarossa's orbital plane.
This orbit implies that Barbarossa+Frey = 0.01024 solar masses. This agrees perfectly with the 0.0103 solar masses predicted for the entire Barbarossa system, according to the 3:1 Neptune precession resonance theory I discussed above.
From this best orbit, I removed E3/E4 and, maintaining the same orbital plane, added D3/D, leaving the other points on this orbit as they are. Then I searched for the shift in the RA & Decl. of D (Frey) which would minimize the standard deviation of the areal speed for the new orbit. The minimum standard deviation was 18.0%, achieved when D was moved 28.25" forward parallel to Barbarossa's path on the celestial sphere, but only 0.3" closer to the path. This suggests that D is not merely a random mistake, but rather that its apparent position is affected by a physical phenomenon. Also, 28.25" * pi = 89", approximately the quantum of deviation observed perpendicular to the path. This movement of D would bring the center of gravity D3/D forward +3.5".
Refractions through a slanted window pane cancel except from objects within the pane. I've found no extremely large proper motions of stars near Barbarossa/Frey in the photos, but according to the USNO-B catalog, Barbarossa's orbital path is enriched in catalog objects with unusually high (automatically calculated) proper motions. This is the phenomenon which caused me to investigate plates of this region.
Please Log in or Create an account to join the conversation.
- Joe Keller
- Offline
- Platinum Member
Less
More
- Thank you received: 0
16 years 8 months ago #20667
by Joe Keller
Replied by Joe Keller on topic Reply from
Ratios Support Barbarossa Cloud Refraction Phenomenon
The temperature of the CMB Planck curve is 2.726 K with an error of about 1 in 3000, as evidenced by the implied significance of the figure (1 in 5000) and also by occasional published statements that it is 2.725. (Elsewhere on this messageboard I explain why the CMB Planck curve must be imperfect due to contamination by far CIRB, causing two different definitions of best fit, which differ by 0.001 deg.) The best fit Planck curve to the sun ("Journey Through the Galaxy", filer.case.edu, 2006) has 5780 K, a figure whose implied accuracy is 1 in 1200.
Thus Tsun / TCMB is 2539.5 with accuracy 1 in 1000.
The 3-yr WMAP result for the CMB dipole is 3.358 mK with error bar 1 in 200.
Thus TCMB / halfdeltaTdipole is 811.79 with accuracy 1 in 200.
The ratio of the ratios, 2539.5/811.79 = 3.1283, equals pi=3.1416, to accuracy 1 in 200. This suggests a simple physical relation between the sun's temperature, the CMB temperature, and the CMB dipole temperature.
Since my discovery of Barbarossa and Frey, I can say more. The two deviation quanta perpendicular to the ecliptic, of centers of mass, on optical infrared photos (the DSS website says they were with an IVN plate and RG715 filter), average (assuming that they are quantized and that the larger deviation should be divided by two) 94.25" +/- range 0.75" = 1/2188 radian with 1% accuracy. The visible light photo (the DSS website says this was with a IIIaF plate and OG590 filter, equivalent to photographic passband R59F) shows perpendicular deviation approx. 88.2" = 1/2339 radian, whose accuracy may be guessed to be 1% also.
PASP 98:1303+, Fig. 1, gives the response functions of these Red & IR plates. Likely the quantized refraction angle ("rainbow angle") in radians, equals the ratio of the "average" frequency of CMB photons, to the "average" frequency of sunlight photons detected by the plate. This assumes that Barbarossa and Frey are seen only by reflected light and that their albedos are independent of wavelength, so that the reflected light approximates the sun's light.
For such ratios to occur with Planck-curve light, the energy-per-unit-wavelength Wien's law peaks would have to be given by
2539.5/2339*28970000 Angstrom/K (i.e., Wien's const.)/5780 K = 5442 A for the Red plate
2539.5/2188*28970000 A/5780 = 5817 A for the IR plate
vs. 28970000 A/5780 = 5012 A for sunlight.
I think that really the effective temperature, corresponds to that "average" energy per photon, which is weighted by energy. That is, if half the energy is in the form of photons of energy E1 apiece and half the energy is in the form of photons of energy E2 apiece, the effective energy per photon is (E1+E2)/2. The integrals for this calculation are in Dwight's "Tables of Integrals" 860.39; here Dwight refers to small integer values of the Riemann zeta function (in, inter alia, Dwight's other book, "Mathematical Tables").
By this definition, the average photon of sunlight is of wavelength 6494 Angstrom. From the size of the quantized refraction, the predicted average photon on the "Red" or "IR" plate is of wavelength 7051 or 7537 A, resp. My crude graphical estimate made by plotting the (wavelength) Planck curve of solar radiation, on the passband graph (Fig. 1, PASP 68:1304) and then roughly plotting the product function, is that the energy-weighted average photon energy for R59F exposed to sunlight, is that of a photon of wavelength 6300 A, and for IsubN (equivalent to IVN + RG715), 7800 A. So, the presumed refraction of IR plate objects D3/D & E3/E4 is roughly explained, but Red plate objects C/C9 seem to be refracted somewhat too much; redness (like classical Kuiper Belt objects) can explain only part of this discrepancy, because the R59F window cuts off at 7000 A.
The temperature of the CMB Planck curve is 2.726 K with an error of about 1 in 3000, as evidenced by the implied significance of the figure (1 in 5000) and also by occasional published statements that it is 2.725. (Elsewhere on this messageboard I explain why the CMB Planck curve must be imperfect due to contamination by far CIRB, causing two different definitions of best fit, which differ by 0.001 deg.) The best fit Planck curve to the sun ("Journey Through the Galaxy", filer.case.edu, 2006) has 5780 K, a figure whose implied accuracy is 1 in 1200.
Thus Tsun / TCMB is 2539.5 with accuracy 1 in 1000.
The 3-yr WMAP result for the CMB dipole is 3.358 mK with error bar 1 in 200.
Thus TCMB / halfdeltaTdipole is 811.79 with accuracy 1 in 200.
The ratio of the ratios, 2539.5/811.79 = 3.1283, equals pi=3.1416, to accuracy 1 in 200. This suggests a simple physical relation between the sun's temperature, the CMB temperature, and the CMB dipole temperature.
Since my discovery of Barbarossa and Frey, I can say more. The two deviation quanta perpendicular to the ecliptic, of centers of mass, on optical infrared photos (the DSS website says they were with an IVN plate and RG715 filter), average (assuming that they are quantized and that the larger deviation should be divided by two) 94.25" +/- range 0.75" = 1/2188 radian with 1% accuracy. The visible light photo (the DSS website says this was with a IIIaF plate and OG590 filter, equivalent to photographic passband R59F) shows perpendicular deviation approx. 88.2" = 1/2339 radian, whose accuracy may be guessed to be 1% also.
PASP 98:1303+, Fig. 1, gives the response functions of these Red & IR plates. Likely the quantized refraction angle ("rainbow angle") in radians, equals the ratio of the "average" frequency of CMB photons, to the "average" frequency of sunlight photons detected by the plate. This assumes that Barbarossa and Frey are seen only by reflected light and that their albedos are independent of wavelength, so that the reflected light approximates the sun's light.
For such ratios to occur with Planck-curve light, the energy-per-unit-wavelength Wien's law peaks would have to be given by
2539.5/2339*28970000 Angstrom/K (i.e., Wien's const.)/5780 K = 5442 A for the Red plate
2539.5/2188*28970000 A/5780 = 5817 A for the IR plate
vs. 28970000 A/5780 = 5012 A for sunlight.
I think that really the effective temperature, corresponds to that "average" energy per photon, which is weighted by energy. That is, if half the energy is in the form of photons of energy E1 apiece and half the energy is in the form of photons of energy E2 apiece, the effective energy per photon is (E1+E2)/2. The integrals for this calculation are in Dwight's "Tables of Integrals" 860.39; here Dwight refers to small integer values of the Riemann zeta function (in, inter alia, Dwight's other book, "Mathematical Tables").
By this definition, the average photon of sunlight is of wavelength 6494 Angstrom. From the size of the quantized refraction, the predicted average photon on the "Red" or "IR" plate is of wavelength 7051 or 7537 A, resp. My crude graphical estimate made by plotting the (wavelength) Planck curve of solar radiation, on the passband graph (Fig. 1, PASP 68:1304) and then roughly plotting the product function, is that the energy-weighted average photon energy for R59F exposed to sunlight, is that of a photon of wavelength 6300 A, and for IsubN (equivalent to IVN + RG715), 7800 A. So, the presumed refraction of IR plate objects D3/D & E3/E4 is roughly explained, but Red plate objects C/C9 seem to be refracted somewhat too much; redness (like classical Kuiper Belt objects) can explain only part of this discrepancy, because the R59F window cuts off at 7000 A.
Please Log in or Create an account to join the conversation.
- Joe Keller
- Offline
- Platinum Member
Less
More
- Thank you received: 0
16 years 8 months ago #20537
by Joe Keller
Replied by Joe Keller on topic Reply from
Please allow me to call attention to my important accurization today (March 2) of my Feb. 22 and Feb. 26, 2008, posts, above. When I accurized the positions today, finding the result when exactly correct Julian epochs are used throughout, I found that practically all discrepancy in center-of-mass Barbarossa/Frey position, disappears when a quantized refraction, ~ +/- n * 90", perpendicular, not to Barbarossa's orbit, but rather to the ecliptic, is assumed.
Please Log in or Create an account to join the conversation.
- Joe Keller
- Offline
- Platinum Member
Less
More
- Thank you received: 0
16 years 8 months ago #13359
by Joe Keller
Replied by Joe Keller on topic Reply from
Barbarossa's Refraction by the "Cloud"
The observed refractions of Barbarossa/Frey, in terms of the approx. 90" quanta perpendicular to the ecliptic (which is at practically the same inclination as Jupiter or Saturn's orbit, at this longitude) are 0, 0, -1, -2, +1, 0. Here, where there is quantization superposed on likely an underlying sinusoidal variation, normal distribution theory might not give very accurate error bars, but using normal distribution theory, the mean refraction is -(1/3) +/- 0.42 (std. error of the mean) quanta.
Assuming the other three visible light images would have had the same quantum of refraction as C/C9, the mean quantum of refraction, for the six images, is 1/2281 radian. Barbarossa's position in the six images averages 11.2 deg below the ecliptic. If the refraction is due to a thin disk or lenticule, containing the known solar system but not Barbarossa, and "n" is the effective index of refraction of the material within this solid disk, then most likely,
n-1 = 1/3 * 1/2281.3 * tan(11.185) = 1/34,612
By comparison, the constant 0.5 * alpha^2, where alpha is the fine structure constant, is 1 / 37,558. Some of my earlier posts on this messageboard explain the suggestive connections between this constant and the so-called CMB. If an outer lenticule exists, similar to this hypothetical inner one, but larger and containing Barbarossa, then refractions of distant stars would cancel, but the refraction of light from Barbarossa would not.
If I use the principal plane of the solar system (approximated as the mean of Jupiter's and Saturn's orbits, weighted by angular momentum) instead of the ecliptic, n-1 becomes 1/30,532. If I use the plane of Earth-moon angular momentum (including rotation)(averaged over one draconitic cycle)(including small corrections for barycentric motion and for average equinox) n-1 becomes 1/36,455, only 3% larger than 0.5*alpha^2 (here the precession of lunar nodes is assumed to cancel, because 1954-2007 contains almost three full cycles). More precisely, averaging the Earth-moon angular momentum vector over the six image times, gives 1/38,010, 1.2% smaller than 0.5*alpha^2.
As mentioned previously, D3/D shows a possible quantized displacement of Frey, relative to Barbarossa, parallel to the ecliptic. Not only is D3/D the image showing the largest (2 quanta) quantized displacement of the c.o.m.; also, D3/D is the image in which Frey's ecliptic latitude differs most from Barbarossa's.
The observed refractions of Barbarossa/Frey, in terms of the approx. 90" quanta perpendicular to the ecliptic (which is at practically the same inclination as Jupiter or Saturn's orbit, at this longitude) are 0, 0, -1, -2, +1, 0. Here, where there is quantization superposed on likely an underlying sinusoidal variation, normal distribution theory might not give very accurate error bars, but using normal distribution theory, the mean refraction is -(1/3) +/- 0.42 (std. error of the mean) quanta.
Assuming the other three visible light images would have had the same quantum of refraction as C/C9, the mean quantum of refraction, for the six images, is 1/2281 radian. Barbarossa's position in the six images averages 11.2 deg below the ecliptic. If the refraction is due to a thin disk or lenticule, containing the known solar system but not Barbarossa, and "n" is the effective index of refraction of the material within this solid disk, then most likely,
n-1 = 1/3 * 1/2281.3 * tan(11.185) = 1/34,612
By comparison, the constant 0.5 * alpha^2, where alpha is the fine structure constant, is 1 / 37,558. Some of my earlier posts on this messageboard explain the suggestive connections between this constant and the so-called CMB. If an outer lenticule exists, similar to this hypothetical inner one, but larger and containing Barbarossa, then refractions of distant stars would cancel, but the refraction of light from Barbarossa would not.
If I use the principal plane of the solar system (approximated as the mean of Jupiter's and Saturn's orbits, weighted by angular momentum) instead of the ecliptic, n-1 becomes 1/30,532. If I use the plane of Earth-moon angular momentum (including rotation)(averaged over one draconitic cycle)(including small corrections for barycentric motion and for average equinox) n-1 becomes 1/36,455, only 3% larger than 0.5*alpha^2 (here the precession of lunar nodes is assumed to cancel, because 1954-2007 contains almost three full cycles). More precisely, averaging the Earth-moon angular momentum vector over the six image times, gives 1/38,010, 1.2% smaller than 0.5*alpha^2.
As mentioned previously, D3/D shows a possible quantized displacement of Frey, relative to Barbarossa, parallel to the ecliptic. Not only is D3/D the image showing the largest (2 quanta) quantized displacement of the c.o.m.; also, D3/D is the image in which Frey's ecliptic latitude differs most from Barbarossa's.
Please Log in or Create an account to join the conversation.
- Joe Keller
- Offline
- Platinum Member
Less
More
- Thank you received: 0
16 years 8 months ago #12833
by Joe Keller
Replied by Joe Keller on topic Reply from
Yesterday I rechecked the 1983 SERC Blue sky survey, and again found no "disappearing dot" near the expected (geocentric) positions of Barbarossa or Frey. That is, I found no dot from 1983 Blue that wasn't also on 1987 Red, 1954 Red and/or 1997 Infrared, except for dots so faint that many of the other dots of similar faintness were absent too.
For a spectral type G2 star (e.g., the sun), B-R=1.06 and R-I=0.34 (astro.pas.rochester.edu, citing Mihalas & Binney's "Galactic Astronomy and also PASP 100:1134). So, if Barbarossa/Frey are colorless in sunlight, they would be about 1.06 mag dimmer (i.e., mag ~+20 instead of ~+19; the detection limit is ~+21) on the 1983 Blue plate than on the 1987 or 1954 Red plates, and about 1.40 mag dimmer than on the 1997 IR plate.
A famous early study of 21 distant solar system bodies, including classical KBOs, "1:2 objects", plutinos (i.e., "2:3 objects"), and centaurs, showed that B-V (data on 19 objects) ranged from 0.71 to 1.24 (mean 0.96) and V-R (data on all 21 objects) ranged from 0.37 to 0.78 (mean 0.625) (Tegler & Romanishin, Nature 407:979). So for Barbarossa/Frey, B-R is likely 1.6 and might unsurprisingly be as high as 2.0.
For a spectral type G2 star (e.g., the sun), B-R=1.06 and R-I=0.34 (astro.pas.rochester.edu, citing Mihalas & Binney's "Galactic Astronomy and also PASP 100:1134). So, if Barbarossa/Frey are colorless in sunlight, they would be about 1.06 mag dimmer (i.e., mag ~+20 instead of ~+19; the detection limit is ~+21) on the 1983 Blue plate than on the 1987 or 1954 Red plates, and about 1.40 mag dimmer than on the 1997 IR plate.
A famous early study of 21 distant solar system bodies, including classical KBOs, "1:2 objects", plutinos (i.e., "2:3 objects"), and centaurs, showed that B-V (data on 19 objects) ranged from 0.71 to 1.24 (mean 0.96) and V-R (data on all 21 objects) ranged from 0.37 to 0.78 (mean 0.625) (Tegler & Romanishin, Nature 407:979). So for Barbarossa/Frey, B-R is likely 1.6 and might unsurprisingly be as high as 2.0.
Please Log in or Create an account to join the conversation.
Time to create page: 0.393 seconds