The guest speaker on supernova, in conjunction with Professor Johnson, said that superluminal neutrinos were detected from a supernova. The reasoning was about how we detected neutrinos from the supernova before we detected its light. So we see neutrinos before we see the star go supernova.
The way I understand this is to think of time flowing at the speed of light. So neutrinos leave the star when it goes supernova, but they travel faster than the flow of time (really the light from the supernova) so they pass time and reach us first. But what if you were at the 'surface' of the star when it blew? Would the neutrinos still reach you first, or would you read both at the same time since neither was traveling any distance (I'm assuming neutrinos and photons have the same source location).
This is like when I first learned that infinity wasn't a number. My question here is, is superluminosity a property over distance? If neutrinos travel faster than light, shouldn't they reach you first despite what your distance from the source is? Even if that distance is 0, unless getting from point A to point A requires traveling some warped path.
Sunday, October 30, 2011
Wednesday, October 19, 2011
Calculating an Astronomical Unit
Main Author: David Vartanyan
Co-authors: John Pharo and Mee Wong-u
Abstract: We will attempt to find the distance between Earth and the sun, an astronomical unit, AU. For our purposes we will assume that an AU is equivalent to the semi-axis of Earth's orbit. Also assume that the sun's angular diameter is 0.5 degrees (half a thumb width approximately). We will also take for simplicity a low-orbit satellite to be in in contact with Earth's surface, radius r_o away from its center.
Introduction:
We extended the arc of the sun in our handout to form an entire circular sun. Then, we studied the periodic movement of Mercury across the surface of the sun. What we were actually seeing was a parallax effect resulting from the movement of the low-orbit satellite as it moved between the poles of the earth. By measuring the ratio of the peaks and troughs of the satellite's orbit to the diameter of the sun, we estimated the ratio of the angular diameter, /alpha, of Mercury's apparent oscillations.
We complete the circle of the sun and measure its diameter as around 20cm. The distance between peaks/troughs of Hg's 'oscillation' is 2mm, so:
and /alpha = .005 degrees, or 8.73e^-5 radians.
There was some controversy as to how we could define the angle subtended by Mercury's 'oscillations'. Initially our group was thinking of defining /beta as this half angle; however /beta gives half the angular diameter of mercurcy as seen from the sun.
Then we tried to define \theta as half the angle subtended by this motion. However we threw out this idea because /theta is equivalent to the angle formed between the line sight of the satellite and a tangent to earth's surface at the location of the satellite, and this fixes Mercury with respect to a distant star.
Rather, since the problem is dependent on the parallax in the Earth-Mercuy-Sun arrangement, we decided to fix the sun relative to Mercury and Earth, assuming without loss of generality that their centers are aligned, and defined /alpha as the twice the angle between the satellite's line of sight and a line connecting all 3 centers.
Calculations:
For our angle measurements (using small angle approximation for /beta and /theta) we have:
From our diagram you can see a triangle formed by these angles or their complements.
We have angle equivalence:
or 
We now explicitly define lengths:
where a_0 is the astronomical unit, a_H_g is the (mean) distance from Mercury to the sun, and \delta a is their difference, the distance from Earth to Mercury.
Substituting in our angle definitions into our angle equivalence and solving for \delta a, we get :
after substituting this into our length definition.
By assuming Kepler's third law for Earth and Mercurcy, we get:
where the constants have canceled in the ratio. T_o is 365 days and T_H_g is 87 days (we can keep period in days rather than seconds since their units cancel). Substituting in our relation between a_o and a_H_g and solving for a_o, we have:
We can calculate the sun's mass to by using Kepler's 3rd Law again but without constructing a ratio:
where M_s is the solar mass and G is the gravitational constant. We have neglected the mass of the earth because it is diminutive to that of the sun. Using the calculated value of the astronomical unit, we calculate M_s to be 4.4*10^29kg. An actual solar mass is about 2*10^30kg.
Defining our solar mass as about 0.2 of an actual solar mass,
we can easily prove by induction that the sun is massless.
Conclusions: The error in our calculated solar mass is due mainly to our approximation of an AU. However, Kepler's Law is also not an exact approach here because we assume that our distance to the sun is along the semi-major axis. But OoM.
Acknowledgements: I used Botox for encoding the equations and a personal scanner to upload my
picture. In drawing the pictures I used a #9 pencil on a 11.5x8.5 sheet of blank paper. The paper was acquired in a ream of 100 sheets from within a box containing 3 reams obtained from Office Depot. My ability to draw rudimentary geometric shapes was acquired over time from unclear sources.
Co-authors: John Pharo and Mee Wong-u
Abstract: We will attempt to find the distance between Earth and the sun, an astronomical unit, AU. For our purposes we will assume that an AU is equivalent to the semi-axis of Earth's orbit. Also assume that the sun's angular diameter is 0.5 degrees (half a thumb width approximately). We will also take for simplicity a low-orbit satellite to be in in contact with Earth's surface, radius r_o away from its center.
Introduction:
We extended the arc of the sun in our handout to form an entire circular sun. Then, we studied the periodic movement of Mercury across the surface of the sun. What we were actually seeing was a parallax effect resulting from the movement of the low-orbit satellite as it moved between the poles of the earth. By measuring the ratio of the peaks and troughs of the satellite's orbit to the diameter of the sun, we estimated the ratio of the angular diameter, /alpha, of Mercury's apparent oscillations.
We complete the circle of the sun and measure its diameter as around 20cm. The distance between peaks/troughs of Hg's 'oscillation' is 2mm, so:
and /alpha = .005 degrees, or 8.73e^-5 radians.
There was some controversy as to how we could define the angle subtended by Mercury's 'oscillations'. Initially our group was thinking of defining /beta as this half angle; however /beta gives half the angular diameter of mercurcy as seen from the sun.
Then we tried to define \theta as half the angle subtended by this motion. However we threw out this idea because /theta is equivalent to the angle formed between the line sight of the satellite and a tangent to earth's surface at the location of the satellite, and this fixes Mercury with respect to a distant star.
Rather, since the problem is dependent on the parallax in the Earth-Mercuy-Sun arrangement, we decided to fix the sun relative to Mercury and Earth, assuming without loss of generality that their centers are aligned, and defined /alpha as the twice the angle between the satellite's line of sight and a line connecting all 3 centers.
Calculations:
For our angle measurements (using small angle approximation for /beta and /theta) we have:
We have angle equivalence:
We now explicitly define lengths:
Substituting in our angle definitions into our angle equivalence and solving for \delta a, we get :
after substituting this into our length definition.
By assuming Kepler's third law for Earth and Mercurcy, we get:
where the constants have canceled in the ratio. T_o is 365 days and T_H_g is 87 days (we can keep period in days rather than seconds since their units cancel). Substituting in our relation between a_o and a_H_g and solving for a_o, we have:
and a_o 
9*10^12 cm. An actual astronomical unit is around 1.5*10^13 cm.
We can calculate the sun's mass to by using Kepler's 3rd Law again but without constructing a ratio:
where M_s is the solar mass and G is the gravitational constant. We have neglected the mass of the earth because it is diminutive to that of the sun. Using the calculated value of the astronomical unit, we calculate M_s to be 4.4*10^29kg. An actual solar mass is about 2*10^30kg.
Defining our solar mass as about 0.2 of an actual solar mass,
we can easily prove by induction that the sun is massless.
Conclusions: The error in our calculated solar mass is due mainly to our approximation of an AU. However, Kepler's Law is also not an exact approach here because we assume that our distance to the sun is along the semi-major axis. But OoM.
Acknowledgements: I used Botox for encoding the equations and a personal scanner to upload my
picture. In drawing the pictures I used a #9 pencil on a 11.5x8.5 sheet of blank paper. The paper was acquired in a ream of 100 sheets from within a box containing 3 reams obtained from Office Depot. My ability to draw rudimentary geometric shapes was acquired over time from unclear sources.
Avatar: The Last E&M-bender
So we had a great idea today during class. I don't remember how, but by some means F=ma came up in our conversation. When I heard this, I thought, what if there's a cartoon like Avatar: The Last Airbender, but instead of the 4 elements wind, water, earth, and fire, we have the four fundamental forces?
We assumed that each force's abilities are reflected in their reflex rates. For example, Gravitots cannot throw Strong Force into the sun before the Strong Force ignites Gravitots into a trillion nuclear explosions. The most practical (and coolest) of the four would be E&MnMs while the villain would undoubtedly be Strong Force. The weak force would be a telepathic handicapped character like Pf Xavier from X-men, but his psionic abilities would be range-limited.
In the second season we'd find out the the Weak Force is E&MnM's long lost-twin, and in the 3rd season all four would realize they're one and form the Unified Grand Force. I can't think of a plot conflict to continue the story beyond this.
We assumed that each force's abilities are reflected in their reflex rates. For example, Gravitots cannot throw Strong Force into the sun before the Strong Force ignites Gravitots into a trillion nuclear explosions. The most practical (and coolest) of the four would be E&MnMs while the villain would undoubtedly be Strong Force. The weak force would be a telepathic handicapped character like Pf Xavier from X-men, but his psionic abilities would be range-limited.
In the second season we'd find out the the Weak Force is E&MnM's long lost-twin, and in the 3rd season all four would realize they're one and form the Unified Grand Force. I can't think of a plot conflict to continue the story beyond this.
Saturday, October 15, 2011
Radiative Transfer: Optical Thickness
2. Main Author: David Vartanyan
Co-authors: Pharaoh John and Mee "not I" Zatarain
Abstract: We will categorize various paths as optically thin or thick. An optically thin path allows photons to transmit through, while an optically thick path absorbs and scatters photons.
The longer the pathlength of a photon through a scattering or absorbing media, the higher chance it will be scattered or absorbed.
a. The path from the center of the sun to us is optically thick since most of the photons will be scattered before they exit the sun. We cannot see the center of the sun since after traveling a certain depth into the sun's surface the sun becomes optically thick.
b. The path from the center of the sun to us for neutrinos is 'thin'. We based this off the fact that neutrinos were once a candidate for dark matter, and thus they qualify as dark, or non-interacting, and would not be scattered or absorbed as they traveled out of the sun.
c. Earth's atmosphere near peak frequency of CMB is optically thin since ground-based telescopes have been used to measure CMB homogeneity, so CMB radio-waves must penetrate earth's atmosphere. Also, we can hear satellite based XM radio-waves like Sirius, confirming earth's atmosphere is 'optically' thin to radio waves.
d. Earth's atmosphere is predominantly nitrogen and oxygen. Photons with energy of 13.7 eV (ionization energy of H) correspond to the first energy transition of an electron in hydrogen. Photons that hit hydrogen will excite the electron to the second energy level, the electron will jump down to the first level and reemit the photon. Hence hydrogen will scatter electrons at 13.7 eV. However, very little of the Earth's atmosphere is hydrogen so photons will pass through the atmosphere, which is correspondingly optically thin.
e. Cloudy days are generally darker, so clouds are optically thick. Clouds are mostly water vapor which scatters photons. Clouds from an above-earth view are white since the tops scatter, and dark from ground perspective since photons don't pass through.
f. Earth's atmosphere is optically thin for near-UV sunburn causing photons at the beach since we've all been sunburned on cloudy days in southern California.
g. Our clothes are optically thin for soft X-rays at airport scanners seeing as all the TSA officials in LAX have had the wonderful opportunity of seeing me naked.
h. Our bodies are mostly optically thick since soft X-rays at scanners don't penetrate through bone. However, we get a colored read-out with some hint of skeletal structure.
i. Our flesh is optically thin for medical X-rays since their purpose is to study bone structure, and we have endoskeletons.
j. Our bones are optically thick for medical X-rays. They are meant to study bone structure not to penetrate through them.
k. Random unidentified astrophysical objects with frequency distribution of blackbodies are optically thick since blackbodies are perfect emitters. Photons hitting them will not pass through but will be scattered.
Optically thin media:
Optically thick media:
Co-authors: Pharaoh John and Mee "not I" Zatarain
Abstract: We will categorize various paths as optically thin or thick. An optically thin path allows photons to transmit through, while an optically thick path absorbs and scatters photons.
The longer the pathlength of a photon through a scattering or absorbing media, the higher chance it will be scattered or absorbed.
a. The path from the center of the sun to us is optically thick since most of the photons will be scattered before they exit the sun. We cannot see the center of the sun since after traveling a certain depth into the sun's surface the sun becomes optically thick.
b. The path from the center of the sun to us for neutrinos is 'thin'. We based this off the fact that neutrinos were once a candidate for dark matter, and thus they qualify as dark, or non-interacting, and would not be scattered or absorbed as they traveled out of the sun.
c. Earth's atmosphere near peak frequency of CMB is optically thin since ground-based telescopes have been used to measure CMB homogeneity, so CMB radio-waves must penetrate earth's atmosphere. Also, we can hear satellite based XM radio-waves like Sirius, confirming earth's atmosphere is 'optically' thin to radio waves.
d. Earth's atmosphere is predominantly nitrogen and oxygen. Photons with energy of 13.7 eV (ionization energy of H) correspond to the first energy transition of an electron in hydrogen. Photons that hit hydrogen will excite the electron to the second energy level, the electron will jump down to the first level and reemit the photon. Hence hydrogen will scatter electrons at 13.7 eV. However, very little of the Earth's atmosphere is hydrogen so photons will pass through the atmosphere, which is correspondingly optically thin.
e. Cloudy days are generally darker, so clouds are optically thick. Clouds are mostly water vapor which scatters photons. Clouds from an above-earth view are white since the tops scatter, and dark from ground perspective since photons don't pass through.
f. Earth's atmosphere is optically thin for near-UV sunburn causing photons at the beach since we've all been sunburned on cloudy days in southern California.
g. Our clothes are optically thin for soft X-rays at airport scanners seeing as all the TSA officials in LAX have had the wonderful opportunity of seeing me naked.
h. Our bodies are mostly optically thick since soft X-rays at scanners don't penetrate through bone. However, we get a colored read-out with some hint of skeletal structure.
i. Our flesh is optically thin for medical X-rays since their purpose is to study bone structure, and we have endoskeletons.
j. Our bones are optically thick for medical X-rays. They are meant to study bone structure not to penetrate through them.
k. Random unidentified astrophysical objects with frequency distribution of blackbodies are optically thick since blackbodies are perfect emitters. Photons hitting them will not pass through but will be scattered.
Optically thin media:
Optically thick media:
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