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User:Marshallsumter/Sources/Astronomy

Source astronomy has its origins in the actions of intelligent life on Earth when they noticed things or entities falling from above and became aware of the sky. Sometimes what they noticed is an acorn or walnut being dropped on them or thrown at them by a squirrel in a tree. Other events coupled with keen intellect allowed these life forms to deduce that some entities falling from the sky are coming down from locations higher than the tops of local trees. It may have taken awhile to realize that the sky is penetrable.

Direct observation and tracking of the origination and trajectories of falling entities such as volcanic bombs presented early intelligent life with vital albeit sometimes dangerous opportunities to compose the science that led to source astronomy.

"Where did it come from?" is the question.

Answering this fundamental question is at the heart of source astronomy. Radiation incoming to the detector may have been altered just after it left the source, just before it arrived at the detector, or anywhere in between. Occasionally, contamination occurs within the detector that can produce spurious conclusions.

"What is the source?"

Minerals
"A time-dependent model of the solar nebula is used to describe the outward transport of hot mineral aggregates from locations in the warm inner regions of the nebula under the influence of photophoresis."

Theoretical source astronomy
Def. an entity from which something comes or is acquired is called a source.

Def. a source or apparent source detected or created at or near the time of the event or events is called a primary source.

Def. a source or apparent source that transforms or transduces anything originating from a primary source is called a secondary source.

Def. a source or apparent source that selects (such as through selective absorption), distills, scatters, or reflects anything from a primary or secondary source is called a tertiary source.

Def. the point of origin of a ray, beam, or stream of small cross section traveling in a line is called a radiation source.

"Def. a natural source in the sky especially at night is called an astronomical source.", after the definition of astronomical source in theoretical astronomy.

Early observations of origination, tracking, and recovery of entities are the proof of concept that entities falling from the sky have an origin or source.

Entities
Def.
 * 1.a: an "independent, separate, or self-contained [astronomical] existence",
 * 1.b: "the [astronomical] existence of a thing as contrasted with its attributes", or
 * 2. "some [astronomical] thing that has separate and distinct existence and objective or conceptual reality",

is called an astronomical entity.

Skies
Def. an "expanse of space that seems to be [overhead] like a dome" is called the sky, or sometimes the heavens.

This definition applies especially well to an individual on top of the Earth's solid crust looking around at what lies above and off to the horizon in all directions. Similarly, it applies to an individual's visual view while floating on a large body of water, where off on the horizon is still water.

A more general definition of 'sky' allows for skies as seen on other worlds.

Is the sky a source or an obstruction between the observer and a source?

The same question may be asked when looking downward at the Earth's crust, regolith, or water below the observer.

The sky may have been thought of at one time as impenetrable. The same assumption may have occurred regarding the Earth below.

Is the sky a mirror?

The sky, also known as the celestial dome, commonly refers to everything that lies a certain distance above the surface of Earth, including the atmosphere and the rest of outer space. In the field of astronomy, the sky is also called the celestial sphere. This is an imaginary dome where the sun, stars, planets, and the moon are seen to be traveling. The celestial sphere is divided into regions called constellations. Usually, the term sky is used from the point of view of the Earth's surface. However, the exact meaning of the term can vary; in some cases, the sky is defined as only the denser portions of the atmosphere, for example.

Celestial spheres
In astronomy and navigation, the celestial sphere is an imaginary sphere of arbitrarily large radius, concentric with the Earth and rotating upon the same axis. All objects in the sky can be thought of as projected upon the celestial sphere. Projected upward from Earth's equator and poles are the celestial equator and the celestial poles. The celestial sphere is a very practical tool for positional astronomy.

The celestial sphere is divided into 89 areas. These are the International Astronomical Union (IAU) 88 modern constellations of astronomy. Although there are only 88 IAU constellations, the sky is actually divided into 89 irregularly shaped boxes as the constellation Serpens is split into two separate sections, Serpens Caput (the snake's head) to the west and Serpens Cauda (the snake's tail) to the east. From each of these areas of the celestial sphere, incoming radiation have been detected above the Earth's atmosphere and occasionally through it.

Celestial equators
The celestial equator (in hours of longitude) passes through these constellations (RA): Pisces (1), Cetus (~2), Taurus (4+), Eridanus (3-), Orion (5), Monoceros (7), Canis Minor (8), Hydra (8 to 15), Sextans (10), Leo (11), Virgo (13), Serpens (16,18), Ophiuchus (17), Aquila (20), and Aquarius (23).

Celestial poles
In the first diagram at right, the Earth is shown as seen from the Sun. The north and south celestial poles, their relation to the Earth's axis of rotation, the plane of the Earth's orbit around the Sun, and Earth's axial tilt are shown.

The second diagram at right shows the path of the celestial north pole around the ecliptic north pole. The beginning of the four "astrological ages" of the historical period are marked with their zodiacal symbols: the Age of Taurus from the Chalcolithic to the Early Bronze Age, the Age of Aries from  the Middle Bronze Age to Classical Antiquity, the Age of Pisces from Late Antiquity to the present, and the Age of Aquarius beginning in the mid 3rd millennium.

The first image at left shows that over the course of an evening, stars appear to rotate about the north celestial pole when viewing in the northern hemisphere of Earth. Polaris, within a degree of the pole, is the single nearly-stationary star just to the right of the centre of this image.

The second image at left shows the star trails during the course of an evening around the south celestial pole as viewed from the southern hemisphere of Earth. The image shows these star rotations around the celestial South Pole over the Very Large Telescope.

The third diagram at right shows how to locate the south celestial pole. Sigma Octantis is currently designated as the south celestial polar star.

Autumnal equinoxes
The vernal equinox is the point of equal Sun light and darkness (no Sun visible). Twelve hours occurs for each. The date is about March 22nd.

Ecliptics
The Earth in its orbit around the Sun causes the Sun to appear on the celestial sphere as moving along the ecliptic (red, in the diagram at right), which is tilted with respect to the celestial equator (blue-white).

The International Astronomical Union's constellations are shown at left in a equirectangular plot of declination vs right ascension of the modern constellations with a dotted line denoting the ecliptic. Constellations are colour-coded by family and year established. Note that the right ascension axis increases from right to left to approximate the view of the night sky.

Zodiacs
The Sun travels through the 13 constellations along the ecliptic, the 12 of the Zodiac and Ophiuchus.

Constellations
The International Astronomical Union provides figures of each constellation and coordinates for the boundaries which can be used to determine the constellational location of a celestial object using its right ascension (RA) and declination (Dec). As the position of these constellations has shifted due to Earth orbit precessions, there is an approximate four minute increase between the right ascension boundaries of epoch 1950 B1950.0 and the most recent calibration of epoch 2000 J2000.0 and a few minutes of change in many declinations.

Milky Way
The center of the galaxy is in the direction of Sagittarius, and the Milky Way "passes" (going westward, Earthview, to the right) through Scorpius, Ara, Norma, Triangulum Australe, Circinus, Centaurus, Musca, Crux, Carina, Vela, Puppis, Canis Major, Monoceros, Orion & Gemini, Taurus, Auriga, Perseus, Andromeda, Cassiopeia, Cepheus & Lacerta, Cygnus, Vulpecula, Sagitta, Aquila, Ophiuchus, Scutum, and back to Sagittarius.

In the diagram at right, spherical triangles are shown for deriving the relationship between equatorial and galactic coordinate systems. The yellow plane is the celestial equator, P and P' are its celestial poles. The green plane is the galactic plane, G and G' are its galactic poles. B is the direction towards the galactic center, which is now used as a starting point for measuring galactic longitude. Prior to 1959, the starting point was the intersection of the galactic and equatorial planes (point C).

Quadrants
In the table of locations by quadrants that follows, ranges of RA, Dec are included, e.g. (23-02, +23 to +52), to help make a fast assessment whether a particular source is in or borders a specific constellation. Questionable locations near a border require consulting the IAU website.

Andromeda
The figure at right shows the sky map of Andromeda. Around the edges of the map are coordinates related to longitude and latitude.

Antlia
The figure at right shows the sky map of Antlia. Around the edges of the map are coordinates related to longitude and latitude.

Apus
The figure at right shows the sky map of Apus. Around the edges of the map are coordinates related to longitude and latitude.

Aquarius
The figure at right shows the sky map of Aquarius. Around the edges of the map are coordinates related to longitude and latitude.

Aquila
The figure at right shows the sky map of Aquila. Around the edges of the map are coordinates related to longitude and latitude.

Ara
The diagram at right is the International Astronomical Union's official sky chart for the constellation Ara with notable sources marked and located.

At left is a three-colors image of NGC 6193 and NGC 6188 obtained with the Curtis-Schmidt telescope at Cerro Tololo Inter-American Observatory (Chile). The red channel is ionized Sulfur, green channel ionized Hydrogen, and the blue channel is double ionized Oxygen.

Aries
At left is the sky chart of the constellation Aries from the IAU.

Auriga
At right is a map of the constellation Auriga, with the surrounding constellations shaded in gray. Vertical coordinates are in declination while horizontal coordinates are right ascension. The green lines outline the constellation's conventional asterism, while the black dots show the location and apparent magnitude of the visible naked eye stars. The heavy blue line bisecting the constellation represents the Galactic Plane of the Milky Way. The lighter blue line to the south of the constellation is the ecliptic.

At left in the near infrared is the planetary nebula IC 2419 in Auriga.

Boötes
The figure at right shows the sky map of Boötes. Around the edges of the map are coordinates related to longitude and latitude.

Caelum
The figure at right shows the sky map of Caelum. Around the edges of the map are coordinates related to longitude and latitude.

Camelopardalis
The official IAU sky chart for the constellation Camelopardalis is at right with notable sources marked and located.

"A bright star [in the image at left] is surrounded by a tenuous shell of gas in this unusual image from the NASA/ESA Hubble Space Telescope. U Camelopardalis, or U Cam for short, is a star nearing the end of its life. As it begins to run low on fuel, it is becoming unstable. Every few thousand years, it coughs out a nearly spherical shell of gas as a layer of helium around its core begins to fuse. The gas ejected in the star’s latest eruption is clearly visible in this picture as a faint bubble of gas surrounding the star."

"U Cam is an example of a carbon star. This is a rare type of star whose atmosphere contains more carbon than oxygen. Due to its low surface gravity, typically as much as half of the total mass of a carbon star may be lost by way of powerful stellar winds."

"Located in the constellation of Camelopardalis (The Giraffe), near the North Celestial Pole, U Cam itself is actually much smaller than it appears in Hubble’s picture. In fact, the star would easily fit within a single pixel at the centre of the image. Its brightness, however, is enough to overwhelm the capability of Hubble’s Advanced Camera for Surveys making the star look much bigger than it really is. The shell of gas, which is both much larger and much fainter than its parent star, is visible in intricate detail in Hubble’s portrait. While phenomena that occur at the ends of stars’ lives are often quite irregular and unstable (see for example Hubble’s images of Eta Carinae, potw1208a), the shell of gas expelled from U Cam is almost perfectly spherical."

"The image was produced with the High Resolution Channel of the Advanced Camera for Surveys [using the 606 nm and 814 nm filters]."

Cancer
The figure at right shows the sky map of Cancer. Around the edges of the map are coordinates related to longitude and latitude.

Canes Venatici
At right is the IAU official sky chart for the constellation Canes Venatici with notable sources marked and located.

The left image is of the Whale galaxy [NGC 4631] in Canes Venatici. The image is a visual record using a backyard telescope. For comparison an ultraviolet image is just below it from the GALEX satellite.

Canis Major
The figure at right shows the sky map of Canis Major. Around the edges of the map are coordinates related to longitude and latitude.

Canis Major [...] is one of the 88 modern constellations, and was included in the 2nd-century astronomer Ptolemy's 48 constellations. [...] Canis Major is a constellation in the southern hemisphere's summer (or northern hemisphere's winter) sky [...] The three-letter abbreviation for the constellation, as adopted by the International Astronomical Union in 1922, is 'CMa'. The official constellation boundaries, as set by Eugène Delporte in 1930, are defined by a polygon of 4 sides. In the equatorial coordinate system, the right ascension coordinates of these borders lie between 06:12.5:0 and 07:27.5:0, while the declination coordinates are between -11.03° and −33.25°. Covering 380 square degrees, it ranks 43rd of the 88 constellations in size.

At left is a visual image of Sirius, the brightest star as seen from Earth, apart from the Sun.

Around 150 AD, the Hellenistic astronomer Claudius Ptolemy described Sirius as reddish, along with five other stars, Betelgeuse, Antares, Aldebaran, Arcturus and Pollux, all of which are clearly of orange or red hue. The discrepancy was first noted by amateur astronomer Thomas Barker, ... who prepared a paper and spoke at a meeting of the Royal Society in London in 1760. The existence of other stars changing in brightness gave credence to the idea that some may change in colour too; Sir John Herschel noted this in 1839, possibly influenced by witnessing Eta Carinae two years earlier. Thomas Jefferson Jackson See resurrected discussion on red Sirius with the publication of several papers in 1892, and a final summary in 1926. He cited not only Ptolemy but also the poet Aratus, the orator Cicero, and general Germanicus as colouring the star red, though acknowledging that none of the latter three authors were astronomers, the last two merely translating Aratus' poem Phaenomena. Seneca, too, had described Sirius as being of a deeper red colour than Mars. However, not all ancient observers saw Sirius as red. The 1st century AD poet Marcus Manilius described it as "sea-blue", as did the 4th century Avienus. It is the standard star for the color white in ancient China, and multiple records from the 2nd century BC up to the 7th century AD all describe Sirius as white in hue.

In 1985, German astronomers Wolfhard Schlosser and Werner Bergmann published an account of an 8th century Lombardic manuscript, which contains De cursu stellarum ratio by St. Gregory of Tours. The Latin text taught readers how to determine the times of nighttime prayers from positions of the stars, and Sirius is described within as rubeola — "reddish". The authors proposed this was further evidence Sirius B had been a red giant at the time.

Canis Minor
At left is the International Astronomical Union official sky chart for the constellation Canis Minor with notable sources marked and located.

The right image is from the Hubble Space Telescope of a region that includes Orion at the right, Monoceras, Canis Major, and Canis Minor. Located within this region is the Cone Nebula.

Capricornus
At right is the IAU sky chart for Capricornus with the notable sources marked and located.

The left image is of the solar twin HIP 102152, "a star located 250 light-years from Earth in the constellation of Capricornus (The Sea Goat)."

"HIP 102152 is more like the Sun than any other solar twin — apart from the fact that it is nearly four billion years older, giving us an unprecedented chance to study how the Sun will look when it ages. It is the oldest solar twin identified to date, and was studied by an international team using ESO’s Very Large Telescope, led by astronomers in Brazil."

"The different colours of the star are caused by the star moving slightly between the two exposures, many years apart."

Carina
At left is the IAU sky chart for the constellation Carina with notable sources located.

"NGC 3199, in the constellation Carina, [...] is the wind-blown partial "ring" around the Wolf-Rayet (W-R) star WR 18 (aka HD 89358), the easternmost (leftmost) of the three bright blue stars near the center of the 2MASS image. NGC 3199 and WR 18 are at a distance of about 3.6 kpc (11,736 light years) from us. W-R stars represent the final evolutionary stages of very massive stars (with ~30 solar masses or greater). The nebula shows an asymmetric appearance, i.e., only one side (the western one) of the shell is bright, both in the optical and the near-infrared. The fainter, eastern side is there, but is much fainter. Some W-R ring nebulae can be seen in 2MASS images, such as the more complete ring around M1-67. But, NGC 3199 is particularly bright in the 2MASS data. Dyson & Ghanbari (1989, A&A, 226, 270) provided an explanation for the ring's appearance through a model where a moving WR 18 is blowing a strong stellar wind into a surrounding uniform interstellar medium. Vigorous mass loss of 10-5 to 10-4 solar masses per year is characteristic of W-R stars, as the star approaches the end of its short life, although not all are surrounded by ring nebulae."

Cassiopeia
At right is a sky chart for the constellation Cassiopeia from the IAU where notable sources are indicated and located.

"This spectacular image [at left] of the supernova remnant Cassiopeia A is the most detailed image ever made of the remains of an exploded star. The one-million-second image shows a bright outer ring (green) ten light years in diameter that marks the location of a shock wave generated by the supernova explosion. A large jet-like structure that protrudes beyond the shock wave can be seen in the upper left. In the accompanying image, specially processed to highlight silicon ions, a counter-jet can be seen on the lower right."

"Surprisingly, the X-ray spectra show that the jet and counter-jet are rich in silicon atoms and relatively poor in iron atoms. This indicates that the jets formed soon after the initial explosion of the star; otherwise, the jets should have contained large quantities of iron from the star's central regions."

"The bright blue fingers located near the shock wave on the lower left are composed almost purely of iron gas. This iron was produced in the central, hottest regions of the star and somehow ejected in a direction almost perpendicular to the jets."

"The bright source at the center of the image is presumed to be a neutron star created during the supernova. Unlike the rapidly rotating neutron stars in the Crab Nebula and Vela supernova remnants that are surrounded by dynamic magnetized clouds of electrons called pulsar wind nebulas, this neutron star is quiet, faint, and so far shows no evidence for pulsed radiation."

"A working hypothesis is that the explosion that created Cassiopeia A produced high-speed jets similar to but less energetic than the hypernova jets thought to produce gamma-ray bursts. During the explosion, the neutron star may have developed an extremely strong magnetic field that helped to accelerate the jets. This strong magnetic field later stifled any pulsar wind activity, so the neutron star today resembles other strong-field neutron stars (a.k.a. "magnetars") in lacking a pulsar wind nebula."

Centaurus
The sky chart for the constellation Centaurus is at right with notable sources indicated and located.

"This image [at left] of the planetary nebula SuWt 2 reveals a bright ring-like structure encircling a bright central star. The central star is actually a close binary system where two stars completely circle each other every five days. The interaction of these stars and the more massive star that sheds material to create the nebula formed the ring structure. The burned out core of the massive companion has yet to be found inside the nebula. The nebula is located 6,500 light-years from Earth in the direction of the constellation Centaurus. This color image was taken on Jan. 31, 1995 with the National Optical Astronomy Observatory's 1.5-meter telescope at the Cerro Tololo Inter-American Observatory in Chile."

The second lower image at left is the Boomerang nebula.

"The Boomerang Nebula is a young planetary nebula and the coldest object found in the Universe so far. The NASA/ESA Hubble Space Telescope image is yet another example of how Hubble's sharp eye reveals surprising details in celestial objects."

"This NASA/ESA Hubble Space Telescope image shows a young planetary nebula known (rather curiously) as the Boomerang Nebula. It is in the constellation of Centaurus, 5000 light-years from Earth. Planetary nebulae form around a bright, central star when it expels gas in the last stages of its life."

"The Boomerang Nebula is one of the Universe's peculiar places. In 1995, using the 15-metre Swedish ESO Submillimetre Telescope in Chile, astronomers Sahai and Nyman revealed that it is the coldest place in the Universe found so far. With a temperature of -272C, it is only 1 degree warmer than absolute zero (the lowest limit for all temperatures). Even the -270C background glow from the Big Bang is warmer than this nebula. It is the only object found so far that has a temperature lower than the background radiation."

"Keith Taylor and Mike Scarrott called it the Boomerang Nebula in 1980 after observing it with a large ground-based telescope in Australia. Unable to see the detail that only Hubble can reveal, the astronomers saw merely a slight asymmetry in the nebula's lobes suggesting a curved shape like a boomerang. The high-resolution Hubble images indicate that 'the Bow tie Nebula' would perhaps have been a better name."

"The Hubble telescope took this image in 1998. It shows faint arcs and ghostly filaments embedded within the diffuse gas of the nebula's smooth 'bow tie' lobes. The diffuse bow-tie shape of this nebula makes it quite different from other observed planetary nebulae, which normally have lobes that look more like 'bubbles' blown in the gas. However, the Boomerang Nebula is so young that it may not have had time to develop these structures. Why planetary nebulae have so many different shapes is still a mystery."

"The general bow-tie shape of the Boomerang appears to have been created by a very fierce 500 000 kilometre-per-hour wind blowing ultracold gas away from the dying central star. The star has been losing as much as one-thousandth of a solar mass of material per year for 1500 years. This is 10-100 times more than in other similar objects. The rapid expansion of the nebula has enabled it to become the coldest known region in the Universe."

"The image was exposed for 1000 seconds through a green-yellow filter. The light in the image comes from starlight from the central star reflected by dust particles."

Cepheus
Starting at the right is the IAU sky chart for the constellation Cepheus.

Two astronomical entities are shown in the images at left: the supernova remnant CTA 1 and the Natal Microcosm, the latter is an image from the Spitzer space telescope.

Cetus
The constellation Cetus is charted by the IAU in the diagram at left.

At right the Chandra X-ray Observatory has confirmed that β Ceti is an X-ray source.

Chamaeleon
The IAU sky chart for the constellation Chamaeleon is at right with notable sources indicated and located.

At left is a ROSAT PSPC false-color image of the Chamaeleon I dark cloud, a well-known site of star formation. The contours show the 100 micron emission from dust in this cloud as measured by the infrared IRAS satellite. This image was made from X-rays with energies between 500 eV and 1100 eV. Since these soft (low energy) X-rays are absorbed by interstellar dust and gas, the observation of a "shadow" provides an indication of the intensity of X-rays coming from behind the absorbing cloud. X-ray "shadows" like this are the X-ray analog of well-known optical dark nebulae such as the Horsehead Nebula, produced by absorption in a dark cloud of optical photons emitted in a bright nebula behind the cloud. In the X-ray case, the entire sky is bright with a diffuse glow known as the diffuse X-ray background (somewhat like the night sky as seen from a modern city, where scattered light from street lamps produces a bright glow in the sky and washes out all but the brightest stars). Observations like this will help us understand the source of the diffuse X-ray background at these energies.

Circinus
Mano de Dios is shown at right in the constellation Circinus, sky chart at left.

Columba
The center of NGC 1808 is imaged at left with the Hubble Space Telescope. The galaxy occurs in the constellation Columba whose sky chart is at right.

Coma Berenices
The GALEX ultraviolet observatory has imaged the NGC 4793 galaxy (at right) in Coma Berenices. The IAU chart for the constellation Coma Berenices is at left.

Corona Australis
"The star R Coronae Australis [in the image at right] lies in one of the nearest and most spectacular star-forming regions. This portrait was taken by the Wide Field Imager (WFI) on the MPG/ESO 2.2-metre telescope at the La Silla Observatory in Chile. The image is a combination of twelve separate pictures taken through red, green and blue filters."

Corona Australis is the constellation charted in the diagram at left.

Corona Borealis
The IAU chart for the constellation Corona Borealis is at left. Abell 2142 is an object in Corona Borealis.

Corvus
The IAU sky chart for the constellation Corvus is at right.

“This NASA Hubble Space Telescope image of the Antennae galaxies (NGC 4038 & 4039) is the sharpest yet of this merging pair of galaxies. During the course of the collision, billions of stars will be formed. The brightest and most compact of these star birth regions are called super star clusters.”

Crater
The International Astronomical Union sky chart for the constellation Crater is at left. Notable sources are located and identified.

Crux
The constellation Crux has borders determined by the IAU and embodied at right in their sky chart.

Cygnus
The IAU sky chart for the constellation Cygnus is at left and includes the currently accepted boundaries.

Delphinus
The IAU has included the currently accepted boundaries for the constellation Delphinus in their sky chart.

Dorado
At left is the IAU sky chart for the constellation Dorado.

Draco
The IAU sky chart for the constellation Draco is at right. Notable sources are identified and located.

Equuleus
At left is the IAU sky chart for the constellation Equuleus. Notable sources are located and named.

Eridanus
The sky chart for the constellation Eridanus is at right.

Fornax
The sky chart for the constellation Fornax from the IAU is on the left.

Gemini
At right is the IAU sky chart for the constellation Gemini with notable sources marked.

At left is a multiwavelength composite that shows the supernova remnant IC 443, also known as the Jellyfish Nebula. Fermi GeV gamma-ray emission is shown in magenta, optical wavelengths as yellow, and infrared data from NASA's Wide-field Infrared Survey Explorer (WISE) mission is shown as blue (3.4 microns), cyan (4.6 microns), green (12 microns) and red (22 microns). Cyan loops indicate where the remnant is interacting with a dense cloud of interstellar gas.

Grus
Grus is a constellation that has notable sources as shown on the sky chart of the IAU at right.

Hercules
At left is the IAU sky chart for the constellation Hercules.

Horologium
At right is the IAU sky chart for the constellation Horologium.

Hydra
The constellation Hydra is close to the ecliptic as shown on the IAU sky chart at right.

Hydrus
The notable sources in the constellation Hydrus are shown in the image at left.

Indus
The IAU sky chart for the constellation Indus is at left.

Lacerta
The sky chart for the constellation Lacerta is shown at right.

Leo
At left is the sky chart for the constellation Leo.

Leo Minor
The sky chart for the constellation Leo Minor is at the right.

Lepus
The constellation Lepus is at left with its notable sources located and indicated.

Libra
At right is the IAU sky chart for the constellation Libra.

Orion
At right is the IAU sky chart for Orion with notable sources indicated.

At left is Abell 520, the "Train Wreck" cluster. The composite view shows inferred mass, visible light and X-ray.

Perseus
At right is the IAU Perseus chart.

At left, baby stars are forming near the eastern rim of the cosmic cloud Perseus, in this infrared image from NASA's Spitzer Space Telescope.

The baby stars are approximately three million years old and are shown as reddish-pink dots to the right of the image. The pinkish color indicates that these infant stars are still shrouded by the cosmic dust and gas that collapsed to form them. These stars are part of the IC348 star cluster, which consists of over 300 known member stars.

The Perseus Nebula can be seen as the large green cloud at the center of the image. Wisps of green are organic molecules called Polycyclic Aromatic Hydrocarbons (PAHs) that have been illuminated by the nearby star formation. Meanwhile, wisps of orange-red are dust particles warmed by the newly forming stars.

The Perseus Nebula is located about 1,043 light-years away in the Perseus constellation.

The image is a three channel false color composite, where emission at 4.5 microns is blue, emission at 8.0 microns is green, and 24-micron emission is red.

Sculptor
At left is the International Astronomical Union sky chart for the constellation Sculptor.

On the right is a visual image of the Sculptor Dwarf Galaxy.

Serpens Caput
At right is the IAU sky chart for the constellation Serpens Caput with notable sources marked.

At left is a Hubble Space Telescope image of Seyfert's Sextet in Serpens Caput. "The NASA/ESA Hubble Space Telescope is witnessing a grouping of galaxies engaging in a slow dance of destruction that will last for billions of years. The galaxies are so tightly packed together that gravitational forces are beginning to rip stars from them and distort their shapes. Those same gravitational forces eventually could bring the galaxies together to form one large galaxy."

"The name of this grouping, Seyfert's Sextet, implies that six galaxies are participating in the action. But only four galaxies are on the dance card. The small face-on spiral with the prominent arms [center] of gas and stars is a background galaxy almost five times farther away than the other four. Only a chance alignment makes it appear as if it is part of the group. The sixth member of the sextet isn't a galaxy at all but a long 'tidal tail' of stars [below, right] torn from one of the galaxies."

Serpens Cauda
The constellation Serpens is split into two separate sections, Serpens Caput (the snake's head) to the west and Serpens Cauda (the snake's tail) to the east. The constellation Ophiuchus is between them.

Taurus
At right is the IAU sky chart for the constellation Taurus. Legend: red oval - Galaxy, yellow circle - Open star cluster, yellow circle with vertical cross-hairs - Globular cluster, green circle inside cross-hairs - Planetary nebula, green square - Bright nebula, black circle - Star, open circle - Variable star.

At left is an APEX telescope image of part of the Taurus Molecular Cloud. It shows a sinuous filament of cosmic dust more than ten light-years long. In it, newborn stars are hidden, and dense clouds of gas are on the verge of collapsing to form yet more stars. The cosmic dust grains are so cold that observations at submillimetre wavelengths, such as these made by the LABOCA camera on APEX, are needed to detect their faint glow. This image shows two regions in the cloud: the upper-right part of the filament shown here is Barnard 211, while the lower-left part is Barnard 213. The submillimetre-wavelength observations from the LABOCA camera on APEX, which reveal the heat glow of the cosmic dust grains, are shown here in orange tones. They are superimposed on a visible-light image of the region, which shows the rich background of stars. The bright star above the filament is φ Tauri.

Triangulum Australe
The figure at right shows the sky map of Triangulum Australe. Around the edges of the map are coordinates related to longitude and latitude, but with the Earth rotating on its axis every 24 hours the celestial coordinates must remain fixed relative to the background light sources in the sky.

"The galaxies of this beautiful interacting pair [on the left] bear some resemblance to musical notes on a stave. Long tidal tails sweep out from the two galaxies: gas and stars were stripped out and torn away from the outer regions of the galaxies. The presence of these tails is the unique signature of an interaction. ESO 69-6 is located in the constellation of Triangulum Australe, the Southern Triangle, about 650 million light-years away from Earth."

Virgo
The Virgo Supercluster is on the left.

Volans
Volans is one of the 12 constellations that were introduced by the Dutch navigators Pieter Dirkszoon Keyser and Frederick de Houtman in the late 16th century. It was first depicted on Petrus Plancius’ globe in 1598. Plancius called the constellation Vliegendenvis (flying fish).

Vulpecula
About 4000 years old and at 1200 ly distant, the Dumbbell nebula on the left is brightened by a very hot star (85,000 K).

Weather
Weather is the state of the atmosphere, to the degree that it is hot or cold, wet or dry, calm or stormy, clear or cloudy. Most weather phenomena occur in the troposphere, just below the stratosphere. Weather refers, generally, to day-to-day temperature and precipitation activity, whereas climate is the term for the average atmospheric conditions over longer periods of time.

On Earth, common weather phenomena include wind, cloud, rain, snow, fog and dust storms. Less common events include natural disasters such as tornadoes, hurricanes, typhoons and ice storms.

Air pollution
Air pollution is the release of chemicals and particulates into the atmosphere. Common gaseous pollutants include carbon monoxide, sulfur dioxide, chlorofluorocarbons (CFCs) and nitrogen oxides produced by industry and motor vehicles. Photochemical ozone and smog are created as nitrogen oxides and hydrocarbons react to sunlight. Particulate matter (PM), or fine dust is characterized by their micrometre size PM10 to PM2.5. Particulate pollution is observed around the globe in varying sizes and compositions and is the focus of many epidemiological studies. PM is generally classified into two main size categories: PM10 and PM2.5. PM10, also known as coarse particulate matter, consists of particles 10 micrometers (μm) and smaller, while PM2.5, also called fine particulate matter, consists of particles 2.5 μm and smaller. Particles 2.5 μm or smaller in size are especially notable as they can be inhaled into the lower respiratory system, and with enough exposure, absorbed into the bloodstream. Particulate pollution can occur directly or indirectly from a number of sources including, but not limited to: agriculture, automobiles, construction, forest fires, chemical pollutants, and power plants.

Telluric sources
Most astronomical observations are conducted by measuring photons (electromagnetic waves) which originate beyond the sky. The molecules in the Earth's atmosphere, however, absorb and emit their own light, especially in the visible and near-IR portion of the spectrum, and any ground-based observation is subject to contamination from these telluric (earth-originating) sources. Water vapor, oxygen, and OH are some of the more important molecules in telluric contamination. Water contamination was particularly pronounced in the Mount Wilson solar Doppler measurements.

It is possible to correct for the effects of telluric contamination in an astronomical spectrum. This is done by preparing a telluric correction function, made by dividing a model spectrum of a star by an observation of an astronomical photometric standard stars. This function can then be multiplied by an astronomical observation at each wavelength point.

While this method can restore the original shape of the spectrum, the regions affected can be prone to high levels of noise due to the low number of counts in that area of the spectrum.

Light pollution
Scientific definitions of light pollution include the following:
 * Degradation of photic habitat by artificial light.
 * Alteration of natural light levels in the outdoor environment owing to artificial light sources.
 * Light pollution is the alteration of light levels in the outdoor environment (from those present naturally) due to man-made sources of light. Indoor light pollution is such alteration of light levels in the indoor environment due to sources of light, which compromises human health.
 * Light pollution is the introduction by humans, directly or indirectly, of artificial light into the environment.

Light pollution obscures the stars in the night sky for city dwellers [and] interferes with astronomical observatories ... Light is particularly problematic for amateur astronomers, whose ability to observe the night sky from their property is likely to be inhibited by any stray light from nearby. Most major optical astronomical observatories are surrounded by zones of strictly enforced restrictions on light emissions.

Sky glow is of particular irritation to astronomers, because it reduces contrast in the night sky to the extent where it may even become impossible to see any but the brightest stars.

To precisely measure how bright the sky gets, night time satellite imagery of the earth is used as raw input for the number and intensity of light sources. These are put into a physical model of scattering due to air molecules and aerosoles to calculate cumulative sky brightness. Maps that show the enhanced sky brightness have been prepared for the entire world.

Some astronomers use narrow-band "nebula filters" which only allow specific wavelengths of light commonly seen in nebulae, or broad-band "light pollution filters" which are designed to reduce (but not eliminate) the effects of light pollution by filtering out spectral lines commonly emitted by sodium- and mercury-vapor lamps, thus enhancing contrast and improving the view of dim objects such as galaxies and nebulae. Unfortunately these light pollution reduction (LPR) filters are not a cure for light pollution. LPR filters reduce the brightness of the object under study and this limits the use of higher magnifications. LPR filters work by blocking light of certain wavelengths, which alters the color of the object, often creating a pronounced green cast. Furthermore, LPR filters only work on certain object types (mainly emission nebulae) and are of little use on galaxies and stars. No filter can match the effectiveness of a dark sky for visual or photographic purposes. Due to their low surface brightness, the visibility of diffuse sky objects such as nebulae and galaxies is affected by light pollution more than are stars. Most such objects are rendered invisible in heavily light polluted skies around major cities. A simple method for estimating the darkness of a location is to look for the Milky Way, which from truly dark skies appears bright enough to cast a shadow.

In addition to skyglow, light trespass can impact observations when artificial light directly enters the tube of the telescope and is reflected from non-optical surfaces until it eventually reaches the eyepiece. This direct form of light pollution causes a glow across the field of view which reduces contrast. Light trespass also makes it hard for a visual observer to become sufficiently dark adapted. The usual measures to reduce this glare, if reducing the light directly is not an option, include flocking the telescope tube and accessories to reduce reflection, and putting a light shield (also usable as a dew shield) on the telescope to reduce light entering from angles other than those near the target. Under these conditions, some astronomers prefer to observe under a black cloth to ensure maximum dark adaptation.

Space weather
Space weather is the concept of changing environmental conditions in near-Earth space or the space from the Sun's atmosphere to the Earth's atmosphere.

Near-earth space is defined as the region from the interface between the solar wind and the Earth into the magnetosphere and down to the ionosphere and thermosphere.

The sun's corona is constantly being lost to space, creating what is essentially a very thin atmosphere throughout the Solar System. The movement of mass ejected from the Sun is known as the solar wind. Inconsistencies in this wind and larger events on the surface of the star, such as coronal mass ejections, form a system that has features analogous to conventional weather systems (such as pressure and wind) and is generally known as space weather. Coronal mass ejections have been tracked as far out in the solar system as Saturn. The activity of this system can affect planetary atmospheres and occasionally surfaces. The interaction of the solar wind with the terrestrial atmosphere can produce spectacular aurorae, and can play havoc with electrically sensitive systems such as electricity grids and radio signals.

Point sources
Def. "[a]n idealized discrete source of radiation that subtends an infinitesimally small angle" is called a point source.

It is "[a] source whose angular extent cannot be measured (< 0".05)."

Absorptions
"[W]hen the medium [behaves] like an amplifier to the incident radiation" it is "possible for negative absorption to arise at radio wavelengths".

Negative absorptions
The necessary and sufficient conditions for negative absorption to occur at radio wavelengths are
 * 1) "the kinetic energy distribution F(η) of the radiating electrons [is] markedly non-thermal with an appreciable excess of high energy electrons such that ∂F/∂η is positive over a finite range of the kinetic energy η" and
 * 2) "the stimulated transition probability [has] a maximum at some finite value of the kinetic energy, the most favorable case occurring when this maximum is a sharp one at the value of η at which ∂F/∂η has a positive maximum."

"These conditions can both be met in principle for the cases in which the dominant radiation process is due [to]
 * 1) [the] Cerenkov effect,
 * 2) gyro radiation by non-relativistic electrons, [and]
 * 3) synchrotron-type radiation by highly relativistic electrons".

Def. "radiation at the fundamental or at the first few harmonics of the gyro frequency by weakly relativistic electrons rotating in a magnetic field" is called gyro radiation.

Def. "radiation by strongly relativistic electrons at high harmonics of the gyro frequency" is called synchrotron radiation.

Def. the frequency of the circular motion of a charged particle in the presence of a uniform magnetic field is called the gyro frequency, here also it is the first or fundamental frequency.

Def. the emission of light (Cerenkov radiation) that occurs when a charged particle passes through an insulator at a speed greater than the speed of light in that medium is called the Cerenkov effect.

Def. the electromagnetic radiation emitted by the accelerating charged particles in a synchrotron that are moving at near the speed of light is called synchrotron radiation.

Def. a component frequency of the signal that is an integer multiple of the fundamental frequency is called a harmonic of the wave.

At right is a diagram which shows the higher harmonics of the fundamental frequency (ff) at the top. Each higher harmonic is an integral multiple of the first: 2ff, 3ff, 4ff, 5ff, 6ff, 7ff, from top to bottom.

Positive absorptions
A rotational transition is an abrupt change in angular momentum in quantum physics. Like all other properties of a quantum particle, angular momentum is quantized, meaning it can only equal certain discrete values, which correspond to different rotational energy states. When a particle loses angular momentum, it is said to have transitioned to a lower rotational energy state. Likewise, when a particle gains angular momentum, a positive rotational transition is said to have occurred. ... [U]nique spectral lines ... result. [For] a net gain or loss of energy during a transition, electromagnetic radiation of a particular frequency must be absorbed or emitted.

"The 111 → 110 rotational transition of formaldehyde (H2CO) [occurs] in absorption in the direction of four dark nebulae. The radiation ... being absorbed appears to be the isotropic microwave background". One of the dark nebulae sampled, per SIMBAD is TGU H1211 P5.

Accelerations
For proton astronomy, “at the high end of the proton energy spectrum (above ≈ 1018 eV) [the Larmor radius] deflection becomes small enough that proton astronomy becomes possible.”

The Larmor radius is the radius of the circular motion of a charged particle in the presence of a uniform magnetic field.

“[F]or a particle of energy E in EeV and charge Z in a magnetic field B in µG [the Larmor radius (RL)] is roughly”

 * $$R_L = 1 kpc \frac{E}{Z B}$$

where
 * $$R_L \ $$ is the Larmor radius,
 * $$E\ $$ is the energy of the particle in EeV
 * $$Z \ $$ is the charge of the particle, and
 * $$B \ $$ is the constant magnetic field.

Diffractions
Def. the "breaking up of an electromagnetic wave as it passes a geometric structure (e.g. a slit), followed by reconstruction of the wave by interference" is called diffraction.

Distillations
Def. the "separation of the volatile parts of a substance from the more fixed; specifically, the operation of driving off gas or vapor from volatile liquids or solids, by heat in a retort or still, and the condensation of the products as far as possible by a cool receiver, alembic, or condenser; rectification; vaporization; condensation" is called distillation.

Emissions
Def. the act of sending or throwing out; the act of putting into circulation is called an emission.

Def. origination by some process, [biological], chemical, or [physical], is called generation.

Holes
Def. a hollow spot in a surface is called a hole.

Def. an opening in a solid, liquid, gas, or plasma is called a hole.

Def. a round or irregular patch on [or apparently on] the surface of [an entity] having a different color, texture etc. and generally round in shape is called a spot.

Def. an opening through which entities such as gases can pass is called a vent.

Lensings
The image at right shows several blue, loop-shaped objects that are multiple images of the same galaxy, duplicated by the gravitational lens effect of the cluster of yellow galaxies near the middle of the photograph. The lens is produced by the cluster's gravitational field that bends light to magnify and distort the image of a more distant object.

Reflections
Def. the property of a propagated wave being thrown back from a surface (such as a mirror) is called reflection.

Def.

 * 1) a method of firing a projectile so that it skips along a surface
 * 2) a glancing rebound
 * 3) a rebound off something wildly in a seemingly random direction is called a ricochet.

Refractions
Def. the turning or bending of any wave, such as a light or sound wave, when it passes from one medium into another of different optical density is called refraction.

Spallations
Def. a reaction in which an entity fragments into many due to stress or impact is called spallation.

Transductions
Def. the process to convert energy from one form to another is called transduction.

Super soft X-ray sources
A super soft X-ray source (SSXS, or SSS) is an astronomical source of very low energy X-rays. Soft X-rays have energies in the 0.09 to 2.5 keV range, whereas hard X-rays are in the 1-20 keV range.

SSXSs are in most cases only detected below 0.5 keV, so that within our own galaxy they are usually hidden by interstellar absorption in the galactic disk. They are readily evident in external galaxies, with ~10 found in the Magellanic Clouds and at least 15 seen in M31.

As of early 2005, more than 100 SSSs have been reported in ~20 external galaxies, the Large Magellanic Cloud (LMC), Small Magellanic Cloud (SMC), and the Milky Way (MW). Those with luminosities below ~3 x 1038 erg/s are consistent with steady nuclear burning in accreting white dwarfs (WD)s or post-novae. There are a few SSS with luminosities ≥1039 erg/s.

Super soft X-rays are believed to be produced by steady nuclear fusion on a white dwarf's surface of material pulled from a binary companion, the so-called close-binary supersoft source (CBSS).

This requires a flow of material sufficiently high to sustain the fusion. Contrast this with the nova, where less flow causes the material to only fuse sporadically. Super soft X-ray sources can evolve into type Ia supernova, where a sudden fusion of material destroys the white dwarf, and neutron stars, through collapse.

Super soft X-ray sources were first discovered by the Einstein Observatory. Further discoveries were made by ROSAT.

Many different classes of objects emit supersoft X-radiation (emission dominantly below 0.5 keV).

Luminous supersoft X-ray sources
Luminous super soft X-ray sources have a characteristic blackbody temperature of a few tens of eV (~20-100 eV) and a bolometric luminosity of ~1038 erg/s (below ~ 3 x 1038 erg/s).

Apparently, luminous SSSs can have equivalent blackbody temperatures as low as ~15 eV and luminosities ranging from 1036 to 1038 erg/s. The numbers of luminous SSSs in the disks of ordinary spiral galaxies such as the MW and M31 are estimated to be on the order of 103.

Milky Way SSXSs
SSXSs have now been discovered in our galaxy and in globular cluster M3. MR Velorum (RX J0925.7-4758) is one of the rare MW super soft X-ray binaries. "The source is heavily reddened by interstellar material, making it difficult to observe in the blue and ultraviolet." The period determined for MR Velorum at ~4.03 d is considerably longer than that of other supersoft systems, which is usually less than a day.

Close-binary supersoft sources
The close-binary supersoft source CBSS model invokes steady nuclear burning on the surface of an accreting white dwarf (WD) as the generator of the prodigious super soft X-ray flux. As of 1999, eight SSXSs have orbital periods between ~4 hr and 1.35 d: RX J0019.8+2156 (MW), RX J0439.8-6809 (LMC), RX J0513.9-6951 (LMC), RX J0527.8-6954 (LMC), RX J0537.7-7034 (LMC), CAL 83 (LMC), CAL 87 LMC), and 1E 0035.4-7230 (SMC).

Symbiotic binaries
A symbiotic binary star is a variable binary star system in which a red giant has expanded its outer envelope and is shedding mass quickly, and another hot star (often a white dwarf) is ionizing the gas. Three symbiotic binaries as of 1999 are SSXSs: AG Dra (BB, MW), RR Tel (WD, MW), and RX J0048.4-7332 (WD, SMC).

Noninteracting white dwarfs
The youngest, hottest WD is very close to 100,000 K, of type DO and is the first single WD recorded as an X-ray source with ROSAT.

Cataclysmic variables
"Cataclysmic variables (CVs) are close binary systems consisting of a white dwarf and a red-dwarf secondary transferring matter via the Roche lobe overflow."

Both fusion- and accretion-powered cataclysmic variables have been observed to be X-ray sources. The accretion disk may be prone to instability leading to dwarf nova outbursts: a portion of the disk material falls onto the white dwarf, the cataclysmic outbursts occur when the density and temperature at the bottom of the accumulated hydrogen layer rise high enough to ignite nuclear fusion reactions, which rapidly burn the hydrogen layer to helium.

Apparently the only SSXS nonmagnetic cataclysmic variable is V Sge: bolometric luminosity of (1 - 10) x 1037, a binary including a blackbody (BB) accretor at T < 80 eV, and an orbital period of 0.514195 d. "

The accretion disk can become thermally stable in systems with high mass-transfer rates (Ṁ). Such systems are called nova-like (NL) stars, because they lack outbursts characteristic of dwarf novae.

VY Scl cataclysmic variables
Among the NL stars is a small group which shows a temporary reduction or cessation of Ṁ from the secondary. These are the VY Scl-type stars or anti-dwarf novae.

V751 Cygni
V751 Cyg (BB, MW) is a VY Scl CV, has a bolometric luminosity of 6.5 x 1036 erg/s, and emits soft X-rays at quiescence. The discovery of a weak soft X-ray source of V751 Cyg at minimum presents a challenge as this is unusual for CVs which commonly display weak hard X-ray emission at quiescence.

The high luminosity (6.5 x 1036 erg/s) is particularly hard to understand in the context of VY Scl stars generally, because observations suggest that the binaries become simple red dwarf + white dwarf pairs at quiescence (the disk mostly disappears). "A high luminosity in soft X-rays poses an additional problem of understanding why the spectrum is of only modest excitation." The ratio He II λ4686/Hβ did not exceed ~0.5 in any of the spectra recorded up to 2001, which is typical for accretion-powered CVs and does not approach the ratio of 2 commonly seen in supersoft binaries (CBSS).

Pushing the edge of acceptable X-ray fits toward lower luminosity suggests that the luminosity should not exceed ~2 x 1033 ergs/s, which gives only ~4 x 1031 ergs/s of reprocessed light in the WD about equal to the secondary's expected nuclear luminosity.

Magnetic cataclysmic variables
X-rays from magnetic cataclysmic variables are common because accretion provides a continuous supply of coronal gas. A plot of number of systems vs. orbit period shows a statistically significant minimum for periods between 2 and 3 hr which can probably be understood in terms of the effects of magnetic braking when the companion star becomes completely convective and the usual dynamo (which operates at the base of the convective envelope) can no longer give the companion a magnetic wind to carry off angular momentum. The rotation has been blamed on asymmetric ejection of planetary nubulae and winds and the fields on in situ dynamos. Orbit and rotation periods are synchronized in strongly magnetized WDs. Those with no detectable field never are synchronized.

With temperatures in the range 11,000 to 15,000 K, all the WDs with the most extreme fields are far too cool to be detectable EUV/X-ray sources, e.g., Grw +70°8247, LB 11146, SBS 1349+5434, PG 1031+234 and GD 229.

Most highly magnetic WDs appear to be isolated objects, although G 23-46 (7.4 MG) and LB 1116 (670 MG) are in unresolved binary systems.

RE J0317-853 is the hottest magnetic WD at 49,250 K, with an exceptionally intense magnetic field of ~340 MG, and implied rotation period of 725.4 s. Between 0.1 and 0.4 keV, RE J0317-853 was detectable by ROSAT, but not in the higher energy band from 0.4 to 2.4 keV. RE J0317-853 is associated with a blue star 16 arcsec from LB 9802 (also a blue WD) but not physically associated. A centered dipole field is not able to reproduce the observations, but an off-center dipole 664 MG at the south pole and 197 MG at the north pole does.

Until recently (1995) only PG 1658+441 possessed an effective temperature > 30,000 K. Its polar field strength is only 3 MG.

The ROSAT Wide Field Camera (WFC) source RE J0616-649 has an ~20 MG field.

PG 1031+234 has a surface field that spans the range from ~200 MG to nearly 1000 MG and rotates with a period of 3h24m.

The magnetic fields in CVs are confined to a narrow range of strengths, with a maximum of 7080 MG for RX J1938.4-4623.

None of the single magnetic stars has been seen as of 1999 as an X-ray source, although fields are of direct relevance to the maintenance of coronae in main sequence stars.

PG 1159 stars
PG 1159 stars are a group of very hot, often pulsating WDs for which the prototype is PG 1159 dominated by carbon and oxygen in their atmospheres.

PG 1159 stars reach luminosities of ~1038 erg/s but form a rather distinct class. RX J0122.9-7521 has been identified as a galactic PG 1159 star.

Novas
There are three SSXSs with bolometric luminosity of ~1038 erg/s that are novae: GQ Mus (BB, MW), V1974 Cyg (WD, MW), and Nova LMC 1995 (WD). Apparently, as of 1999 the orbital period of Nova LMC 1995 if a binary was not known.

U Sco, a recurrent nova as of 1999 unobserved by ROSAT, is a WD (74-76 eV), Lbol ~ (8-60) x 1036 erg/s, with an orbital period of 1.2306 d.

Planetary nebulas
In the SMC, 1E 0056.8-7154 is a WD with bolometric luminosity of 2 x 1037 that has a planetary nebula associated with it.

Super soft active galactic nuclei
Supersoft active galactic nuclei reach luminosities up to 1045 erg/s.

Large amplitude outbursts
Large amplitude outbursts of super soft X-ray emission have been interpreted as tidal disruption events.

Greens
Chlorophyll is a common source of green on Earth's surface by reflection of sunlight. Malachite is a green mineral also by reflection of sunlight that occurs in rocks at or near the interface between Earth's atmosphere and crust.

There is a green emission continuum.

Hypotheses

 * 1) Some intervening material between the detector and the alleged source may turn out to be the source.

title: User:Marshallsumter/Sources/Astronomy


text:
User:Egm4313.s12.team8.lucas/R2.1

=R2.1=

Problem Statement
Part i) Find the non-homogeneous L2-ODE-CC in standard form and the solution for the ODE in terms of the initial conditions ($$\displaystyle y(0)=1$$,  $$\displaystyle y'(0)=0$$) and the general excitation where $$\displaystyle r(x)=0$$, and plot the solution.

Part ii) Generate 3 non-standard, non-homoegeneous L2-ODE-CC that yield the same two roots of the characteristic equation as in Part i.

Part i
The standard form of a standard non-homogeneous L2-ODE-CC is: $$\displaystyle y''+ay'+by=r(x)$$

First, we use the given lambas to find the characteristic equation for the homogeneous solution:

Since,

then,

So,

The answer will be in the form of:

Differentiating the homogeneous solution gives:

By substituting Eqn. 1a into Eqn. 1b, we get:

Contributions
Solved and typed by: Egm4313.s12.team8.lucas 04:14, 8 February 2012 (UTC)

Proofread by:

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/

title: User:Egm4313.s12.team8.lucas/R2.1


text:
Panspermia/A critique of Panspermia according to Hoyle

(Review Paper) Cited in

Points Made
The Hoyle-Wickramasinghe panspermia theory proposes that life on earth was originated approximately 4 billion years ago by comets containing microbial life. Some of these microorganisms continue to be found. Nowadays Panspermia is considered as a serious alternative to life being originated solely on earth. Research performed by Becquerel and some other scientists in 1924 suggested that bacteria would not survive the radioactive environment of space, especially ultraviolet radiation. This theory was later proven wrong when it was discovered that microorganisms are not so easily destroyed by ultraviolet radiation but are rather rendered dormant which could easily be reversed when environmental conditions become favorable for the microorganism.

Methods
Research performed demonstrated that bacteria can survive the extreme conditions that exist in space. It can withstand radiation as well as extremely low temperatures.These bacteria are able to do this by using a condensed thin coating of carbonaceous material around them as a shield from potentially harmful radiation. During the 1970’s inorganic dust had properties identical to particles of bacterial composition. These bacteria were primarily found in interstellar dust and can trace the origin of some particles on earth. Replication of these particles is not possible in the interstellar environment; Therefore Hoyle- Wickramasinghe suggested that microbial life follows the amplification process that stars, comets and planetary systems go through on the cosmic life cycle. Comets are then considered to amplify microbial life during this life cycle that can last a few billion years.

.

Results
Microbial life on earth can be dated back to around 3.8 billion years (3.8 Gy) when comets and asteroid were bombarding the earth. This microbial life can be found or earth in new habitats such as geothermal vents, ocean floor, radioactive dumps and Antarctic soil. The microorganisms recovered from the Earth’s crust prove that microbial life can survive the pressure of the ocean depths. Recent discoveries continue to support that microbial life was transferred in the cosmic life cycle and therefore transferred to earth.

Micro-fossils of cyanobacteria were found embedded in the CI1 carbonaceous meteorites which further raises the possibility of life on comets.Micro-fossils of cyanobacteria were found embedded in the CI1 carbonaceous meteorites which further raises the possibility of life on comets ..


text:
User:Egm4313.s12.team2/Report2

=R2.1=

Problem Statement
Given two roots and the initial conditions:
 * $$ \lambda _1 = -2 \ $$, $$ \lambda _2 = +5 \ $$
 * $$ y(0) = 1 \ $$, $$ y'(0) = 0 \ $$

Find the non-homogeneous L2-ODE-CC in standard form and the solution in terms of the initial conditions and the general excitation r(x). Consider $$ r(x) = 0 \ $$. Plot the solution.

Soluion-Part 1
Given:
 * $$ \lambda_1 = -2, \lambda_2 = 5 \ $$
 * $$ y(0) = 1, y'(0) = 0 \ $$

With these roots, the original,standard ODE can be found by finding the characteristic equation.
 * $$ (\lambda - (-2))(\lambda - 5) = 0 \ $$
 * $$ \lambda^2 - 3\lambda - 10 = 0 \ $$

The solution is the sum of the homogenous solution and the particular solution. Since r(x) is unknown, yp will represent the particular solution.
 * $$ y(t) = y_h + y_p \ $$

The roots are real and distinct so the homogenous solution takes the form:
 * $$ y_h = C_1 e^{-2t} + C_2 e^{5t} \ $$

By taking the derivative and substituting the initial conditions, the following system is derived to calculate the coefficients.
 * $$ y(0) = 1 = C_1 + C_2 \ $$
 * $$ y'(0) = 0 -2C_1 + 5C_2 \ $$
 * $$ C_1 = 1- C_2 \ $$
 * $$ 0 = -2 + 2C_2 + 5C_2 \ $$
 * $$ 2 = 7C_2 \ $$
 * $$C_1 = \frac{5}{7}, C_2 = \frac{2}{7} \ $$

The solution to the nonhomogenous L2-ODE is
 * $$ y(t) = \frac{5}{7} e^{-2t} + \frac{2}{7} e^{5t} + y_p \ $$

If r(x)=0, Yp=0 then y(t)= yh. The plot is shown below.

Solution-Part 2
Given:
 * $$ \lambda_1 = 3,\lambda_2 = -6 \ $$

The characteristic equation is
 * $$ (\lambda - 3)(\lambda - (-6)) = 0 \rightarrow \lambda^2 + 3\lambda -18 = 0 \ $$

The standard L2-ODE is
 * $$ y'' + 3y' - 18y = r(x) \ $$

Non-standard
 * $$ m(y'' + 3y' -18y) = r(x) \ $$

3 non-standard equations can be found by multiplying the standard form by a constant,m. Choosing m=3,5,7 respectively.
 * $$ 3y'' + 9y' - 54y = r(x) \ $$
 * $$ 5y'' + 15y' - 90y = r(x) \ $$
 * $$ 7y'' + 21y' - 126y = r(x) \ $$

--John Plunkett 17:30, 7 February 2012 (UTC)

=R2.2=

Problem Statement
Given initial conditions $$ y(0) = 1 \ $$, $$ y'(0) = 0 \ $$ and excitation $$ r(x) = 0 \ $$ , find the solution to $$ y'' - 10y' + 25y = r(x) \ $$

Solution

 * $$ y'' - 10y' + 25y = r(x) \ $$
 * $$ \lambda^2 - 10\lambda + 25 = 0 \ $$
 * $$ (\lambda - 5)(\lambda - 5) = 0 \ $$
 * $$ \lambda = 5 \ $$

Double real root
 * $$ \lambda = -\frac 1 2 a $$
 * $$ y = (C_1 + C_2 x)e^{- \frac {ax} 2} \ $$
 * $$ y = C_1 e^{5x} + C_2 xe^{5x} \ $$

Apply initial conditions
 * $$ 1 = C_1 e^{5(0)} + C_2 (0)e^{5(0)} \ $$
 * $$ 1 = C_1 e \ $$
 * $$ C_1 = 1 \ $$
 * $$ y' = 5C_1 e^{5x} + C_2 (1*e^{5x}) + x*5xe^{5x} \ $$
 * $$ y' = 5C_1 e^{5x} + C_2 e^{5x} + 5x^2 e^{5x} \ $$
 * $$ y' = 5(1) e^{5(0)x} + C_2 e^{5(0)} + 5(0)^2 e^{5(0)} \ $$
 * $$ 0 = 5+ C_2 + 0 \ $$
 * $$ C_2 = -5 \ $$
 * $$ y = (1) e^{5x} + (-5) xe^{5x} \ $$

$$ y = e^{5x}-5xe^{5x} \ $$ --Corey Williams 17:55, 6 February 2012 (UTC)

Matlab Code
x = 0:0.01:1; >> ft = exp(5*x) - 5*x.*exp(5*x); >> plot(x,ft)

Plot
--Corey Williams 15:10, 7 February 2012 (UTC)

=R2.3=

Problem Statement
K 2011 p59 #3 and 4: Find a general solution, check by substitution.

Problem 3

 * $$ y'' + 6y' + 8.96y = 0 \ $$
 * $$ r^2 + 6r + 8.96 = 0 \ $$
 * $$ r = (\frac{-6 \pm \sqrt{36 - 4(8.96)}}{2}) \ $$
 * $$ r = -3 \pm 0.2 \ $$
 * $$ r = -2.8, -3.2 $$

Therefore, the general solution takes the form of: $$ y_g = C_1 e^{-2.8x} + C_2 e^{-3.2x} \ $$
 * $$ y' = -2.8C_1 e^{-2.8x} - 3.2C_2 e^{-3.2x} \ $$

 * $$ y'' = 7.84C_1 e^{-2.8x} + 10.24C_2 e^{-3.2x} \ $$

Substitution back into the original equation should result in 0 = 0
 * $$ 0 = (7.84C_1 e^{-2.8x} + 10.24C_2 e^{-3.2x}) + 6 (-2.8C_1 e^{-2.8x} - 3.2C_2 e^{-3.2x}) + 8.96 (C_1 e^{-2.8x} + C_2 e^{-3.2x}) \ $$

Problem 4

 * $$ y'' + 4y' +(\pi^2 + 4)y = 0 \ $$
 * $$ r^2 + 4r = (\pi^2 + 4) = 0 \ $$
 * $$ r = (\frac{-4 \pm \sqrt{16 - 4(\pi^2 + 4)}}{2}) \ $$
 * $$ r = -2\pm \sqrt{4 - (\pi^2 + 4)} \ $$
 * $$ r = -2\pm\pi i \ $$

Therefore, the general solution takes the form of: $$ y_g = Ae^{-2t}cos(\pi t) + Be^{-2t}sin(\pi t) \ $$
 * $$ y' = A[-2e^{-2t}cos(\pi t) - \pi e^{-2t}sin(\pi t)] + B[-2e^{-2t}sin(\pi t) + \pi e^{-2t}cos(\pi t)] \ $$

 * $$ y'' = A[4e^{-2t}cos(\pi t) + 2\pi e^{-2t}sin(\pi t) + 2\pi e^{-2t}sin(\pi t) - \pi^2 e^{-2t}cos(\pi t)] \ $$

 * $$ + B[4e^{-2t}sin(\pi t) - 2\pi e^{-2t}cos(\pi t) - 2\pi e^{-2t}cos(\pi t) - \pi^2 e^{-2t}sin(\pi t)] \ $$

Substitution back into the original equation should result in 0 = 0
 * $$ 0 = A[4e^{-2t}cos(\pi t) + 2\pi e^{-2t}sin(\pi t) + 2\pi e^{-2t}sin(\pi t) - \pi^2 e^{-2t}cos(\pi t)] \ $$

 * $$ + B[4e^{-2t}sin(\pi t) - 2\pi e^{-2t}cos(\pi t) - 2\pi e^{-2t}cos(\pi t) - \pi^2 e{^-2t}sin(\pi t)]  \ $$

 * $$ + (4A[-2e^{-2t}cos(\pi t) - \pi e^{-2t}sin(\pi t)] + B[-2e^{-2t}sin(\pi t) + \pi e^{-2t}cos(\pi t)] + (\pi^2+4)(Ae^{-2t}cos(\pi t) + Be^{-2t}sin(\pi t)) \ $$

 * $$ 0 = 0 \ $$

Substitution of the solution and its derivatives into the original differential equation returns a value of zero, thereby proving that the solution is correct.

=R2.4=

Problem Statement
K 2011 p59 #5 and 6: Find a general solution, check by substitution.

Problem 5
$$ y'' +2\pi y' +\pi y = 0 \ $$ $$ r^2 +2\pi r +\pi ^2 = 0 \ $$ $$r = \frac{-2\pi \pm \sqrt{4\pi ^2 - 4\pi^2} }{2} \ $$ $$ r= \frac{-2\pi}{2} = -\pi\ $$ Therefore, the general solution takes the form of:

$$ y_g = C_1e^{-\pi t} + C_2te^{-\pi t} \ $$ $$ y' = -\pi C_1e^{-\pi t} + C_2[e^{-\pi t} -\pi te^{-\pi t}] \ $$ $$ y'' = \pi ^2 C_1e^{-\pi t} + C_2[-\pi e^{-\pi t} - \pi(e^{-\pi t}-\pi te^{-\pi t})] \ $$ $$ y'' = \pi ^2 C_1e^{-\pi t} + C_2[-\pi ^2te^{-\pi t} - 2\pi e^{-\pi t}] \ $$

Substitution back into the original equation should result as 0=0

$$0 = \pi^2C_1e^{-\pi t} + \pi^2C_2te^{-\pi t} -2\pi C_2e^{-\pi t} - 2\pi^2C_1e^{-\pi t} + 2\pi C_2e^{-\pi t} -2\pi^2C_2e^{-\pi t}+\pi^2 C_1e^{-\pi t} + \pi^2C_2te^{-\pi t} \ $$ $$ 0 = 0 \ $$ Thus the substitution checks out, and the form of the general solution is proven.

--Stephen Leshko 20:10, 30 January 2012 (UTC)

Problem 6

 * $$ 10y'' -32y' +25.6y = 0 \ $$

Which can be written in standard form as:
 * $$ y'' -3.2y' +2.56y = 0 \ $$

The characteristic equation is then:
 * $$ \lambda ^2 - 3.2\lambda +2.56 = 0 \ $$
 * $$ \lambda_1,\lambda_2 = \frac{3.2 \pm \sqrt{(-3.2)^2 - 4(2.56)}}{2} \ $$

Therefore the roots $$ \lambda_1, \lambda_2 = 1.6 \ $$ are actually a double root at $$ 1.6 $$ The form for the solution of the differential equation with double-real roots is of the following:
 * $$ y_h(t) = C_1e^{1.6t} + C_2te^{1.6t} \ $$

And then in order to check the solution, find the first and second derivatives to plug back into the original equation and check their equality to zero.
 * $$ y'_h(t) = 1.6C_1 e^{1.6t} + C_2[e^{1.6t} +1.6te^{1.6t}] \ $$
 * $$ y''_h(t) = 2.56C_1e^{1.6t} + C_2[1.6e^{1.6t} + 1.6e^{1.6t} + 2.56te^{1.6t}] \ $$

 * $$ 10[1.6C_1e^{1.6t} +C_2[3.2e^{1.6t} +2.56te^{1.6t}]] - 32[1.6C_1e^{1.6t} +C_2[e^{1.6t} + 1.6te^{1.6t}]] + 25.6[C_1e^{1.6t} +C_2te^{1.6t}] = 0 \ $$

 * $$ 25.6C_1e^{1.6t} + 32C_2e^{1.6t} + 25.6C_2te^{1.6t} - 51.2C_1e^{1.6t} -32C_2e^{1.6t} - 51.2C_2te^{1.6t} + 25.6C_1e^{1.6t} + 25.6C_2te^{1.6t} = 0 \ $$

The solution then checks out

--Stephen Leshko 21:28, 5 February 2012 (UTC)

=R2.5=

Problem Statement
K 2011 p59 #16 and 17: Find an ODE $$ y'' +ay' + by = 0 \ $$ for the given basis.

Problem 16
Basis: $$ e^{2.6x}, e^{-4.3x} \ $$

 * $$ \lambda_1,\lambda_2 =2.6, -4.3 = \frac{-a \pm \sqrt{a^2 - 4b}}{2} \ $$

Therefore multiplying each $$ \lambda \ $$ by 2 yields:
 * $$ 5.2 = -a + \sqrt{a^2 - 4b} \ $$ ............. (1)
 * $$ -8.6 = -a - \sqrt{a^2 -4b} \ $$ ............. (2)

Adding the equations together gives:
 * $$ -3.4 = -2a \ $$ so $$ a = 1.7 \ $$

Plug $$ a \ $$ back into equation (1) to get:
 * $$ 5.2 = -1.7 + \sqrt{1.7^2 - 4b} \ $$

Solving for $$ b \ $$:
 * $$ b = \frac {(5.2 +1.7)^2 - 1.7^2}{-4} = -11.18 \ $$

So then the differential equation would be: $$ y'' +1.7y' - 11.18 y = 0 \ $$

Problem 17
Basis: $$ e^{-\sqrt{5}x} ,xe^{-\sqrt{5}x} \ $$

From the basis it can be seen that $$ -\sqrt{5} \ $$ is a double root of the characteristic equation
 * $$ \lambda_1, \lambda_2 = - \sqrt{5} \ $$

So then the characteristic equation must have root such that:
 * $$ (\lambda + \sqrt{5})^2 = 0 \ $$

Multiplying out gives:
 * $$ \lambda ^2 + 2 \sqrt{5} \lambda + 5 = 0 \ $$

Which corresponds to the differential equation: $$ y'' + \sqrt{20} y' + 5y = 0 \ $$ --Stephen Leshko 21:31, 5 February 2012 (UTC)

=R2.6=

Problem Statement
Realize spring-dashpot-mass systems in series, as shown in Figure 2.6, with a double root $$ \lambda = -3 $$. Find the values for the parameters k, c, m.

Figure 2.6

Solution
The equation of motion for spring-dashpot-mass system in series is: $$ m(y''_k+\frac{k}{c}y'_k)+ky_k=f(t)\ $$ Distribute m: $$ my''_k+\frac{mk}{c}y'_k+ky_k=f(t)\ $$ The standard L2-ODE-CC with excitation is: $$ Ay''+By'+Cy=r(x)\ $$ The homogeneous equation is: $$ Ay''_h+By'_h+Cy_h=0\ $$ Given the double root: $$ \lambda=-3\ $$

The characteristic equation is: $$ [\lambda-(-3)]^2=(\lambda+3)^2=\lambda^2+6\lambda+9=0\ $$ The homogeneous L2-ODE-CC for this characteristic equation is: $$ y''+6y'+9y=0\ $$ Compare coefficients with: $$ my''_k+\frac{mk}{c}y'_k+ky_k=f(t)\ $$ The result is: $$ m=1\ $$

$$ k=9\ $$

$$ 6=\frac{mk}{c}\ $$

$$ 6=\frac{mk}{c}\rightarrow c=\frac{mk}{6}=\frac{(1)(9)}{6}\ $$

$$ c=\frac{3}{2}\ $$

--Jonathon Ladin 17:15, 07 February 2012 (UTC)

=R2.7=

=R2.8=

Problem Statement
K2011 pg. 59 #5,13 Find the general solution and check by substitution.

Problem 5
Given the following L2-0DE:
 * $$ y'' + y' + 3.25y = 0 \ $$

The corresponding characteristic equation is
 * $$ \lambda^2 + \lambda +3.25 = 0\ $$

The roots are
 * $$ \lambda_1, \lambda_2 = \frac{-1 \pm \sqrt{1-4*(3.25)}}{2} = -\frac{1}{2} \pm j\sqrt{3} \ $$

These roots yield the following solution.
 * $$ y(t) = C_1 e^{-\frac{t}{2}}cos(\sqrt{3}t) + C_2 e^{-\frac{t}{2}}sin(\sqrt{3}t) \ $$

After taking the derivatives and substituting them into the differential equation, the coefficients of the sine and cosine terms sum to zero. Cosine terms:
 * $$ 0 = (3.25C_1 - .5C_1 + \sqrt{3} C_2 + .25C_1 - 3C_1 -.5\sqrt{3}C_2 -.5\sqrt{3}C_2) e^{-.5t}cos(\sqrt{3}t) \ $$
 * $$ 0 = (3.25C_2 - \sqrt{3}C_1 - .5C_2 + .5\sqrt{3}C_1 +.5\sqrt{3}C_1 + .25C_2 -3C_2) e^{-.5t}sin(\sqrt{3}t) \ $$

Problem 13
Given the following L2-ODE:
 * $$ y'' + .54y' + (\pi+.0729)y = 0 \ $$

The Characteristic equation is
 * $$ \lambda^2 +.54\lambda + (\pi +.0729) = 0 \ $$

The roots are
 * $$ \lambda_1, \lambda_2 = \frac{-.54 \pm \sqrt{(.54)^2 - 4(\pi + .0729)}}{2} = -.27 \pm j(1.772)  \ $$

The roots can be used to find the following general, homogenous solution
 * $$ y(t) = C_1 e^{-.27t} cos(1.772t) + C_2 e^{-.27t} sin(1.772t) \ $$

After taking the derivatives and substituting them into the differential equation, the coefficients of the sine and cosine terms sum to zero.

Cosine terms:
 * $$ 0 = (3.2145C_1 - .1458C_1 + .9569C_2 + .0729C_1 - 3.14C_1 - .47844C_2 - .47844C_2) e^{-.27t} cos(1.772t) \ $$

Sine terms:
 * $$ 0 = (3.2145C_2 - .95688C_1 - .1458C_2 + .47844C_1 + .47844C_1 + .0729C_2 - 3.14C_2) e^{-.27t} sin(1.772t) \ $$

--Egm4313.s12.team2.plunkett.jp|John Plunkett 19:06, 8 February 2012 (UTC)

=R2.9=

Problem Statement
Given the initial conditions: Y(0)=1, Y'(0)=0 and the excitation r(x)=0  Find and plot the solution to the differential equation corresponding to $$ \lambda ^2+ 4 \lambda + 13 = r(x) $$

Solution
The differential equation that corresponds to $$ \lambda ^2 + 4 \lambda +13 = r(x) $$ is $$ y'' + 4y' + 13y = r(x) $$  Recognizing that the equation is equal to the excitation r(x) which was declared to be zero in the problem statement we can proceed to find the roots of the equation via the quadratic formula  $$ \lambda_1,\lambda_2 = \frac{-4 \pm \sqrt{(4)^2 - 4(13)(1)}}{2} \ = -2 \pm \frac{ /sqrt{16-52} } {2}$$ $$ \lambda_1,\lambda_2 = -2 \pm 3i $$

The general solution to a homogenous ODE with the complex roots $$ r_1, r_2=\alpha \pm \beta i$$takes on the general form:  $$ y(t)=C_1 e ^{\alpha t} cos(\beta t) + C_2 e ^{\alpha t} sin(\beta t) \ $$

Given $$ \lambda_1,\lambda_2 = -2 \pm 3i $$ we can substitute $$ -2$$for $$/alpha $$ and  $$3$$ for  $$\beta$$ into the general form and use the initial values given in the problem statement to begin to solve for $$ C_1 $$and $$ C_2 $$.

$$ y(t)=C_1 e ^{-2t} cos(3t) + C_2 e ^{-2t} sin(3t) \ $$  $$ y(0)=C_1 \cancelto{1} {e ^{-2(0)}cos(3(0))} + \cancelto{0} {C_2 e ^{-2(0)} sin(3(0))}= 1$$$$ C_1=1$$

$$ y'(t)= (-2)(1) e ^{-2t} cos(3t) -(3)(1)e ^{-2t} sin(3t)-(-2) C_2 e ^{-2t} sin(3t) +(3) C_2 e ^{-2t}cos(3t) \ $$  $$ y'(0)= (-2)(1)\cancelto{1} {e ^{-2(0)} cos(3(0))} - \cancelto{0} {3)(1)e ^{-2(0)} sin(3(0))} -\cancelto{0} {(-2) C_2 e ^{-2(0)} sin(3(0))} +(3) C_2 \cancelto{1} { ^{-2(0)}cos(3(0))}=0$$ $$ y'(0)=-2+3C_2=0 \ $$ <br\>

$$ C_2 = \frac{2} {3} \ $$ <br\><br\> Therefore the the solution particular to the given initial conditions is:<br\> $$ y(t)=e ^{-2t} cos(3t) +\frac{2} {3} e ^{-2t} sin(3t) \ $$ <br\> --Nicolas Ellis 05:13, 7 February 2012 (UTC)

Effort Distribution
--Corey Williams 20:40, 3 February 2012 (UTC)

title: User:Egm4313.s12.team2/Report2


text:
University of Florida/Egm4313/s12.team7/Report2

==Background Theory: Solving ODEs with Constant Coefficients ==

Consider a second-order homogeneous linear ODE with constant coefficients a and b.

To solve this problem, note that the solution to a first-order linear ODE of the form:

is an exponential function, yielding a solution of the form:

Based upon this, we will expand the solution to apply to second-linear homogeneous ODEs. To do so, we must first find equations describing $$y'$$ and $$y''$$, and so we will take the derivatives (with respect to x) of ($$).

We will now substitute ($$), ($$), and ($$) into ($$) to obtain the relationship:

 * Simplifying, we have:

Since ($$) follows the same form as the quadratic equation, we can solve for $$\lambda _{1}$$ and $$\lambda _{2}$$ as follows:

Referring back to algebra, we know that the solution to these two equations can be one of three cases:

Case 1
Two real roots if...

These two roots give us two solutions:

The corresponding general solution then takes the form of the following:
 * $$y_{1} = e^{\lambda _{1} x}$$ || || and || || $$y_{2} = e^{\lambda _{2} x}$$
 * }
 * }

Case 2
A real double root if...

This yields only one solution:
 * $$y_{1} = e^{\lambda x}$$

In order to form a basis, a set of two linearly independent solutions, we must find another solution. To do so, we will use the method of reduction of order, where:
 * $$y_{2} = uy_{1}$$
 * $$y'_{2} = u'y_{1} + uy'_{1}$$
 * and
 * $$y_{2} = uy_{1} + 2u'y'_{1} + uy''_{1}$$
 * yeilding
 * $$ (uy_{1} + 2u'y'_{1} + uy_{1}) + a(u'y_{1} + uy'_{1}) + buy_{1} = 0 $$
 * $$ uy_{1} + u'(2y'_{1} + ay_{1}) + u(y_{1} + a'y_{1} + by_{1}) = 0 $$

Since $$y_{1}$$ is a solution of (1-1), the last set of parentheses is zero. Similarly, the first parentheses is zero as well, because
 * $$2y'_{1} = -ae^{ax/2} = -ay_{1}$$

From integration, we get the solution:
 * $$ u = c_{1}x + c_{2} $$

If we set $$c_{1} = 1$$ and $$c_{2} = 0$$
 * $$y_{2} = xy_{1}$$

Thus, the general solution is:

Case 3
Complex conjugate roots if...

In this case, the roots of ($$) are complex:
 * $$ y_{1} = e^{-ax/2} cos( \omega x) $$
 * and
 * $$ y_{2} = e^{-ax/2} sin( \omega x) $$

Thus, the corresponding general solution is of the form:

Contributions
Typed by - Mark James 2/6/2012 12:08 PM UTC Reviewed By - Jennifer Melroy 2/6/2012 2:15 UTC Edited By - N/A

Problem Statement
For this problem, the task was to derive the linear second-order non-homogeneous ODE and its solution given the two roots, initial conditions, and general excitation. However, for this problem it was assumed that r(x) = 0. Then, three non-standard and non-homogeneous ODE's were generated from the two given roots.

Roots:

Initial conditions:

Background Theory
To solve this problem, it is necessary to know the general form for a linear second-order ODE:

and that the characteristic equation for this type of ODE can be written as

Solution
Substitution both roots into the characteristic equation,

and distributing fully yields

The characteristic equation can be used to form the second-order ODE: Therefore, the solution for the homogeneous function is in the form of

and its derivative is

Algebraic substitution yields:

Plugging in initial conditions,

and

Algebraic substitution and solving for each constant term yields:

Therefore, since r(x)=0,

and the graph of y(x) is

Equation 1 (m = 7)
Therefore,

Equation 2 (m = 2)
Therefore,

Equation 3 (m = 20)
Therefore,

Contributions
Typed by - Maxwell Shuman 2/6/2012 15:10 PM UTC Reviewed By - Jennifer Melroy 2/6/2012 2:15 UTC Edited By - N/A

Problem Statement
Find and plot the solution for

With the initial conditions of

Background Theory
From the background theory at the beginning of this report, using the quadratic formula we find that

In this case

This would make this problem under the category of Case 2, in which the general solution becomes equation ($$)

Solution
To find ʎ we use

and we find that

Because the problem is in the category of Case 2, the general solution is

Differentiating this general solution gives us

Using equation($$) and plugging in the inital condition of y(0)= 1, we find that

Using equation ($$) and plugging in the initial condition of y'(0)=0 and $$c_1=1$$, we find that

The general solution for this problem will be

Contributions
Typed by ---Dalwin Marrero 21:59, 6 February 2012 (UTC) Reviewed By - Jennifer Melroy 2/6/2012 2:15 UTC Edited By - N/A

Problem Statement
For this problem, we were tasked with finding the general solution to the homogeneous linear ODE given below:

Solution
Following the process that yields ($$), we find the equation:

To find the form of the general solution, we must solve for the values of $$\lambda$$ using the quadratic formula:
 * [[File:P2_2_3.png]]

Since the solutions to $$\lambda$$ are complex numbers, our solution takes the form of ($$):

$$c_{1}$$ and $$c_{2}$$ are constant coefficients of unknown value. Had we been given initial values of $$y(x)$$ and $$y'(x)$$, we would solve for those values as well.

Confirmation of Solution
To confirm that the solution is true, we will substitute $$y$$, $$y'$$, and $$y''$$ into ($$). If the final result is a tautology then the solution is confirmed. If the result is a contradiction, a statement that is never true, then our solution is disproved.

Substituting ($$), ($$), and ($$) into ($$), we are left with:

Combining similar terms allows us to clean up this solution and check our answer:

Final Solution
Since the outcome of our check is a tautology, the solution is confirmed for all values of A and B and we affirm that the final solution is:
 * $$y = c_{1}e^{-2.8x} + c_{2}e^{-3.2x}$$

Problem Statement
For this problem, we were tasked with finding the general solution to the homogeneous linear ODE given below:

Solution
Following the process that yields ($$), we find the equation:

To find the form of the general solution, we must solve for the values of $$\lambda$$ using the quadratic formula:

Since the solutions to $$\lambda$$ are complex numbers, our solution takes the form of ($$):

A and B are constant coefficients of unknown value. Had we been given initial values of $$y(x)$$ and $$y'(x)$$, we would solve for those values as well.

Confirmation of Solution
To confirm that the solution is true, we will substitute $$y$$, $$y'$$, and $$y''$$ into ($$). If the final result is a tautology then the solution is confirmed. If the result is a contradiction, a statement that is never true, then our solution is disproved.

Substituting ($$), ($$), and ($$) into ($$), we are left with:

Combining similar terms allows us to clean up this solution and check our answer:

Final Solution
Since the outcome of our check is a tautology, the solution is confirmed for all values of A and B and we affirm that the final solution is:
 * $$y = e^{-2x}(Acos(\pi x)+ Bsin(\pi x)) $$

Contributions
Typed by - Mark James 2/6/2012 12:08 PM UTC Reviewed By - Jennifer Melroy 2/6/2012 2:15 UTC Edited By - N/A

Problem Statement
For this problem, we were asked to find the general solutions for problem 5 on page 59 of (need to site book; do not know how to)

Solution
Following the process that yields ($$), we find the equation:

To find the form of the general solution, we must solve for the values of $$\lambda$$. Instead of going through the quadratic formula, we can analyze the discriminant.

Thus we see that the discriminant follows Case 2 and the general solution must follow the form of ($$), and we have the following:

where $$c_{1}$$ and $$ c_{2}$$ are unknown constant coefficients. If this were an initial value problem with values of $$y(x)$$ and $$y'(x)$$, we would solve for those values as well.

Confirmation of Solution
To confirm that the solution is true, we will substitute $$y$$, $$y'$$, and $$y''$$ and check to see if the final result is a tautology or a contradiction.

Substituting, we are left with:

After combining similar terms, we have the form:

Final Solution
Since the above result is a tautology for all values of $$c_{1}$$ and $$ c_{2}$$,the general solution is:

Problem Statement
For this problem, we were asked to find the general solutions for problem 6 on page 59 of (need to site book; do not know how to)

Solution
For this problem, we have to divide by 10 to put the equation in standard form, making it

Following the process that yields ($$), we find the equation:

To find the form of the general solution, we must solve for the values of $$\lambda$$. Instead of going through the quadratic formula, we can analyze the discriminant.

Thus we see that the discriminant follows Case 2 and the general solution must follow the form of ($$), and we have the following:

where $$c_{1}$$ and $$ c_{2}$$ are unknown constant coefficients. If this were an initial value problem with values of $$y(x)$$ and $$y'(x)$$, we would solve for those values as well.

Confirmation of Solution
To confirm that the solution is true, we will substitute $$y$$, $$y'$$, and $$y''$$ and check to see if the final result is a tautology or a contradiction.

Substituting, we are left with:

After combining similar terms, we have the form:

Final Solution
Since the above result is a tautology for all values of $$c_{1}$$ and $$ c_{2}$$,the general solution is:

Contributions
Typed by - Mark James 2/6/2012 12:08 PM UTC Reviewed By - Jennifer Melroy 2/6/2012 2:15 UTC Edited By - N/A

Background Theory
A basis is the root of a ODE. They take the form of

They can be used to identify which background case can be applied.

It can also be noted that

Problem Statement
For this problem find the ordinary differential equation in the from of ($$)

using the following basis  the fit the form of ($$)

Solution
For this solution, it is assumed that

Since($$) and ($$) match the given roots in Case 1, the general solution in the form of ($$) can be found by substituting ($$) and ($$). This results in

From ($$), $$\lambda _{1} $$ and $$\lambda _{2} $$ can be found

Using the relationship in ($$), and proprieties of quadratic equations it can be assumed

Multiplying ($$) out give the form

Since ($$) is in the form of ($$), the relationship in ($$) can be used to get the form of

Problem Statement
For this problem find the ordinary differential equation in the from of ($$)

using the following basis  the fit the form of ($$)

Solution
For this solution, it is assumed that

Since($$) and ($$) match the given roots in Case 2, the general solution in the form of ($$) can be found by substituting ($$) and ($$). This results in

From ($$), $$\lambda$$ can be found

Using the relationship in ($$), and proprieties of quadratic equations it can be assumed

Multiplying ($$) out give the form

Since ($$) is in the form of ($$), the relationship in ($$) can be used to get the form of

Contributions
Solved and Typed by - Jennifer Melroy 2/4/2012 9:20 PM UTC Reviewed By - Edited By - N/A

Problem Statement
Given this system: It was found that the equation for this system is

The parameters of this equation can be found by the characteristic equation for a double real root if given a value for $$ \lambda $$. Find the values of m, k , and c with $$ \lambda= -3 $$.

Solution
With $$ \lambda=-3 $$ then the characteristic equation becomes

If we relate these equations we can say that the coefficients in ($$) and ($$) are the same. With that said then,

We are fortunate that with this method m and k are solved without any need for calculation.

This leaves c to be solved using ($$).

So m = 1, k = 9 and c = 1.5

Contributions
Typed by - Yamil Herrera 2/6/2012 12:08 PM UTC Reviewed By - Jennifer Melroy 2/6/2012 2:15 UTC Edited By - N/A

Background
Taylor series is a method used to find the value of a equation using a set of infinite sums of the its derivatives at a certain point a. Taylor Series:

The MacLaurin Series is the most basic form of the Taylor series by making a=0. Maclaurin Series:

Problem Statement
Develop a MacLaurin series for the following equations: $$e^{t}$$, cos(t), and sin(t).

Solution
The first step for finding the solution is to solve for f(a) and its derivatives. For $$y=e^{t}$$ (where y = f(a)) the first three derivatives come out to:

Once the derivatives are found, make t=0, and plug back into ($$) which results in the the equation:

Doing the same thing for y= cos(t)the derivatives come out to:

Again, plugging in zero and placing the ys into ($$) you get:

Repeating the same method for y= sin(t) as the previous two equations the derivatives come out as:

Using ($$) you get the equation:

Problem Statement
For this problem, find the general solution for the following equation:

Solution
From the equation ($$) our a=1 and b=3.25

With this information we can decide what kind of root we have.

So with a=1 and b=3.25

$$(1)^2-4(3.25)= -12 $$

Since ($$) is less than zero, we can use Case 3 from our Background Theory. Our general solution will be:

To find $$\omega$$ we need to use the characteristic equation ($$) and solve for the root $$\lambda$$

where Using ($$) we can solve for $$ \omega $$

Now that we have $$ \omega $$ we can plug it in to get our general solution from ($$)

Confirmation of Solution
To confirm that the solution is true, we will find $$y'$$ and $$y''$$ and plug it into ($$). If the result agrees and equals zero then it it considered true.

Substituting ($$), ($$), and ($$) into ($$), we are left with:

Combining similar terms allows us to clean up this solution and check our answer:

Problem Statement
For this problem, find the general solution for the following equation:

Solution
From the equation ($$) our a=0.54 and b=0.0729+ $$ \pi $$

With this information we can decide what kind of root we have.

So with a=0.54 and b=0.0729+ $$ \pi $$3.25

$$(0.54)^2-4(0.0729+ \pi)= -12.57 $$

Since ($$) is less than zero, we can use Case 3 from our Background Theory. Our general solution will be:

To find $$\omega$$ we need to use the characteristic equation ($$) and solve for the root $$\lambda$$

where Using ($$) we can solve for $$ \omega $$

Now that we have $$ \omega $$ we can plug it in to get our general solution from ($$)

Confirmation of Solution
To confirm that the solution is true, we will find $$y'$$ and $$y''$$ and plug it into ($$). If the result agrees and equals zero then it it considered true.

Substituting ($$), ($$), and ($$) into ($$), we are left with:

Combining similar terms allows us to clean up this solution and check our answer:

Contributions
Typed by - Yamil Herrera 2/6/2012 12:08 PM UTC Reviewed By - Jennifer Melroy 2/6/2012 2:15 UTC Edited By - N/A

Problem Statement
For this problem we were to find and plot a solution to the following problem  We also had to superpose three figures: one for this problem, and problems.

Background Theory
In the problem we were given the initial conditions $$y(0)=1$$ , $$y'(0)=0$$, and that there is no excitation, having $$r(x)=0 $$.

There are a few steps that we have to go through before we can solve this problem. We first have to solve for $$\lambda_{1}$$ and $$\lambda_{2}$$. To find out what $$\lambda_{1}$$ and $$\lambda_{2}$$ equal we can apply pythagorean theorem, We will know if this problem has complex roots or not by applying $$ a^2-4b $$. If the answer we get is greater than zero we will use case one. If it is equal to zero we will use vase two and if it is less than zero we will use case 3. , so we will use case 3. The standard form for case 3 is ($$) We will also have to find out what $$a$$ and $$w$$ are equal to and we use this equation to find out what they are

Part 1
This first step in solving this problem is to factor in the fact that there is no excitation.

Once we have does this we need to solve for $$\lambda_{1}$$ and $$\lambda_{2}$$. When apply ($$) we get. After reducing it down we end up with Now that we know what $$\lambda_{1}$$ and $$\lambda_{2}$$ are equal to, we apply ($$) to find out what $$a$$ and $$w$$ are. We find that $$a=2$$ and $$w=3$$ Knowing that we have complex roots, we can now plug the known values of $$a$$ and $$w$$, into ($$). Once we do this we have The next step we have to do is to take the derivative of ($$). We end up getting We know have to find what $$C_{1}$$ and $$C_{2}$$ are equal to. From what we already know about our initial conditions we can apply those and what we found $$a$$ and $$w$$ to be into both ($$)

and ($$).

We have now found out what all the variables are equal to and we can plug them into ($$), getting out final answer

When we graph this solution it looks like

Part 2
For the second part we had to graph the equations from problems 2.1, 2.6 and 2.9.

Contributions
Typed by - Ron D'Amico 2/7/2012 10:39 PM UTC Reviewed By - Jennifer Melroy 2/6/2012 2:15 UTC Edited By - N/A


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User:Egm4313.s12.team14.kimiagarov/HW 2

Problem set 2.3 (Page59)

= Problem 3 =

Given

 * $$\displaystyle{y''+6y'+8.96y = 0}$$

Find
General solution to the differential equation

Solution
Characteristic equation:

 * $$\displaystyle{\lambda^2+6\lambda+8.96 = 0}$$

Using the quadratic equation:

 * $$\displaystyle{x = \frac{-b\plusmn\sqrt{b^2-4ac}}{2a}\,}$$

where:

 * $$\displaystyle{a=1, b=6, c=8.96}$$

Place values into quadratic equation:

 * $$\displaystyle{x = \frac{-6\plusmn\sqrt{6^2-4*1*8.96}}{2*1}\,}$$

 * $$\displaystyle{x = \frac{-6\plusmn\sqrt{36-35.84}}{2}\,}$$

 * $$\displaystyle{x = \frac{-6\plusmn0.4}{2}\,}$$

 * $$\displaystyle{x = \frac{-6.4}{2}\,,\frac{-5.6}{2}}$$

 * $$\displaystyle{x = -3.2,-2.8}$$

$$\displaystyle{y = c_1e^{-3.2x}+c_2e^{-2.8x}}$$

= Problem 4 =

Given

 * $$\displaystyle{y''+4y'+(\pi^2+4)y = 0}$$

Find
General solution to the differential equation

Solution
Characteristic equation:

 * $$\displaystyle{\lambda^2+4\lambda+(\pi^2+4) = 0}$$

Using the quadratic equation:

 * $$\displaystyle{\lambda = \frac{-b\plusmn\sqrt{b^2-4ac}}{2a}\,}$$

where:

$$\displaystyle{a=1, b=4, c=(\pi^2+4)}$$

Place values into quadratic equation:

 * $$\displaystyle{\lambda = \frac{-4\plusmn\sqrt{4^2-4*1*(\pi^2+4)}}{2*1}\,}$$

 * $$\displaystyle{\lambda = \frac{-4\plusmn\sqrt{16-4\pi^2-16)}}{2}\,}$$

 * $$\displaystyle{\lambda = \frac{-4\plusmn2\pi i}{2}\,}$$

 * $$\displaystyle{\lambda = -2\plusmn\pi i}$$

$$\displaystyle{y = e^{-2x}(C_1\cos\pi x + C_2\sin\pi x)}$$

title: User:Egm4313.s12.team14.kimiagarov/HW 2