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	<title>Singularities in Space</title>
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	<description>The Consequences of the Geometry of Spacetime</description>
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		<title>An introduction to the formation of black holes</title>
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		<pubDate>Fri, 12 Jun 2009 15:22:15 +0000</pubDate>
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		<description><![CDATA[Before even beginning to comprehend the absurdity that is the black hole, a good understanding of Einstein&#8217;s theory of relativity is required. In short, Einstein was able to prove that gravity is a consequence of the geometry and curvature of space. By doing so, Einstein has shown that the gravity surrounding an object is due [...]<img alt="" border="0" src="http://stats.wordpress.com/b.gif?host=jmckennonblackholestudies.wordpress.com&amp;blog=8119166&amp;post=3&amp;subd=jmckennonblackholestudies&amp;ref=&amp;feed=1" width="1" height="1" />]]></description>
			<content:encoded><![CDATA[<p>Before even beginning to comprehend the absurdity that is the black hole, a good understanding of Einstein&#8217;s theory of relativity is required. In short, Einstein was able to prove that gravity is a consequence of the geometry and curvature of space. By doing so, Einstein has shown that the gravity surrounding an object is due to the mass of the object, and thus massive objects have a greater gravitational influence than less massive objects.</p>
<p>Another consequence of Einstein&#8217;s TOR, was the proof that gravity bends light, and actually can &#8220;slow down&#8221; time (this of course, is severely dependent on the frame of reference from which the gravitational influence is being observed, but that is an entirely different matter) . For now, it is just necessary to understand a few of the many consequences of Einstein&#8217;s famous theory.</p>
<p> </p>
<p>Einstein&#8217;s famous field equation is as follows:</p>
<p><img src='http://s0.wp.com/latex.php?latex=R_%7Bab%7D+-+%5Cfrac+%7B1%7D%7B2%7D+R_%7BG_%7Bab%7D%7D%3D+k+T_%7Bab%7D&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='R_{ab} - &#92;frac {1}{2} R_{G_{ab}}= k T_{ab}' title='R_{ab} - &#92;frac {1}{2} R_{G_{ab}}= k T_{ab}' class='latex' />.</p>
<p>The left hand side of this equation is known as the Einstein tensor, which is a specific, divergence free combination of the Ricci tensor and the metric tensor. The right side of this equation is known as the energy-momentum tensor. In this case, k is fixed as a constant equating to <img src='http://s0.wp.com/latex.php?latex=%5Cfrac+%7B8+%5CPi+G%7D%7Bc%5E4%7D&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='&#92;frac {8 &#92;Pi G}{c^4}' title='&#92;frac {8 &#92;Pi G}{c^4}' class='latex' />, where G is Isaac Newton&#8217;s gravitational constant, and c is the speed of light.</p>
<p>When Einstein first published his theory on November 25, 1915, he was not able to directly specify exact solutions to his equations of relativity, only approximate ones. This left the door open for Karl Schwarzschild, who was the director of Astrophysical Observatory at the time, to address some of the mathematical properties of Einstein&#8217;s equations. Schwarzschild first attempted to find a solution to the simplest (at least calculation-wise) of Einstein&#8217;s equations;  the case of a spherically symmetrical star witha given mass density ( in laymen&#8217;s terms : a sphere with a given mass). Schwarzschild began by finding a solution to Einstein&#8217;s equations for the space outside of the star. He was looking for a solution in which the lefthand side of Einstein&#8217;s equation is zero, and shows spherical symmetry. This means that the tensor of curvature will have the same structure in all directions of space, solely depending on the distance from the center of the spherical object.  Since the case at hand here deals with gravitation due to a sphere, two results can be expected : time being warped by the gravitational influence,  and the influences&#8217; forced curvature of the space around it. Using this knowledge as a stepping stone, Schwarzschild was able to proceed further with his investigation of the properties of Einstein&#8217;s equations.  Saving the long derivation for another time, Schwarzschild obtained his solution as:</p>
<p><img src='http://s0.wp.com/latex.php?latex=c%5E2+d+%5Ctau+%5E2%3D+%281-+%5Cfrac+%7Br_s%7D%7Br%7D+c%5E2+dt%5E2+-+%5Cfrac+%7Bdr%5E2%7D%7B1-%7Br_s%7D%7Br%7D%7D-r%5E2%28d+%5Ctheta+%2Bsin%5E2+%5Ctheta+d+%5Cpsi%5E2%29&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='c^2 d &#92;tau ^2= (1- &#92;frac {r_s}{r} c^2 dt^2 - &#92;frac {dr^2}{1-{r_s}{r}}-r^2(d &#92;theta +sin^2 &#92;theta d &#92;psi^2)' title='c^2 d &#92;tau ^2= (1- &#92;frac {r_s}{r} c^2 dt^2 - &#92;frac {dr^2}{1-{r_s}{r}}-r^2(d &#92;theta +sin^2 &#92;theta d &#92;psi^2)' class='latex' />.</p>
<p>Where <img src='http://s0.wp.com/latex.php?latex=%5Ctau&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='&#92;tau' title='&#92;tau' class='latex' /> is the proper time (measured by a clock traveling with the particle) in seconds. c is the speed of light in meters per second. t is the time coordinate (from a stationary clock at infinity) in seconds. r is the radial coordinate (circumference of a circle centered on the star divided by <img src='http://s0.wp.com/latex.php?latex=2+%5CPi&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='2 &#92;Pi' title='2 &#92;Pi' class='latex' />) in meters. <img src='http://s0.wp.com/latex.php?latex=%5Ctheta&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='&#92;theta' title='&#92;theta' class='latex' /> is the colaltitude angle in radians. <img src='http://s0.wp.com/latex.php?latex=r_%7Bs%7D&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='r_{s}' title='r_{s}' class='latex' /> is the Schwarzschild radius in meters, which is equal to <img src='http://s0.wp.com/latex.php?latex=r_%7Bs%7D%3D%5Cfrac+%7B2+G+M%7D%7Bc%5E2%7D&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='r_{s}=&#92;frac {2 G M}{c^2}' title='r_{s}=&#92;frac {2 G M}{c^2}' class='latex' /> where G is the gravitational constant.</p>
<p>Upon examining Schwarzschild&#8217;s solution (known as the Schwarzschild metric), it is easy to notice that singularities occur at <img src='http://s0.wp.com/latex.php?latex=r%3Dr_%7Bs%7D&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='r=r_{s}' title='r=r_{s}' class='latex' /> and <img src='http://s0.wp.com/latex.php?latex=r%3D0&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='r=0' title='r=0' class='latex' />. After further examiniation, it is seen that <img src='http://s0.wp.com/latex.php?latex=r%3Dr_%7Bs%7D&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='r=r_{s}' title='r=r_{s}' class='latex' /> is a coordinate singularity, which comes from a poor choice in coordinates or coordinate conditions. The case r=0 however, is a different case.</p>
<p>If you were to look at a clock, orbiting near the sphere under examination here, the clock will be operating with little influence from gravitation. As the clock gets closer to the sphere, the ticking of the clock slows, and thus, time is essentially slowed due to the gravitational influence of the sphere. If t stands for a small slice of time on the clock far away from the sphere, then the time difference between the clock close to the sphere and the clock far from the sphere, <img src='http://s0.wp.com/latex.php?latex=%5Ctau&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='&#92;tau' title='&#92;tau' class='latex' /> at radius r is given by:<br />
<img src='http://s0.wp.com/latex.php?latex=%28%5Cfrac+%7B%5Ctau%7D%7Bt%7D+%29%5E2%3D%5Cfrac+%7Br-R_s%7D%7Br%7D&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='(&#92;frac {&#92;tau}{t} )^2=&#92;frac {r-R_s}{r}' title='(&#92;frac {&#92;tau}{t} )^2=&#92;frac {r-R_s}{r}' class='latex' />. This equation shows that as you approach the sphere, the flow of time slows. But when <img src='http://s0.wp.com/latex.php?latex=r%3DR_s&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='r=R_s' title='r=R_s' class='latex' />, time stops, which is a rather peculiar result. It is a tough idea to wrap ones&#8217; brain around. The solution does show that at the Schwarzschild radius, time does appear to freeze up, but this, of course, depends on the frame of reference as I will discuss shortly.</p>
<p>If a person were to spend a couple of hours orbiting near the Schwarzschild radius, and then return to their space ship far from the sphere, they will have noticed that millions of years have passed.</p>
<p>Returning to my previous statement that gravity causes rays of light to bend, it is possible to calculate the path of a ray of light as it passes by an object with a large mass at a given distance. If there were no gravitational influence on the ray of light, it would be traveling in a straight line. But, due to spacetimes&#8217; curvature, the actual path that a ray of light takes turns out to be a geodesic line as opposed to a straight line. If a ray of light on its&#8217; way to Earth passes by a massive galaxy, the light will be deflected by a certain angle (given by <img src='http://s0.wp.com/latex.php?latex=%5Ctheta%3D+2%28%5Cfrac+%7BR_s%7D%7BR_c%7D&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='&#92;theta= 2(&#92;frac {R_s}{R_c}' title='&#92;theta= 2(&#92;frac {R_s}{R_c}' class='latex' /> where <img src='http://s0.wp.com/latex.php?latex=R_s&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='R_s' title='R_s' class='latex' /> is the Schwarzschild radius, and <img src='http://s0.wp.com/latex.php?latex=R_c&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='R_c' title='R_c' class='latex' /> is the shortest distance between the light and the massive object&#8217;s center). Now, if this light were to pass by a massive galaxy, the light could reach Earth on two different paths, creating a type of gravitational lensing effect(depending on the shape of the massive galaxy or object it passes by).</p>
<p>With some of the theory of relativity out of the way, a roundabout discussion of the formation of black holes can now begin.</p>
<p>A question to consider: What happens if an objects&#8217; circumference is smaller than its&#8217; Schwarzschild circumference? That question, will lead us on a new journey towards the understanding of black holes.</p>
<p>In the early 20th century, an astronmer by the name of Fritz Zwicky,  began to consider the possibility of objects in space that have a considerably greater matter density than that of a regular star. This came shortly after the discovery of neutrons in the nucleus of an atom; particles with a neutral charge, bound to protons in the nucleus. Unbound neutrons are not stable on their own. When a free neutron is on its&#8217; own, it quickly decays. When a neutron decays into a proton, the decay also creates an electron and a neutrino. Since a neutron&#8217;s mass is roughly 0.1 times larger than a proton, the energy released by this decay goes into the electron and the neutrino. A free proton however, cannot become a neutron (easily proven, due to the fact that the mass of a proton is less than the mass of a neutron). In a nucleus however, this type of reaction is entirely possible, as the masses of these particles are slightly different inside of a nucleus as opposed to being outside of it. This process is  fairly common amongst the cosmos, and thus plays an important role in the functioning of stars.</p>
<p>If you were to invent a sort of &#8220;super pressurizer&#8221;(this does not exist, but for sake of explanation, let&#8217;s pretend it does) and begin compressing a gas, say hydrogen, you would notice several interesting events. As shown in Harald Fritzsch&#8217;s <em>The Curvature of Spacetime, </em>the atoms of the hydrogen will move together as the pressure increases, eventually liquefying as the particles in the nucleus get closer and closer to one another. As more and more pressure is put on the gas eventually reaching about 1million tons per cubic centimeter, the electrons and protons will become so tightly packed against one another, that the protons and electrons will fuse together into neutrons, leaving us with a gas entirely composed of neutrons. This unique behavior leads us to the cosmos.</p>
<p>In order to save time, let&#8217;s consider stars as simple systems, with the life of the star solely dependent on its mass (which is a pretty true statement on its own anyways). Currently, it is known to us that stars get the majority of their energy from nuclear fusion at their core. With such monstrous amounts of energy being released due to fusion, gravity remains as the saving force that keeps them intact.  In the case of our sun, the solar matter is ever-trying to explode, and gravity is ever opposing this action. This gravitational opposition creates implosion which keeps the sun as we know it today. In billions of years, the fuel for this nuclear reaction (hydrogen, in the case of our sun) will be completely used up. When this happens, the nuclear process will gradually slow, and gravity will begin to take over. With less force opposing gravity, the matter of the sun will begin to compress, eventually becoming what is known as a &#8220;white dwarf&#8221; roughly the size of Earth. As time progresses, the temperature of the sun will drop, and nothing will remain but a black dwarf; a small sphere entirely composed of ashes.</p>
<p>As we know, the greater an objects mass, the greater the gravitational influence will be on that object. In the universe, there are many stars with masses much larger than that of the sun. These stars will have much hotter interiors, and produce much more pressure for the radiation that is opposing the force of gravity. These stars have much shorter lives than other stars with smaller masses.</p>
<p>The great U.S. astrophysicist, Subrahmanyan Chandrasekhar calculated thatin a white dwarf, the atomic matter can only fight the pull of gravity if its mass is less than 1.4 solar masses. In the case of stars with masses larger than this, the atoms in the center would crumble due to the immense pressure. In our universe however, there are countless stars with solar masses larger than 1.4 solar masses. When these stars begin to cool down, a completely different situation occurs. Going back to the idea of a gas composed entirely of neutrons, the stars with masses larger than 1.4 solar masses will undergo a similar fate. Gravity on the star will become stronger and stronger as the matter becomes more and more compressed, ultimately becoming a huge nucleus composed of neutrons. This nucleus will have an immense mass, but will be very tiny. This nucleus will also have a rather tiny radius due to the pressure caused by gravity. While it is unknown exactly whether or not these neutron stars do exist, there are a number of indirect observations that hint towards the existence of such stars. </p>
<p>In the late 1960&#8242;s,  astronomers discovered the first pulsar. These pulsars are objects that send out electromagnetic signals at a very precise rate (so precise, we could actually use these signals to keep track of time). The regularity of these signals is due to the rotation of the pulsars. By observing these signals, the radius of these pulsars can be concluded to be about 10km, very similar to the radius of these neutron stars. Which, leads us to an indirect proof of the existence of neutron stars.</p>
<p>In the event that the mass limit for dwarf stars is exceeded, we have seen what happens. But what would happen in the event that a neutron star with a large mass ,(larger than 2 solar masses, as calculated by John Wheeler in the early 1950&#8242;s) begins to cool down? In short, the star will inevitably collapse under the immense pressure from gravity. The star will completely bypass the state of a neutron star, and the neutrons left in the nucleus will fuse, forming a nuclues consisting of solely quarks. The behavior of quarks is largely unknown, but one factor is certain: quarks do not have the ability to resist the force of gravity. This quark-nucleus will too, collapse. This event is known as a gravitational collapse. J. Robert Oppenheimer was the first to examine this problem in the late 1930&#8242;s. Oppenheimer believed that nothing known to man can stop this gravitational collapse. He was right.</p>
<p>To show this, let&#8217;s look at an example of a star with a mass of 10 solar masses at the end of its radioactive phase. For sake of experiment, lets also assume that it is of perfect spherical symmetricity. By making this assumption, the spacetime out side of the star can be explained by the Schwarzschild solution. Using Einstein&#8217;s TOR, we know that the flow of time at the center of the star is slower than at the regions far from the center. As the collapse begins, the star is able to increase or decrease its radius as it collapses, without altering the Schwarzschild metric. The spherical symmetry remains intact throughout the collapse. In order to illustrate what happens in this scenario, let&#8217;s pretend we&#8217;re in a space shuttle quite a bit away from the center of the star, watching the collapse. The collapse begins slowly at first, but the diameter of the star decreases quicker and quicker. After a short time (hours to be exact) the circumference of the star will have decreased to just about four times the Schwarzschild value. The Schwarzschild radius, in this case, is about 30 kilometers (10 times the value of the sun&#8217;s SR). Since the stars circumference is only 4 times the Schwarzschild value, the flow of time on the surface of the star will be about 15% slower than over at the orbiting shuttle. Shortly after, the circumference of the star is only two times the Schwarzschild value, causing time to be 41% slower on the surface of the star compared to time flowing in the shuttle. As the collapse continues, the shuttle observers will see the collapse slowing up more and more dramatically. Now, slightly before the collapse begins, say an astronaut from the shuttle gets in a little pod and flies over to the surface of the star to observe the collapse more closely. In both the pod and the space ship sit a very accurate clocks, that both send radio signals to one another to keep track of time. The clocks advance synchronously to begin with. As the collapse progresses, the clock on the pod will produce these radio signals at slower and slower intervals due to the gravitational warp. Now, back to the collapse. The circumference of the star is two times its&#8217; Schwarzschild radius, and the time warp is very prevalent (41%).The difference between the circumference of the star  and the Schwarzschild circumference will be decreasing by a factor of two every 0.00007 seconds (Fritzsch 192).  After barely one one thousandth of one second, the difference has now decreased by a factor of 20,000.  The collapse continues,  and the circumference shrinks and shrinks, but never quite reaches the Schwarzschild circumference, approaching it asymptotically. Over in the space shuttle, the warp in time increases by a factor of two every 0.00014 seconds. This behaves the same way as the circumference, as the relative time on the space shuttle becomes more and more warped due to the collapse, eventually appearing to freeze up to the observers aboard the shuttle. As the circumference gets closer and closer to the Schwarzschild value, the collapse will appear to have stopped to the observers aboard the shuttle. To the outside world, the star wil have vanished, as time has become so warped that the light radiated from the star has become a low, radio frequency signal, unseen to the human eye.</p>
<p>Since the Schwarzschild circumference is reached at time <img src='http://s0.wp.com/latex.php?latex=t%3D+%5Cinfty&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='t= &#92;infty' title='t= &#92;infty' class='latex' />, at that time, no light whatsoever will be being emitted from the star. The immense gravitation of the star traps the light that is trying to be emitted, not letting it escape. Since light is no longer being emitted, the only influence the star has on the environment around it is by means of gravity. This gravitational impact on the surrounding environment is the only way to determine the location of this &#8220;black&#8221; star. These black stars have been dubbed as &#8220;black holes,&#8221; a phrase coined by John Wheeler.</p>
<p>In order for a particle to exit the atmosphere of a planet or star, a certain escape velocity must be reached. This escape velocity is determined by the planet/stars mass and diameter. The larger the mass, the larger the escape velocity must be to exit the atmosphere. The smaller the radius of the star, the higher the escape velocity must be as well, due to the stronger gravitational effect for objects with smaller radii. If the gravitational influence on a planet/star is so strong that light cannot escape, the photons will be deflected and fall back down to the surface. At the Schwarzschild radius, this effect occurs. This distance is also known as the event horizon. Particles, such as photons, move with a velocity of 186000 km/sec (the speed of light). In order to contain these particles, gravity must be extremely powerful. This leaves the idea that there may be any number of invisible stars due to the fact that their gravitational pull is so strong, not even light can escape. But back to our buddy the astronaut, who has been following the collapse of this star very closely.  For him, time never slows down; the warp of time is never seen by him due to his location relative to the center of the star. As long as the astronaut remains at a distance further than the Schwarzschild radius, the point of no return,  he will be able to return to the space ship and share his findings. How close he gets to the Schwarzschild radius determines how much time has actually passed on the space shuttle as opposed to the time in his pod. The astronaut may have only been in  the vicinity of the event horizon for a few days, but for the space shuttle, fifty years may have passed. Let&#8217;s assume the astronaut decides to stay, following the collapse. As he gets closer and closer to the event horizon, the effects ot the curvature of spacetime become increasingly more powerful. The force of the pull at his feet will be much stronger than the force at his head. This force will literally tear him to shreds, causing a painful death by gravitation. Now, let&#8217;s&#8217; assume the pod is capable of withstanding this gravitational force, and have the pod be at the surface of the collapsing star nearing the event horizon. Inside the pod, there is no effect of time warping. Since the probe is in its&#8217; own relative time, the probe is able to fly towards the center of the star, passing the event horizon without any special effect. Seen from the pod itself, time passes just as it does here on Earth. In the space shuttle, this progression would take an infinite amount of time, and would never been seen by the shuttle. Immediately after the pod passes the critical point, it has no chance of ever escaping. No radio waves or light is able to escape once it passes the event horizon, so the fate of the pod has been sealed. As the circumference becomes smaller and smaller, the  entire entity of stellar matter will have condensed to a single point, which of course, seems entirely impossible. This point will have an infinite matter density, and becomes what is known as a singularity in mathematics. Since the curvature of spacetime describes gravitation, as the singularity is approached, space and time actually switch roles. This shows that beyond the force of gravity keeping everything in the black hole, once inside the black hole a person would actually have to travel backwards in time to escape the black hole. Much beyond this point however, is largely un proven in the world of astrophysics. Many formulas and equations have been developed and used to predict what would happen at this point, but, we will never know what actually occurs at this point. It is from here, that my research will continue.</p>
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		<pubDate>Wed, 10 Jun 2009 15:59:52 +0000</pubDate>
		<dc:creator>jmckennon</dc:creator>
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