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		<title>Q: How can photons have energy and momentum, but no mass?</title>
		<link>http://www.askamathematician.com/?p=3685</link>
		<comments>http://www.askamathematician.com/?p=3685#comments</comments>
		<pubDate>Wed, 08 Sep 2010 22:56:02 +0000</pubDate>
		<dc:creator>Physicist</dc:creator>
				<category><![CDATA[-- By the Physicist]]></category>
		<category><![CDATA[Relativity]]></category>

		<guid isPermaLink="false">http://www.askamathematician.com/?p=3685</guid>
		<description><![CDATA[Physicist: Classically (according to Newton) kinetic energy is given by and the momentum is given by , where m is the mass and v is the velocity.  But if you plug in the mass and velocity for light you get .  But that&#8217;s no good.  If light didn&#8217;t carry energy, it wouldn&#8217;t be able to [...]]]></description>
			<content:encoded><![CDATA[<p><strong>Physicist</strong>: Classically (according to Newton) kinetic energy is given by <img src='http://s.wordpress.com/latex.php?latex=E%3D%5Cfrac%7B1%7D%7B2%7Dmv%5E2&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='E=\frac{1}{2}mv^2' title='E=\frac{1}{2}mv^2' class='latex' />and the momentum is given by <img src='http://s.wordpress.com/latex.php?latex=P%3Dmv&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='P=mv' title='P=mv' class='latex' />, where m is the mass and v is the velocity.  But if you plug in the mass and velocity for light you get <img src='http://s.wordpress.com/latex.php?latex=E%3D%5Cfrac%7B1%7D%7B2%7D0c%5E2%3D0&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='E=\frac{1}{2}0c^2=0' title='E=\frac{1}{2}0c^2=0' class='latex' />.  But that&#8217;s no good.  If light didn&#8217;t carry energy, it wouldn&#8217;t be able to heat stuff up.</p>
<p>The difficulty comes from the fact that Newton&#8217;s laws paint an incomplete (and ultimately incorrect) picture.  When relativity came along it was revealed that there&#8217;s a fundamental difference in the physics of the massive and the massless.  Relativity makes the (experimentally backed) assumptions that: #1) it doesn&#8217;t matter whether, or how fast, you&#8217;re moving (all physical laws stay the same) and #2) the speed of light is invariant (always the same to everyone).</p>
<p>Any object with mass travels slower than light and so may as well be stationary (#1).</p>
<p>Anything with zero mass always travels at the speed of light.  But since the speed-of-light is always the speed-of-light to everyone (#2) there&#8217;s no way for these objects to ever be stationary (unlike massive stuff).  Vive la différence des lois!  It&#8217;s not important here, but things (like light) that travel at the speed of light never experience the passage of time.  Isn&#8217;t that awesome?</p>
<p>The point is: light and ordinary matter are very different, and the laws that govern them are just as different.</p>
<p style="text-align: center;">
<div id="attachment_3711" class="wp-caption aligncenter" style="width: 419px"><a href="http://www.askamathematician.com/wp-content/uploads/2010/09/lightandmatter.jpg"><img class="size-full wp-image-3711  " title="yet surprisingly similar" src="http://www.askamathematician.com/wp-content/uploads/2010/09/lightandmatter.jpg" alt="" width="409" height="223" /></a><p class="wp-caption-text">Light and Matter: different</p></div>
<p>That being said, in 1905 <a href="http://en.wikipedia.org/wiki/Special_relativity#Relativistic_mechanics" onclick="pageTracker._trackPageview('/outgoing/en.wikipedia.org/wiki/Special_relativity_Relativistic_mechanics?referer=');">Einstein managed to write a law that works whenever</a>: <img src='http://s.wordpress.com/latex.php?latex=E%5E2%3DP%5E2c%5E2%2Bm%5E2c%5E4&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='E^2=P^2c^2+m^2c^4' title='E^2=P^2c^2+m^2c^4' class='latex' />.  The same year (the <span style="text-decoration: underline;">same</span> <em>freaking</em> year) he figured out that light is both a particle and a wave and that the energy of a photon isn&#8217;t governed by it&#8217;s mass or it&#8217;s velocity (like matter), but instead is governed entirely by f, it&#8217;s frequency: E=hf, where h is <a href="http://en.wikipedia.org/wiki/Planck_constant" onclick="pageTracker._trackPageview('/outgoing/en.wikipedia.org/wiki/Planck_constant?referer=');">Planck&#8217;s constant</a>.</p>
<p>For light m=0, so E=Pc (energy and momentum are proportional).  Notice that you can never have zero momentum, since something with zero mass and zero energy isn&#8217;t something, it&#8217;s nothing.  This is just another way of saying that light can never be stationary.</p>
<p>Also!  Say you have an object with mass m, that isn&#8217;t moving (P=0).  Then you get: E=mc<sup>2</sup> (awesome)!</p>
<p> &nbsp; </p>
<p><em>Unrelated tangen</em>t: It took a little while, but the laws governing the massive and the massless are even more inter-related than the &#8216;Stein originally thought.  He figured out that the energy of a photon is related to it&#8217;s frequency (E=hf), but why are photons so special?  Why do they get to have frequencies?  They&#8217;re not special.  Years later (1924) de Broglie drew the most natural line from Einstein&#8217;s various equations from light to matter.  <img src='http://s.wordpress.com/latex.php?latex=mc%5E2%3DE%3Dhf&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='mc^2=E=hf' title='mc^2=E=hf' class='latex' />  So for a given amount of matter you can find it&#8217;s frequency.  Holy crap!  Everything has a frequency!</p>
<p>On the off chance that anyone out there got unduly excited about that last statement: the frequencies never go out of wack, you can&#8217;t tune them, more importantly they are utterly unimportant on the Human scale, or even the single-cell scale, and don&#8217;t ever buy a bracelet or anything else with &#8220;quantum&#8221; in the name.</p>
<div id="attachment_3715" class="wp-caption aligncenter" style="width: 309px"><a href="http://www.askamathematician.com/wp-content/uploads/2010/09/q-ray-bracelet.jpg"><img class="size-full wp-image-3715" title="q-ray-bracelet" src="http://www.askamathematician.com/wp-content/uploads/2010/09/q-ray-bracelet.jpg" alt="" width="299" height="299" /></a><p class="wp-caption-text">No, no, no, no, no, no, no, no, no.</p></div>
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		<title>Q: If you were on the inside of the Sun falling in, the matter closer to the surface doesn&#8217;t affect your acceleration, but the matter closer to the core does. Why is that?</title>
		<link>http://www.askamathematician.com/?p=3649</link>
		<comments>http://www.askamathematician.com/?p=3649#comments</comments>
		<pubDate>Thu, 02 Sep 2010 18:04:50 +0000</pubDate>
		<dc:creator>Physicist</dc:creator>
				<category><![CDATA[-- By the Physicist]]></category>
		<category><![CDATA[Astronomy]]></category>
		<category><![CDATA[Math]]></category>
		<category><![CDATA[Physics]]></category>

		<guid isPermaLink="false">http://www.askamathematician.com/?p=3649</guid>
		<description><![CDATA[The original question was: Plait talks about the &#8220;physics of solid bodies&#8221; and why, specifically, if you were on the inside of the Sun falling in, the matter &#8220;behind&#8221; you- closer to the surface- doesn&#8217;t affect your acceleration at all, and all that matters is the matter &#8220;in front&#8221; of you- closer to the core. [...]]]></description>
			<content:encoded><![CDATA[<p><em>The original question was</em>: Plait talks about the &#8220;physics of solid bodies&#8221; and why, specifically, if you were on the inside of the Sun falling in, the matter &#8220;behind&#8221; you- closer to the surface- doesn&#8217;t affect your acceleration at all, and all that matters is the matter &#8220;in front&#8221; of you- closer to the core. Why is that?</p>
<p><strong>Physicist</strong>: The short, uninteresting answer is that the gravity from any layer above you cancels itself out.  If you take any sample layer above you, and you happen to be closer to one side, then you&#8217;ll find that the side you&#8217;re closer to has more pull on you, but there&#8217;s less of it.  Conversely, the far side has less pull, but there&#8217;s more of it.  For a sphere (but not a ring) these forces cancel exactly.  So as you fall in you can ignore all the layers above you.</p>
<div id="attachment_3672" class="wp-caption aligncenter" style="width: 290px"><a href="http://www.askamathematician.com/wp-content/uploads/2010/09/balanced.jpg"><img class="size-full wp-image-3672" title="balanced" src="http://www.askamathematician.com/wp-content/uploads/2010/09/balanced.jpg" alt="" width="280" height="316" /></a><p class="wp-caption-text">Pick a layer.  Anything inside will experience exactly the same amount of pull in every direction, and so, no pull at all.</p></div>
<p><strong>Answer gravy</strong>: One of the greatest tools in the physicist&#8217;s tool kit is &#8220;<a href="http://en.wikipedia.org/wiki/Gaussian_surface" onclick="pageTracker._trackPageview('/outgoing/en.wikipedia.org/wiki/Gaussian_surface?referer=');">Gaussian Surfaces</a>&#8220;.  They let you shortcut really difficult math problems using pictures and a little reasoning.  Even better, you <a href="http://www.mensa.org/" onclick="pageTracker._trackPageview('/outgoing/www.mensa.org/?referer=');">come across smarter than perhaps you deserve</a>, which is a big plus.</p>
<p>A Gaussian surface is nothing more than an invisible bubble that you draw in space.  The &#8220;<a href="http://en.wikipedia.org/wiki/Inverse-square_law" onclick="pageTracker._trackPageview('/outgoing/en.wikipedia.org/wiki/Inverse-square_law?referer=');">inverse square law</a>&#8221; of gravity can actually be rewritten as &#8220;the total amount of gravity pointing into the bubble is proportional to the amount of matter inside the bubble&#8221;.  The arrangement of matter (both inside and outside the bubble) certainly changes how gravity points into (or out) of the bubble, but the <em>total</em> amount of gravity pointing through depends only on the amount of matter inside.</p>
<div id="attachment_3655" class="wp-caption aligncenter" style="width: 459px"><a href="http://www.askamathematician.com/wp-content/uploads/2010/09/situations.jpg"><img class="size-full wp-image-3655" title="Here's a fun experiment you can do at home!  Stand back and squint.  See if you can see the googly-eyed spitting clown." src="http://www.askamathematician.com/wp-content/uploads/2010/09/situations.jpg" alt="" width="449" height="356" /></a><p class="wp-caption-text">(upper left) when the mass is symmetrical and in the middle, then the gravity is exactly the same everywhere on the surface.  (upper right) if the matter is off to the side, then gravity will be stronger, or weaker, or point in different directions at different points on the surface, but the total stays the same.  (bottom) matter outside of the Gaussian surface can affect how gravity pokes through, but it can&#39;t affect the total. </p></div>
<p>Now say that your bubble is an exact fit around a sphere of matter.  Everything is perfectly symmetric, so there&#8217;s no reason for gravity to be any stronger or weaker anywhere and, given the amount of mass inside the sphere, you can figure out how strong the gravity is.  Now say you add more matter, but uniformly, on top of your original sphere.</p>
<div id="attachment_3661" class="wp-caption aligncenter" style="width: 460px"><a href="http://www.askamathematician.com/wp-content/uploads/2010/09/layers.jpg"><img class="size-full wp-image-3661" title="layers" src="http://www.askamathematician.com/wp-content/uploads/2010/09/layers.jpg" alt="" width="450" height="210" /></a><p class="wp-caption-text">In both situations the total amount of matter inside the bubble is the same, and everything is nice and symmetrical, so the gravity along the surface of the bubble is the same.</p></div>
<p>The matter inside the sphere has remained the same, so the pull at the surface of that sphere remains the same.  As a result, so long as the matter above is at least fairly symmetrical (which is the case for any planet or star you can think of), you can ignore the layers above the surface of the bubble.</p>
<p>Specifically, as you fall farther and farther into the Sun (or Earth, or whatever else is round) you can figure out how much gravity you&#8217;re feeling by using a Gaussian surface, for which you only need the matter below you.  The layers above will exert no pull on you, and you will exert no net pull on them (for every action/force there is an equal and opposite reaction/force).</p>
<p><em>This part has nothing to do with the question</em>: You can use Gaussian surfaces to prove some surprising things.  Specifically: Dyson spheres work, and black holes have no more gravity than the stars they came from.</p>
<p>From the last argument (above) you know that the layers above you have no net gravitational effect on you.  But what if you fall a little way into a planet, and suddenly find that the inside of it is completely hollow?  One you&#8217;re inside <span style="text-decoration: underline;">all</span> the layers are layers above you.  So there&#8217;s no gravity at all inside of a large hollow sphere (at least, none caused by the sphere).  If you built a really huge sphere around a star you&#8217;d have a &#8220;Dyson&#8217;s sphere&#8221;.  The sphere doesn&#8217;t pull the star, and the star doesn&#8217;t pull the sphere.  It&#8217;s stable no matter where the star is inside the ball.  So long as no one shoves anything, everything will just float neutrally right where it is.</p>
<div id="attachment_3665" class="wp-caption aligncenter" style="width: 460px"><a href="http://www.askamathematician.com/wp-content/uploads/2010/09/dyson.jpg"><img class="size-full wp-image-3665  " title="jk!  It was &quot;Freeman Dyson&quot;.  No fate, y'all!" src="http://www.askamathematician.com/wp-content/uploads/2010/09/dyson.jpg" alt="" width="450" height="174" /></a><p class="wp-caption-text">(left) the set up for a Dyson Sphere.  The perfectly spherical shell has no gravitational effect on anything inside the sphere, including the star.  (center) Miles Dyson, inventor of the Dyson sphere, and future inventor of Skynet.  (right) an artist&#39;s interpretation of a Dyson sphere.</p></div>
<p>Now, put a Gaussian surface around a star.  There&#8217;s a certain amount of matter in the star, and that tells you how much gravity is pointing through the surface.  If the star shrinks, who cares?  Same amount of mass = same amount of gravity.</p>
<div id="attachment_3668" class="wp-caption aligncenter" style="width: 472px"><a href="http://www.askamathematician.com/wp-content/uploads/2010/09/collapse.jpg"><img class="size-full wp-image-3668" title="collapse" src="http://www.askamathematician.com/wp-content/uploads/2010/09/collapse.jpg" alt="" width="462" height="226" /></a><p class="wp-caption-text">The gravity through the outer Gaussian surface stays the same, since both contain the same amount of matter.  The gravity through the inner Gaussian surface increases dramatically after the star collapses, because it contains all of the star&#39;s mass, instead of just a small part of it.</p></div>
<p>But if you draw a small Gaussian surface around the core of the star you&#8217;ll find that the gravity along the surface is small, because there is (relatively) little mass inside of it.  If for some reason you found yourself in the center of the Sun, you&#8217;d be floating in zero G&#8217;s.  Point of fact; you&#8217;d also be on fire.</p>
<p>Now when the Star collapses, all of the matter is drawn into a tiny region.  <span style="text-decoration: underline;">Both</span> spheres (see diagram on the right) now contain all of the star&#8217;s matter, and thus the same <span style="text-decoration: underline;">total</span> amount of gravity pokes through them.  The only difference is that the inner sphere is smaller, so the gravity has to be more intense to get the same total as the outer sphere.<br />
Black holes do have very intense gravity, but only in the region where the star used to be.</p>
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		<title>Q: How do surge protectors work?</title>
		<link>http://www.askamathematician.com/?p=3608</link>
		<comments>http://www.askamathematician.com/?p=3608#comments</comments>
		<pubDate>Sun, 29 Aug 2010 13:32:44 +0000</pubDate>
		<dc:creator>Physicist</dc:creator>
				<category><![CDATA[-- By the Physicist]]></category>
		<category><![CDATA[Engineering]]></category>
		<category><![CDATA[Physics]]></category>

		<guid isPermaLink="false">http://www.askamathematician.com/?p=3608</guid>
		<description><![CDATA[Physicist: To control power in a house or an outlet you&#8217;d generally use a fuse.  But fuses are slow, they need time to heat up.  A surge (or the faster &#8220;spike&#8221;) happen too fast, so reacting to a surge is no good.  Dealing with a surge properly has to be built into the nature of [...]]]></description>
			<content:encoded><![CDATA[<p><strong>Physicist</strong>: To control power in a house or an outlet you&#8217;d generally use a fuse.  But fuses are slow, they need time to heat up.  A surge (or the faster &#8220;spike&#8221;) happen too fast, so <em>reacting</em> to a surge is no good.  Dealing with a surge properly has to be built into the nature of the machine.  There are two (general) ways to do this: &#8220;choking&#8221; and &#8220;shunting&#8221;.</p>
<p>Shunting is what a sink does when it over-flows.  You never worry about water getting to the ceiling above your sink, because the moment the water gets as high as the edge the sink starts overflowing and stops filling up.  The ceiling is protected by the innate nature of the sink (which doesn&#8217;t need to react, it just &#8220;does&#8221;).</p>
<div id="attachment_3638" class="wp-caption aligncenter" style="width: 461px"><a href="http://www.askamathematician.com/wp-content/uploads/2010/08/shunt.jpg"><img class="size-full wp-image-3638" title="shunt" src="http://www.askamathematician.com/wp-content/uploads/2010/08/shunt.jpg" alt="" width="451" height="214" /></a><p class="wp-caption-text">If the water in the sink gets too high it simply falls over the edge.  Similarly, if the voltage in the load line gets too high, the resistance of the varistor (variable resistor) drops from very, very high to nearly zero. This allows a connection to ground, and electricity can then flow out of the circut, instead of through it.</p></div>
<p>The zero-finesse surge protector is just a varistor placed between the power line and the ground line.  The ground line (in this metaphor) is the floor around the sink, where all the overflowing water gets dumped.</p>
<p>While there is some fancy quantum mechanics tied up in varistors (I&#8217;m talkin&#8217; <a href="http://en.wikipedia.org/wiki/Electronic_band_structure" onclick="pageTracker._trackPageview('/outgoing/en.wikipedia.org/wiki/Electronic_band_structure?referer=');">valence and conduction bands</a> here), what really gets my physics juices jumping is waves and frequencies.</p>
<p>The other type of surge protector is essentially a &#8220;<a href="http://en.wikipedia.org/wiki/Band-pass_filter" onclick="pageTracker._trackPageview('/outgoing/en.wikipedia.org/wiki/Band-pass_filter?referer=');">band pass filter</a>&#8221; centered around 60 hz (the frequency of the current in the electrical grid).  You can think of this as the &#8220;radio&#8221; of the surge protector tuning into the &#8220;station&#8221; of the wall outlet (and tuning out everything else).</p>
<p>Now, I hope this doesn&#8217;t come as a shock to anyone, but most surges (e.g., lightning strikes) are fast.  It turns out that because of the <a href="http://www.askamathematician.com/?p=3270">Uncertainty Principle</a>, things that happen really fast are necessarily spread out over a lot of frequencies (small <em>time</em> uncertainty means large <em>frequency</em> uncertainty).</p>
<div id="attachment_3644" class="wp-caption aligncenter" style="width: 355px"><a href="http://www.askamathematician.com/wp-content/uploads/2010/08/surge.jpg"><img class="size-full wp-image-3644" title="surge" src="http://www.askamathematician.com/wp-content/uploads/2010/08/surge.jpg" alt="" width="345" height="620" /></a><p class="wp-caption-text">The frequency spectrum for a wall socket (top) and a lightning bolt (bottom).  Keep in mind that this is not a graph in time, but in frequency.  If you played a single note on a piano you&#39;d get something like the top graph, and if you banged on all the keys at once you&#39;d get something like the bottom graph.</p></div>
<p>So while a power surge may have a lot of energy overall, the amount of energy right around 60 Hz, where it can get into the circuit, should be fairly small.  You can build chokes to limit the frequencies that get past the surge protector by using carefully tuned <a href="http://en.wikipedia.org/wiki/RLC_circuit" onclick="pageTracker._trackPageview('/outgoing/en.wikipedia.org/wiki/RLC_circuit?referer=');">LRC circuits</a>, but generally (since most of the energy is in frequencies much higher than 60 Hz) you can just build a &#8220;low pass filter&#8221;.</p>
<p>Which is just fancy talk for &#8220;an inductor&#8221;.</p>
<p>Which is just fancy talk for &#8220;a <a href="http://en.wikipedia.org/wiki/Inductor" onclick="pageTracker._trackPageview('/outgoing/en.wikipedia.org/wiki/Inductor?referer=');">coil of wire</a>&#8220;.</p>
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		<title>Relativity and Quantum Mechanics: the elevator pitch</title>
		<link>http://www.askamathematician.com/?p=3616</link>
		<comments>http://www.askamathematician.com/?p=3616#comments</comments>
		<pubDate>Tue, 24 Aug 2010 19:45:08 +0000</pubDate>
		<dc:creator>Physicist</dc:creator>
				<category><![CDATA[-- By the Physicist]]></category>
		<category><![CDATA[Quantum Theory]]></category>
		<category><![CDATA[Relativity]]></category>

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		<description><![CDATA[Physicist: A woman on the subway, about two stations away from her stop, asked us &#8220;what are relativity and quantum mechanics?&#8221; So, this is a two-stop elevator pitch for the two most pivotal sciences since slicedbreadology. Relativity: Speed is just distance over time (as in &#8220;miles per hour&#8221;). Normally when you change you&#8217;re own speed, [...]]]></description>
			<content:encoded><![CDATA[<p><strong>Physicist</strong>: A woman on the subway, about two stations away from her stop, asked us &#8220;what are relativity and quantum mechanics?&#8221;<br />
So, this is a two-stop <a href="http://en.wikipedia.org/wiki/Elevator_pitch" onclick="pageTracker._trackPageview('/outgoing/en.wikipedia.org/wiki/Elevator_pitch?referer=');">elevator pitch</a> for the two most pivotal sciences since slicedbreadology.</p>
<div id="attachment_3618" class="wp-caption aligncenter" style="width: 297px"><a href="http://www.askamathematician.com/wp-content/uploads/2010/08/Elevator.gif"><img class="size-full wp-image-3618 " title="Elevator" src="http://www.askamathematician.com/wp-content/uploads/2010/08/Elevator.gif" alt="" width="287" height="287" /></a><p class="wp-caption-text">Elevators: Wonderful, mechanical rooms, quietly skirting the ever-thinning line between broom closet and robot.</p></div>
<p><em>Relativity</em>: Speed is just distance over time (as in &#8220;miles per hour&#8221;). Normally when you change you&#8217;re own speed, the speeds of everything else changes (for your point of view).  For example, if you&#8217;re walking slowly down the street everyone else will be moving quickly (and, for the sake of this example, in the same direction), but if you pick up the pace and walk normally, then everyone else will barely be moving at all.</p>
<p>But the speed of light is different. No matter how you move, it will always stay the same. Since that particular speed refuses to change, distance and time have to change instead.  Relativity is the study of how distance and time change with speed, and the consequences that follow from those changes.</p>
<p><em>Quantum Mechanics</em>: When you look at very, very small objects, like individual particles, you begin to find that they don&#8217;t behave right. If particles were like ordinary objects, but smaller like tiny billiard balls, then you&#8217;d expect them to act like ordinary (but tiny) objects.  Instead, they &#8220;ooze&#8221; from place to place, move through impassable barriers, exist in several places at the same time, and interfere with each other. It&#8217;s impossible even to say exactly where they are.<br />
All of these are impossible (or at least very unlikely) behaviors for solid &#8220;particle-ish&#8221; objects.  But all of these behaviors are explained, and even expected, if all of matter is actually some kind of wave.<br />
So, quantum mechanics is the (more-accurate-than-every-other-science) study of the universe from the perspective that everything, at the lowest levels, is made up of some kind of waves.</p>
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		<title>Q: Why are orbits elliptical?  Why is the Sun in one focus, and what&#8217;s in the other?</title>
		<link>http://www.askamathematician.com/?p=3501</link>
		<comments>http://www.askamathematician.com/?p=3501#comments</comments>
		<pubDate>Thu, 19 Aug 2010 03:34:55 +0000</pubDate>
		<dc:creator>Physicist</dc:creator>
				<category><![CDATA[-- By the Physicist]]></category>
		<category><![CDATA[Astronomy]]></category>
		<category><![CDATA[Physics]]></category>

		<guid isPermaLink="false">http://www.askamathematician.com/?p=3501</guid>
		<description><![CDATA[Physicist: This question always bothered me too.  The short answer is: it falls out of the math.  Specifically, the math of first year physics and second year calculus.  The fact that the Sun is in one focus is just one of those things.  It&#8217;s nothing special.  Even less special is the other focus, which contains [...]]]></description>
			<content:encoded><![CDATA[<p><strong>Physicist</strong>: This question always bothered me too.  The short answer is: it falls out of the math.  Specifically, the math of first year physics and second year calculus.  The fact that the Sun is in one focus is just one of those things.  It&#8217;s nothing special.  Even less special is the other focus, which contains nothing at all.</p>
<div id="attachment_3593" class="wp-caption aligncenter" style="width: 460px"><a href="http://www.askamathematician.com/wp-content/uploads/2010/08/otherfocus.jpg"><img class="size-full wp-image-3593 " title="otherfocus" src="http://www.askamathematician.com/wp-content/uploads/2010/08/otherfocus.jpg" alt="" width="450" height="463" /></a><p class="wp-caption-text">Ellipses and their foci have a lot of useful properties.  It so happens that an orbiting object traces out an ellipse, with the thing it orbits around at one of the focuses.  Coincidence?  Yes.</p></div>
<p>I can&#8217;t find a good intuitive reason why orbits are elliptical.  In fact, I can&#8217;t even find a mathematical derivation.  So, because it should be found somewhere, I&#8217;ll leave the derivation floating in the answer gravy.</p>
<p><strong>Answer gravy</strong>: The force of gravity is usually written as <img src='http://s.wordpress.com/latex.php?latex=ma%3DF%3D-%5Cfrac%7BGMm%7D%7BR%5E2%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='ma=F=-\frac{GMm}{R^2}' title='ma=F=-\frac{GMm}{R^2}' class='latex' />.  You can rewrite this using vector notation as <img src='http://s.wordpress.com/latex.php?latex=m%5Cddot%7B%5Cvec%7Bx%7D%7D%3D-%5Cfrac%7BGMm%5Cvec%7Bx%7D%7D%7B%7C%5Cvec%7Bx%7D%7C%5E3%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='m\ddot{\vec{x}}=-\frac{GMm\vec{x}}{|\vec{x}|^3}' title='m\ddot{\vec{x}}=-\frac{GMm\vec{x}}{|\vec{x}|^3}' class='latex' />, where the dot on top is a time derivative.  To keep the notation both standard and confusing, <img src='http://s.wordpress.com/latex.php?latex=%5Cvec%7Bx%7D%3D%28x%2Cy%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\vec{x}=(x,y)' title='\vec{x}=(x,y)' class='latex' />.</p>
<img src='http://s.wordpress.com/latex.php?latex=%5Cbegin%7Barray%7D%7Bll%7Dm%5Cddot%7B%5Cvec%7Bx%7D%7D%3D-%5Cfrac%7BGMm%5Cvec%7Bx%7D%7D%7B%7C%5Cvec%7Bx%7D%7C%5E3%7D%5C%5C%5CRightarrow%5Cddot%7B%5Cvec%7Bx%7D%7D%3D-%5Cfrac%7BGM%7D%7B%7C%5Cvec%7Bx%7D%7C%5E3%7D%5Cvec%7Bx%7D%5C%5C%5CRightarrow%5Cdot%7B%5Cvec%7Bx%7D%7D%5Ccdot%5Cddot%7B%5Cvec%7Bx%7D%7D%3D-%5Cfrac%7BGM%7D%7B%7C%5Cvec%7Bx%7D%7C%5E3%7D%5Cvec%7Bx%7D%5Ccdot%5Cdot%7B%5Cvec%7Bx%7D%7D%5C%5C%5CRightarrow%5Cdot%7B%5Cvec%7Bx%7D%7D%5Ccdot%5Cddot%7B%5Cvec%7Bx%7D%7D%3D-%5Cfrac%7BGM%7D%7B%28%5Cvec%7Bx%7D%5Ccdot%5Cvec%7Bx%7D%29%5E%7B3%2F2%7D%7D%5Cvec%7Bx%7D%5Ccdot%5Cdot%7B%5Cvec%7Bx%7D%7D%26%5Cleft%5C%7B%5Cvec%7Bx%7D%5Ccdot%5Cvec%7Bx%7D%3D%7C%5Cvec%7Bx%7D%7C%5E2%5Cright.%5C%5C%5CRightarrow%5Cfrac%7Bd%7D%7Bdt%7D%5Cleft%5B%5Cfrac%7B1%7D%7B2%7D%20%5Cdot%7B%5Cvec%7Bx%7D%7D%5Ccdot%5Cdot%7B%5Cvec%7Bx%7D%7D%5Cright%5D%3D%5Cfrac%7Bd%7D%7Bdt%7D%20%5Cleft%5B%5Cfrac%7BGM%7D%7B%28%5Cvec%7Bx%7D%5Ccdot%5Cvec%7Bx%7D%29%5E%7B1%2F2%7D%7D%20%5Cright%5D%26%5Cleft%5C%7B%5Cfrac%7Bd%7D%7Bdt%7D%5Cleft%28%5Cvec%7Bx%7D%5Ccdot%5Cvec%7Bx%7D%5Cright%29%3D2%5Cvec%7Bx%7D%5Ccdot%5Cdot%7B%5Cvec%7Bx%7D%7D%5Cright.%5C%5C%5CRightarrow%20%5Cfrac%7Bd%7D%7Bdt%7D%5Cleft%5B%7C%5Cdot%7B%5Cvec%7Bx%7D%7D%7C%5E2%20%5Cright%5D%3D%5Cfrac%7Bd%7D%7Bdt%7D%5Cleft%5B%5Cfrac%7B2GM%7D%7B%7C%5Cvec%7Bx%7D%7C%7D%5Cright%5D%5C%5C%5CRightarrow%20%7C%5Cdot%7B%5Cvec%7Bx%7D%7D%7C%5E2%3D%5Cfrac%7B2GM%7D%7B%7C%5Cvec%7Bx%7D%7C%7D%2Bc%5Cend%7Barray%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\begin{array}{ll}m\ddot{\vec{x}}=-\frac{GMm\vec{x}}{|\vec{x}|^3}\\\Rightarrow\ddot{\vec{x}}=-\frac{GM}{|\vec{x}|^3}\vec{x}\\\Rightarrow\dot{\vec{x}}\cdot\ddot{\vec{x}}=-\frac{GM}{|\vec{x}|^3}\vec{x}\cdot\dot{\vec{x}}\\\Rightarrow\dot{\vec{x}}\cdot\ddot{\vec{x}}=-\frac{GM}{(\vec{x}\cdot\vec{x})^{3/2}}\vec{x}\cdot\dot{\vec{x}}&amp;\left\{\vec{x}\cdot\vec{x}=|\vec{x}|^2\right.\\\Rightarrow\frac{d}{dt}\left[\frac{1}{2} \dot{\vec{x}}\cdot\dot{\vec{x}}\right]=\frac{d}{dt} \left[\frac{GM}{(\vec{x}\cdot\vec{x})^{1/2}} \right]&amp;\left\{\frac{d}{dt}\left(\vec{x}\cdot\vec{x}\right)=2\vec{x}\cdot\dot{\vec{x}}\right.\\\Rightarrow \frac{d}{dt}\left[|\dot{\vec{x}}|^2 \right]=\frac{d}{dt}\left[\frac{2GM}{|\vec{x}|}\right]\\\Rightarrow |\dot{\vec{x}}|^2=\frac{2GM}{|\vec{x}|}+c\end{array}' title='\begin{array}{ll}m\ddot{\vec{x}}=-\frac{GMm\vec{x}}{|\vec{x}|^3}\\\Rightarrow\ddot{\vec{x}}=-\frac{GM}{|\vec{x}|^3}\vec{x}\\\Rightarrow\dot{\vec{x}}\cdot\ddot{\vec{x}}=-\frac{GM}{|\vec{x}|^3}\vec{x}\cdot\dot{\vec{x}}\\\Rightarrow\dot{\vec{x}}\cdot\ddot{\vec{x}}=-\frac{GM}{(\vec{x}\cdot\vec{x})^{3/2}}\vec{x}\cdot\dot{\vec{x}}&amp;\left\{\vec{x}\cdot\vec{x}=|\vec{x}|^2\right.\\\Rightarrow\frac{d}{dt}\left[\frac{1}{2} \dot{\vec{x}}\cdot\dot{\vec{x}}\right]=\frac{d}{dt} \left[\frac{GM}{(\vec{x}\cdot\vec{x})^{1/2}} \right]&amp;\left\{\frac{d}{dt}\left(\vec{x}\cdot\vec{x}\right)=2\vec{x}\cdot\dot{\vec{x}}\right.\\\Rightarrow \frac{d}{dt}\left[|\dot{\vec{x}}|^2 \right]=\frac{d}{dt}\left[\frac{2GM}{|\vec{x}|}\right]\\\Rightarrow |\dot{\vec{x}}|^2=\frac{2GM}{|\vec{x}|}+c\end{array}' class='latex' />
<p>c is an &#8220;<a href="http://en.wikipedia.org/wiki/Constant_of_integration" onclick="pageTracker._trackPageview('/outgoing/en.wikipedia.org/wiki/Constant_of_integration?referer=');">integration constant</a>&#8220;, it can be any number.  Jumping over to <a href="http://en.wikipedia.org/wiki/Polar_coordinate_system#Converting_between_polar_and_Cartesian_coordinates" onclick="pageTracker._trackPageview('/outgoing/en.wikipedia.org/wiki/Polar_coordinate_system_Converting_between_polar_and_Cartesian_coordinates?referer=');">polar coordinates</a> <img src='http://s.wordpress.com/latex.php?latex=%5Cleft%28%5Cbegin%7Barray%7D%7Bl%7Dx%3DR%5Ccos%7B%28%5Ctheta%29%7D%5C%5Cy%3DR%5Csin%7B%28%5Ctheta%29%7D%5Cend%7Barray%7D%5Cright%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\left(\begin{array}{l}x=R\cos{(\theta)}\\y=R\sin{(\theta)}\end{array}\right)' title='\left(\begin{array}{l}x=R\cos{(\theta)}\\y=R\sin{(\theta)}\end{array}\right)' class='latex' /> you can rewrite the usual velocity in terms of how fast you&#8217;re moving toward or away from the Sun <img src='http://s.wordpress.com/latex.php?latex=%28%5Cdot%7BR%7D%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='(\dot{R})' title='(\dot{R})' class='latex' /> and how fast you&#8217;re going around <img src='http://s.wordpress.com/latex.php?latex=%28%5Cdot%7B%5Ctheta%7D%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='(\dot{\theta})' title='(\dot{\theta})' class='latex' />.</p>
<img src='http://s.wordpress.com/latex.php?latex=%5Cbegin%7Barray%7D%7Bll%7D%5CRightarrow%5Cdot%7BR%7D%5E2%2BR%5E2%5Cdot%7B%5Ctheta%7D%5E2%3D%5Cfrac%7B2GM%7D%7BR%7D%2Bc%26%5Cleft%5C%7B%5Cbegin%7Barray%7D%7Bll%7D%7C%5Cvec%7Bx%7D%7C%3DR%5C%5C%7C%5Cdot%7B%5Cvec%7Bx%7D%7D%7C%5E2%3D%5Cdot%7BR%7D%5E2%2BR%5E2%5Cdot%7B%5Ctheta%7D%5E2%5Cend%7Barray%7D%5Cright.%5C%5C%5CRightarrow%5Cleft%28%5Cfrac%7BdR%7D%7Bd%5Ctheta%7D%5Cdot%7B%5Ctheta%7D%20%5Cright%29%5E2%2BR%5E2%5Cdot%7B%5Ctheta%7D%5E2%3D%5Cfrac%7B2GM%7D%7BR%7D%2Bc%26%5Cleft%5C%7B%5Cfrac%7BdR%7D%7Bdt%7D%3D%5Cfrac%7BdR%7D%7Bd%5Ctheta%7D%5Cfrac%7Bd%5Ctheta%7D%7Bdt%7D%5Cright.%5C%5C%5CRightarrow%5Cleft%28%5Cleft%28%5Cfrac%7BdR%7D%7Bd%5Ctheta%7D%5Cright%29%5E2%2BR%5E2%5Cright%29%5Cdot%7B%5Ctheta%7D%5E2%3D%5Cfrac%7B2GM%7D%7BR%7D%2Bc%5C%5C%5CRightarrow%5Cleft%28%5Cleft%28%5Cfrac%7BdR%7D%7Bd%5Ctheta%7D%5Cright%29%5E2%2BR%5E2%5Cright%29%5Cfrac%7BL%5E2%7D%7BR%5E4%7D%3D%5Cfrac%7B2GM%7D%7BR%7D%2Bc%26%5Cleft%5C%7BR%5E2%5Cdot%7B%5Ctheta%7D%3DL%5Cright.%5C%5C%5CRightarrow%5Cleft%28%5Cfrac%7B1%7D%7BR%5E2%7D%5Cfrac%7BdR%7D%7Bd%5Ctheta%7D%5Cright%29%5E2%2B%5Cfrac%7B1%7D%7BR%5E2%7D%3D%5Cfrac%7B2GM%7D%7BL%5E2R%7D%2B%5Cfrac%7Bc%7D%7BL%5E2%7D%5C%5C%5CRightarrow%5Cleft%28%5Cfrac%7B1%7D%7BR%5E2%7D%5Cfrac%7BdR%7D%7Bd%5Ctheta%7D%5Cright%29%5E2%2B%5Cfrac%7B1%7D%7BR%5E2%7D%3D2%5Calpha%5Cfrac%7B1%7D%7BR%7D%20%2BC%26%5Cleft%5C%7B%5Cbegin%7Barray%7D%7Bl%7DC%3D%5Cfrac%7Bc%7D%7BL%5E2%7D%5C%5C%5Calpha%3D%5Cfrac%7BGM%7D%7BL%5E2%7D%5Cend%7Barray%7D%5Cright.%5Cend%7Barray%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\begin{array}{ll}\Rightarrow\dot{R}^2+R^2\dot{\theta}^2=\frac{2GM}{R}+c&amp;\left\{\begin{array}{ll}|\vec{x}|=R\\|\dot{\vec{x}}|^2=\dot{R}^2+R^2\dot{\theta}^2\end{array}\right.\\\Rightarrow\left(\frac{dR}{d\theta}\dot{\theta} \right)^2+R^2\dot{\theta}^2=\frac{2GM}{R}+c&amp;\left\{\frac{dR}{dt}=\frac{dR}{d\theta}\frac{d\theta}{dt}\right.\\\Rightarrow\left(\left(\frac{dR}{d\theta}\right)^2+R^2\right)\dot{\theta}^2=\frac{2GM}{R}+c\\\Rightarrow\left(\left(\frac{dR}{d\theta}\right)^2+R^2\right)\frac{L^2}{R^4}=\frac{2GM}{R}+c&amp;\left\{R^2\dot{\theta}=L\right.\\\Rightarrow\left(\frac{1}{R^2}\frac{dR}{d\theta}\right)^2+\frac{1}{R^2}=\frac{2GM}{L^2R}+\frac{c}{L^2}\\\Rightarrow\left(\frac{1}{R^2}\frac{dR}{d\theta}\right)^2+\frac{1}{R^2}=2\alpha\frac{1}{R} +C&amp;\left\{\begin{array}{l}C=\frac{c}{L^2}\\\alpha=\frac{GM}{L^2}\end{array}\right.\end{array}' title='\begin{array}{ll}\Rightarrow\dot{R}^2+R^2\dot{\theta}^2=\frac{2GM}{R}+c&amp;\left\{\begin{array}{ll}|\vec{x}|=R\\|\dot{\vec{x}}|^2=\dot{R}^2+R^2\dot{\theta}^2\end{array}\right.\\\Rightarrow\left(\frac{dR}{d\theta}\dot{\theta} \right)^2+R^2\dot{\theta}^2=\frac{2GM}{R}+c&amp;\left\{\frac{dR}{dt}=\frac{dR}{d\theta}\frac{d\theta}{dt}\right.\\\Rightarrow\left(\left(\frac{dR}{d\theta}\right)^2+R^2\right)\dot{\theta}^2=\frac{2GM}{R}+c\\\Rightarrow\left(\left(\frac{dR}{d\theta}\right)^2+R^2\right)\frac{L^2}{R^4}=\frac{2GM}{R}+c&amp;\left\{R^2\dot{\theta}=L\right.\\\Rightarrow\left(\frac{1}{R^2}\frac{dR}{d\theta}\right)^2+\frac{1}{R^2}=\frac{2GM}{L^2R}+\frac{c}{L^2}\\\Rightarrow\left(\frac{1}{R^2}\frac{dR}{d\theta}\right)^2+\frac{1}{R^2}=2\alpha\frac{1}{R} +C&amp;\left\{\begin{array}{l}C=\frac{c}{L^2}\\\alpha=\frac{GM}{L^2}\end{array}\right.\end{array}' class='latex' />
<p>L is the <a href="http://hyperphysics.phy-astr.gsu.edu/hbase/amom.html" onclick="pageTracker._trackPageview('/outgoing/hyperphysics.phy-astr.gsu.edu/hbase/amom.html?referer=');">angular momentum</a> of the planet in question, and it&#8217;s constant.  It may seem silly but, with the advantage of foresight, it&#8217;s better to solve this problem in terms of 1/R instead of R.</p>
<img src='http://s.wordpress.com/latex.php?latex=%5Cbegin%7Barray%7D%7Bll%7D%5CRightarrow%20%5Cleft%28-%5Cfrac%7BdS%7D%7Bd%5Ctheta%7D%5Cright%29%5E2%2BS%5E2%3D2%5Calpha%20S%20%2BC%26%5Cleft%5C%7B%5Cbegin%7Barray%7D%7Bl%7DS%3D%5Cfrac%7B1%7D%7BR%7D%5C%5C%5Cfrac%7BdS%7D%7Bd%5Ctheta%7D%3D-%5Cfrac%7B1%7D%7BR%5E2%7D%5Cfrac%7BdR%7D%7Bd%5Ctheta%7D%5Cend%7Barray%7D%5Cright.%5C%5C%5CRightarrow-%5Cfrac%7BdS%7D%7Bd%5Ctheta%7D%3D%5Csqrt%7B-S%5E2%2B2%5Calpha%20S%2BC%7D%5C%5C%5CRightarrow%20d%5Ctheta%3D%5Cfrac%7B-dS%7D%7B%5Csqrt%7B-S%5E2%2B2%5Calpha%20S%2BC%7D%7D%5C%5C%5CRightarrow%5Cint%20d%5Ctheta%3D-%5Cint%5Cfrac%7BdS%7D%7B%5Csqrt%7B-S%5E2%2B2%5Calpha%20S%2BC%7D%7D%5C%5C%5CRightarrow%5Ctheta%2BD%3D-%5Cint%5Cfrac%7BdS%7D%7B%5Csqrt%7BC%2B%5Calpha%5E2-%28S-%5Calpha%29%5E2%7D%7D%5C%5C%3D%5Cint%5Cfrac%7B%5Csqrt%7BC%2B%5Calpha%5E2%7D%5Csin%7B%28u%29%7Ddu%7D%7B%5Csqrt%7BC%2B%5Calpha%5E2-%28C%2B%5Calpha%5E2%29%5Ccos%5E2%7B%28u%29%7D%7D%7D%26%5Cleft%5C%7B%5Cbegin%7Barray%7D%7Bl%7DS-%5Calpha%3D%5Csqrt%7BC%2B%5Calpha%5E2%7D%5Ccos%7B%28u%29%7D%5C%5CdS%3D-%5Csqrt%7BC%2B%5Calpha%5E2%7D%5Csin%7B%28u%29%7Ddu%5Cend%7Barray%7D%5Cright.%5C%5C%3D%5Cint%5Cfrac%7B%5Csin%7B%28u%29%7Ddu%7D%7B%5Csqrt%7B1-%5Ccos%5E2%7B%28u%29%7D%7D%7D%5C%5C%3D%5Cint%5Cfrac%7B%5Csin%7B%28u%29%7Ddu%7D%7B%5Csqrt%7B%5Csin%5E2%7B%28u%29%7D%7D%7D%5C%5C%3D%5Cint%20du%5C%5C%5CRightarrow%5Ctheta%2BD%3Du%5C%5C%5CRightarrow%5Ccos%7B%28%5Ctheta%2BD%29%7D%3D%5Ccos%7B%28u%29%7D%5C%5C%5CRightarrow%5Csqrt%7BC%2B%5Calpha%5E2%7D%5Ccos%7B%28%5Ctheta%2BD%29%7D%3D%5Csqrt%7BC%2B%5Calpha%5E2%7D%5Ccos%7B%28u%29%7D%5C%5C%5CRightarrow%5Csqrt%7BC%2B%5Calpha%5E2%7D%5Ccos%7B%28%5Ctheta%2BD%29%7D%3DS-%5Calpha%5C%5C%5CRightarrow%5Cfrac%7B1%7D%7BR%7D%3DS%3D%5Calpha%2B%5Csqrt%7BC%2B%5Calpha%5E2%7D%5Ccos%7B%28%5Ctheta%2BD%29%7D%5C%5C%5CRightarrow%20R%3D%5Cfrac%7B1%7D%7B%5Calpha%2B%5Csqrt%7BC%2B%5Calpha%5E2%7D%5Ccos%7B%28%5Ctheta%2BD%29%7D%7D%5C%5C%5CRightarrow%20R%3D%5Cfrac%7BP%7D%7B1%2B%5Cepsilon%5Ccos%7B%28%5Ctheta%2BD%29%7D%7D%26%5Cleft%5C%7B%5Cbegin%7Barray%7D%7Bl%7DP%3D%5Cfrac%7B1%7D%7B%5Calpha%7D%3D%5Cfrac%7BL%5E2%7D%7BGM%7D%5C%5C%5Cepsilon%3D%5Csqrt%7B%5Cfrac%7BC%7D%7B%5Calpha%5E2%7D%2B1%7D%5Cend%7Barray%7D%5Cright.%5Cend%7Barray%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\begin{array}{ll}\Rightarrow \left(-\frac{dS}{d\theta}\right)^2+S^2=2\alpha S +C&amp;\left\{\begin{array}{l}S=\frac{1}{R}\\\frac{dS}{d\theta}=-\frac{1}{R^2}\frac{dR}{d\theta}\end{array}\right.\\\Rightarrow-\frac{dS}{d\theta}=\sqrt{-S^2+2\alpha S+C}\\\Rightarrow d\theta=\frac{-dS}{\sqrt{-S^2+2\alpha S+C}}\\\Rightarrow\int d\theta=-\int\frac{dS}{\sqrt{-S^2+2\alpha S+C}}\\\Rightarrow\theta+D=-\int\frac{dS}{\sqrt{C+\alpha^2-(S-\alpha)^2}}\\=\int\frac{\sqrt{C+\alpha^2}\sin{(u)}du}{\sqrt{C+\alpha^2-(C+\alpha^2)\cos^2{(u)}}}&amp;\left\{\begin{array}{l}S-\alpha=\sqrt{C+\alpha^2}\cos{(u)}\\dS=-\sqrt{C+\alpha^2}\sin{(u)}du\end{array}\right.\\=\int\frac{\sin{(u)}du}{\sqrt{1-\cos^2{(u)}}}\\=\int\frac{\sin{(u)}du}{\sqrt{\sin^2{(u)}}}\\=\int du\\\Rightarrow\theta+D=u\\\Rightarrow\cos{(\theta+D)}=\cos{(u)}\\\Rightarrow\sqrt{C+\alpha^2}\cos{(\theta+D)}=\sqrt{C+\alpha^2}\cos{(u)}\\\Rightarrow\sqrt{C+\alpha^2}\cos{(\theta+D)}=S-\alpha\\\Rightarrow\frac{1}{R}=S=\alpha+\sqrt{C+\alpha^2}\cos{(\theta+D)}\\\Rightarrow R=\frac{1}{\alpha+\sqrt{C+\alpha^2}\cos{(\theta+D)}}\\\Rightarrow R=\frac{P}{1+\epsilon\cos{(\theta+D)}}&amp;\left\{\begin{array}{l}P=\frac{1}{\alpha}=\frac{L^2}{GM}\\\epsilon=\sqrt{\frac{C}{\alpha^2}+1}\end{array}\right.\end{array}' title='\begin{array}{ll}\Rightarrow \left(-\frac{dS}{d\theta}\right)^2+S^2=2\alpha S +C&amp;\left\{\begin{array}{l}S=\frac{1}{R}\\\frac{dS}{d\theta}=-\frac{1}{R^2}\frac{dR}{d\theta}\end{array}\right.\\\Rightarrow-\frac{dS}{d\theta}=\sqrt{-S^2+2\alpha S+C}\\\Rightarrow d\theta=\frac{-dS}{\sqrt{-S^2+2\alpha S+C}}\\\Rightarrow\int d\theta=-\int\frac{dS}{\sqrt{-S^2+2\alpha S+C}}\\\Rightarrow\theta+D=-\int\frac{dS}{\sqrt{C+\alpha^2-(S-\alpha)^2}}\\=\int\frac{\sqrt{C+\alpha^2}\sin{(u)}du}{\sqrt{C+\alpha^2-(C+\alpha^2)\cos^2{(u)}}}&amp;\left\{\begin{array}{l}S-\alpha=\sqrt{C+\alpha^2}\cos{(u)}\\dS=-\sqrt{C+\alpha^2}\sin{(u)}du\end{array}\right.\\=\int\frac{\sin{(u)}du}{\sqrt{1-\cos^2{(u)}}}\\=\int\frac{\sin{(u)}du}{\sqrt{\sin^2{(u)}}}\\=\int du\\\Rightarrow\theta+D=u\\\Rightarrow\cos{(\theta+D)}=\cos{(u)}\\\Rightarrow\sqrt{C+\alpha^2}\cos{(\theta+D)}=\sqrt{C+\alpha^2}\cos{(u)}\\\Rightarrow\sqrt{C+\alpha^2}\cos{(\theta+D)}=S-\alpha\\\Rightarrow\frac{1}{R}=S=\alpha+\sqrt{C+\alpha^2}\cos{(\theta+D)}\\\Rightarrow R=\frac{1}{\alpha+\sqrt{C+\alpha^2}\cos{(\theta+D)}}\\\Rightarrow R=\frac{P}{1+\epsilon\cos{(\theta+D)}}&amp;\left\{\begin{array}{l}P=\frac{1}{\alpha}=\frac{L^2}{GM}\\\epsilon=\sqrt{\frac{C}{\alpha^2}+1}\end{array}\right.\end{array}' class='latex' />
<p>The choice of P and ε may seem arbitrary (and it is), but it has some historical relevance.  P is called the &#8220;semi-latus recturn&#8221; and it basically describes the size of the orbit.  ε is called the &#8220;eccentricity&#8221;, and it describes how lopsided the orbit is.  ε=0 means the orbit is a circle, 0&lt;ε&lt;1 means the orbit is elliptical, and 1≤ε means that the orbit is open (not actually orbiting).  For reference, the Earth&#8217;s eccentricity is ε=0.01671123 and Halley&#8217;s comet&#8217;s is ε=0.967.</p>
<p>D just describes what direction the far side of the ellipse points in, so it&#8217;s not actually important to the overall shape.</p>
<p>It turns out that this last equation relating R and θ is all you need to define an ellipse, such that the center of the system, (0,0), is at one of the foci.  Here&#8217;s a proof:</p>
<p>An ellipse with a focus at (0,0) can be written <img src='http://s.wordpress.com/latex.php?latex=%5Cfrac%7B%28x%2BF%29%5E2%7D%7BA%5E2%7D%2B%5Cfrac%7By%5E2%7D%7BB%5E2%7D%3D1&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\frac{(x+F)^2}{A^2}+\frac{y^2}{B^2}=1' title='\frac{(x+F)^2}{A^2}+\frac{y^2}{B^2}=1' class='latex' /> where F is the distance from the center of the ellipse to the focus and <img src='http://s.wordpress.com/latex.php?latex=F%5E2%3DA%5E2-B%5E2&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='F^2=A^2-B^2' title='F^2=A^2-B^2' class='latex' />.</p>
<img src='http://s.wordpress.com/latex.php?latex=%5Cbegin%7Barray%7D%7Bll%7DR%3D%5Cfrac%7BP%7D%7B1%2B%5Cepsilon%5Ccos%7B%28%5Ctheta%29%7D%7D%5C%5C%5CRightarrow%20R%2B%5Cepsilon%20R%5Ccos%7B%28%5Ctheta%29%7D%3DP%5C%5C%5CRightarrow%20%5Csqrt%7Bx%5E2%2By%5E2%7D%2B%5Cepsilon%20x%3DP%5Cquad%5Cquad%5Cquad%5Cleft%5C%7B%5Cbegin%7Barray%7D%7Bl%7Dx%3DR%5Ccos%7B%28%5Ctheta%29%7D%5C%5Cy%3DR%5Csin%7B%28%5Ctheta%29%7D%5Cend%7Barray%7D%5Cright.%5C%5C%5CRightarrow%20%5Csqrt%7Bx%5E2%2By%5E2%7D%3DP-%5Cepsilon%20x%5C%5C%5CRightarrow%20x%5E2%2By%5E2%3DP%5E2-2P%5Cepsilon%20x%2B%5Cepsilon%5E2x%5E2%5C%5C%5CRightarrow%20%281-%5Cepsilon%5E2%29x%5E2%2B2P%5Cepsilon%20x%2By%5E2%3DP%5E2%5C%5C%5CRightarrow%20x%5E2%2B2%5Cfrac%7BP%5Cepsilon%7D%7B1-%5Cepsilon%5E2%7D%20x%2B%5Cfrac%7By%5E2%7D%7B1-%5Cepsilon%5E2%7D%3D%5Cfrac%7BP%5E2%7D%7B1-%5Cepsilon%5E2%7D%5C%5C%5CRightarrow%20x%5E2%2B2%5Cfrac%7BP%5Cepsilon%7D%7B1-%5Cepsilon%5E2%7D%20x%2B%5Cleft%28%5Cfrac%7BP%5Cepsilon%7D%7B1-%5Cepsilon%5E2%7D%5Cright%29%5E2%2B%5Cfrac%7By%5E2%7D%7B1-%5Cepsilon%5E2%7D%3D%5Cfrac%7BP%5E2%7D%7B1-%5Cepsilon%5E2%7D%2B%5Cleft%28%5Cfrac%7BP%5Cepsilon%7D%7B1-%5Cepsilon%5E2%7D%5Cright%29%5E2%5C%5C%5CRightarrow%20%5Cleft%28x%2B%5Cfrac%7BP%5Cepsilon%7D%7B1-%5Cepsilon%5E2%7D%5Cright%29%5E2%2B%5Cfrac%7By%5E2%7D%7B1-%5Cepsilon%5E2%7D%3D%5Cfrac%7BP%5E2%281-%5Cepsilon%5E2%29%7D%7B%281-%5Cepsilon%5E2%29%5E2%7D%2B%5Cfrac%7BP%5E2%5Cepsilon%5E2%7D%7B%281-%5Cepsilon%5E2%29%5E2%7D%5C%5C%5CRightarrow%20%5Cleft%28x%2B%5Cfrac%7BP%5Cepsilon%7D%7B1-%5Cepsilon%5E2%7D%5Cright%29%5E2%2B%5Cfrac%7By%5E2%7D%7B1-%5Cepsilon%5E2%7D%3D%5Cfrac%7BP%5E2%7D%7B%281-%5Cepsilon%5E2%29%5E2%7D%5C%5C%5CRightarrow%20%5Cfrac%7B%5Cleft%28x%2B%5Cfrac%7BP%5Cepsilon%7D%7B1-%5Cepsilon%5E2%7D%5Cright%29%5E2%7D%7B%5Cleft%28%5Cfrac%7BP%5E2%7D%7B%281-%5Cepsilon%5E2%29%5E2%7D%5Cright%29%7D%2B%5Cfrac%7By%5E2%7D%7B%5Cleft%28%5Cfrac%7BP%5E2%7D%7B1-%5Cepsilon%5E2%7D%5Cright%29%7D%3D1%5Cend%7Barray%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\begin{array}{ll}R=\frac{P}{1+\epsilon\cos{(\theta)}}\\\Rightarrow R+\epsilon R\cos{(\theta)}=P\\\Rightarrow \sqrt{x^2+y^2}+\epsilon x=P\quad\quad\quad\left\{\begin{array}{l}x=R\cos{(\theta)}\\y=R\sin{(\theta)}\end{array}\right.\\\Rightarrow \sqrt{x^2+y^2}=P-\epsilon x\\\Rightarrow x^2+y^2=P^2-2P\epsilon x+\epsilon^2x^2\\\Rightarrow (1-\epsilon^2)x^2+2P\epsilon x+y^2=P^2\\\Rightarrow x^2+2\frac{P\epsilon}{1-\epsilon^2} x+\frac{y^2}{1-\epsilon^2}=\frac{P^2}{1-\epsilon^2}\\\Rightarrow x^2+2\frac{P\epsilon}{1-\epsilon^2} x+\left(\frac{P\epsilon}{1-\epsilon^2}\right)^2+\frac{y^2}{1-\epsilon^2}=\frac{P^2}{1-\epsilon^2}+\left(\frac{P\epsilon}{1-\epsilon^2}\right)^2\\\Rightarrow \left(x+\frac{P\epsilon}{1-\epsilon^2}\right)^2+\frac{y^2}{1-\epsilon^2}=\frac{P^2(1-\epsilon^2)}{(1-\epsilon^2)^2}+\frac{P^2\epsilon^2}{(1-\epsilon^2)^2}\\\Rightarrow \left(x+\frac{P\epsilon}{1-\epsilon^2}\right)^2+\frac{y^2}{1-\epsilon^2}=\frac{P^2}{(1-\epsilon^2)^2}\\\Rightarrow \frac{\left(x+\frac{P\epsilon}{1-\epsilon^2}\right)^2}{\left(\frac{P^2}{(1-\epsilon^2)^2}\right)}+\frac{y^2}{\left(\frac{P^2}{1-\epsilon^2}\right)}=1\end{array}' title='\begin{array}{ll}R=\frac{P}{1+\epsilon\cos{(\theta)}}\\\Rightarrow R+\epsilon R\cos{(\theta)}=P\\\Rightarrow \sqrt{x^2+y^2}+\epsilon x=P\quad\quad\quad\left\{\begin{array}{l}x=R\cos{(\theta)}\\y=R\sin{(\theta)}\end{array}\right.\\\Rightarrow \sqrt{x^2+y^2}=P-\epsilon x\\\Rightarrow x^2+y^2=P^2-2P\epsilon x+\epsilon^2x^2\\\Rightarrow (1-\epsilon^2)x^2+2P\epsilon x+y^2=P^2\\\Rightarrow x^2+2\frac{P\epsilon}{1-\epsilon^2} x+\frac{y^2}{1-\epsilon^2}=\frac{P^2}{1-\epsilon^2}\\\Rightarrow x^2+2\frac{P\epsilon}{1-\epsilon^2} x+\left(\frac{P\epsilon}{1-\epsilon^2}\right)^2+\frac{y^2}{1-\epsilon^2}=\frac{P^2}{1-\epsilon^2}+\left(\frac{P\epsilon}{1-\epsilon^2}\right)^2\\\Rightarrow \left(x+\frac{P\epsilon}{1-\epsilon^2}\right)^2+\frac{y^2}{1-\epsilon^2}=\frac{P^2(1-\epsilon^2)}{(1-\epsilon^2)^2}+\frac{P^2\epsilon^2}{(1-\epsilon^2)^2}\\\Rightarrow \left(x+\frac{P\epsilon}{1-\epsilon^2}\right)^2+\frac{y^2}{1-\epsilon^2}=\frac{P^2}{(1-\epsilon^2)^2}\\\Rightarrow \frac{\left(x+\frac{P\epsilon}{1-\epsilon^2}\right)^2}{\left(\frac{P^2}{(1-\epsilon^2)^2}\right)}+\frac{y^2}{\left(\frac{P^2}{1-\epsilon^2}\right)}=1\end{array}' class='latex' />
<p>Put it all together, and you&#8217;ll find that this is definitely an ellipse with a focus at the point (0,0), the location being orbited around (like the Sun for instance).</p>
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			<wfw:commentRss>http://www.askamathematician.com/?feed=rss2&amp;p=3501</wfw:commentRss>
		<slash:comments>2</slash:comments>
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		<item>
		<title>Q: What would happen if everyone in the world jumped at the same time?</title>
		<link>http://www.askamathematician.com/?p=3479</link>
		<comments>http://www.askamathematician.com/?p=3479#comments</comments>
		<pubDate>Sun, 15 Aug 2010 17:20:02 +0000</pubDate>
		<dc:creator>Physicist</dc:creator>
				<category><![CDATA[-- By the Physicist]]></category>
		<category><![CDATA[Physics]]></category>

		<guid isPermaLink="false">http://www.askamathematician.com/?p=3479</guid>
		<description><![CDATA[Physicist: Sounds like a party! It would create a spherical wave that would descend through the Earth, focus at the core, and then expand again hitting the entire surface again, at more or less the same time, a little under 20 minutes later. In some areas of the Earth there&#8217;s a chance that someone might [...]]]></description>
			<content:encoded><![CDATA[<p><strong>Physicist</strong>: Sounds like a party!</p>
<p>It would create a spherical wave that would descend through the  Earth, focus at the core, and then expand again hitting the entire surface again, at more or less the same time, a little under 20 minutes later.  In some areas of the Earth there&#8217;s a chance that someone might feel or hear a gentle bump.  But probably not.  Also, don&#8217;t worry about the core. It&#8217;s already suffering from much worse than a bunch of people jumping around.</p>
<p>A more interesting question might be &#8220;can you weaponize hopping?&#8221;</p>
<div id="attachment_3489" class="wp-caption alignnone" style="width: 441px"><a href="http://www.askamathematician.com/wp-content/uploads/2010/08/800px-Jumping_Giulia_Salar_de_Uyuni_Bolivia_Luca_Galuzzi_2006.jpg"><img class="size-full wp-image-3489" title="800px-Jumping_Giulia_Salar_de_Uyuni_Bolivia_Luca_Galuzzi_2006" src="http://www.askamathematician.com/wp-content/uploads/2010/08/800px-Jumping_Giulia_Salar_de_Uyuni_Bolivia_Luca_Galuzzi_2006.jpg" alt="" width="431" height="288" /></a><p class="wp-caption-text">Good clean fun?  Or the most deadly stealth weapon since HAARP?</p></div>
<p>There is some precedent for shock waves alone doing damage.  Both <a href="http://en.wikipedia.org/wiki/Mercury_%28planet%29#Impact_basins_and_craters" onclick="pageTracker._trackPageview('/outgoing/en.wikipedia.org/wiki/Mercury_28planet_29_Impact_basins_and_craters?referer=');">Mercury</a> and Saturn&#8217;s moon <a href="http://en.wikipedia.org/wiki/Mimas_%28moon%29" onclick="pageTracker._trackPageview('/outgoing/en.wikipedia.org/wiki/Mimas_28moon_29?referer=');">Mimas</a> have suffered massive impacts (the Caloris and Herschel impacts respectively) that created shock waves that moved through/around them and focused on the far side, causing geological scale damage (Allegedly.  It&#8217;s hard to say for sure what caused the damage).</p>
<p>Depending on how high the people of the world are willing to jump (without injury), we can generate energy on the order of approximately 1-2 kilotons of TNT.  For comparison, the Little Boy nuclear bomb detonated over Hiroshima had an estimated yield of around 15 kilotons.</p>
<p>Now say that the entire world decides that Paris has created more than enough high art, and needs to be dealt with.  By carefully timing when everyone jumps off of their kitchen tables, so that the waves thus created by everyone arrive at the Eiffel tower at the same moment, a fair amount of damage could be done (maybe).  Due to fluctuations in density and material throughout the planet waves have a tendency to get scattered.  As a result, the best you can do is a &#8220;fuzzy focus&#8221;, like trying to burn something with a smoked-glass magnifying lens.  So here&#8217;s a guess:</p>
<p>I&#8217;d bet that if everyone on the Earth jumped in the right sequence then you could (mostly) focus the waves at some point on the surface and create an earthquake of no more than 5.0 on the <a href="http://en.wikipedia.org/wiki/Richter_magnitude_scale#Richter_magnitudes" onclick="pageTracker._trackPageview('/outgoing/en.wikipedia.org/wiki/Richter_magnitude_scale_Richter_magnitudes?referer=');">Richter scale</a>.</p>
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		<title>Q: How can electrons &#8220;jump&#8221; between places without covering the intervening distance?</title>
		<link>http://www.askamathematician.com/?p=3432</link>
		<comments>http://www.askamathematician.com/?p=3432#comments</comments>
		<pubDate>Thu, 12 Aug 2010 13:34:02 +0000</pubDate>
		<dc:creator>Physicist</dc:creator>
				<category><![CDATA[-- By the Physicist]]></category>
		<category><![CDATA[Physics]]></category>
		<category><![CDATA[Quantum Theory]]></category>

		<guid isPermaLink="false">http://www.askamathematician.com/?p=3432</guid>
		<description><![CDATA[Physicist: Frequently in quantum mechanics you&#8217;ll find that particles are restricted to only a certain set of states or locations, and yet somehow they can move from one to the next. It&#8217;s like moving between islands without crossing any water. &#8220;Classically&#8221; (19th century, pre-relativity, pre-quantum) this is impossible. If you see a particle in one [...]]]></description>
			<content:encoded><![CDATA[<p><strong>Physicist</strong>: Frequently in quantum mechanics you&#8217;ll find that particles are restricted to only a certain set of states or locations, and yet somehow they can move from one to the next. It&#8217;s like moving between islands without crossing any water.<br />
&#8220;Classically&#8221; (19th century, pre-relativity, pre-quantum) this is impossible. If you see a particle in one place and then see it again, but in a new place, then of course it must have traversed the distance from one to the other.<br />
But here&#8217;s the essential difference between quantum mechanics (correct) and classical physics (wrong): particles aren&#8217;t solid objects that have a genuine position, instead they&#8217;re waves that are &#8220;smeared out&#8221;.</p>
<p>A standing wave (like a guitar string, or an electron orbital) usually has &#8220;nodes&#8221; where the wave is always zero.</p>
<div id="attachment_3451" class="wp-caption alignnone" style="width: 226px"><a href="http://www.askamathematician.com/wp-content/uploads/2010/08/vibration-setup-8181a.jpg"><img class="size-full wp-image-3451" title="vibration-setup-8181a" src="http://www.askamathematician.com/wp-content/uploads/2010/08/vibration-setup-8181a.jpg" alt="" width="216" height="355" /></a><p class="wp-caption-text">The string remains stationary in the middle, but the wave has no problem getting past that point.  This image stolen, unrepentantly, from http://people.rit.edu/andpph/text-string-vibrations.html.</p></div>
<p>But of course there&#8217;s a big difference between the wave being zero at a node, and the wave being unable to get to the other side of that node. It&#8217;s the difference between the string in the above picture, and what you would have if you nailed that string to a piece of wood at the node and left the bottom half dangling.</p>
<p>For you calculus buffs out there, the difference is hidden away in the derivatives.</p>
<p>If you have a particle (wave) like this and you measure where it is many times you&#8217;ll find that it&#8217;s on the top about half the time, on the bottom about half the time, and never ever in the middle (ever).</p>
<div id="attachment_3467" class="wp-caption alignnone" style="width: 442px"><a href="http://www.askamathematician.com/wp-content/uploads/2010/08/coldpizza.png"><img class="size-full wp-image-3467" title="coldpizza" src="http://www.askamathematician.com/wp-content/uploads/2010/08/coldpizza.png" alt="" width="432" height="432" /></a><p class="wp-caption-text">The shape of the standing waves formed inside a microwave oven.  The peaks and troughs move up and down, but the lines in between are stationary and zero.  This is why microwaves have rotating plates, to keep food from sitting in the nodes (and staying cold) or from staying in the peak areas (and burning).</p></div>
<p>One of my personal fave examples of &#8220;large scale&#8221; quantum weirdness is microwave ovens.  A microwave oven creates a standing wave of photons with a &#8220;plus-shaped&#8221; (+) node, which leaves the center of the chamber especially cold.  The chance of finding a photon anywhere on the plus-shaped-node is zero.  So there are four different cells that it should be impossible for the microwave photons to move between, but they don&#8217;t seem to mind at all.  It&#8217;s not like particles exist or something.</p>
<p>So thinking of things like electrons as particles leads to incorrect conclusions.  Thinking of them as waves is really  the way to go.</p>
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		<title>Q: Why do we only see one rainbow at a time?</title>
		<link>http://www.askamathematician.com/?p=3429</link>
		<comments>http://www.askamathematician.com/?p=3429#comments</comments>
		<pubDate>Wed, 11 Aug 2010 19:29:31 +0000</pubDate>
		<dc:creator>Physicist</dc:creator>
				<category><![CDATA[-- By the Physicist]]></category>
		<category><![CDATA[Physics]]></category>

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		<description><![CDATA[Physicist: A rainbow is an angle-dependent illusion.  They don&#8217;t actually exist anywhere, they just appear to.  Since they can only exist at a particular angle (with respect to you and the Sun), you can never see them anywhere else in the sky. The next time you see a rainbow, draw an imaginary line from the [...]]]></description>
			<content:encoded><![CDATA[<p><strong>Physicist</strong>: A <span style="color: #ff0000;">r</span><span style="color: #ff6600;">a</span><span style="color: #ffff00;">i</span><span style="color: #00ff00;">n</span><span style="color: #0000ff;">b</span><span style="color: #800080;">o</span><span style="color: #cc99ff;">w</span> is an angle-dependent illusion.  They don&#8217;t actually exist anywhere, they just appear to.  Since they can only exist at a particular angle (with respect to you and the Sun), you can never see them anywhere else in the sky.</p>
<div id="attachment_3434" class="wp-caption alignnone" style="width: 410px"><a href="http://www.askamathematician.com/wp-content/uploads/2010/08/rnbw4-1.gif"><img class="size-full wp-image-3434" title="rnbw4-1" src="http://www.askamathematician.com/wp-content/uploads/2010/08/rnbw4-1.gif" alt="" width="400" height="300" /></a><p class="wp-caption-text">All rainbows form a circle 84° across and exactly opposite from the Sun, no matter where you are.</p></div>
<p>The next time you see a rainbow, draw an imaginary line from the sun through you.  You&#8217;ll notice that this line goes exactly through the middle of the rainbow.  This comes about because of how rainbows are created.  No matter where you are, the red line is 138 degrees from the sun and the purple line is 140 degrees from the Sun.</p>
<p>For very small drops of water the surface tension overwhelms all other forces, and the drop is pulled into a nearly perfect sphere (which has nice optical properties).  When light encounters the surface of water (or anything at all really) it splits into its component colors.  As the light travels into the drop the various colors mostly end up getting scrambled so much that almost anywhere you look you&#8217;ll see a more or less even combination of them (white light).  However, light that enters the drop, reflects off the back, and leaves again at about 42° from the direction it came from produces colors that stay separated.</p>
<div id="attachment_3439" class="wp-caption alignnone" style="width: 476px"><a href="http://www.askamathematician.com/wp-content/uploads/2010/08/rays.jpg"><img class="size-full wp-image-3439" title="Even if you could find the pot of gold you'd have Warwick Davis following you around for six movies.  Best not to bother." src="http://www.askamathematician.com/wp-content/uploads/2010/08/rays.jpg" alt="" width="466" height="381" /></a><p class="wp-caption-text">The path taken by light through a water drop with one reflection (upper left) and with two reflections (upper right).  At 42° the light is not recombined into white light, and thus forms a rainbow.  The recombined white light makes the sky under rainbows brighter than the sky above.</p></div>
<p>This happens again at around 52°, but this time the effect is caused by two reflections inside the water drop.  As a result of there being two reflection (instead of one), the colors in secondary rainbows are reversed compared to the primary rainbow at 42°.</p>
<div id="attachment_3443" class="wp-caption alignnone" style="width: 410px"><a href="http://www.askamathematician.com/wp-content/uploads/2010/08/rainbowbrightwuzhere.jpg"><img class="size-full wp-image-3443" title="rainbowbrightwuzhere" src="http://www.askamathematician.com/wp-content/uploads/2010/08/rainbowbrightwuzhere.jpg" alt="" width="400" height="300" /></a><p class="wp-caption-text">Pretty rainbow, bright sky underneath, and reverse-color-order secondary rainbow above.</p></div>
<p>Entirely unimportant fun fact: the only way to see a complete, full circle, rainbow is to be flying (or falling).  Otherwise whatever you&#8217;re standing on will cast a shadow that will cut off the bottom of the circle.</p>
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		<title>Q: Why does putting spin on a ball change how it moves through the air?</title>
		<link>http://www.askamathematician.com/?p=3188</link>
		<comments>http://www.askamathematician.com/?p=3188#comments</comments>
		<pubDate>Mon, 09 Aug 2010 03:26:53 +0000</pubDate>
		<dc:creator>Physicist</dc:creator>
				<category><![CDATA[-- By the Physicist]]></category>
		<category><![CDATA[Physics]]></category>

		<guid isPermaLink="false">http://www.askamathematician.com/?p=3188</guid>
		<description><![CDATA[Physicist: This question is especially perplexing after a first year physics course, where every question starts with &#8220;ignoring air resistance&#8230;&#8221;. There are a couple of different way to approach the answer, but I like this one best. Way back in the day a dude named D&#8217;Alembert came up with the unsurprisingly named &#8220;D&#8217;Alembert&#8217;s paradox&#8221; which [...]]]></description>
			<content:encoded><![CDATA[<p><strong>Physicist</strong>: This question is especially perplexing after a first year physics course, where every question starts with &#8220;ignoring air resistance&#8230;&#8221;.  There are a couple of different way to approach the answer, but I like this one best.</p>
<p>Way back in the day a dude named D&#8217;Alembert came up with the unsurprisingly named &#8220;D&#8217;Alembert&#8217;s paradox&#8221; which essentially says that in air and water an object traveling at subsonic speeds experiences neither drag nor lift.  That&#8217;s obviously not the case which is why this is &#8220;D&#8217;Alembert&#8217;s paradox&#8221; and not &#8220;D&#8217;Alembert&#8217;s well known fact&#8221;.<br />
The resolution is to take into account vorticity and the rotation of air flow around the object in question.<br />
Rotation of air allows the air to flow faster over the top or the bottom of the ball. But a funny thing happens when air moves quickly; the transverse air pressure (pressure to the sides) drops.  This effect is called the <a href="http://m.youtube.com/#/watch?client=mv-google&amp;xl=xl_blazer&amp;v=P-xNXrELCmU" onclick="pageTracker._trackPageview('/outgoing/m.youtube.com/_/watch?client=mv-google_amp_xl=xl_blazer_amp_v=P-xNXrELCmU&amp;referer=');">Bernoulli effect</a> or, in certain cases, the Venturi effect.</p>
<div id="attachment_3408" class="wp-caption alignnone" style="width: 458px"><a href="http://www.askamathematician.com/wp-content/uploads/2010/08/spinlift.jpg"><img class="size-full wp-image-3408" title="If all equations looked like this, then purveyors of mathematics would merely be very dull as opposed to stunningly dull." src="http://www.askamathematician.com/wp-content/uploads/2010/08/spinlift.jpg" alt="" width="448" height="155" /></a><p class="wp-caption-text">If you combine the flow of air rotating around a ball, with the flow of air past a ball, then the result is a faster flow past one side and a slower flow past the other side.  In this example there is less pressure on the top than the bottom.</p></div>
<p>So how can you induce air to rotate around a ball in the air? Base balls, golf balls, and especially tennis balls are designed to have rough surfaces that grab the air a little bit.  So if the ball spins as it moves it&#8217;ll drag the air around it a little, which leads to air rotating around the ball, which makes the air on one side move faster than the air on the other, which leads to a lower pressure on one side compared to the other, which creates a net force, which pushes the ball around.<br />
So if you want the ball to curve downward you make it spin as though it were rolling away from you.  If you want it to curve up (well&#8230; Stay in the air longer, <a href="http://www.youtube.com/watch?v=WhXZ186FNIk" onclick="pageTracker._trackPageview('/outgoing/www.youtube.com/watch?v=WhXZ186FNIk&amp;referer=');">no Human can make a ball actually curve upward</a> in flight) you&#8217;d want the ball to spin as though it were rolling toward you.  And of course, you can also make the ball veer left or right using the corresponding spin.</p>
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		<title>Quantum mech, choices, and time travel too!</title>
		<link>http://www.askamathematician.com/?p=3366</link>
		<comments>http://www.askamathematician.com/?p=3366#comments</comments>
		<pubDate>Fri, 06 Aug 2010 12:32:05 +0000</pubDate>
		<dc:creator>Physicist</dc:creator>
				<category><![CDATA[-- By the Physicist]]></category>
		<category><![CDATA[Philosophical]]></category>
		<category><![CDATA[Physics]]></category>
		<category><![CDATA[Quantum Theory]]></category>

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		<description><![CDATA[Physicist: Recently I sent a series of emails back and forth with a reader that seem interesting enough to post. Conversations (near a chalkboard especially) are the best way to learn just about anything. &#160; Q: I wake up at 6am. I brush my teeth get dressed and go downstairs. I eat my breakfast at [...]]]></description>
			<content:encoded><![CDATA[<p><strong>Physicist</strong>: Recently I sent a series of emails back and forth with a reader that seem interesting enough to post. Conversations (near a chalkboard especially) are the best way to learn just about anything.</p>
<p> &#160; </p>
<p><strong>Q:</strong><br />
I wake up at 6am.<br />
I brush my teeth get dressed and go downstairs.<br />
I eat my breakfast at 7am and I&#8217;m just about to leave out the door.<br />
when i notice there is an apple and an orange in a fruitbowl on a table next to me<br />
at exactly 7:02am.  I CHOOSE to take the apple.</p>
<p>I then make a choice whether I should choose to take my car or save gas and take the bus.<br />
I CHOOSE to take the bus.</p>
<p>SUDDENLY.  when i walk into the office, a strange event occurs and time starts moving backwards.<br />
It goes back and back and back until finally it&#8217;s 6am in the morning and i wake up,<br />
brush my teeth get dressed and go downstairs to eat my breakfast.</p>
<p>Here&#8217;s my tough quantum mechanics question for you,<br />
at exactly 7:02am  will I still choose to take the apple?</p>
<p>And if this process were repeated over and over and over again for about a million times.  will the choice I make ALWAYS BE the apple?</p>
<p> &#160; </p>
<p><strong>A:</strong><br />
I&#8217;m working on a (to long) <a href="http://www.askamathematician.com/?p=2903">post about Bell&#8217;s theorem</a>. The thought experiment you propose, about going back in time, is one of the better ways to understand it.</p>
<p>To actually answer your question: if choosing the apple is based on some quantum mechanical process in your brain (and there&#8217;s a good chance that at least some part of it is), then every time that choice is made the result is random. Time travel or not.<br />
Part of the weirdness comes from the fact that every possible thing that can happen does. So (even when you time travel) some versions of you take the apple and some versions don&#8217;t.</p>
<p> &#160; </p>
<p><strong>Q:</strong><br />
Ok, remember how I took the bus in the thought experiment?</p>
<p>My question is, does quantum mechanics also apply in a reversal of time?</p>
<p>For instance, lets say that time started to slowly reverse.<br />
Will I always get onto the bus backwards and head home.</p>
<p>OR.  </p>
<p>will my car magically appear (even though i didn&#8217;t take it)<br />
and will I backwards drive home in that?</p>
<p>So the concluding question is,<br />
do quantum principles apply in a reversal of time as it does when time moves forwards?</p>
<p> &#160; </p>
<p><strong>A:</strong><br />
You&#8217;ll often hear &#8220;everything that can happen does&#8221; so if a particle can take two different paths it will actually take both.<br />
If I understand your question correctly, the answer is yes.  It turns out that &#8220;everything that could have happened did&#8221;.  The &#8220;branching&#8221; goes both forward and backward in time.  This is demonstrated by things like the &#8220;Franson experiment&#8221; that demonstrates the interference of a single photon with an earlier version of itself.<br />
Driving a car, for example, will leave telltale signs that later make it impossible for you to have actually taken a bus.  Chair fibers, leaving tire tracks, you&#8217;ll remember it, etc.<br />
But if, in every way, you could have done either one, then you did both (no magically appearing cars).<br />
This is actually the backbone of the <a href="http://en.wikipedia.org/wiki/Path_integral_formulation" onclick="pageTracker._trackPageview('/outgoing/en.wikipedia.org/wiki/Path_integral_formulation?referer=');">Feynman path integral</a> technique.</p>
<p> &#160; </p>
<p><strong>Q:</strong><br />
So are you telling me that just next to us, could exist a place where the Nazis won the second world war?<br />
A place where there exists a flying spaghetti monster? (to quote richard dawkins)</p>
<p>Or even a place out there in a dimension somewhere where there exists an all knowing omnipresent, omnipotent, all encompassing being who &#8220;watches over us&#8221; etc..</p>
<p> &#160; </p>
<p><strong>A:</strong><br />
Sure.  BUT, it&#8217;s impossible to interact with things that are even a little bit different.  For example, a stream of identical photons (lasers) will all interact with each other strongly. You can see evidence of this in effects like <a href="http://en.wikipedia.org/wiki/Speckle_pattern" onclick="pageTracker._trackPageview('/outgoing/en.wikipedia.org/wiki/Speckle_pattern?referer=');">speckling</a>. Non-coherent (regular) light is made up of all kinds of different photons, and the best way to figure out how they&#8217;ll behave is to assume that they&#8217;ll ignore each other. This is sort of a metaphor, and sort of a concrete example.<br />
So while, yes, there are almost certainly universes where the Nazis won, it doesn&#8217;t matter.  It&#8217;ll never have any impact on our universe whatsoever.<br />
A good rule of thumb is: if there is any conceivable way, whatsoever, for anything to tell the difference between universes, then they can&#8217;t interact (from the perspective of that thing that can tell the difference).</p>
<p> &#160; </p>
<p><strong>Q:</strong><br />
Couldn&#8217;t that then solve the entire God dilemma? I mean if in only one of these infinite dimensions there existed an all encompassing all knowing all powerful entity, wouldn&#8217;t this entity then transcend all dimensions? (since he is all encompassing)</p>
<p> &#160; </p>
<p><strong>A:</strong><br />
If you want to consider God, then it&#8217;s best not to do it in any kind of physics based context. That being said:<br />
Remember that if two universes are even slightly dissimilar, they won&#8217;t interact at all.  By &#8220;slightly dissimilar&#8221; I mean something like a single electron being conspicuously out of place.<br />
So any existing Gods that follow the most basic laws of logic and quantum mechanics will be stuck in their native worlds.<br />
If you&#8217;re not worried about Gods that follow physical laws, then, again, physics is literally the worst possible forum.<br />
Also, you have to be careful with this kind of reasoning. You can make up just about anything and claim that it should exist in every version of the universe.<br />
The rule &#8220;anything that can happen does&#8221; carries a bit more heft that it seems to at first. If something  <em>can&#8217;t</em> happen, then it doesn&#8217;t happen in <em>any</em> version of the universe.<br />
For example, spaghetti can neither fly nor think, so the <a href="http://www.venganza.org/" onclick="pageTracker._trackPageview('/outgoing/www.venganza.org/?referer=');">FSM</a> (pasta be upon him) can&#8217;t exist in any universe, no matter how much anyone dresses like a pirate.</p>
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