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		<title>The Big Bang Theory &#8211; Part II</title>
		<link>http://www.thulasidas.com/2008-11/the-big-bang-theory-part-ii.htm</link>
		<comments>http://www.thulasidas.com/2008-11/the-big-bang-theory-part-ii.htm#comments</comments>
		<pubDate>Wed, 19 Nov 2008 23:35:22 +0000</pubDate>
		<dc:creator>Manoj</dc:creator>
				<category><![CDATA[Debates]]></category>
		<category><![CDATA[philosophy]]></category>
		<category><![CDATA[physics]]></category>
		<category><![CDATA[SFN - Science Forums]]></category>
		<category><![CDATA[Ashtekar]]></category>
		<category><![CDATA[big bang theory]]></category>
		<category><![CDATA[cosmology]]></category>
		<category><![CDATA[paradigm shift]]></category>
		<category><![CDATA[quantum gravity]]></category>
		<category><![CDATA[relativity]]></category>
		<category><![CDATA[spac-time]]></category>
		<category><![CDATA[space time continuum]]></category>
		<category><![CDATA[universe]]></category>

		<guid isPermaLink="false">http://www.thulasidas.com/?p=615</guid>
		<description><![CDATA[This post is a continuation of my earlier musings on the Big Bang theory. This one looks at the foundational assumptions of quantum gravity. In management speak, it is a high level overview, which sounds like I understand it. In a physicist's lingo, it is merely a layman description or a hand-waving argument. In other words, the management types out there may like it better than the smart ones. You be the judge! <a href="http://www.thulasidas.com/2008-11/the-big-bang-theory-part-ii.htm">Continue reading <span class="meta-nav">&#8594;</span></a>]]></description>
			<content:encoded><![CDATA[<p>After reading a <a href="http://arxiv.org/abs/0810.0514" target="_blank">paper by Ashtekar</a> on quantum gravity and thinking about it, I realized what my trouble with the Big Bang theory was. It is more on the fundamental assumptions than the details. I thought I would summarize my thoughts here, more for my own benefit than anybody else&#8217;s.</p>
<p>Classical theories (including SR and QM) treat space as continuous nothingness; hence the term space-time continuum. In this view, objects exist in continuous space and interact with each other in continuous time.</p>
<p>Although this notion of space time continuum is intuitively appealing, it is, at best, incomplete. Consider, for instance, a spinning body in empty space. It is expected to experience centrifugal force. Now imagine that the body is stationary and the whole space is rotating around it. Will it experience any centrifugal force?</p>
<p>It is hard to see why there would be any centrifugal force if space is empty nothingness.</p>
<p>GR introduced a paradigm shift by encoding gravity into space-time thereby making it dynamic in nature, rather than empty nothingness. Thus, mass gets enmeshed in space (and time), space becomes synonymous with the universe, and the spinning body question becomes easy to answer. Yes, it will experience centrifugal force if it is the universe that is rotating around it because it is equivalent to the body spinning. And, no, it won&#8217;t, if it is in just empty space. But &#8220;empty space&#8221; doesn&#8217;t exist. In the absence of mass, there is no space-time geometry.</p>
<p>So, naturally, before the Big Bang (if there was one), there couldn&#8217;t be any space, nor indeed could there be any &#8220;before.&#8221; Note, however, that the Ashtekar paper doesn&#8217;t clearly state why there had to be a big bang. The closest it gets is that the necessity of BB arises from the encoding of gravity in space-time in GR. Despite this encoding of gravity and thereby rendering space-time dynamic, GR still treats space-time as a smooth continuum &#8212; a flaw, according to Ashtekar, that QG will rectify.</p>
<p>Now, if we accept that the universe started out with a big bang (and from a small region), we have to account for quantum effects. Space-time has to be quantized and the only right way to do it would be through quantum gravity. Through QG, we expect to avoid the Big Bang singularity of GR, the same way QM solved the unbounded ground state energy problem in the hydrogen atom.</p>
<p>What I described above is what I understand to be the physical arguments behind modern cosmology. The rest is a mathematical edifice built on top of this physical (or indeed philosophical) foundation. If you have no strong views on the philosophical foundation (or if your views are consistent with it), you can accept BB with no difficulty. Unfortunately, I do have differing views.</p>
<p>My views revolve around the following questions. </p>
<ul>
<li><a href="/2008-11/what-is-space.htm" target="_blank">What is space?</a></li>
<li><a href="/2008-11/why-the-speed-of-light.htm" target="_blank">Why is the speed of light important in it?</a></li>
<li>Where does the <a href="/2008-08/uncertainly-principle.htm" target="_blank">Heisenberg Uncertainty Principle</a> come from?</li>
</ul>
<p>These posts may sound like useless philosophical musings, but I do have some concrete (and in my opinion, important) results, listed below. </p>
<ul>
<li><a href="/2008-11/are-radio-sources-and-gamma-ray-bursts-luminal-booms.htm" target="_blank">Are GRBs and Radio Sources Luminal Booms?</a> (An article published in IJMP-D, which became one of the &#8220;Top Accessed Articles&#8221; of the journal.  <img src='http://www.thulasidas.com/wp-includes/images/smilies/icon_smile.gif' alt=':-)' class='wp-smiley' /> )</li>
<li><a href="/2008-11/light-travel-time-effects-and-cosmological-features.htm" target="_blank">Light Travel Time Effects and Cosmological Features</a> (Trying to get this one published.)</li>
</ul>
<p>There is much more work to be done on this front. But for the next couple of years, with my new book contract and pressures from my quant career, I will not have enough time to study GR and cosmology with the seriousness they deserve. I hope to get back to them once the current phase of spreading myself too thin passes. </p>
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		<title>Tsunami</title>
		<link>http://www.thulasidas.com/2007-11/tsunami.htm</link>
		<comments>http://www.thulasidas.com/2007-11/tsunami.htm#comments</comments>
		<pubDate>Sun, 04 Nov 2007 23:07:09 +0000</pubDate>
		<dc:creator>Manoj</dc:creator>
				<category><![CDATA[Debates]]></category>
		<category><![CDATA[physics]]></category>
		<category><![CDATA[SFN - Science Forums]]></category>
		<category><![CDATA[environment]]></category>

		<guid isPermaLink="false">http://theunrealuniverse.com/blog/?p=27</guid>
		<description><![CDATA[The Asian Tsunami two and a half years ago unleashed tremendous amount energy on the coastal regions around the Indian ocean. What do you think would've have happened to this energy if there had been no water to carry it away from the earthquake? <a href="http://www.thulasidas.com/2007-11/tsunami.htm">Continue reading <span class="meta-nav">&#8594;</span></a>]]></description>
			<content:encoded><![CDATA[<p>The Asian Tsunami two and a half years ago unleashed tremendous amount energy on the coastal regions around the Indian ocean. What do you think would&#8217;ve have happened to this energy if there had been no water to carry it away from the earthquake? I mean, if the earthquake (of the same kind and magnitude) had taken place on land instead of the sea-bed as it did, presumably this energy would&#8217;ve been present. How would it have manifested? As a more violent earthquake? Or a longer one?</p>
<p>I picture the earthquake (in cross-section) as a cantilever spring being held down and then released. The spring then transfers the energy to the tsunami in the form of potential energy, as an increase in the water level. As the tsunami radiates out, it is only the potential energy that is transferred; the water doesn&#8217;t move laterally, only vertically. As it hits the coast, the potential energy is transferred into the kinetic energy of the waves hitting the coast (water moving laterally then).</p>
<p>Given the magnitude of the energy transferred from the epicenter, I am speculating what would&#8217;ve happened if there was no mechanism for the transfer. Any thoughts?</p>
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		<title>Universe &#8211; Size and Age</title>
		<link>http://www.thulasidas.com/2007-03/universe-size-and-age.htm</link>
		<comments>http://www.thulasidas.com/2007-03/universe-size-and-age.htm#comments</comments>
		<pubDate>Sun, 25 Mar 2007 22:40:46 +0000</pubDate>
		<dc:creator>Manoj</dc:creator>
				<category><![CDATA[Debates]]></category>
		<category><![CDATA[physics]]></category>
		<category><![CDATA[SFN - Science Forums]]></category>
		<category><![CDATA[big bang theory]]></category>
		<category><![CDATA[galaxy]]></category>
		<category><![CDATA[relativity]]></category>
		<category><![CDATA[speed of light]]></category>
		<category><![CDATA[universe size]]></category>

		<guid isPermaLink="false">http://www.thulasidas.com/?p=484</guid>
		<description><![CDATA[I posted this question that was bothering me when I read that they found a galaxy at about 13 billion light years away. My understanding of that statement is: At distance of 13 billion light years, there was a galaxy 13 billion years ago, so that we can see the light from it now. Wouldn't that mean that the universe is at least 26 billion years old? It must have taken the galaxy about 13 billion years to reach where it appears to be, and the light from it must take another 13 billion years to reach us. <a href="http://www.thulasidas.com/2007-03/universe-size-and-age.htm">Continue reading <span class="meta-nav">&#8594;</span></a>]]></description>
			<content:encoded><![CDATA[<p>I posted this question that was bothering me when I read that they found a galaxy at about 13 billion light years away. My understanding of that statement is: At distance of 13 billion light years, there was a galaxy 13 billion years ago, so that we can see the light from it now. Wouldn&#8217;t that mean that the universe is at least 26 billion years old? It must have taken the galaxy about 13 billion years to reach where it appears to be, and the light from it must take another 13 billion years to reach us.</p>
<p>In answering my question, Martin and Swansont (who I assume are academic phycisists) point out my misconceptions and essentially ask me to learn more. All shall be answered when I&#8217;m assimilated, it would appear! <img src='http://www.thulasidas.com/wp-includes/images/smilies/icon_smile.gif' alt=':-)' class='wp-smiley' /> </p>
<p>This debate is published as a prelude to my post on the Big Bang theory, coming up in a day or two.</p>
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<td style="font-size:14pt">Mowgli</td>
<td class="smallfont" align="right">03-26-2007 10:14 PM</td>
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<div><strong>Universe &#8211; Size and Age</strong></div>
<div>I was reading a post in <a href="http://www.space.com/scienceastronomy/distant_galaxy_040216.html" target="_blank">http://www.space.com/</a> stating that they found a galaxy at about 13 billion light years away.  I am trying to figure out what that statement means.  To me, it means that 13 billion years ago, this galaxy was where we see it now.  Isn&#8217;t that what 13b LY away means?  If so, wouldn&#8217;t that mean that the universe has to be at least 26 billion years old?  I mean, the whole universe started from one singular point; how could this galaxy be where it was 13 billion years ago unless it had at least 13 billion years to get there?  (Ignoring the inflationary phase for the moment&#8230;) I have heard people explain that the space itself is expanding.  What the heck does that mean?  Isn&#8217;t it just a fancier way of saying that the speed of light was smaller some time ago?</div>
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<td style="font-size:14pt"><span style="color: #ccffcc;">swansont</span></td>
<td class="smallfont" align="right">03-27-2007 09:10 AM</td>
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<div>Originally Posted by <strong>Mowgli</strong><br />
(Post 329204)</div>
<div style="font-style:italic">I mean, the whole universe started from one singular point; how could this galaxy be where it was 13 billion years ago unless it had at least 13 billion years to get there?  (Ignoring the inflationary phase for the moment&#8230;)</div>
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<p><span style="color: #ccffcc;">Ignoring all the rest, how would this mean the universe is 26 billion years old?</span></p>
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<div>Originally Posted by <strong>Mowgli</strong><br />
(Post 329204)</div>
<div style="font-style:italic">I have heard people explain that the space itself is expanding.  What the heck does that mean?  Isn&#8217;t it just a fancier way of saying that the speed of light was smaller some time ago?</div>
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<p><span style="color: #ccffcc;">The speed of light is an inherent part of atomic structure, in the fine structure constant (alpha).  If c was changing, then the patterns of atomic spectra would have to change.  There hasn&#8217;t been any confirmed data that shows that alpha has changed (there has been the occasional paper claiming it, but you need someone to repeat the measurements), and the rest is all consistent with no change.</span></div>
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<td style="font-size:14pt"><span style="color: #99ccff;">Martin</span></td>
<td class="smallfont" align="right"><span style="color: #99ccff;">03-27-2007 11:25 AM</span></td>
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<div><span style="color: #99ccff;">To confirm or reinforce what swansont said, there are speculation and some fringe or nonstandard cosmologies that involve c changing over time (or alpha changing over time), but the changing constants thing just gets more and more ruled out.I&#8217;ve been watching for over 5 years and the more people look and study evidence the LESS likely it seems that there is any change.  They rule it out more and more accurately with their data.So it is probably best to ignore the &#8220;varying speed of light&#8221; cosmologies until one is thoroughly familiar with standard mainstream cosmology.</span><span style="color: #99ccff;">You have misconceptions Mowgli</span></p>
<ul>
<li><span style="color: #99ccff;">General Relativity (the 1915 theory) trumps Special Rel (1905)</span></li>
<li><span style="color: #99ccff;">They don&#8217;t actually contradict if you understand them correctly, because SR has only a very limited local applicability, like to the spaceship passing by:-)</span></li>
<li><span style="color: #99ccff;">Wherever GR and SR SEEM to contradict, believe GR. It is the more comprehensive theory.</span></li>
<li><span style="color: #99ccff;">GR does not have a speed limit on the rate that very great distances can increase.  the only speed limit is on LOCAL stuff (you can&#8217;t catch up with and pass a photon)</span></li>
<li><span style="color: #99ccff;">So we can and DO observe stuff that is receding from us faster than c.  (It&#8217;s far away, SR does not apply.)</span></li>
<li><span style="color: #99ccff;">This was explained in a Sci Am article I think last year</span></li>
<li><span style="color: #99ccff;">Google the author&#8217;s name Charles Lineweaver and Tamara Davis.</span></li>
<li><span style="color: #99ccff;">We know about plenty of stuff that is presently more than 14 billion LY away.</span></li>
<li><span style="color: #99ccff;">You need to learn some cosmology so you wont be confused by these things.</span></li>
<li><span style="color: #99ccff;">Also a &#8220;singularity&#8221; does not mean a single point.  that is a popular mistake because the words SOUND the same.</span></li>
<li><span style="color: #99ccff;">A singularity can occur over an entire region, even an infinite region.</span></li>
</ul>
<p><span style="color: #99ccff;">Also the &#8220;big bang&#8221; model doesn&#8217;t look like an explosion  of matter whizzing away from some point. It shouldn&#8217;t be imagined like that.  The best article explaining common mistakes people have is this Lineweaver and Davis thing in Sci Am.  I think it was Jan or Feb 2005 but I could be a year off. Google it. Get it from your local library or find it online. Best advice I can give.</span></div>
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<td style="font-size:14pt">Mowgli</td>
<td class="smallfont" align="right">03-28-2007 01:30 AM</td>
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<div>To swansont on why I thought 13 b LY implied an age of 26 b years:When you say that there is a galaxy at 13 b LY away, I understand it to mean that 13 billion years ago my time, the galaxy was at the point where I see it now (which is 13 b LY away from me).  Knowing that everything started from the same point, it must have taken the galaxy at least 13 b years to get where it was 13 b years ago.  So 13+13.  I&#8217;m sure I must be wrong.To Martin: You are right, I need to learn quite a bit more about cosmology.  But a couple of things you mentioned surprise me &#8212; how do we observe stuff that is receding from as FTL?  I mean, wouldn&#8217;t the relativistic Doppler shift formula give imaginary 1+z?  And the stuff beyond 14 b LY away &#8211; are they &#8220;outside&#8221; the universe?I will certainly look up and read the authors you mentioned.  Thanks.</div>
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<td style="font-size:14pt"><span style="color: #ccffcc;">swansont</span></td>
<td class="smallfont" align="right">03-28-2007 03:13 AM</td>
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<div>Originally Posted by <strong>Mowgli</strong><br />
(Post 329393)</div>
<div style="font-style:italic">To swansont on why I thought 13 b LY implied an age of 26 b years:When you say that there is a galaxy at 13 b LY away, I understand it to mean that 13 billion years ago my time, the galaxy was at the point where I see it now (which is 13 b LY away from me).  Knowing that everything started from the same point, it must have taken the galaxy at least 13 b years to get where it was 13 b years ago.  So 13+13.  I&#8217;m sure I must be wrong.</div>
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<p><span style="color: #ccffcc;">That would depend on how you do your calibration.  Looking only at a Doppler shift and ignoring all the other factors, if you know that speed correlates with distance, you get a certain redshift and you would probably calibrate that to mean 13b LY if that was the actual distance.  That light would be 13b years old.</span></p>
<p><span style="color: #ccffcc;">But as Martin has pointed out, space is expanding; the cosmological redshift is different from the Doppler shift.  Because the intervening space has expanded, AFAIK the light that gets to us from a galaxy 13b LY away is not as old, because it was closer when the light was emitted.  I would think that all of this is taken into account in the measurements, so that when a distance is given to the galaxy, it&#8217;s the actual distance.</span></div>
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<td style="font-size:14pt"><span style="color: #99ccff;">Martin</span></td>
<td class="smallfont" align="right">03-28-2007 08:54 AM</td>
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<div>Originally Posted by <strong>Mowgli</strong><br />
(Post 329393)</div>
<div style="font-style:italic">I will certainly look up and read the authors you mentioned.</div>
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<p><span style="color: #99ccff;">This post has 5 or 6 links to that Sci Am article by Lineweaver and Davis</span></p>
<p><a href="http://scienceforums.net/forum/showthread.php?p=142965#post142965" target="_blank">http://scienceforums.net/forum/showt&#8230;965#post142965</a></p>
<p><span style="color: #99ccff;">It is post #65 on the Astronomy links sticky thread</span></p>
<p><span style="color: #99ccff;">It turns out the article was in the March 2005 issue.</span></p>
<p><span style="color: #99ccff;">I think it&#8217;s comparatively easy to read&#8212;well written. So it should help.</span></p>
<p><span style="color: #99ccff;">When you&#8217;ve read the Sci Am article, ask more questions&#8212;your questions might be fun to try and answer:-)</span></div>
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		<title>Twin Paradox &#8211; Take 2</title>
		<link>http://www.thulasidas.com/2007-03/twin-paradox-take-2.htm</link>
		<comments>http://www.thulasidas.com/2007-03/twin-paradox-take-2.htm#comments</comments>
		<pubDate>Wed, 21 Mar 2007 23:08:53 +0000</pubDate>
		<dc:creator>Manoj</dc:creator>
				<category><![CDATA[Debates]]></category>
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		<guid isPermaLink="false">http://theunrealuniverse.com/blog/?p=28</guid>
		<description><![CDATA[Posting an old discussion on SFN. The Twin Paradox is usually explained away by arguing that the traveling twin feels the motion because of his acceleration/deceleration, and therefore ages slower. [...] <a href="http://www.thulasidas.com/2007-03/twin-paradox-take-2.htm">Continue reading <span class="meta-nav">&#8594;</span></a>]]></description>
			<content:encoded><![CDATA[<p>The Twin Paradox is usually explained away by arguing that the traveling twin feels the motion because of his acceleration/deceleration, and therefore ages slower.</p>
<p>But what will happen if the twins both accelerate symmetrically? That is, they start from rest from one space point with synchronized clocks, and get back to the same space point at rest by accelerating away from each other for some time and decelerating on the way back. By the symmetry of the problem, it seems that when the two clocks are together at the end of the journey, at the same point, and at rest with respect to each other, they have to agree.</p>
<p>Then again, during the whole journey, each clock is in motion (accelerated or not) with respect to the other one. In SR, every clock that is in motion with respect to an observer&#8217;s clock is supposed run slower. Or, the observer&#8217;s clock is always the fastest. So, for each twin, the other clock must be running slower. However, when they come back together at the end of the journey, they have to agree. This can happen only if each twin sees the other&#8217;s clock running faster at some point during the journey. What does SR say will happen in this imaginary journey?</p>
<p>(Note that the acceleration of each twin can be made constant. Have the twins cross each other at a high speed at a constant linear deceleration. They will cross again each other at the same speed after sometime. During the crossings, their clocks can be compared.)</p>
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		<title>Superluminal Laser Dots</title>
		<link>http://www.thulasidas.com/2006-08/superluminal-laser-dots.htm</link>
		<comments>http://www.thulasidas.com/2006-08/superluminal-laser-dots.htm#comments</comments>
		<pubDate>Wed, 02 Aug 2006 23:37:42 +0000</pubDate>
		<dc:creator>Manoj</dc:creator>
				<category><![CDATA[Debates]]></category>
		<category><![CDATA[physics]]></category>
		<category><![CDATA[SFN - Science Forums]]></category>

		<guid isPermaLink="false">http://theunrealuniverse.com/blog/?p=30</guid>
		<description><![CDATA[A discussion in the Science Forums on the appearance of a laser dot on a ceiling. It is thought that if you pointed a laser dot on a ceiling and move the laser gun fast enough, the dot could move superluminally. Could it really? <a href="http://www.thulasidas.com/2006-08/superluminal-laser-dots.htm">Continue reading <span class="meta-nav">&#8594;</span></a>]]></description>
			<content:encoded><![CDATA[<p align="center"><img src="/img/laser.gif" border="0" alt="" /></p>
<p>In the animation above, the laser pointer is at the bottom-center of the figure. I&#8217;m turning it so fast that the laser dot on the ceiling should travel at 10c (which is indicated by the white circle moving across the ceiling.) But it takes a while for the laser to reach the ceiling. The light is indicated by the small red dots which move towards the ceiling (10 times slower than the white circle). The dot appears when the light hits the ceiling.</p>
<p>As you can see, the light from the laser first hits the ceiling at a point close to the top (indicated by the black dot), and subsequently, light on either side starts hitting the ceiling, making two dots (in yellow and green). This is the how one laser pointer creates two dots appears at two places at the same time. Note how the dots slow down considerably as they move away from the center. Light travel time effects dominate at shallow angles.</p>
<p>But CPL.Luke is right. If the ceiling was a spherical shell and the laser was at its center, there would be only one dot moving at 10c. (At least, that&#8217;s what I get when I try to work it out.) In effect, by having a spherical ceiling, you are cutting out the shallow angles; the laser is always perpendicular to the ceiling. In this case you can treat the laser as a solid rod, but with a constant delay equal to r/c (which differentiates to zero, thus not affecting the speed of the dot).</p>
<p><strong>If a superliminal object did exist, would it appear as two identical objects to us?</strong></p>
<p>Okay, I will start with an animation first to make it interesting. I will post the notations and algebra in the next post.</p>
<p>In the figure below, we have a (purely hypothetical) superluminal object &#8211; the white circle flying across the animation at ten times the speed of light. As it flies by, it emits light. We consider the light rays (the red lines with small red circles at the end) coming towards the observer at the bottom-center of the animation. As we can see, the first ray of light that reaches the observer is emitted at a point close to the point of closest approach to the observer, indicated by a black dot that appears when the ray reaches him, say at time=<a onclick="do_texpopup('t_o', 'math'); return false;" href="javascript:;"></a>. The rays emitted <em>before</em> this first ray reach the observer <em>after</em> <a onclick="do_texpopup('t_o', 'math'); return false;" href="javascript:;"></a>. This reversal of the order in which the rays reach the observer gives rise to the perception of two objects moving away from the black dot. (If the object doesn&#8217;t change during its flight, the two &#8220;phantom&#8221; objects are identical to each other.)</p>
<p align="center"><img src="/img/grb-small.gif" border="0" alt="" /></p>
<p>Now, my question is, if we see two objects in a symmetric formation in the night sky, can we be sure that they are really two, and not our perception of one object in motion? Of course we can if we say that nothing can really travel faster than light. Assuming hypothetically that we didn&#8217;t know about SR and its constraint on the speed, is there any way we could work out the &#8220;real&#8221; speed from our observation of the rate of angular separation? My feeling is that there are at least two configurations (one superluminal object going in one direction or two objects &#8211; superluminal or otherwise &#8211; going in opposite directions) which will result in the same observation.</p>
<p><strong>Algebra behind the animation</strong></p>
<p>This post gives the algebra behind the animation.  First, let&#8217;s define the notations used using the following figure.</p>
<p align="center"><img title="animation" src="/img/notat.gif" alt="animation" /></p>
<p>Here, the object is traveling along the thick horizontal line at a speed <img src="http://l.wordpress.com/latex.php?latex=%5Cbeta&#038;bg=FFFFFF&#038;fg=000000&#038;s=0" title="\beta" style="vertical-align:-20%;" class="tex" alt="\beta" />.  The black dot in the animation (where the object first appears to the observer) is B&#8217;. B is the point of closest approach.  Let&#8217;s set the time <img src="http://l.wordpress.com/latex.php?latex=t%3D0&#038;bg=FFFFFF&#038;fg=000000&#038;s=0" title="t=0" style="vertical-align:-20%;" class="tex" alt="t=0" /> when the object is at the point B. The line of flight (at its closest point B) is at a distance of y from the observer at O.  A is a typical point at a distance x from B.  <img src="http://l.wordpress.com/latex.php?latex=%5Ctheta&#038;bg=FFFFFF&#038;fg=000000&#038;s=0" title="\theta" style="vertical-align:-20%;" class="tex" alt="\theta" /> is the angle between the line of flight and the observer&#8217;s line of sight.  <img src="http://l.wordpress.com/latex.php?latex=%5Cphi&#038;bg=FFFFFF&#038;fg=000000&#038;s=0" title="\phi" style="vertical-align:-20%;" class="tex" alt="\phi" /> is that the angle that the object subtends at the observer&#8217;s position O with respect to the normal.  Let&#8217;s set <img src="http://l.wordpress.com/latex.php?latex=c%3D1&#038;bg=FFFFFF&#038;fg=000000&#038;s=0" title="c=1" style="vertical-align:-20%;" class="tex" alt="c=1" /> to simplify the algebra, so that  <img src="http://l.wordpress.com/latex.php?latex=t_o&#038;bg=FFFFFF&#038;fg=000000&#038;s=0" title="t_o" style="vertical-align:-20%;" class="tex" alt="t_o" />, the observer&#8217;s time is <img src="http://l.wordpress.com/latex.php?latex=t%20-%20y&#038;bg=FFFFFF&#038;fg=000000&#038;s=0" title="t - y" style="vertical-align:-20%;" class="tex" alt="t - y" />. (A- is another representative point where <img src="http://l.wordpress.com/latex.php?latex=t%2C%20x&#038;bg=FFFFFF&#038;fg=000000&#038;s=0" title="t, x" style="vertical-align:-20%;" class="tex" alt="t, x" /> and <img src="http://l.wordpress.com/latex.php?latex=%5Cphi&#038;bg=FFFFFF&#038;fg=000000&#038;s=0" title="\phi" style="vertical-align:-20%;" class="tex" alt="\phi" /> are negative.)</p>
<p>With these notations, we can write down the following equation for the real position of the object at time <img src="http://l.wordpress.com/latex.php?latex=t&#038;bg=FFFFFF&#038;fg=000000&#038;s=0" title="t" style="vertical-align:-20%;" class="tex" alt="t" />:</p>
<p><img src="http://l.wordpress.com/latex.php?latex=x%20%3D%20y%5Ctan%5Cphi%20%3D%20%5Cbeta%20t&#038;bg=FFFFFF&#038;fg=000000&#038;s=0" title="x = y\tan\phi = \beta t" style="vertical-align:-20%;" class="tex" alt="x = y\tan\phi = \beta t" /></p>
<p>Or,<br />
<img src="http://l.wordpress.com/latex.php?latex=t%20%3D%20%5Cfrac%7By%5Ctan%5Cphi%7D%7B%5Cbeta%7D&#038;bg=FFFFFF&#038;fg=000000&#038;s=0" title="t = \frac{y\tan\phi}{\beta}" style="vertical-align:-20%;" class="tex" alt="t = \frac{y\tan\phi}{\beta}" /></p>
<p>A photon emitted by the object at A (at time <img src="http://l.wordpress.com/latex.php?latex=t&#038;bg=FFFFFF&#038;fg=000000&#038;s=0" title="t" style="vertical-align:-20%;" class="tex" alt="t" />) will reach O after traversing the hypotenuse.  A photon emitted at B will reach the observer at <img src="http://l.wordpress.com/latex.php?latex=t%20%3D%20y&#038;bg=FFFFFF&#038;fg=000000&#038;s=0" title="t = y" style="vertical-align:-20%;" class="tex" alt="t = y" />, since we have chosen <img src="http://l.wordpress.com/latex.php?latex=c%20%3D%201&#038;bg=FFFFFF&#038;fg=000000&#038;s=0" title="c = 1" style="vertical-align:-20%;" class="tex" alt="c = 1" />.  We have defined the observer&#8217;s time <img src="http://l.wordpress.com/latex.php?latex=t_o&#038;bg=FFFFFF&#038;fg=000000&#038;s=0" title="t_o" style="vertical-align:-20%;" class="tex" alt="t_o" /> such that <img src="http://l.wordpress.com/latex.php?latex=t%20%3D%20t_o%20%2B%20y&#038;bg=FFFFFF&#038;fg=000000&#038;s=0" title="t = t_o + y" style="vertical-align:-20%;" class="tex" alt="t = t_o + y" />, then we have:</p>
<p><img src="http://l.wordpress.com/latex.php?latex=t_o%20%3D%20t%20%2B%20%5Cfrac%7By%7D%7B%5Ccos%5Cphi%7D%20-%20y&#038;bg=FFFFFF&#038;fg=000000&#038;s=0" title="t_o = t + \frac{y}{\cos\phi} - y" style="vertical-align:-20%;" class="tex" alt="t_o = t + \frac{y}{\cos\phi} - y" /></p>
<p>which gives the relation between <img src="http://l.wordpress.com/latex.php?latex=t_o&#038;bg=FFFFFF&#038;fg=000000&#038;s=0" title="t_o" style="vertical-align:-20%;" class="tex" alt="t_o" /> and <img src="http://l.wordpress.com/latex.php?latex=%5Cphi&#038;bg=FFFFFF&#038;fg=000000&#038;s=0" title="\phi" style="vertical-align:-20%;" class="tex" alt="\phi" />.</p>
<p><img src="http://l.wordpress.com/latex.php?latex=t_o%20%3D%20y%5Cleft%28%20%5Cfrac%7B%5Ctan%5Cphi%7D%5Cbeta%20%2B%20%5Cfrac%7B1%7D%7B%5Ccos%5Cphi%7D%20-%201%5Cright%29&#038;bg=FFFFFF&#038;fg=000000&#038;s=0" title="t_o = y\left( \frac{\tan\phi}\beta + \frac{1}{\cos\phi} - 1\right)" style="vertical-align:-20%;" class="tex" alt="t_o = y\left( \frac{\tan\phi}\beta + \frac{1}{\cos\phi} - 1\right)" /></p>
<p>Expanding the equation for <img src="http://l.wordpress.com/latex.php?latex=t_o&#038;bg=FFFFFF&#038;fg=000000&#038;s=0" title="t_o" style="vertical-align:-20%;" class="tex" alt="t_o" /> to second order, we get:</p>
<p><img src="http://l.wordpress.com/latex.php?latex=t_o%20%3D%20y%5Cleft%28%5Cfrac%5Cphi%5Cbeta%20%2B%20%5Cfrac%7B%5Cphi%5E2%7D%7B2%7D%5Cright%29&#038;bg=FFFFFF&#038;fg=000000&#038;s=0" title="t_o = y\left(\frac\phi\beta + \frac{\phi^2}{2}\right)" style="vertical-align:-20%;" class="tex" alt="t_o = y\left(\frac\phi\beta + \frac{\phi^2}{2}\right)" /><br />
(Call this equation Q.)</p>
<p>The minimum value of <img src="http://l.wordpress.com/latex.php?latex=t_o&#038;bg=FFFFFF&#038;fg=000000&#038;s=0" title="t_o" style="vertical-align:-20%;" class="tex" alt="t_o" /> occurs at <img src="http://l.wordpress.com/latex.php?latex=%5Cphi_%7B0%7D%3D-1%2F%5Cbeta&#038;bg=FFFFFF&#038;fg=000000&#038;s=0" title="\phi_{0}=-1/\beta" style="vertical-align:-20%;" class="tex" alt="\phi_{0}=-1/\beta" /> (which defines the position of the black dot in the animation, the point B&#8217;)  and it is <img src="http://l.wordpress.com/latex.php?latex=%7Bt_o%7D_%7Bmin%7D%20%3D%20-y%2F2%5Cbeta%5E2&#038;bg=FFFFFF&#038;fg=000000&#038;s=0" title="{t_o}_{min} = -y/2\beta^2" style="vertical-align:-20%;" class="tex" alt="{t_o}_{min} = -y/2\beta^2" />.  To the observer, the object first appears at the position <img src="http://l.wordpress.com/latex.php?latex=%5Cphi%3D-1%2F%5Cbeta&#038;bg=FFFFFF&#038;fg=000000&#038;s=0" title="\phi=-1/\beta" style="vertical-align:-20%;" class="tex" alt="\phi=-1/\beta" />.  Then it appears to stretch and split, rapidly at first, and slowing down later.</p>
<p>The quadratic equation Q above can be recast as:<br />
<img src="http://l.wordpress.com/latex.php?latex=1%2B%5Cfrac%7B2%5Cbeta%5E2%7D%7By%7Dt_o%20%3D%20%5Cleft%281%2B%5Cbeta%5Cphi%5Cright%29%5E2&#038;bg=FFFFFF&#038;fg=000000&#038;s=0" title="1+\frac{2\beta^2}{y}t_o = \left(1+\beta\phi\right)^2" style="vertical-align:-20%;" class="tex" alt="1+\frac{2\beta^2}{y}t_o = \left(1+\beta\phi\right)^2" /><br />
which will be more useful later in the derivation. (Call this equation U.)</p>
<p>The angular separation between the objects flying away from each otheris the difference between the roots of the quadratic equation Q:</p>
<p style="text-align: left;"><img src="http://l.wordpress.com/latex.php?latex=%5CPhi%20%5C%2C%3D%5C%2C%20%5Cphi_1-%5Cphi_2%20&#038;bg=FFFFFF&#038;fg=000000&#038;s=0" title="\Phi \,=\, \phi_1-\phi_2 " style="vertical-align:-20%;" class="tex" alt="\Phi \,=\, \phi_1-\phi_2 " /><br />
<img src="http://l.wordpress.com/latex.php?latex=%5C%2C%3D%5C%2C%20%5Cfrac%7B2%7D%7B%5Cbeta%7D%5Csqrt%7B1%2B%5Cfrac%7B2%5Cbeta%5E2%7D%7By%7Dt_o%7D%20&#038;bg=FFFFFF&#038;fg=000000&#038;s=0" title="\,=\, \frac{2}{\beta}\sqrt{1+\frac{2\beta^2}{y}t_o} " style="vertical-align:-20%;" class="tex" alt="\,=\, \frac{2}{\beta}\sqrt{1+\frac{2\beta^2}{y}t_o} " /><br />
<img src="http://l.wordpress.com/latex.php?latex=%5C%2C%3D%5C%2C%20%5Cfrac%7B2%7D%7B%5Cbeta%7D%5Cleft%281%2B%5Cbeta%5Cphi%5Cright%29&#038;bg=FFFFFF&#038;fg=000000&#038;s=0" title="\,=\, \frac{2}{\beta}\left(1+\beta\phi\right)" style="vertical-align:-20%;" class="tex" alt="\,=\, \frac{2}{\beta}\left(1+\beta\phi\right)" /><br />
making use of the &#8220;useful&#8221; equation U above.  Thus, we have the angular separation either in terms of the observer&#8217;s time (<img src="http://l.wordpress.com/latex.php?latex=%5CPhi%28t_o%29&#038;bg=FFFFFF&#038;fg=000000&#038;s=0" title="\Phi(t_o)" style="vertical-align:-20%;" class="tex" alt="\Phi(t_o)" />) or the angular position of the object (<img src="http://l.wordpress.com/latex.php?latex=%5CPhi%28%5Cphi%29&#038;bg=FFFFFF&#038;fg=000000&#038;s=0" title="\Phi(\phi)" style="vertical-align:-20%;" class="tex" alt="\Phi(\phi)" />) as illustrated in the next figure, which illustrates how the angular separation  is expressed either in terms of the observer&#8217;s time (<img src="http://l.wordpress.com/latex.php?latex=%5CPhi%28t_o%29&#038;bg=FFFFFF&#038;fg=000000&#038;s=0" title="\Phi(t_o)" style="vertical-align:-20%;" class="tex" alt="\Phi(t_o)" />) or the angular position of the object (<img src="http://l.wordpress.com/latex.php?latex=%5CPhi%28%5Cphi%29&#038;bg=FFFFFF&#038;fg=000000&#038;s=0" title="\Phi(\phi)" style="vertical-align:-20%;" class="tex" alt="\Phi(\phi)" />).</p>
<p align="center"><img title="Phi" src="/img/phiphi.gif" alt="Phi" /></p>
<p>The rate at which the angular separation occurs is:</p>
<p><img src="http://l.wordpress.com/latex.php?latex=%5Cfrac%7Bd%5CPhi%7D%7Bdt_o%7D%20%5C%2C%3D%5C%2C%20%5Cfrac%7B2%5Cbeta%7D%7By%5Csqrt%7B1%2B%5Cfrac%7B2%5Cbeta%5E2%7D%7By%7Dt_o%7D%7D&#038;bg=FFFFFF&#038;fg=000000&#038;s=0" title="\frac{d\Phi}{dt_o} \,=\, \frac{2\beta}{y\sqrt{1+\frac{2\beta^2}{y}t_o}}" style="vertical-align:-20%;" class="tex" alt="\frac{d\Phi}{dt_o} \,=\, \frac{2\beta}{y\sqrt{1+\frac{2\beta^2}{y}t_o}}" /><br />
<img src="http://l.wordpress.com/latex.php?latex=%5C%2C%3D%5C%2C%20%5Cfrac%7B2%5Cbeta%7D%7By%5Cleft%281%2B%5Cbeta%5Cphi%5Cright%29%7D&#038;bg=FFFFFF&#038;fg=000000&#038;s=0" title="\,=\, \frac{2\beta}{y\left(1+\beta\phi\right)}" style="vertical-align:-20%;" class="tex" alt="\,=\, \frac{2\beta}{y\left(1+\beta\phi\right)}" /></p>
<p>Again, making use of the useful equation U.  Defining the apparent age of the formation <img src="http://l.wordpress.com/latex.php?latex=t_%7Bage%7D%20%3D%20t_o%20-%20%7Bt_o%7D_%7Bmin%7D&#038;bg=FFFFFF&#038;fg=000000&#038;s=0" title="t_{age} = t_o - {t_o}_{min}" style="vertical-align:-20%;" class="tex" alt="t_{age} = t_o - {t_o}_{min}" /> and knowing <img src="http://l.wordpress.com/latex.php?latex=%7Bt_o%7D_%7Bmin%7D%20%3D%20-y%2F2%5Cbeta%5E2&#038;bg=FFFFFF&#038;fg=000000&#038;s=0" title="{t_o}_{min} = -y/2\beta^2" style="vertical-align:-20%;" class="tex" alt="{t_o}_{min} = -y/2\beta^2" />, we can write:</p>
<p><img src="http://l.wordpress.com/latex.php?latex=%5Cfrac%7Bd%5CPhi%7D%7Bdt_o%7D%20%5C%2C%3D%5C%2C%20%5Cfrac%7B2%5Cbeta%7D%7By%5Csqrt%7B1%2B%5Cfrac%7B2%5Cbeta%5E2%7D%7By%7Dt_o%7D%7D&#038;bg=FFFFFF&#038;fg=000000&#038;s=0" title="\frac{d\Phi}{dt_o} \,=\, \frac{2\beta}{y\sqrt{1+\frac{2\beta^2}{y}t_o}}" style="vertical-align:-20%;" class="tex" alt="\frac{d\Phi}{dt_o} \,=\, \frac{2\beta}{y\sqrt{1+\frac{2\beta^2}{y}t_o}}" /><br />
<img src="http://l.wordpress.com/latex.php?latex=%5C%2C%3D%5C%2C%20%5Cfrac%7B2%5Cbeta%7D%7By%5Csqrt%7B1-%5Cfrac%7Bt_o%7D%7B%7Bt_o%7D_%7Bmin%7D%7D%7D%7D&#038;bg=FFFFFF&#038;fg=000000&#038;s=0" title="\,=\, \frac{2\beta}{y\sqrt{1-\frac{t_o}{{t_o}_{min}}}}" style="vertical-align:-20%;" class="tex" alt="\,=\, \frac{2\beta}{y\sqrt{1-\frac{t_o}{{t_o}_{min}}}}" /><br />
<img src="http://l.wordpress.com/latex.php?latex=%5C%2C%3D%5C%2C%20%5Csqrt%7B%5Cfrac%7B4%5Cbeta%5E2%7D%7By%5E2%7D%5C%2C%5Ctimes%5C%2C%5Cfrac%7B-%7Bt_o%7D_%7Bmin%7D%7D%7Bt_o-%7Bt_o%7D_%7Bmin%7D%7D%7D&#038;bg=FFFFFF&#038;fg=000000&#038;s=0" title="\,=\, \sqrt{\frac{4\beta^2}{y^2}\,\times\,\frac{-{t_o}_{min}}{t_o-{t_o}_{min}}}" style="vertical-align:-20%;" class="tex" alt="\,=\, \sqrt{\frac{4\beta^2}{y^2}\,\times\,\frac{-{t_o}_{min}}{t_o-{t_o}_{min}}}" /><br />
<img src="http://l.wordpress.com/latex.php?latex=%5C%2C%3D%5C%2C%5Csqrt%7B%5Cfrac%7B2%7D%7By%5C%2C%20t_%7Bage%7D%7D%7D&#038;bg=FFFFFF&#038;fg=000000&#038;s=0" title="\,=\,\sqrt{\frac{2}{y\, t_{age}}}" style="vertical-align:-20%;" class="tex" alt="\,=\,\sqrt{\frac{2}{y\, t_{age}}}" /></p>
<p>Note that in order to go from the angular rate to the speed (even the apparent speed), we need to estimate <img src="http://l.wordpress.com/latex.php?latex=y&#038;bg=FFFFFF&#038;fg=000000&#038;s=0" title="y" style="vertical-align:-20%;" class="tex" alt="y" />, which is model-based.</p>
<p>Now, can you tell me why the object doesn&#8217;t appear at two places if <img src="http://l.wordpress.com/latex.php?latex=%5Cbeta%20%26lt%3B%201%20&#038;bg=FFFFFF&#038;fg=000000&#038;s=0" title="\beta &lt; 1 " style="vertical-align:-20%;" class="tex" alt="\beta &lt; 1 " />? <img src='http://www.thulasidas.com/wp-includes/images/smilies/icon_smile.gif' alt=':-)' class='wp-smiley' /> </p>
<p>In fact, there is much more to this puzzle.</p>
<ul>
<li>The reason the dot on the right doesn&#8217;t move faster than c is closely related to way the relativistic speed limit is derived.</li>
<li>One should also consider that saying that the dot is on the ceiling at a particular point at a certain instant of time is not good enough. When will the observer (presumably at the laser gun) see it? There is one more leg of light travel (from the dot back to the gun) that needs to included. This consideration may be behind the definition of simultaneity (using the round trip travel of light) in SR.</li>
<li>To the left of the point of separation (the black point), the flow of time is reversed. In other words, if you changed the laser color as you scanned from the left edge of the figure to the center, the change in the color of the dot would appear in the reverse order (in time).</li>
<li>The Doppler shift also is reversed in this region. This is why I was intrigued by the left-handed material, where the group velocity and the Doppler shift are reversed.</li>
</ul>
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