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	<title>Coherent Light &#187; lightspeed</title>
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		<title>a constant c?</title>
		<link>http://www.1310nm.net/blogs/2008/05/15/a-constant-c/</link>
		<comments>http://www.1310nm.net/blogs/2008/05/15/a-constant-c/#comments</comments>
		<pubDate>Thu, 15 May 2008 18:16:39 +0000</pubDate>
		<dc:creator>John Dixon</dc:creator>
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		<category><![CDATA[lightspeed]]></category>
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		<guid isPermaLink="false">http://www.1310nm.net/blogs/?p=21</guid>
		<description><![CDATA[It all started with a discussion around the tea-table a few days ago, about the fact that a lightyear is a measurement of distance and not of time, because a light year is the distance traveled by a photon over the period of a year. (d=s*t for those who like the maths, or the area &#8230; </p><p><a class="more-link block-button" href="http://www.1310nm.net/blogs/2008/05/15/a-constant-c/">Continue reading &#187;</a>]]></description>
			<content:encoded><![CDATA[<p>It all started with a discussion around the tea-table a few days ago, about the fact that a lightyear is a measurement of distance and not of time, because a light year is the distance traveled by a photon over the period of a year. (<code>d=s*t</code> for those who like the maths, or the area under a speed-time graph). Then someone else remembered that 2008 is a leap year, and contains an extra day..<br />
So the question is does light travel slower on leap years?<br />
Here I set out to prove that it does.<br />
<span id="more-21"></span><br />
<strong>So let’s take a look at this more closely..</strong><br />
Firstly a light-year is a measure of distance, so this is a fixed length (since it’s based on SI units), and can be represented by <em>d</em>. Secondly we have a value <code>t</code> that represents the time this takes (1 year). Finally we have a <code>c</code> that is a representation of the speed of light. So we take our old <code>d=s*t</code> equation, and juggle it to get <code><img src='http://s0.wp.com/latex.php?latex=s%3D%5Cfrac%7Bd%7D%7Bt%7D&#038;bg=ffffff&#038;fg=000&#038;s=0' alt='s=&#92;frac{d}{t}' title='s=&#92;frac{d}{t}' class='latex' /></code>. Then we plug in our variables instead, so we have <code><img src='http://s0.wp.com/latex.php?latex=c%3D%5Cfrac%7Bd%7D%7Bt%7D&#038;bg=ffffff&#038;fg=000&#038;s=0' alt='c=&#92;frac{d}{t}' title='c=&#92;frac{d}{t}' class='latex' /></code>.</p>
<p>Now this is where we need to look more closely at the problem, since a year is not a standard length. Every four years we have an extra day added to cope with some differences in measurement of astronomical (<a href="http://en.wikipedia.org/wiki/Sidereal_year">sidereal</a> and <a href="http://en.wikipedia.org/wiki/Solar_year">solar</a>, I believe) time. So in effect we are doing this calculation over four years, but with two differing values of <code>t</code>, <code><img src='http://s0.wp.com/latex.php?latex=t_n&#038;bg=ffffff&#038;fg=000&#038;s=0' alt='t_n' title='t_n' class='latex' /></code> for three years and <code><img src='http://s0.wp.com/latex.php?latex=t_l&#038;bg=ffffff&#038;fg=000&#038;s=0' alt='t_l' title='t_l' class='latex' /></code> for the leap year. This will give us two values of <code>c</code>, <code><img src='http://s0.wp.com/latex.php?latex=c_n&#038;bg=ffffff&#038;fg=000&#038;s=0' alt='c_n' title='c_n' class='latex' /></code> for non leap years, and <code><img src='http://s0.wp.com/latex.php?latex=c_l&#038;bg=ffffff&#038;fg=000&#038;s=0' alt='c_l' title='c_l' class='latex' /></code> for leap years.<br />
A quick search of wikipedia shows that <code>d</code> is 9,460,730,472,580,800 m<br />
We also have two values of <code>t</code>, <code><img src='http://s0.wp.com/latex.php?latex=t_n&#038;bg=ffffff&#038;fg=000&#038;s=0' alt='t_n' title='t_n' class='latex' /></code> which is 365 days, and <code><img src='http://s0.wp.com/latex.php?latex=t_l&#038;bg=ffffff&#038;fg=000&#038;s=0' alt='t_l' title='t_l' class='latex' /></code> of 366 days.<br />
We’ll convert the time into seconds (so it’s an SI unit), using the 60*60*24 (seconds, minutes, hours) or 86400 seconds a day. <code><img src='http://s0.wp.com/latex.php?latex=t_n&#038;bg=ffffff&#038;fg=000&#038;s=0' alt='t_n' title='t_n' class='latex' /></code> is 31,536,000 seconds, and <code><img src='http://s0.wp.com/latex.php?latex=t_l&#038;bg=ffffff&#038;fg=000&#038;s=0' alt='t_l' title='t_l' class='latex' /></code> is 31,622,400.</p>
<p><strong>So what are the speeds of light?</strong><br />
So quickly plugging the values into my calculator, gives <em><img src='http://s0.wp.com/latex.php?latex=c_n&#038;bg=ffffff&#038;fg=000&#038;s=0' alt='c_n' title='c_n' class='latex' /></em> a result of 9,460,730,472,580,800 / 31,536,000, or 299,997,795.3 m/s.<br />
Similarly <code><img src='http://s0.wp.com/latex.php?latex=c_l&#038;bg=ffffff&#038;fg=000&#038;s=0' alt='c_l' title='c_l' class='latex' /></code> produces a result of 9,460,730,472,580,800 / 31,622,400, or 299,178,129.192 m/s.<br />
This looks right, since a leap year is longer than other years, and light would therefore have to travel slower to cover the same distance.</p>
<p><strong>How can we check this?</strong><br />
A physicist might do this with a torch, a tape measure and a stop watch, but since we are good mathematicians, we check the results, by seeing if we can use these answers to get back to the original figures, but by using a different question.<br />
Since we know the ratio of the lengths of time, we should be able to use the ratio of the speeds to reproduce this.<br />
i.e. <code><img src='http://s0.wp.com/latex.php?latex=%5Cfrac%7Bc_l%7D%7Bc_n%7D%3D%5Cfrac%7Bt_l%7D%7Bt_n%7D&#038;bg=ffffff&#038;fg=000&#038;s=0' alt='&#92;frac{c_l}{c_n}=&#92;frac{t_l}{t_n}' title='&#92;frac{c_l}{c_n}=&#92;frac{t_l}{t_n}' class='latex' /></code>. If we juggle this to get a value on one side, we should get <code><img src='http://s0.wp.com/latex.php?latex=%28%5Cfrac%7Bc_l%7D%7Bc_n%7D+%2A+t_n%3D%5Cfrac%7Bt_l%7D%7Bt%7D&#038;bg=ffffff&#038;fg=000&#038;s=0' alt='(&#92;frac{c_l}{c_n} * t_n=&#92;frac{t_l}{t}' title='(&#92;frac{c_l}{c_n} * t_n=&#92;frac{t_l}{t}' class='latex' /></code><br />
So here are the numbers… (299997795.3/299178129.192) * 31,536,000 which produces an answer of 31,622,400. So this is self-consistent at least, as 31,622,400 is the number of seconds in a leap year, exactly the number we&#8217;re looking for.</p>
<p><strong>So how much slower is light in a leap year?</strong><br />
Well, that’s easy, we simply boil down the ratio of <code><img src='http://s0.wp.com/latex.php?latex=%5Cfrac%7Bt_n%7D%7Bt_l%7D&#038;bg=ffffff&#038;fg=000&#038;s=0' alt='&#92;frac{t_n}{t_l}' title='&#92;frac{t_n}{t_l}' class='latex' /></code>, or <img src='http://s0.wp.com/latex.php?latex=%5Cfrac%7B31%2C536%2C000%7D%7B31%2C622%2C400%7D&#038;bg=ffffff&#038;fg=000&#038;s=0' alt='&#92;frac{31,536,000}{31,622,400}' title='&#92;frac{31,536,000}{31,622,400}' class='latex' />. Lo and behold, it becomes <img src='http://s0.wp.com/latex.php?latex=%5Cfrac%7B365%7D%7B366%7D&#038;bg=ffffff&#038;fg=000&#038;s=0' alt='&#92;frac{365}{366}' title='&#92;frac{365}{366}' class='latex' />, or 99.726776% of the speed of light in other years.</p>
<p><strong>So doesn’t this make life complicated?</strong><br />
The short answer is yes, it does, so science has conveniently forgotten the fact that the speed of light is variable (it tends to conflict with some of Einstein’s theories), and has instead chosen a different method of approximating the speed of light. This basically amounts to creating an average speed over four years, or 1461 days. Doing this calculation, we have <code><img src='http://s0.wp.com/latex.php?latex=c+%3D+%5Cfrac%7B%284d%29%7D%7B%28%283t_n%29%2Bt_l%29%7D&#038;bg=ffffff&#038;fg=000&#038;s=0' alt='c = &#92;frac{(4d)}{((3t_n)+t_l)}' title='c = &#92;frac{(4d)}{((3t_n)+t_l)}' class='latex' /></code>. Which gives the following math to calculate. ( 4 * 9,460,730,472,580,800) / ((3 * 31,536,000) + 31,622,400) This equates to 299,792,458 m/s, which is the average speed of light often quoted by scientists. Astronomers also fudge the time, by creating the Julian year, which comprised 365.25 days (and a <a href="http://www.thefreedictionary.com/smidgen">smidgen</a>) to make the calculation consistent.</p>
<p><strong>I’ve heard that the speed of light is slower in denser environments, could this be a reason?</strong><br />
It’s a well known fact that if you increase the density of objects, light slows down as it passes through them. This is especially true for black holes, which have infinite density, and light never makes it out, as it slows to a stop. However even I have to admit that the notion that the universe becomes more dense to slow light for a period of time every four years is preposterous!}</p>
<p>So just remember, that even though the scientists would have you believe that <code>c</code> is a constant, it isn’t!</p>
<div class="notice_block">This post is under constant revision due to the rapidly moving nature of time.</div>
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