tag:blogger.com,1999:blog-54971381097658583832014-10-28T05:36:28.285-07:00ArcsecondMarkkimarkkonnenhttp://www.blogger.com/profile/11122368349695914223noreply@blogger.comBlogger45125tag:blogger.com,1999:blog-5497138109765858383.post-48141200888672863682008-10-05T18:16:00.000-07:002008-10-05T18:18:33.175-07:00Moving to WordpressThe LaTeX feature I installed here is no longer working, so I'm moving to Wordpress, where LaTeX support is built in. The new location for the blog is<br /><a href="http://arcsecond.wordpress.com/">http://arcsecond.wordpress.com/</a>Markkimarkkonnenhttp://www.blogger.com/profile/11122368349695914223noreply@blogger.com2tag:blogger.com,1999:blog-5497138109765858383.post-84607714439695067252008-09-29T19:45:00.000-07:002008-09-30T01:46:07.931-07:00Re-examing the Inner Product (Euclidean Space)I'm auditing Kip Thorne's <a href="http://www.pma.caltech.edu/Courses/ph136/yr2008/text.html">"Applications of Classical Physics"</a> this year. Today was the first day of class, and Kip gave a preamble which, if I may be allowed to paraphrase in a wild, inaccurate, and completely unfair manner, (which is basically to say every word of this is existed only in my head) went something approximately like this:<br /><blockquote>Welcome to class. Here is the website. Please call me "Kip". I'm 68 years old and my signature lucky ponytail is a thing of the past. There's not enough time left for ridiculous formalities. I am about to retire. I'm going to start writing books and making movies and various other things that famous retired physicists can do. I'm too old to waterski, so this is the next best option. This is will be last class I ever teach. There will be no grades. I would prefer you actually learn some stuff. Now, let's suppose the laws of physics are frame-independent, and see what restrictions this places on force law dealing with a classical field in Minkowski space..."</blockquote><br /><br />Awesome.<br /><br />One thing that interested me was our classes' definition of the inner product between two vectors. Since we wanted to do everything in a frame-independent manner, we couldn't simply define the inner product by<br /><br /><img src="http://www.forkosh.dreamhost.com/mimetex.cgi?%5Cmathbf%7BA%7D%20%5Ccdot%20%5Cmathbf%7BB%7D%20=%20A_x%20B_x%20+%20A_y%20B_y%20+%20A_z%20B_z" align="middle" border="0" /><br /><br />because that presumes the existence of basis vectors. The other common definition is<br /><br /><img src="http://www.forkosh.dreamhost.com/mimetex.cgi?%5Cmathbf%7BA%7D%20%5Ccdot%20%5Cmathbf%7BB%7D%20=%20A%20B%20%5Ccos%7B%5Ctheta%7D" align="middle" border="0" /><br /><br />which is good, but we wanted something that could generalize to the Minkowski space of special relativity. So the definition we came in two parts. First, we defined the inner product of a vector with itself. Then use that definition to bridge to the inner product of arbitrary vectors.<br /><br />For a vector <img src="http://www.forkosh.dreamhost.com/mimetex.cgi?%5Cmathbf%7BA%7D" align="middle" border="0" />, define<br /><br /><img src="http://www.forkosh.dreamhost.com/mimetex.cgi?%5Cmathbf%7BA%7D%20%5Ccdot%20%5Cmathbf%7BA%7D%20=%20-%20%5CDelta%20s%5E2" align="middle" border="0" /><br /><br />where <img src="http://www.forkosh.dreamhost.com/mimetex.cgi?%5CDelta%20s%5E2" align="middle" border="0" /> is the square of the physical invariant interval between two events. To measure this interval, have an unaccelerated clock move from the origin to the event. (<img src="http://www.forkosh.dreamhost.com/mimetex.cgi?%5CDelta%20s" align="middle" border="0" /> should be real, meaning the square of the interval is positive, and <img src="http://www.forkosh.dreamhost.com/mimetex.cgi?-%5CDelta%20s%5E2" align="middle" border="0" /> is negative. So the inner product of the vector between timelike events with itself is negative). In the case of spacelike events, you should instead find the inertial reference frame in which the events are simultaneous, and lay down a measuring stick between them to get the interval.<br /><br />Now we know how to take the inner product of a vector with itself. Define the inner product of two vectors <img src="http://www.forkosh.dreamhost.com/mimetex.cgi?%5Cmathbf%7BA%7D" align="middle" border="0" /> and <img src="http://www.forkosh.dreamhost.com/mimetex.cgi?%5Cmathbf%7BB%7D" align="middle" border="0" /> by<br /><br /><img src="http://www.forkosh.dreamhost.com/mimetex.cgi?%5Cmathbf%7BA%7D%20%5Ccdot%20%5Cmathbf%7BB%7D%20=%20%5Cfrac%7B1%7D%7B4%7D%5Cleft%28%20%28%5Cmathbf%7BA%7D%20+%20%5Cmathbf%7BB%7D%29%5E2%20-%20%28%5Cmathbf%7BA%7D%20-%20%5Cmathbf%7BB%7D%29%5E2%20%5Cright%29" align="middle" border="0" /><br /><br />This is more subtle than it might initially appear. You can't just go willy-nilly with the algebra and start simplifying that right hand side out. You don't know any properties of this inner product, so you cannot, for example, write<br /><br /><img src="http://www.forkosh.dreamhost.com/mimetex.cgi?%28%5Cmathbf%7BA%7D%20+%20%5Cmathbf%7BB%7D%29%20%5Ccdot%20%28%5Cmathbf%7BA%7D%20+%20%5Cmathbf%7BB%7D%29%20=%20A%5E2%20+%20%5Cmathbf%7BA%7D%20%5Ccdot%20%5Cmathbf%7BB%7D%20+%20%5Cmathbf%7BB%7D%20%5Ccdot%20%5Cmathbf%7BA%7D%20+%20B%5E2" align="middle" border="0" /><br /><br />Instead you have to work with the actual sums. It was claimed in class that this definition of the inner product is bilinear in the arguments, which wasn't obvious to me. So I asked about it, and Kip suggested I try to work it out for myself in Euclidean space using the Pythagorean theorem. So here goes.<br /><br />Break down the vector <img src="http://www.forkosh.dreamhost.com/mimetex.cgi?%5Cmathbf%7BB%7D" align="middle" border="0" /> into a component parallel to <img src="http://www.forkosh.dreamhost.com/mimetex.cgi?%5Cmathbf%7BA%7D" align="middle" border="0" /> and a component perpendicular to <img src="http://www.forkosh.dreamhost.com/mimetex.cgi?%5Cmathbf%7BA%7D" align="middle" border="0" /><br /><br /><img src="http://www.forkosh.dreamhost.com/mimetex.cgi?%5Cmathbf%7BB%7D%20=%20B_%5Cparallel%20+%20B_%5Cperp" align="middle" border="0" /><br /><br />The vector <img src="http://www.forkosh.dreamhost.com/mimetex.cgi?%5Cmathbf%7BA%7D%20+%20%5Cmathbf%7BB%7D" align="middle" border="0" /> now becomes <img src="http://www.forkosh.dreamhost.com/mimetex.cgi?A%20+%20B_%5Cparallel%20%5Cmathbf%7B%5Chat%7B%5Cparallel%7D%7D%20+%20B_%5Cperp%20%5Cmathbf%7B%5Chat%7B%5Cperp%7D%7D" align="middle" border="0" />, where <img src="http://www.forkosh.dreamhost.com/mimetex.cgi?%5Cmathbf%7B%5Chat%7B%5Cparallel%7D%7D" align="middle" border="0" /> and <img src="http://www.forkosh.dreamhost.com/mimetex.cgi?%5Cmathbf%7B%5Chat%7B%5Cperp%7D%7D" align="middle" border="0" /> are unit vectors in the directions parallel and perpendicular to <img src="http://www.forkosh.dreamhost.com/mimetex.cgi?%5Cmathbf%7BA%7D" align="middle" border="0" />.<br /><br />Because these two vectors make a right angle, the Pythagorean theorem applies.<br /><br /><img src="http://www.forkosh.dreamhost.com/mimetex.cgi?%5Cleft%7C%20%5Cmathbf%7BA%7D%20+%20%5Cmathbf%7BB%7D%20%5Cright%7C%5E2%20=%20%28A%20+%20B_%5Cparallel%29%5E2%20+%20B_%5Cperp%5E2" align="middle" border="0" /><br /><br />similarly,<br /><br /><img src="http://www.forkosh.dreamhost.com/mimetex.cgi?%5Cleft%7C%20%5Cmathbf%7BA%7D%20-%20%5Cmathbf%7BB%7D%20%5Cright%7C%5E2%20=%20%28A%20-%20B_%5Cparallel%29%5E2%20+%20B_%5Cperp%5E2" align="middle" border="0" /><br /><br />This is great, because the right hand sides of those equations are just numbers. Expand that out, subtract the two equations, and divide by four to obtain<br /><br /><img src="http://www.forkosh.dreamhost.com/mimetex.cgi?%5Cfrac%7B1%7D%7B4%7D%20%5Cleft%7C%20%5Cmathbf%7BA%7D%20+%20%5Cmathbf%7BB%7D%20%5Cright%7C%5E2%20-%20%5Cleft%7C%20%5Cmathbf%7BA%7D%20-%20%5Cmathbf%7BB%7D%20%5Cright%7C%5E2%20=%20A%20B_%5Cparallel" align="middle" border="0" /><br /><br />That's just the normal definition of the dot product, and is obviously linear in both vectors. Now, how about Minkowski space?Markkimarkkonnenhttp://www.blogger.com/profile/11122368349695914223noreply@blogger.com3tag:blogger.com,1999:blog-5497138109765858383.post-87117268857241950522008-09-29T18:21:00.000-07:002008-09-29T19:19:45.806-07:00Why Am I Posting This Crap?Working with physical models inevitably leads to mathematical problems. In physics education, many professors build up enough tools and techniques to shove these problems aside, and then quickly move on to the real physics. That's simply a reflection of the fact that they care deeply about the physics itself, and I appreciate their efforts to keep students from getting bogged down in details. But taken to the extreme, this attitude produces calculating-maniacs with tons of knowledge and little mathematical intuition. Take at look at the canonical book on math methods for physics (Arfken), and you'll see that it's simply an encyclopedic dictionary of methods.<br /><br />I'm only now becoming mature enough to go back and re-examine many of the things I learned earlier with an eye towards appreciation rather than computation. This practice intrigues me. The purpose of a blog is ostensibly to create content for public consumption. That's not what I'm doing. I'm working through things for myself. Not because I feel a dire need to understand. Because I can get enraptured by the absurd power of a little solid thought to unearth something I never saw before, and yet later becomes laughably obvious.<br /><br />According to Google Analytics, in the last month, 58 total people have visited the site in 172 separate visits, and viewed 268 pages. They've spent a total of 5 hours, 41 minutes reading the posts. That's only a small fraction of the cumulative time it took me to write the posts. Also, a lot of the statistics I quoted are just me, logging in from various computers, trying to crank up the scores so I'll feel more important.<br /><br />Nikita, Julia, and Kangway have actually been participating and giving me feedback, which is pretty much incredible, so thank you for doing that. But looking at the recent posts I've written, I realized a few things:<br /><ul><li>not that many people care about things such as the geometric interpretation of a vector product rule<br /></li><li>people who do care were better students the first time around than I was, and mostly already know</li><li>people who would care, but don't already know this stuff, most likely don't have the assumed background to learn much from my posts.<br /></li><li>even if the content of the posts is "just right" for you, I skip a bunch of steps and gloss over the parts of the arguments that are less interesting to me, because they're too tedious to draw/<img src="http://www.forkosh.dreamhost.com/mimetex.cgi?%5CLaTeX" align="middle" border="0" /> for the blog. Hence, following the posts from scratch would be a rough road. It's not the sort of reading people are looking for when browsing through blog posts. They want quick, to-the-point, easily-comprehensible and entertaining content. This blog is the antithesis.<br /></li></ul>My goal is simply to explore nuances that interest me, however mundane. That explains a few of the points above. When I start writing a post, I have to think about how much background knowledge to assume the reader has. I want to write about the parts that I think are interesting, which means essentially the parts (and only the parts) that I didn't completely understand before I started working on the post. I generally assume that everything I am totally comfortable with, the reader is as well. That means people will frequently be flabbergasted by the strange logical jumps I take and the way I write certain details off as trivial. On the other hand, everything that I had to work out and think about for a while, I feel the need to explain in more detail than many readers would be interested in wading through.<br /><br />Today I finished a post about vector products and determinants. I was aware of the geometric interpretations from a few sources, but I had never put the pieces together for myself. By the end, I was amazed. Not only had I built up the idea of a determinant, motivated geometrically, but I had also shown how to use the determinant to solve a system of equations. I didn't even know that would pop out of the argument. If you had asked me to demonstrate how a determinant works before I started writing, I probably wouldn't even have known. Additionally, the path that led me there was remarkably short, and since I had worked through it myself, it walks through a logical (to my mind) problem-solving process.<br /><br />Math books tend to define and develop the determinant as a number characterizing a linear operator on an n-dimensional vector space. The treatment is more abstract. The proofs showing how to calculate it and how it solves systems of equations frequently rely on manipulating sums of indices and leave me feeling unsatisfied. I believe them, and see that they're correct, but I have difficulty visualizing them as well as the author of the book. I don't knock their approach. I think it's probably the best way. But this supplementary cogitation on the blog has bolstered my appreciation.<br /><br />Physics books, on the other hand, will normally skip over the entire journey and roll the answer out right away, then tell you to start inverting some matrices. Save the math questions for your linear algebra class!<br /><br />Writing this blog is my way of taking control of my own educational process. I think about whatever I want, at my own pace, in my own way. Why then post it on the internet? Anything I write has probably been written somewhere else, and probably with more insight and clarity than I'm providing. If my goals are personal education, why paste it all over this blog, advertising it for people to see?<br /><br />If I were to keep these developments to myself, I wouldn't think them through fully. I'd work until I saw the gist of the argument, then say to myself, "the rest is just mopping up details from here. better move on to something more important."<br /><br />If I don't take the time to mop up the details, the bits of understanding I did build flit away. So by assuming the responsibility to clean things up to the point where someone could potentially read through the post and work out the results for themself, I force myself to clean up the ideas in my own head. That's what I'm after.Markkimarkkonnenhttp://www.blogger.com/profile/11122368349695914223noreply@blogger.com32tag:blogger.com,1999:blog-5497138109765858383.post-30503033730830696102008-09-29T14:20:00.000-07:002008-09-29T15:46:19.977-07:00What does a determinant have to do with a cross product? pt. 2In the <a href="http://arcsecond.blogspot.com/2008/09/what-does-determinant-have-to-do-with.html">previous post</a>, I showed that the quantity<br /><br /><img src="http://www.forkosh.dreamhost.com/mimetex.cgi?%5Cmathbf%7BA%7D%20%5Ccdot%20%5Cmathbf%7BB%7D%20%5Ctimes%20%5Cmathbf%7BC%7D" align="middle" border="0" /><br /><br />and the determinant<br /><!-- \left| \begin{array}{ccc} A_x & A_y & A_z \\ B_x & B_y & B_z \\ C_x & C_y & C_z \end{array} \right| --><br /><br /><img src="http://www.forkosh.dreamhost.com/mimetex.cgi?%5Cleft%7C%20%5Cbegin%7Barray%7D%7Bccc%7D%20A_x%20&%20A_y%20&%20A_z%20%5C%5C%20B_x%20&%20B_y%20&%20B_z%20%5C%5C%20C_x%20&%20C_y%20&%20C_z%20%5Cend%7Barray%7D%20%5Cright%7C" align="middle" border="0" /><br /><br />are both tests that the three vectors <img src="http://www.forkosh.dreamhost.com/mimetex.cgi?%5Cmathbf%7BA%7D" align="middle" border="0" /> , <img src="http://www.forkosh.dreamhost.com/mimetex.cgi?%5Cmathbf%7BB%7D" align="middle" border="0" />, and <img src="http://www.forkosh.dreamhost.com/mimetex.cgi?%5Cmathbf%7BC%7D" align="middle" border="0" /> are independent. Now we'll go about showing that they're actually the same test - that they're the same number.<br /><br />You may be familiar with the geometric interpretation of <img src="http://www.forkosh.dreamhost.com/mimetex.cgi?%5Cmathbf%7BA%7D%20%5Ccdot%20%5Cmathbf%7BB%7D%20%5Ctimes%20%5Cmathbf%7BC%7D" align="middle" border="0" /> as the volume of a parallelipiped (box) with sides <img src="http://www.forkosh.dreamhost.com/mimetex.cgi?%5Cmathbf%7BA%7D" align="middle" border="0" /> , <img src="http://www.forkosh.dreamhost.com/mimetex.cgi?%5Cmathbf%7BB%7D" align="middle" border="0" /> , and <img src="http://www.forkosh.dreamhost.com/mimetex.cgi?%5Cmathbf%7BC%7D" align="middle" border="0" />.<br /><br />Just a quick explanation - <img src="http://www.forkosh.dreamhost.com/mimetex.cgi?%5Cmathbf%7BB%7D%20%5Ctimes%20%5Cmathbf%7BC%7D" align="middle" border="0" /> is a vector perpendicular to both, whose length is the area of the parallelogram they form. See stolen image below:<br /><br /><img src="http://chortle.ccsu.edu/VectorLessons/vch12/parallelogramArea.gif" /><br /><br />If I take that and dot it with <img src="http://www.forkosh.dreamhost.com/mimetex.cgi?%5Cmathbf%7BA%7D" align="middle" border="0" /> , I get the projection of <img src="http://www.forkosh.dreamhost.com/mimetex.cgi?%5Cmathbf%7BA%7D" align="middle" border="0" /> into the perpendicular direction, times the area of the box. That is, I get the height of the box times the area of its base. That's the volume. Wikipedia visualization:<br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://upload.wikimedia.org/wikipedia/commons/thumb/3/3e/Parallelepiped_volume.svg/780px-Parallelepiped_volume.svg.png"><img style="margin: 0pt 10px 10px 0pt; float: left; cursor: pointer; width: 320px;" src="http://upload.wikimedia.org/wikipedia/commons/thumb/3/3e/Parallelepiped_volume.svg/780px-Parallelepiped_volume.svg.png" alt="" border="0" /></a><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br />There's one subtlety, which is that the volume of the box records its "handedness" in the sign of <img src="http://www.forkosh.dreamhost.com/mimetex.cgi?%5Cmathbf%7BA%7D%20%5Ccdot%20%5Cmathbf%7BB%7D%20%5Ctimes%20%5Cmathbf%7BC%7D" align="middle" border="0" />. By handedness, I mean that if we looked at a mirror image of the box, its volume would be multiplied by negative one. If you do that funny thing with your right hand, and find that <img src="http://www.forkosh.dreamhost.com/mimetex.cgi?%5Cmathbf%7BB%7D%20%5Ctimes%20%5Cmathbf%7BC%7D" align="middle" border="0" /> points the same sort of direction that <img src="http://www.forkosh.dreamhost.com/mimetex.cgi?%5Cmathbf%7BA%7D" align="middle" border="0" /> does, you have a righthanded box. Otherwise it's left-handed.<br /><br />This interpretation of <img src="http://www.forkosh.dreamhost.com/mimetex.cgi?%5Cmathbf%7BA%7D%20%5Ccdot%20%5Cmathbf%7BB%7D%20%5Ctimes%20%5Cmathbf%7BC%7D" align="middle" border="0" /> as the volume turns out to be the same as the determinant. To see why, we'll make a list of properties that the area of the box has, and then show that those properties imply the normal calculation of the determinant.<br /><ul><li>If the sides of the box are orthonormal (all perpendicular to each other, and all unit length), its volume is one (or minus one for a left-handed box).<br /></li><li>The volume of the box is linear in a vector. By this I mean that if we multiplied any side of the box by some constant, we'd multiply the volume of the entire box by that constant. Also, if we make two boxes that share two vectors but differ in the third, the sum of the volumes is the same as the volume of the box whose third side is the sum of the vectors. Notationally:</li></ul> <img src="http://www.forkosh.dreamhost.com/mimetex.cgi?%28%20%5Calpha%20%5Cmathbf%7BA%7D%20+%20%5Cbeta%20%5Cmathbf%7BA%27%7D%29%20%5Ccdot%20%28%5Cmathbf%7BB%7D%20%5Ctimes%20%5Cmathbf%7BC%7D%29%20=%20%5Calpha%20%5Cmathbf%7BA%7D%20%5Ccdot%20%5Cmathbf%7BB%7D%20%5Ctimes%20%5Cmathbf%7BC%7D%20+%20%5Cbeta%20%5Cmathbf%7BA%27%7D%20%5Ccdot%20%5Cmathbf%7BB%7D%20%5Ctimes%20%5Cmathbf%7BC%7D" align="middle" border="0" /><br /> and similarly for <img src="http://www.forkosh.dreamhost.com/mimetex.cgi?%5Cmathbf%7BB%7D" align="middle" border="0" /> and <img src="http://www.forkosh.dreamhost.com/mimetex.cgi?%5Cmathbf%7BC%7D" align="middle" border="0" /><br /><ul><li>If you switch the roles of any two vectors, you'll multiply the volume of the box by minus one. This is obvious if you flip<img src="http://www.forkosh.dreamhost.com/mimetex.cgi?%5Cmathbf%7BB%7D" align="middle" border="0" /> and <img src="http://www.forkosh.dreamhost.com/mimetex.cgi?%5Cmathbf%7BC%7D" align="middle" border="0" />, because then the angle between them gets measured the opposite way, so its sine is the opposite of what it used to be. The other possibilities you can visualize if you just play around with it a little. It works.</li></ul>With these properties, we can uniquely determine the value of the volume of the box from its components, and show that it's the determinant. It's a little long for three dimensions, so you can work through that on your own if you want, after seeing it in two dimensions.<br /><br />Imagine we have two 2-D vectors in component form, and we put them in a matrix. Then we ask for the determinant. The first property says the determinant of the identity is one.<br /><br /><!-- \left| \begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array} \right| = 1 --><br /><img src="http://www.forkosh.dreamhost.com/mimetex.cgi?%5Cleft%7C%20%5Cbegin%7Barray%7D%7Bcc%7D%201%20&%200%20%5C%5C%200%20&%201%20%5Cend%7Barray%7D%20%5Cright%7C%20=%201" align="middle" border="0" /><br /><br />Using linearity in the vectors, we find<br /><br /><!-- \left| \begin{array}{cc} A_x & 0 \\ 0 & B_y \end{array} \right| = A_x B_y \left| \begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array} \right| = A_x B_y --><br /><br /><img src="http://www.forkosh.dreamhost.com/mimetex.cgi?%5Cleft%7C%20%5Cbegin%7Barray%7D%7Bcc%7D%20A_x%20&%200%20%5C%5C%200%20&%20B_y%20%5Cend%7Barray%7D%0A%5Cright%7C%20=%20A_x%20B_y%20%5Cleft%7C%20%5Cbegin%7Barray%7D%7Bcc%7D%201%20&%200%20%5C%5C%200%20&%201%0A%5Cend%7Barray%7D%20%5Cright%7C%20=%20A_x%20B_y" align="middle" border="0" /><br /><br />Now we switch the vectors in the identity and multiply the determinant by -1.<br /><br /><img src="http://www.forkosh.dreamhost.com/mimetex.cgi?%5Cleft%7C%20%5Cbegin%7Barray%7D%7Bcc%7D%200%20&%201%20%5C%5C%201%20&%200%20%5Cend%7Barray%7D%20%5Cright%7C%20=%20-1" align="middle" border="0" /><br /><!-- \left| \begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array} \right| = -1 --><br />and<br /><!-- \left| \begin{array}{cc} 0 & A_y \\ B_x & 0 \end{array} \right| = -A_y B_x --><br /><img src="http://www.forkosh.dreamhost.com/mimetex.cgi?%5Cleft%7C%20%5Cbegin%7Barray%7D%7Bcc%7D%200%20&%20A_y%20%5C%5C%20B_x%20&%200%20%5Cend%7Barray%7D%20%5Cright%7C%20=%20-A_y%20B_x" align="middle" border="0" /><br /><br />Finally use the linearity within a vector, which states<br /><img src="http://www.forkosh.dreamhost.com/mimetex.cgi?%5Cleft%7C%20%5Cbegin%7Barray%7D%7Bcc%7D%20A_x%20&%20A_y%20%5C%5C%20B_x%20&%20B_y%20%5Cend%7Barray%7D%20%5Cright%7C%20=%3Cbr%3E%5Cleft%7C%0A%5Cbegin%7Barray%7D%7Bcc%7D%20A_x%20&%200%20%5C%5C%20B_x%20&%20B_y%20%5Cend%7Barray%7D%20%5Cright%7C%20+%0A%5Cleft%7C%20%5Cbegin%7Barray%7D%7Bcc%7D%200%20&%20A_y%20%5C%5C%20B_x%20&%20B_y%20%5Cend%7Barray%7D%0A%5Cright%7C%20=%3Cbr%3E%5Cleft%7C%20%5Cbegin%7Barray%7D%7Bcc%7D%20A_x%20&%200%20%5C%5C%20B_x%20&%200%0A%5Cend%7Barray%7D%20%5Cright%7C%20+%20%5Cleft%7C%20%5Cbegin%7Barray%7D%7Bcc%7D%20A_x%20&%200%20%5C%5C%200%20&%0AB_y%20%5Cend%7Barray%7D%20%5Cright%7C%20+%20%5Cleft%7C%20%5Cbegin%7Barray%7D%7Bcc%7D%200%20&%20A_y%20%5C%5C%20B_x%0A&%200%20%5Cend%7Barray%7D%20%5Cright%7C%20+%20%5Cleft%7C%20%5Cbegin%7Barray%7D%7Bcc%7D%200%20&%20A_y%20%5C%5C%200%0A&%20B_y%20%5Cend%7Barray%7D%20%5Cright%7C%20%3Cbr%3E%0A=%20A_x%20B_y%20-%20A_y%20B_x" align="middle" border="0" /><br /><!-- \left| \begin{array}{cc} A_x & A_y \\ B_x & B_y \end{array} \right| = \left| \begin{array}{cc} A_x & 0 \\ B_x & B_y \end{array} \right| + \left| \begin{array}{cc} 0 & A_y \\ B_x & B_y \end{array} \right| = \left| \begin{array}{cc} A_x & 0 \\ B_x & 0 \end{array} \right| + \left| \begin{array}{cc} A_x & 0 \\ 0 & B_y \end{array} \right| + \left| \begin{array}{cc} 0 & A_y \\ B_x & 0 \end{array} \right| + \left| \begin{array}{cc} 0 & A_y \\ 0 & B_y \end{array} \right| = A_x B_y - A_y B_x --><br /><br />where we could drop the determinants of matrices with an entire column zero because they are dependent vectors.<br /><br />That is the determinant for a 2-by-2 matrix. If you want to do it for 3-by-3, you can use the exact same process of breaking it down into 27 matrices with three nonzero entries each, and then see by permutations of the identity which ones are zero, and where the plus and minus signs go. Out will fall the normal expression for a 3-by-3 determinant.<br /><br />So that's where that comes from.<br /><br /><h4>Why it Matters</h4>In the <a href="http://arcsecond.blogspot.com/2008/09/what-does-determinant-have-to-do-with.html">previous post</a>, we were trying to solve the problem of taking three vectors and finding a linear combination of them that yielded a fourth vector. That is, find <img src="http://www.forkosh.dreamhost.com/mimetex.cgi?a" align="middle" border="0" />, <img src="http://www.forkosh.dreamhost.com/mimetex.cgi?b" align="middle" border="0" />, <img src="http://www.forkosh.dreamhost.com/mimetex.cgi?c" align="middle" border="0" /> such that<br /><br /><img src="http://www.forkosh.dreamhost.com/mimetex.cgi?a%20%5Cmathbf%7BA%7D%20+%20b%20%5Cmathbf%7BB%7D%20+%20c%20%5Cmathbf%7BC%7D%20=%20%5Cmathbf%7BD%7D" align="middle" border="0" /><br /><br />where <img src="http://www.forkosh.dreamhost.com/mimetex.cgi?%5Cmathbf%7BA%7D" align="middle" border="0" />, <img src="http://www.forkosh.dreamhost.com/mimetex.cgi?%5Cmathbf%7BB%7D" align="middle" border="0" />, <img src="http://www.forkosh.dreamhost.com/mimetex.cgi?%5Cmathbf%7BC%7D" align="middle" border="0" />, and <img src="http://www.forkosh.dreamhost.com/mimetex.cgi?%5Cmathbf%7BD%7D" align="middle" border="0" /> are all arbitary vectors.<br /><br />We worked far enough just to test if the problem had a solution, but we found this result:<br /><br /><img src="http://www.forkosh.dreamhost.com/mimetex.cgi?a%20=%20%5Cfrac%7BD_%5Cperp%20B%20C%20%5Csin%7B%5Ctheta%7D%7D%7B%5Cmathbf%7BA%7D%20%5Ccdot%20%5Cmathbf%7BB%7D%20%5Ctimes%20%5Cmathbf%7BC%7D%7D" /><br /><br />From there, we need to find <img src="http://www.forkosh.dreamhost.com/mimetex.cgi?D_%5Cperp" align="middle" border="0" /> by dotting it with a unit vector perpendicular to both <img src="http://www.forkosh.dreamhost.com/mimetex.cgi?%5Cmathbf%7BB%7D" align="middle" border="0" /> and <img src="http://www.forkosh.dreamhost.com/mimetex.cgi?%5Cmathbf%7BC%7D" align="middle" border="0" />.<br /><br /><img src="http://www.forkosh.dreamhost.com/mimetex.cgi?D_%5Cperp%20=%20%5Cmathbf%7BD%7D%20%5Ccdot%20%5Cfrac%7B%5Cmathbf%7BB%7D%20%5Ctimes%20%5Cmathbf%7BC%7D%7D%7BB%20C%20%5Csin%7B%5Ctheta%7D%7D" align="middle" border="0" /><br /><br />Plug this into the equation for <img src="http://www.forkosh.dreamhost.com/mimetex.cgi?a" align="middle" border="0" />, and the stuff about <img src="http://www.forkosh.dreamhost.com/mimetex.cgi?B%20C%20%5Csin%7B%20%5Ctheta%20%7D" align="middle" border="0" /> cancels, leaving<br /><br /><img src="http://www.forkosh.dreamhost.com/mimetex.cgi?a%20=%20%5Cfrac%7B%5Cmathbf%7BD%7D%20%5Ccdot%20%5Cmathbf%7BB%7D%20%5Ctimes%20%5Cmathbf%7BC%7D%7D%7B%5Cmathbf%7BA%7D%20%5Ccdot%20%5Cmathbf%7BB%7D%20%5Ctimes%20%5Cmathbf%7BC%7D%7D" align="middle" border="0" /><br /><br />Those are determinants. That's why they matter. They let you solve systems of linear equations, such as the system of linear equations that results from the problem we just solved.Markkimarkkonnenhttp://www.blogger.com/profile/11122368349695914223noreply@blogger.com76tag:blogger.com,1999:blog-5497138109765858383.post-46262115445987567782008-09-27T02:14:00.000-07:002008-09-27T16:12:37.440-07:00Debate Debate<span style="font-weight: bold;">Relying Exclusively of Volatile Eruptions of the Reptile Brain Mark:</span> You should watch the debate.<br /><br /><span style="font-weight: bold;">Preposterously Idealized Self SErving Rational Mark:</span> Will it be interesting?<br /><br /><span style="font-weight: bold;">REVERB Mark:</span> You better be interested! It's the Presidential Debate!<br /><br /><span style="font-weight: bold;">PISSER Mark</span>: I don't follow that.<br /><br /><span style="font-weight: bold;">REVERB Mark</span>: You're a functioning, adult member of the society now. You live in a democracy, and that's a great privilege. It's your responsibility to take part in the governance of this nation, and you didn't even vote in the last election! Now you better turn on that debate.<br /><br /><span style="font-weight: bold;">PISSER Mark:</span> Given my demographic, occupation, community, and personal interests it's not difficult to guess which candidate I support. I doubt that watching the debate will change that.<br /><br /><span style="font-weight: bold;">REVERB Mark</span>: That's not the point. Even if you know who you're going to vote for, you still need to stay well informed.<br /><br /><span style="font-weight: bold;">PISSER Mark</span>: Why?<br /><br /><span style="font-weight: bold;">REVERB Mark</span>: Because you're more privileged than any but the tiniest fraction of human beings alive, and yet you take it all for granted. You live in a nation where you personally can have real, meaningful participation in the political process, and it's your responsibility to exercise that right.<br /><br /><span style="font-weight: bold;">PISSER Mark</span>: I admit I frequently take the perks of life in America for granted, but I don't see why following politics is a responsibility, rather than a personal choice. I'd rather do something else.<br /><br /><span style="font-weight: bold;">REVERB Mark</span>: What do you think politics is, then?<br /><br /><span style="font-weight: bold;">PISSER Mark:</span> A hobby.<br /><br /><span style="font-weight: bold;">REVERB Mark</span>: A what?<br /><br /><span style="font-weight: bold;">PISSER Mark</span>: A hobby. A pastime. Like being a sports fan. You can read about sports in the newspaper, listen to experts talk about it on the radio. Then you can choose which team you like, and "support" them by buying bumper stickers and paying for admission to their games. All the same applies with politics, except the bumper sticker becomes a sign on your front lawn and the ticket to the game becomes a vote. They're both just ways for people to have something to get riled up about without doing much harm.<br /><br /><span style="font-weight: bold;">REVERB Mark</span>: Not an apt analogy at all. Sports are just games. It doesn't really matter who wins the World Series. But it sure matters who wins the Presidency!<br /><br /><span style="font-weight: bold;">PISSER Mark</span>: A little bit, maybe. My job prospects might be a little dimmer with one candidate in charge than the other. Their differences on social issues could affect how pissed off the people around me are. The foreign policy they enact will cause me to read different headlines in the news.<br /><br /><span style="font-weight: bold;">REVERB Mark</span>: That's not just news items. All that stuff is really happening out there, to real people. It's not just all about you. To some people it makes a huge difference who's President.<br /><br /><span style="font-weight: bold;">PISSER Mark</span>: But we're discussing my personal decision on whether to watch the debate.<br /><br /><span style="font-weight: bold;">REVERB Mark</span>: Look, I know what you're thinking. You're thinking, "my vote is just one gumball in Candy Land. Doesn't make a bit of difference. The vote in my state always goes to the same party, and one more vote cast either way can't change the result." But you know what? One hundred thousand votes can change the result. And what if one hundred thousand people just like you are using the same logic?<br /><br /><span style="font-weight: bold;">PISSER Mark</span>: Are you saying that if I go to vote it will change one hundred thousand lazy people's minds?<br /><br /><span style="font-weight: bold;">REVERB Mark</span>: You know that's not what I'm saying. I'm saying you shouldn't vote because you think your particular vote is going to make all the difference. You should vote because it's a statement that you give a rat's ass about the future of this country! Not just for yourself, but for everyone you share this country with. And since the US is a leading world superpower, everyone you share this Earth with. And before you vote, you need to be informed. So you need to watch the debate.<br /><br /><span style="font-weight: bold;">PISSER Mark</span>: Watching just one debate won't make me informed. Becoming a competent participant in democracy is a significant commitment.<br /><br /><span style="font-weight: bold;">REVERB Mark</span>: One that every adult American ought to make.<br /><br /><span style="font-weight: bold;">PISSER Mark</span>: Have you met many adult Americans? On the whole, they're not the ones I want making any important decisions. Basically, ability to choose a good Presidential candidate will undoubtedly fall on a curve like this:<br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://4.bp.blogspot.com/_2y3Souq-IUU/SN4RTwiAzKI/AAAAAAAABB4/l1r7QlNsdhQ/s1600-h/debatePlot.jpg"><img style="margin: 0pt 10px 10px 0pt; float: left; cursor: pointer;" src="http://4.bp.blogspot.com/_2y3Souq-IUU/SN4RTwiAzKI/AAAAAAAABB4/l1r7QlNsdhQ/s320/debatePlot.jpg" alt="" id="BLOGGER_PHOTO_ID_5250653246703324322" border="0" /></a><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br />So let the people who are innately interested in the whole process, and can do a good job of it, pour their efforts into politics.<br /><br /><span style="font-weight: bold;">REVERB Mark</span>: All the more reason for you personally to get involved. You're way at the top of that curve. Remember high school? You don't want to common man making the decisions - then YOU better get involved and do something about it.<br /><br /><span style="font-weight: bold;">PISSER Mark</span>: That's not really my point. I don't want to invest huge amounts of time and energy into becoming an "informed citizen", which by the way makes the dubious presupposition that it's possible to make a good choice from behind a computer screen, because it doesn't interest me much. Even if I did watch the debate, I don't know enough about foreign policy or economics to decide who I agree with. I'd inevitably fall back on judging demeanor and rhetoric. I could do more good by putting that effort into something I'll enjoy.<br /><br /><span style="font-weight: bold;">REVERB Mark</span>: Like what? What can you do that's more important than your civic duty?<br /><br /><span style="font-weight: bold;">PISSER Mark</span>: I don't know... plant a tree or something, I guess.<br /><br /><span style="font-weight: bold;">REVERB Mark</span>: Oh come on. You've never planted a tree in your life.<br /><br /><span style="font-weight: bold;">PISSER Mark</span>: I did, actually. I planted the free pine tree seedling they gave me third grade on Arbor Day. I think dad accidentally ran over it with the lawn mower.<br /><br /><span style="font-weight: bold;">REVERB Mark</span>: Not the point. That's not what you <i>will</i> do with that extra time. What you will do is read stupid websites or play flash games, or something inane like that.<br /><br /><span style="font-weight: bold;">PISSER Mark</span>: Or maybe I'll study for a while. Or work on that project I had planned, or...<br /><br /><span style="font-weight: bold;">REVERB Mark</span>: Oh please. You've already got three tabs open and they're Pacman, SomethingAwful.com, and then something more awful than that.<br /><br /><span style="font-weight: bold;">PISSER Mark</span>: sigh. Where is it streaming from?Markkimarkkonnenhttp://www.blogger.com/profile/11122368349695914223noreply@blogger.com1tag:blogger.com,1999:blog-5497138109765858383.post-67411480511088623722008-09-26T21:19:00.000-07:002008-09-26T23:06:54.548-07:00What does a determinant have to do with a cross product? pt. 1You're probably aware of a nifty mnemonic for calculating a cross product.<br /><br /><img src="http://www.forkosh.dreamhost.com/mimetex.cgi?%5Cmathbf%7BB%7D%20%5Ctimes%20%5Cmathbf%7BC%7D%20=%20%5Cleft%7C%20%5Cbegin%7Barray%7D%7Bccc%7D%20%5Cmathbf%7B%5Chat%7Bx%7D%7D%20&%20%5Cmathbf%7B%5Chat%7By%7D%7D%20&%20%5Cmathbf%7B%5Chat%7Bz%7D%7D%20%5C%5C%20B_x%20&%20B_y%20&%20B_z%20%5C%5C%20C_x%20&%20C_y%20&%20C_z%20%5Cend%7Barray%7D%20%5Cright%7C" align="middle" border="0" /><br /><br />If you take a dot product of the above expressions with another vector <img src="http://www.forkosh.dreamhost.com/mimetex.cgi?%5Cmathbf%7BA%7D" align="middle" border="0" />, you'll get<br /><br /><img src="http://www.forkosh.dreamhost.com/mimetex.cgi?%5Cmathbf%7BA%7D%20%5Ccdot%20%5Cmathbf%7BB%7D%20%5Ctimes%20%5Cmathbf%7BC%7D%20=%20%5Cleft%7C%20%5Cbegin%7Barray%7D%7Bccc%7D%20A_x%20&%20A_y%20&%20A_z%20%5C%5C%20B_x%20&%20B_y%20&%20B_z%20%5C%5C%20C_x%20&%20C_y%20&%20C_z%20%5Cend%7Barray%7D%20%5Cright%7C" align="middle" border="0" /><br /><br />What a crazy magic coincidence that vector products and determinants can be calculated the same way!<br /><br />I won't say magic coincidences don't exist, but I will say that when you find one it may behoove you to look a little deeper. There is a deeper connection.<br /><br />I'll hint at it today by posing a certain problem, then showing that the problem has a solution whenever either expression, the determinant or the vector product, is nonzero. So is it two different conditions that show the problem can be solved, or the same condition in different guises? We'll deal with that later.<br /><br />Imagine you are given four random vectors in space, <img src="http://www.forkosh.dreamhost.com/mimetex.cgi?%5Cmathbf%7BA%7D%20,%20%5Cmathbf%7BB%7D,%20%5Cmathbf%7BC%7D,%20%5Cmathbf%7BD%7D" align="middle" border="0" />. Then you are asked to express the fourth as a linear combination of the first three. That is, find <img src="http://www.forkosh.dreamhost.com/mimetex.cgi?a,%20b,%20c" align="middle" border="0" /> such that <img src="http://www.forkosh.dreamhost.com/mimetex.cgi?a%20%5Cmathbf%7BA%7D%20+%20b%20%5Cmathbf%7BB%7D%20+%20c%20%5Cmathbf%7BC%7D%20=%20%5Cmathbf%7BD%7D" align="middle" border="0" /><br /><br />I won't solve the problem. I'm just asking for a test to see if it's soluble (besides throwing water on it). Especially, I want to see if it's soluble for all possible <img src="http://www.forkosh.dreamhost.com/mimetex.cgi?%5Cmathbf%7BD%7D" align="middle" border="0" />, not just special cases.<br /><br /><br /><h4>Using A Basis</h4>Set up a standard Cartesian basis in three dimensions, and write out the vector equation in components<br /><img src="http://www.forkosh.dreamhost.com/mimetex.cgi?a%20A_x%20+%20b%20B_x%20+%20c%20C_x%20=%20D_x" align="middle" border="0" /><br /><img src="http://www.forkosh.dreamhost.com/mimetex.cgi?a%20A_y%20+%20b%20B_y%20+%20c%20C_y%20=%20D_y" align="middle" border="0" /><br /><img src="http://www.forkosh.dreamhost.com/mimetex.cgi?a%20A_z%20+%20b%20B_z%20+%20c%20C_z%20=%20D_z" align="middle" border="0" /><br /><br />which is a system of linear equations that can be written in matrix form as<br /><br /><img src="http://www.forkosh.dreamhost.com/mimetex.cgi?%5Cleft%28%20%5Cbegin%7Barray%7D%7Bccc%7D%20A_x%20&%20A_y%20&%20A_z%20%5C%5C%20B_x%20&%20B_y%20&%20B_z%20%5C%5C%20C_x%0A&%20C_y%20&%20C_z%20%5Cend%7Barray%7D%20%5Cright%29%20%5Cleft%28%20%5Cbegin%7Barray%7D%7Bc%7D%20a%20%5C%5C%20b%20%5C%5C%20c%0A%5Cend%7Barray%7D%20%5Cright%29%20=%20%5Cleft%28%20%5Cbegin%7Barray%7D%7Bc%7D%20D_x%20%5C%5C%20D_y%20%5C%5C%20D_z%20%5Cend%7Barray%7D%20%5Cright%29" align="middle" border="0" /><br /><br />(Aside: If you know how to tell <img src="http://www.forkosh.dreamhost.com/mimetex.cgi?%5CLaTeX" align="middle" border="0" /> to make the vertical columns the same size in those matrices, drop me a comment.)<br /><br />If you've taken linear algebra, you'll recognize that the problem can be solved only if the above matrix has a non-zero determinant. That same determinant is the one we were interested in earlier, since it's also the value of <img src="http://www.forkosh.dreamhost.com/mimetex.cgi?%5Cmathbf%7BA%7D%20%5Ccdot%20%5Cmathbf%7BB%7D%20%5Ctimes%20%5Cmathbf%7BC%7D" align="middle" border="0" /><br /><br /><br /><h4>Without a Basis</h4>Vectors <img src="http://www.forkosh.dreamhost.com/mimetex.cgi?%5Cmathbf%7BB%7D" align="middle" border="0" /> and <img src="http://www.forkosh.dreamhost.com/mimetex.cgi?%5Cmathbf%7BC%7D" align="middle" border="0" /> define a plane. They're two independent vectors (I'm assuming they're independent - if they're dependent, you might as well just get rid of one of them for the purposes of solving this problem, but it's clear that in general that won't work) in a two-dimensional space, so they form a basis for that space.<br /><br />Break <img src="http://www.forkosh.dreamhost.com/mimetex.cgi?%5Cmathbf%7BD%7D" align="middle" border="0" /> down into two parts, the part in the plane of <img src="http://www.forkosh.dreamhost.com/mimetex.cgi?%5Cmathbf%7BB%7D" align="middle" border="0" /> and <img src="http://www.forkosh.dreamhost.com/mimetex.cgi?%20%5Cmathbf%7BC%7D" align="middle" border="0" />, which we'll call <img src="http://www.forkosh.dreamhost.com/mimetex.cgi?%5Cmathbf%7BD_%7C%7C%7D" align="middle" border="0" /> and the part perpendicular to the plane, <img src="http://www.forkosh.dreamhost.com/mimetex.cgi?D_%5Cperp" align="middle" border="0" />.<br /><br /><img src="http://www.forkosh.dreamhost.com/mimetex.cgi?%5Cmathbf%7BB%7D" align="middle" border="0" /> and <img src="http://www.forkosh.dreamhost.com/mimetex.cgi?%5Cmathbf%7BC%7D" align="middle" border="0" /> are irrelevant to the perpendicular part of the equation<img src="http://www.forkosh.dreamhost.com/mimetex.cgi?a%20%5Cmathbf%7BA%7D%20+%20b%20%5Cmathbf%7BB%7D%20+%20c%20%5Cmathbf%7BC%7D%20=%20%5Cmathbf%7BD%7D" align="middle" border="0" />, so we're left with<br /><br /><img src="http://www.forkosh.dreamhost.com/mimetex.cgi?a%20A_%5Cperp%20=%20D_%5Cperp" align="middle" border="0" /><br /><br />We need to find <img src="http://www.forkosh.dreamhost.com/mimetex.cgi?A_%5Cperp" align="middle" border="0" />, so we better dot <img src="http://www.forkosh.dreamhost.com/mimetex.cgi?%5Cmathbf%7BA%7D" align="middle" border="0" /> with a unit vector in the perpendicular direction.<br /><br /><img src="http://www.forkosh.dreamhost.com/mimetex.cgi?A_%5Cperp%20=%20%5Cmathbf%7BA%7D%20%5Ccdot%20%5Cfrac%7B%5Cmathbf%7BB%7D%20%5Ctimes%20%5Cmathbf%7BC%7D%7D%7BB%20C%20%5Csin%7B%5Ctheta%7D%7D" align="middle" border="0" /><br /><br />Combining the previous two equations lets us solve for <img src="http://www.forkosh.dreamhost.com/mimetex.cgi?a" align="middle" border="0" />.<br /><br /><img src="http://www.forkosh.dreamhost.com/mimetex.cgi?a%20=%20%5Cfrac%7BD_%5Cperp%20B%20C%20%5Csin%7B%5Ctheta%7D%7D%7B%5Cmathbf%7BA%7D%20%5Ccdot%20%5Cmathbf%7BB%7D%20%5Ctimes%20%5Cmathbf%7BC%7D%7D" align="middle" border="0" /><br /><br />Unless <img src="http://www.forkosh.dreamhost.com/mimetex.cgi?%5Cmathbf%7BD%7D" align="middle" border="0" /> is by extraordinary chance in the plane of <img src="http://www.forkosh.dreamhost.com/mimetex.cgi?%5Cmathbf%7BB%7D" align="middle" border="0" /> and <img src="http://www.forkosh.dreamhost.com/mimetex.cgi?%5Cmathbf%7BC%7D" align="middle" border="0" /> (and we're interested in the general case, not special ones), we need <img src="http://www.forkosh.dreamhost.com/mimetex.cgi?a" align="middle" border="0" /> to be finite, meaning <img src="http://www.forkosh.dreamhost.com/mimetex.cgi?%5Cmathbf%7BA%7D%20%5Ccdot%20%5Cmathbf%7BB%7D%20%5Ctimes%20%5Cmathbf%7BC%7D" align="middle" border="0" /> needs to be nonzero. This condition also has built into it the requirement that <img src="http://www.forkosh.dreamhost.com/mimetex.cgi?%5Cmathbf%7BB%7D" align="middle" border="0" /> and <img src="http://www.forkosh.dreamhost.com/mimetex.cgi?%5Cmathbf%7BC%7D" align="middle" border="0" /> be independent, so it's a sufficient condition for the problem to have a solution.<br /><br /><h4>Summary for today</h4><br />The equation<br /><br /><img src="http://www.forkosh.dreamhost.com/mimetex.cgi?a%20%5Cmathbf%7BA%7D%20+%20b%20%5Cmathbf%7BB%7D%20+%20c%20%5Cmathbf%7BC%7D%20=%20%5Cmathbf%7BD%7D" align="middle" border="0" /><br /><br />can be solved for any <img src="http://www.forkosh.dreamhost.com/mimetex.cgi?%5Cmathbf%7BD%7D" align="middle" border="0" /> whatsoever if<br /><br /><img src="http://www.forkosh.dreamhost.com/mimetex.cgi?%5Cleft%7C%20%5Cbegin%7Barray%7D%7Bccc%7D%20A_x%20&%20A_y%20&%20A_z%20%5C%5C%20B_x%20&%20B_y&%20B_z%20%5C%5C%20C_x%20&%20C_y%20&%20C_z%20%5Cend%7Barray%7D%20%5Cright%7C%20%5Cneq%200" align="middle" border="0" /><br /><br />also, the same is true if<br /><br /><img src="http://www.forkosh.dreamhost.com/mimetex.cgi?%5Cmathbf%7BA%7D%20%5Ccdot%20%5Cmathbf%7BB%7D%20%5Ctimes%20%5Cmathbf%7BC%7D%20%5Cneq%200" align="middle" border="0" /><br /><br />That doesn't prove the determinant and the vector product are identical, but it's hinting at why it's true. Tomorrow I'll probably post a little about what determinants are, and the geometrical interpretation of the vector product.Markkimarkkonnenhttp://www.blogger.com/profile/11122368349695914223noreply@blogger.com1tag:blogger.com,1999:blog-5497138109765858383.post-6848228167648732862008-09-22T14:23:00.000-07:002008-09-22T15:43:09.803-07:00A X (B X C)Any intro to physics course will teach you the identity<br /><br /><img src="http://www.forkosh.dreamhost.com/mimetex.cgi?%5Cmathbf%7BA%7D%20%5Ctimes%20%28%5Cmathbf%7BB%7D%20%5Ctimes%20%5Cmathbf%7BC%7D%29%20=%20%5Cmathbf%7BB%7D%28%5Cmathbf%7BA%7D%20%5Ccdot%20%5Cmathbf%7BC%7D%29%20-%20%5Cmathbf%7BC%7D%28%5Cmathbf%7BA%7D%20%5Ccdot%20%5Cmathbf%7BB%7D%29" align="middle" border="0" /><br /><br />It's easy to prove by explicitly writing out each vector in component form, cranking through both sides of the equation, finally getting the same thing on each. But there's some insight to be gained in looking at the identity from a purely geometric point of view. Warning: the following is a bit dense, but also short. Go one sentence at a time and you should be fine.<br /><br />Together, <img src="http://www.forkosh.dreamhost.com/mimetex.cgi?%5Cmathbf%7BB%7D" align="middle" border="0" /> and <img src="http://www.forkosh.dreamhost.com/mimetex.cgi?%5Cmathbf%7BC%7D" align="middle" border="0" /> are a basis for a plane. <img src="http://www.forkosh.dreamhost.com/mimetex.cgi?%5Cmathbf%7BB%7D%20%5Ctimes%20%5Cmathbf%7BC%7D" align="middle" border="0" /> is perpendicular to the plane. Take that perpendicular vector product, cross it with <img src="http://www.forkosh.dreamhost.com/mimetex.cgi?%5Cmathbf%7BA%7D" align="middle" border="0" />, and you get a vector perpendicular to the perpendicular to the plane. That's back in the plane again. So we know that the double cross product is in the plane somewhere. But any vector in the plane can be written as a sum of <img src="http://www.forkosh.dreamhost.com/mimetex.cgi?%5Cmathbf%7BB%7D" align="middle" border="0" /> and <img src="http://www.forkosh.dreamhost.com/mimetex.cgi?%5Cmathbf%7BC%7D" align="middle" border="0" /> (that's what it means to be a basis), so we only need to find the right coefficients. Those coefficients come from requiring that we find a combination of <img src="http://www.forkosh.dreamhost.com/mimetex.cgi?%5Cmathbf%7BB%7D" align="middle" border="0" /> and <img src="http://www.forkosh.dreamhost.com/mimetex.cgi?%5Cmathbf%7BC%7D" align="middle" border="0" /> perpendicular to <img src="http://www.forkosh.dreamhost.com/mimetex.cgi?%5Cmathbf%7BA%7D" align="middle" border="0" />. Just dot the expression <img src="http://www.forkosh.dreamhost.com/mimetex.cgi?%5Cmathbf%7BB%7D%28%5Cmathbf%7BA%7D%20%5Ccdot%20%5Cmathbf%7BC%7D%29%20-%20%5Cmathbf%7BC%7D%28%5Cmathbf%7BA%7D%20%5Ccdot%20%5Cmathbf%7BB%7D%29" align="middle" border="0" /><br />with <img src="http://www.forkosh.dreamhost.com/mimetex.cgi?%5Cmathbf%7BA%7D" align="middle" border="0" />, and you'll get zero.<br /><br />That proves both sides of the equation<br /><br /><img src="http://www.forkosh.dreamhost.com/mimetex.cgi?%5Cmathbf%7BA%7D%20%5Ctimes%20%28%5Cmathbf%7BB%7D%20%5Ctimes%20%5Cmathbf%7BC%7D%29%20=%0A%5Cmathbf%7BB%7D%28%5Cmathbf%7BA%7D%20%5Ccdot%20%5Cmathbf%7BC%7D%29%20-%20%5Cmathbf%7BC%7D%28%5Cmathbf%7BA%7D%20%5Ccdot%0A%5Cmathbf%7BB%7D%29" align="middle" border="0" /><br /><br />point in the same direction, but doesn't prove they have the same length. It's not crazy difficult to prove the they have the same length without using a coordinate system, but I don't have a neat insightful trick for it, either.<br /><br />I started by noting that you could break <img src="http://www.forkosh.dreamhost.com/mimetex.cgi?%5Cmathbf%7BA%7D" align="middle" border="0" /> down into perpendicular and parallel components to the plane defined by the other two vectors, and that as far as the identity is concerned the perpendicular part of <img src="http://www.forkosh.dreamhost.com/mimetex.cgi?%5Cmathbf%7BA%7D" align="middle" border="0" /> doesn't matter (plug it in and you'll see). Then, because both sides of the equation are proportional to the lengths of all three vectors, you might as well divide by their lengths to make them unit vectors. All that's left is to find the lengths of both sides of the equation in terms of trig functions of the angles between the vectors. You have tree angles between the coplanar vectors. Write one as a sum of the other two, expand out all the trig functions, and fiddle around with it a bit until you see that it's actually a trig identity. That gets the final part of the theorem in there.Markkimarkkonnenhttp://www.blogger.com/profile/11122368349695914223noreply@blogger.com0tag:blogger.com,1999:blog-5497138109765858383.post-76602458651441007852008-09-22T10:29:00.001-07:002008-09-22T14:14:56.233-07:00Space ElevatorsSpace elevators are <a href="http://www.timesonline.co.uk/tol/news/uk/science/article4799369.ece">in the news again</a>, so I'll write a post covering the "intro to physics" aspects of what a space elevator is and why it's hard to build. This doesn't address the actual difficulties of building the elevator - which are in the engineering. It's just to get the overall conceptual framework.<br /><br /><h3>Geosynchronous Orbits</h3><br />A space elevator is a theoretical long tether that reaches all the way up into outer space and just hangs there all by itself. In theory, once you put one up, you can use it to climb up into outer space very cheaply compared to chemical rockets.<br /><br />How could a long tether "just hang there?" Because it's in orbit around the Earth. A space elevator is a special type of satellite, specifically, one in a geosynchronous orbit.<br />The idea here is that it's possible to have a satellite always directly above the same spot on Earth. This isn't the normal state of things. Most satellites are in low-Earth orbit and whip around the planet in fast circles every few hours. But the higher up you put a satellite, the longer it will take to orbit. If you put it high enough, it will take 24 hours to orbit. That way, the Earth can spin underneath it at just the same pace it's orbiting, and by happy coincidence the satellite appears to stay in the same spot, to an observer on the surface of the Earth.<br /><br />There are a lot of restrictions on where you can place such an orbit. First off, the shape and size of an orbit, along with the gravity of the planet and the laws of physics, determine how long an orbit takes. You can't just blast your rockets a little softer and expect to orbit slowly. If you do that, you won't be going fast enough to orbit any more, and gravity will pull you back to the ground.<br /><br />If you want an orbit that takes 24 hours and is "smooth" (always has the same angular velocity), you're restricted to making it a circular orbit at a height that works out to be 36,000 km above the surface of the Earth. The radius of the Earth is only 6,000 km, so a space elevator is long enough to wrap around the Earth multiple times, and together they look roughly like a flying sparrow with a two-meter long hair trailing out its butt. Hold a hula hoop around a soccer ball. That is roughly the correct scale for how high up this circular orbit would be. By comparison, most orbits for satellites and space shuttle trips and things like that would be loops hovering about half a centimeter from the ball's surface.<br /><br />Second, a geosynchronous orbit has to be over the equator. A satellite in a circular orbit doesn't have to be over the equator, of course. Take that soccer ball and hula hoop from before. Spin and twirl the hula hoop however you like. As long as the Earth is still at the center, you have a valid orbit. But it will only stay over the same spot if it's at the equator. Otherwise, the satellite wanders north and south again over the course of its day. If you try to cheat by simply sliding the hula hoop up some, so it's parallel to the equator but over say, the 45<sup>o</sup> line of latitude, it's no longer a possible physical orbit. The soccer ball isn't at the center any more, so its gravity would inevitably drag the orbit back down.<br /><br /><h3>Tethers</h3><br />Imagine a satellite in geosynchronous orbit. Now imagine you drop a little fishing line from the bottom of the satellite, reaching down towards Earth. In response, the rest of the satellite will have to rise up a tiny little bit to keep the center of mass at the same height. Now drop that line lower and lower. The rest of the satellite continues creeping up a little bit higher, and the line keeps reaching lower, until eventually you have a tether that reaches all the way from Earth to geosynchronous orbit. That's a space elevator.<br /><br />The first problem that creates is that the fishing line is under enormous tension. If it's made of any ordinary material, it will rip apart. To see why, imagine a spot on the line a few thousand kilometers above the Earth. If you cut the line there, the stuff below it would clearly not be in geosynchronous orbit - it isn't high enough up. So all that stuff would fall.<br /><br />Since that material below the cut point <i>isn't</i> falling in an operational space elevator, something must be holding it up. That something is tension in the line. The higher up the space elevator you go, the more material there is to hold up, so the tension gets greater and greater. By the time you reach geosynchronous orbit, the line has to support the weight of tens of thousands of kilometers of line beneath it. That's a lot of weight, and any normal material would imply rip apart. So to build a space elevator you need something really strong. So far, there are no long ropes that are anywhere near strong enough, although some people speculate that it's possible to build one out of carbon nanotubes.<br /><br /><h3>Geosynchronous Orbits II</h3><br /><br />The gravitational potential of two point masses is given by<br /><br /><img src="http://www.forkosh.dreamhost.com/mimetex.cgi?%20U%20=%20-%20%5Cfrac%7BG%20m_e%20m_s%7D%7Br%7D" align="middle" border="0" /><br /><br /><img src="http://www.forkosh.dreamhost.com/mimetex.cgi?G" align="middle" border="0" /> is a constant, <img src="http://www.forkosh.dreamhost.com/mimetex.cgi?m_e" align="middle" border="0" /> and <img src="http://www.forkosh.dreamhost.com/mimetex.cgi?m_s" align="middle" border="0" /> are the objects' masses, and <img src="http://www.forkosh.dreamhost.com/mimetex.cgi?r" align="middle" border="0" /> is the distance between them. We're giving the Earth (which in real life takes up lots of space) a single definite location at one point. Gauss' law allows us to do this for a body with spherical symmetry, but really it doesn't matter for the calculation I have in mind, because the distance <img src="http://www.forkosh.dreamhost.com/mimetex.cgi?r" align="middle" border="0" /> for a space elevator will turn out to be much greater than the radius of the Earth.<br /><br />Using a reference frame in which the center of mass of the Earth does not move, and assuming this is an inertial frame, the kinetic energy of the system is<br /><br /><img src="http://www.forkosh.dreamhost.com/mimetex.cgi?T%20=%20%5Cfrac%7Bm_s%20v%5E2%7D%7B2%7D" align="middle" border="0" /><br /><br />Where <img src="http://www.forkosh.dreamhost.com/mimetex.cgi?v" align="middle" border="0" /> is the magnitude of the satellite's velocity in this frame. We'll use ordinary spherical coordinates, so that the position of the satellite is given by <img src="http://www.forkosh.dreamhost.com/mimetex.cgi?%28r,%5Cphi,%5Ctheta%29" align="middle" border="0" />. We're only interested in circular orbits, where <img src="http://www.forkosh.dreamhost.com/mimetex.cgi?r" align="middle" border="0" /> is a constant, and <img src="http://www.forkosh.dreamhost.com/mimetex.cgi?%5Ctheta%20=%20%5Cfrac%7B%5Cpi%7D%7B2%7D" align="middle" border="0" />. Then the velocity is<br /><br /><img src="http://www.forkosh.dreamhost.com/mimetex.cgi?v%20=%20r%20%5Comega%20%5Chat%7B%5Cphi%7D%20+%20%5Cdot%7Br%7D%20%5Chat%7Br%7D" align="middle" border="0" /><br /><br />where<br /><br /><img src="http://www.forkosh.dreamhost.com/mimetex.cgi?%5Comega%20=%20%5Cfrac%7Bd%20%5Cphi%7D%7Bdt%7D" align="middle" border="0" />.<br /><br />But we need a circular orbit, for which <img src="http://www.forkosh.dreamhost.com/mimetex.cgi?%5Cdot%7Br%7D%20=%200" align="middle" border="0" /><br /><br />The Lagrangian is<br /><br /><img src="http://www.forkosh.dreamhost.com/mimetex.cgi?%5Ccal%7BL%7D%20=%20U%20-%20T%20=%20%5Cfrac%7BG%20m_e%20m_s%7D%7Br%7D%20-%20%5Cfrac%7Bm_s%20r%5E2%20%5Comega%5E2%7D%7B2%7D" align="middle" border="0" /><br /><br />Then the Euler-Lagrange equations give<br /><br /><img src="http://www.forkosh.dreamhost.com/mimetex.cgi?%5Cfrac%7Bd%20%5Ccal%7BL%7D%7D%7Bdr%7D%20=%20%5Cfrac%7Bd%7D%7Bdt%7D%20%5Cleft%28%20%5Cfrac%7B%5Ccal%7BL%7D%7D%7Bd%20%5Cdot%7Br%7D%7D%20%5Cright%29%20=%20%5Cfrac%7BG%20m_e%20m_s%7D%7Br%5E2%7D%20-%20m_s%20r%20%5Comega%5E2%20=%200" align="middle" border="0" /><br /><br />which yields<br /><br /><img src="http://www.forkosh.dreamhost.com/mimetex.cgi?r%20=%20C%20%5Comega%5E%7B%5Cfrac%7B-2%7D%7B3%7D%7D" align="middle" border="0" /><br /><br />where <img src="http://www.forkosh.dreamhost.com/mimetex.cgi?C" align="middle" border="0" /> is a constant depending on the gravitational constant and the mass of the Earth.<br /><br />Setting <img src="http://www.forkosh.dreamhost.com/mimetex.cgi?%5Comega%20=%20%5Cfrac%7B2%20%5Cpi%7D%7B24%20hours%7D" align="middle" border="0" /> and plugging through the numbers gives <i>r</i>=4.2*10<sup>24</sup> km. That's the height of a geosynchronous orbit from the center of the Earth.<br /><br /><br /><h3>Tension II</h3><br /><br />The force on a differential element of the cable, <img src="http://www.forkosh.dreamhost.com/mimetex.cgi?dl" align="middle" border="0" />, is<br /><br /><img src="http://www.forkosh.dreamhost.com/mimetex.cgi?F_tot%20=%20F_g%20+%20dT%20=%20dm%20*%20a" align="middle" border="0" /><br /><br />where <img src="http://www.forkosh.dreamhost.com/mimetex.cgi?F_g" align="middle" border="0" /> is the force of gravity on the segment and <img src="http://www.forkosh.dreamhost.com/mimetex.cgi?dT" align="middle" border="0" /> is the change in tension from below to above that point. <img src="http://www.forkosh.dreamhost.com/mimetex.cgi?dm" align="middle" border="0" /> is the mass of the length element, and <img src="http://www.forkosh.dreamhost.com/mimetex.cgi?a" align="middle" border="0" /> is its acceleration. The positive direction is defined to be away from the center of the Earth. The element is accelerating in a circular orbit, with acceleration<br /><br /><img src="http://www.forkosh.dreamhost.com/mimetex.cgi?a%20=%20-%20%5Comega%5E2%20r" align="middle" border="0" /><br /><br />This gives<br /><br /><img src="http://www.forkosh.dreamhost.com/mimetex.cgi?dT%20=%20%5Cfrac%7BG%20m_e%20*%20dm%7D%7Br%5E2%7D%20-%20dm%20*%20%5Comega%5E2%20r" align="middle" border="0" /><br /><br />If the mass of the element, <img src="http://www.forkosh.dreamhost.com/mimetex.cgi?dm" align="middle" border="0" />, is written in terms of the cross-sectional area of the cable at a given length <img src="http://www.forkosh.dreamhost.com/mimetex.cgi?A%28l%29" align="middle" border="0" />, and the density <img src="http://www.forkosh.dreamhost.com/mimetex.cgi?%5Crho" align="middle" border="0" />, we can then write the tension as<br /><br /><img src="http://www.forkosh.dreamhost.com/mimetex.cgi?T%28r%29%20=%20%5Cint_%7Br_e%7D%5Er%20A%28l%29%20%5Crho%20%5Cleft%28%20%5Cfrac%7BG%20m_e%7D%7Bl%5E2%7D%20-%20%5Comega%5E2%20l%20%5Cright%29%20dl" align="middle" border="0" /><br /><br />The difference between <i>r</i> and <i>l</i> is that <i>l</i> is a dummy variable for integrating over. It begins at <i>r<sub>e</sub></i>, the radius of the Earth, and goes to whatever radius we're considering.<br /><br />The tension is zero at the surface of the Earth, then increases until the integrand becomes zero at geosynchronous orbit. Then the tension slowly slinks back down to zero at the far end of the elevator.<br /><br />To determine whether or not the cable will rip, it's not the total tension that matters, but whether the tension is greater than the cable's tensile strength. And the cable's strength is proportional to it's cross section <img src="http://www.forkosh.dreamhost.com/mimetex.cgi?A%28r%29" align="middle" border="0" />. This suggests a strategy for building the elevator - down near the ground where the tension is small, make the cable very thin. It doesn't need to be strong there. Up near geosynchronous orbit where the tension is very high, make the cable thick to deal with the high stress. This plan could be called "tapering" the cable.<br /><br />The ideal taper for the cable is one in which the tension per unit area is constant all along the cable. To make the problem workable, assume the tension doesn't drop to zero at the surface of the Earth (in which case the "ideal" cable would be zero thickness there, and all the way up), but instead goes to some finite value <img src="http://www.forkosh.dreamhost.com/mimetex.cgi?T_0" align="middle" border="0" />.<br /><br />Then we have:<br /><br /><img src="http://www.forkosh.dreamhost.com/mimetex.cgi?%5Cfrac%7BT_0%7D%7BA_0%7D%20=%20%5Cfrac%7BT%28r%29%7D%7BA%28r%29%7D" align="middle" border="0" /><br /><br />this can be combined with the expression for the tension to give a differential equation for the cross-section length<br /><br /><img src="http://www.forkosh.dreamhost.com/mimetex.cgi?%5Cfrac%7BdA%7D%7Bdr%7D%20=%20%5Cfrac%7BA_0%20%5Crho%7D%7BT_0%7D%20%5Cleft%28%20%5Cfrac%7BG%20m_e%7D%7Br%5E2%7D%20-%20%5Comega%20r%5E2%5Cright%29" align="middle" border="0" /><br /><br />which you can solve to find<br /><br /><img src="http://www.forkosh.dreamhost.com/mimetex.cgi?A%28r%29%20=%20C%20e%5E%7B%5Cfrac%7B-A_0%20%5Crho%7D%7BT_0%7D%20%5Cleft%28%20%5Cfrac%7BG%20m_e%7D%7Br%7D%20+%20%5Cfrac%7B%5Comega%5E2%20r%5E2%7D%7B2%7D%20%5Cright%29%7D" align="middle" border="0" /><br /><br />where <i>C</i> is just a constant set so that the cable has the appropriate width at the surface of the Earth.Markkimarkkonnenhttp://www.blogger.com/profile/11122368349695914223noreply@blogger.com6tag:blogger.com,1999:blog-5497138109765858383.post-80417151340836691792008-09-20T23:18:00.000-07:002008-09-20T23:49:53.282-07:00Depth Perception, part IIThe goal is to understand why a mirror allows depth perception, but a television or painting does not. In the first part of the answer, I demonstrated that depth perception relies on seeing the same object with two eyes. Eyes detect to angle to the object, but the two eyes will detect different angles. The bigger the difference between these angles, the closer the object.<br /><br />Here's a diagram of a television and a mirror. A pair of eyes looks at each, and sees a purple ball.<br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://1.bp.blogspot.com/_2y3Souq-IUU/SNXuK04iw4I/AAAAAAAABBo/jW-HD5QkyKU/s1600-h/depthperception2.jpg"><img style="margin: 0px auto 10px; display: block; text-align: center; cursor: pointer;" src="http://1.bp.blogspot.com/_2y3Souq-IUU/SNXuK04iw4I/AAAAAAAABBo/jW-HD5QkyKU/s320/depthperception2.jpg" alt="" id="BLOGGER_PHOTO_ID_5248362810532086658" border="0" /></a><br />Notice that for the television, the light rays converge at the screen. But the light rays bounce off the mirror in different places, converging only at the ball.<br /><br />Although the ball is actually in front of the mirror, there's a problem. The light you detect is the light right at your eyes. You can't detect the light's entire path. So if you want to reconstruct the path of the light to see where light beams intersect, you just have to guess. That's what your brain does, and it simply guesses that the light doesn't bend or bounce or anything like that. It just goes back straight to the source. So the situation for the mirror is that the light actually travels the solid line, but your brain thinks it travels the dotted line.<br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://2.bp.blogspot.com/_2y3Souq-IUU/SNXuLKLSRZI/AAAAAAAABBw/uK3YQlbnsOA/s1600-h/depthperception3.jpg"><img style="margin: 0px auto 10px; display: block; text-align: center; cursor: pointer;" src="http://2.bp.blogspot.com/_2y3Souq-IUU/SNXuLKLSRZI/AAAAAAAABBw/uK3YQlbnsOA/s320/depthperception3.jpg" alt="" id="BLOGGER_PHOTO_ID_5248362816247842194" border="0" /></a><br />An object can appear to have a depth well back behind the mirror. Objects on a television screen will always appear to be right at the television screen, since that's where the lines leading to those objects converge.<br /><br />The essential difference is that the television or painting is generating the light you see. The light from any given point of the painting radiates out in all directions, always showing the same object.<br /><br />The mirror doesn't generate its light - it reflects it. So if two observers are in different places and look at the same spot on a mirror, they see different objects there, not the same.Markkimarkkonnenhttp://www.blogger.com/profile/11122368349695914223noreply@blogger.com0tag:blogger.com,1999:blog-5497138109765858383.post-16762129328700666302008-09-19T00:53:00.000-07:002008-09-19T02:15:01.544-07:00Depth Perception, part 1This post will take a look at how depth perception works. The next post will build on this background to answer <a href="http://arcsecond.blogspot.com/2008/09/new-problem-depth-perception.html">yesterday's question</a>.<br /><br />First experiment:<br /><ol><li>Close your eyelids. </li><li>Now open them.</li></ol> If you did the first part of that experiment right away, you shouldn't have known what to do next, because you wouldn't be able to read the instructions. Close your eyes and you can't see any more. Conclusion: you see with your eyes. (Actually, this just shows that eyes are "necessary but not sufficient". But hey, let's not split corneas here.)<br /><br />Second experiment: Cut open a cow's eye and diagram it. You should get this:<br /><img src="http://upload.wikimedia.org/wikipedia/commons/thumb/1/1e/Schematic_diagram_of_the_human_eye_en.svg/508px-Schematic_diagram_of_the_human_eye_en.svg.png" /><br /><br />The salient point is that your eye is designed to give your brain the information "lots of light is coming from this direction, not so much from that direction, and none at all from a third direction". Overall, there are lots and lots of directions. When you put them all together, you get an image.<br /><br />That's what one eye does. The way it does it is optics. First, light comes into the eye and passes through the lens. All light that comes in parallel, regardless of whether it enters the left or right side of the eye, then gets focused to the same part of the retina. That's the glory of the lens. So all the light falling on a certain part of the retina came from the same direction in space, not from entering the same part of your eye.<br /><br />This directional knowledge lets your build an image. Here is a picture you can look at with your eyes:<br /><img src="http://www.ashleymills.com/ae/images/lenna_pixels.png" /><br />To see each pixel, you must look in a slightly different direction. Putting each pixel with its appropriate direction allows you to build the image back up in your mind.<br /><br />So eyes only care about the direction light comes in. However, you have two eyes, and they're in different places. That means light from the same source will enter from different angles. Here's a gratuitous graphic:<br /><img src="http://www.vision3d.com/images/bb.jpeg" /><br /><br />Okay, so light from an object, like a baseball you want to catch (actually, depth perception is not the tool baseball outfielders use to flag down a fly ball, but that's another story), enters your left eye from a direction (&theta<sub>l</sub>, &phi<sub>l</sub>), and enters your right eye from a different direction (&theta<sub>r</sub>, &phi<sub>r</sub>). How far away is the object? Here, &theta is the polar angle, the angle between the line from the top of your head to your feet and the path of the light. &phi is the azimuthal angle, telling how far to the right or left the direction is. Some people use a different convention. Mine is like this:<br /><img src="http://www.tf.uni-kiel.de/matwis/amat/elmat_en/kap_3/illustr/spherical_coordinates.gif" /><br />You stand with your head pointed towards <i>z</i> and your face towards <i>x</i>.<br /><br />Your eyes are separated (obviously) by a distance d. Also, we'll make the approximation that the object is far enough away from your face that it's essentially the same distance from each eye. In that case, the polar angle &theta will be the same for each eye.<br /><br />The three lines, one connecting your eyes, the other two from each eye to the object, form a triangle like this:<br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://3.bp.blogspot.com/_2y3Souq-IUU/SNNo8gVjrOI/AAAAAAAABBg/stu6O-oekpM/s1600-h/depthperception1.jpg"><img style="margin: 0px auto 10px; display: block; text-align: center; cursor: pointer;" src="http://3.bp.blogspot.com/_2y3Souq-IUU/SNNo8gVjrOI/AAAAAAAABBg/stu6O-oekpM/s320/depthperception1.jpg" alt="" id="BLOGGER_PHOTO_ID_5247653379498552546" border="0" /></a><br />Boom! The math is right in there! I actually just spent ten minutes searching for and installing this thing that should let me start <img src="http://www.forkosh.dreamhost.com/mimetex.cgi?%5CLaTeX" align="middle" border="0" />'ing up my blog, and then I didn't even use it (except to make that sexy <img src="http://www.forkosh.dreamhost.com/mimetex.cgi?%5CLaTeX" align="middle" border="0" /> logo).<br /><br />Anyway, yup. <img src="http://www.forkosh.dreamhost.com/mimetex.cgi?d%20%5Capprox%20%5Cfrac%7Bw%7D%7B%5CDelta%7D" align="middle" border="0" /> where <i>d</i> is the distance to the object, and <i>w</i> is the width of your eyes. &Delta is the difference between the azimuthal angles, and &phi is that azimuthal angle (either one. the difference is small). I made two simplifying assumptions. The first is that the difference in the angles is small, which just means the thing isn't right up in your face. The second is that the object is close to right in front of you, which is necessary to be able to use depth perception on it.<br /><br />Nice.<br /><br />Now there's a problem, which is that you only get so much accuracy in determining the angles. As &delta becomes smaller and smaller, small errors result in big problems for <i>d</i>. So depth perception only works for objects close to you, or a medium distance away. For very far objects, you need to use other clues. The same problem occurs for astronomers attempting to find the distance to a star by <a href="http://en.wikipedia.org/wiki/Parallax">parallax.</a>Markkimarkkonnenhttp://www.blogger.com/profile/11122368349695914223noreply@blogger.com11tag:blogger.com,1999:blog-5497138109765858383.post-21503870819116116672008-09-18T02:48:00.000-07:002008-09-18T02:52:31.336-07:00New Problem: Depth PerceptionNo matter how good your cameraman, or the equipment he's using, it's impossible to create the illusion of depth on a television. You might be able to tell some things are further away than others, but they won't ever look truly 3-D the way the real world does when you have two eyes open.<br /><br />On the other hand, a mirror can give you a perfect illusion of depth (I am not using "perfect" in a technical sense here). A mirror image world is just as depth-rich to your binocular vision as the genuine article.<br /><br />So why is that? Why can't a television gives depth perception? And what can a mirror do that a television can't?Markkimarkkonnenhttp://www.blogger.com/profile/11122368349695914223noreply@blogger.com1tag:blogger.com,1999:blog-5497138109765858383.post-19079144737533139052008-09-16T01:12:00.001-07:002008-09-16T01:15:25.754-07:00MeditationA lot of people have asked me about my vipassana meditation retreat last month. The only ones who've received answers are the people who asked me in person, since they're much harder to shirk.<br /><br />The meditation camp was a psychologically intense experience. It was important enough to me that if I want to explain it, I think I should put the effort into explaining it well.<br /><br />I think it will take me about two weeks to write up all my thoughts as a memoir of my time there. I'll post the entire story at once when I'm done. In the mean time, thanks to everyone for your interest - it shows me that putting the effort into writing the story will be worthwhile.Markkimarkkonnenhttp://www.blogger.com/profile/11122368349695914223noreply@blogger.com0tag:blogger.com,1999:blog-5497138109765858383.post-8109583088695388312008-09-15T23:02:00.000-07:002008-09-16T01:17:49.671-07:00Answer: Circling BugsThe mathematical answer to <a href="http://arcsecond.blogspot.com/2008/09/circling-bugs.html">yesterday's problem</a> about the circling bugs is much prettier than the previous one about the demon and lake. We might as well just solve the problem for "n" bugs, then let n=4 for the square. It is a bit of a special case, though, which you might be able to solve completely by intuition.<br /><br />We'll let n bugs be distributed uniformly around a unit circle, and set the units of time so their walking speed is one. Each begins chasing its nearest neighbor. Because the situation is exactly the same for every bug, the shape cannot distort. If it starts out a square, it stays a square. It can only get larger or smaller, translate, or rotate.<br /><br />Also by symmetry, it can't translate. Which way would it go? If you let it translate in the direction of one bug, you're being unfair to all the others, but the problem is perfectly equable.<br /><br />The shape will rotate and change size. Begin by imagining the case of infinitely many bugs, all point-sized, that form a complete circle. Then if they chase the bug in front of them, they'll simply march around the circle without it ever changing size at all.<br /><br />If there are finite bugs, the shape must shrink as it rotates. In the case of a circle, the bug can never get any closer to the next guy he's chasing, because whoever he's chasing is running directly away from him at the same speed. But for a finite-sided object, the chased bug isn't running directly away any more. He's running at an angle, so the distance between the bugs decreases. Hence, the shape shrinks.<br /><br />Down to business: find the equation for the motion of a bug.<br /><br />Let's write an equation for its velocity in polar coordinates. If the position of the bug is (r,&theta), then the position of the next bug is (r,&theta+2&pi/n). Skipping a step or two, the velocity of the bug is (-sin[&pi/n],cos[&pi/n]). You get that by taking the geometric fact that the exterior angle of a regular n-gon is 2&pi/n, and noticing the deviation of the bug's path from a pure circular motion is exactly half that.<br /><br />The bugs all start a distance of one from the center, so they will all meet there after a time (and distance walked) 1/sin(&pi/n). If there are lots of bugs, we can make a small-angle approximation to get n/&pi. More bugs, longer to meet.<br /><br />Integrating the velocity above (and remembering that things are a little more complicated in polar coordinates, because d&theta/dt = v<sub>&theta</sub>/r(t)), we have, for a bug that begins at (1,0)<br />p(t) = (1-sin[&pi/n]*t,-cot[&pi/n]*ln[1-sin(&pi/n)t])<br /><a href="http://mathworld.wolfram.com/LogarithmicSpiral.html">A logarithmic spiral</a>.<br /><br />For the case n=4, our formula says the bugs will walk a distance 2<sup>1/2</sup> before colliding. In this special case, the motion of the bug "running away" is at a right angle. He isn't running away at all, so much as trying to execute a circular orbit. So, from a bug's perspective, the distance between the bugs shrinks just as fast as it would if the one he were chasing were simply stationary.Markkimarkkonnenhttp://www.blogger.com/profile/11122368349695914223noreply@blogger.com0tag:blogger.com,1999:blog-5497138109765858383.post-241320170141722192008-09-14T04:01:00.000-07:002008-09-14T04:07:45.322-07:00Circling BugsHere's another problem I liked from "How Would You Move Mount Fuji". I'll state the original problem, then the generalization, and hope that when I try to solve it tomorrow I can do a little better than I did yesterday with the boat and demon. I think this one is actually a famous problem.<br /><br />There are four beetles located the the four corners of a square. At the same moment, they all start crawling directly towards the next beetle clockwise around from them. That beetle they're aiming for is also crawling, so each beetle continually adjusts its course so that it aims straight towards the next beetle.<br /><br />Will they collide in finite time? How far will they walk before they collide?<br /><br />Generalizations: Let there be "n" beetles at the corners of a regular n-gon. Same questions. Also, if you drew a path marking the route of a single beetle, how many degrees around the center of the square would it subtend (i.e. how many times does the beetle wrap around the center)? What is their exact path?Markkimarkkonnenhttp://www.blogger.com/profile/11122368349695914223noreply@blogger.com2tag:blogger.com,1999:blog-5497138109765858383.post-7521263404035013772008-09-13T20:42:00.001-07:002008-09-14T03:54:26.456-07:00(Kind of) Answer: Lake and DemonThis post is about <a href="http://arcsecond.blogspot.com/2008/09/lake-and-demon.html">yesterday's problem</a>. I have to admit - the question is tougher than I thought it was (it's another of those I asked before solving). The question in the book is simply, "Imagine the demon goes four times as fast as your boat, how can you escape?" That question is purely conceptual. I'll answer it, then go as far as I can right now towards, "what is the minimum ratio of the speed of your boat to speed of the demon for you to be able to guarantee escape?" (Obviously less than 1/4). <br /><br />I'll set the units of distance so the radius of the lake is 1, then set the units of time so the speed of your boat is also 1. The speed of the demon is D.<br /><br />At first glance, you can realize that if D < Pi, you can just shoot for the shore opposite of where the demon is standing, and you'll get there first. But you can escape even if your ratio is worse than this.<br /><br />Qualitatively, here is the plan: You start rowing in the direction directly away from the demon. He picks a direction and starts running that way, anticipating where you'll land. But it's to your advantage to keep the demon 180 degrees around the lake from you - as far as possible. To do this, just angle your boat a bit so whichever way the demon runs, you stay directly opposite him.<br /><br />As long as you're still close to the center of the lake, this is easy to do. Your angular speed can be quite high because it's a small circle around the center. The demon's angular speed is low because he's so far out. Hence, you can keep him 180 degrees away from you.<br /><br />What's the maximum distance from the center of the lake where you can still force the demon to be 180 degrees away from you? That happens when your maximum angular velocities match, which is at a distance 1/D.<br /><br />After that, you still have to be able to make it to shore. If you were going on a straight shot, your speed would have to be great enough so that the time for you to get to shore is less than the time for the demon to run halfway around the lake. Algebraically,<br />(1 - 1/D) < Pi/D<br />D < 1 + Pi<br /><br />The new strategy takes us from D < Pi to D < 1+Pi (slightly greater than 4), but is there something even better? Maybe it's worth your time to curve away from the demon despite the fact you can't keep him a full 180 degrees away from you.<br /><br />A moment's thought will tell you undoubtedly yes - you can do better than moving out to the cutoff circle and then sprinting straight for shore. Suppose you are making that beeline for shore. The demon has picked a direction and is running towards your expected landing spot. If you turn your heading a very slight amount away from his side of the lake, you now have two components to your velocity - a "radial" component towards the edge of the lake and a "circumferential" component around in a circle. By the Pythagorean theorem, the squares of these speeds add up to the maximum speed of your boat. So if the component of circumferential speed is small, the change in the radial speed is second-order. Explicitly:<br /><br />S<sub>r</sub><sup>2</sup> + S<sub>c</sub><sup>2</sup> = 1<br />S<sub>r</sub> = (1 - S<sub>c</sub><sup>2</sup>)<sup>1/2</sup><br /><br />using the binomial theorem, for small S<sub>c</sub>, this is<br /><br />S<sub>r</sub> = 1 - 1/2*S<sub>c</sub><sup>2</sup><br /><br />So you can pay a very small price to your radial velocity in return for some circumferential velocity. Losing radial velocity means it takes you longer to get to shore, while gaining circumferential velocity means it takes the demon longer to get to your landing spot. There's a positive effect to veering off course, and a negative effect. But since you can make the ratio of the negative effect to the positive effect arbitrarily small, it'll always be worth your while to veer away from a straight shot to shore at least a little bit, assuming your goal is to land with the demon as far away from you as possible.<br /><br />Let's go back to the beginning and try a more quantitative approach.<br /><br />If we want to escape the demon, we should always go at full speed. The demon's strategy is simple - he also goes at full speed in whatever direction takes him closer to you. If both directions are equal (because you're 180 degrees apart), he picks one at random. So the only thing we have to choose is our bearing, based on our current location and the demon's current location. What should the criterion be? I don't have a proof this is optimal, but if your goal is to land on shore as far from the demon as possible, let's make the criterion:<br /><br /><b>Choose the heading that maximizes the quantity:<br />(dI/dt) + (dr/dt*I)/(1-r)<br />Where "I" indicates the distance from the demon's position to the "intercept point", the point on the shore closest to your boat, and "r" is your distance from the center of the lake.</b><br /><br />I derived this criterion with the following reasoning:<br />Imagine you were going to make a beeline to shore from wherever you currently are. If we divide the demon's distance to the intercept point "I" by his speed "D", we get the demon's expected time to intercept, I/D. On the other hand, your expected time to get there is just your distance from shore, 1-r, because your speed is 1. Finally, if we divide those two times by each other, we get a ratio I/(D*(1-r)). If that ratio is bigger than one, you've made it - you could escape with a bee-line trajectory now. Unfortunately, if D>Pi, that ratio at the beginning of the problem is less than one. But by choosing a clever trajectory, we can try to push that ratio up to one, and then escape. So at every moment, we're going to choose the heading so that this ratio is as high as we can get it a short time "dt" from now. To do that, just take the derivative of the ratio with respect to time and maximize it.<br /><br />If you take the derivative, you'll noticed I dropped the constant "1/(D*(1-r))". This yields the condition above. As long as the derivative of that ratio is positive, we'll assume you still have a chance. Once you can't push the ratio any higher he's got you, since if you were going to escape, the ratio would go to infinity right before you got away, and if he's going to catch you, the ratio goes to zero.<br /><br />Choose "@" to be the angle between your heading and the heading straight for shore. I'll let you work through the calculus for yourself should you so desire. The goal is to get "@" as a function of "I" and "r". That is, choose the best course whatever the current situation is. I found:<br /><br />tan(@) = (1-r)/(I*r)<br /><br />Reality check: I expect that as you near the shore, you should aim more and more towards it. As r->1, @->0.<br /><br />I have to be careful with this formula - it only applies when the demon is NOT directly opposite you. If he's directly opposite you, then angling your boat brings him closer no matter which way you turn - a factor not accounted for in my calculation. So this formula only applies once the boat is already past the cutoff circle and racing for shore.<br /><br />Now we've worked out a strategy, all that's left is integration. Unfortunately, the integration isn't very easy. We'll start things off by introducing a polar coordinate system (r,@). The center of the lake is (0,0). The demon starts out at (1,0), and the boat, having already completed the first stage of its maneuvers, begins at (1/D,Pi).<br /><br />Let's say the demon chooses to run up towards the top of the circle (increasing @). Then his position as a function of time is (1,D*t).<br /><br />Your position is trickier. It depends on your bearing, which depends on both your own position again and the demon's position. If you write out these equations, you'll see that they're pretty hopelessly coupled. I get<br /><br />dr/dt = I*r/((1-r)^2 + (I*r)^2)^1/2<br />d@/dt = (1-r)/(r*((1-r)^2 + (I*r)^2)^1/2)<br /><br />where I = @ - D*t, and "@" and "r" are your polar coordinates as functions of time.<br /><br />So from there, I'm basically analytically screwed. If you'd like to do the numerical analysis and figure something out, please let me know what you find.<br /><br />The sad part is, even doing the integration doesn't actually find an upper bound for D. It only finds an upper bound assuming my particular strategy is the best possible strategy. Just because it seems like a good one in my head doesn't mean there isn't one that's better, and I honestly have no idea how to go about proving either:<br />A) such-and-such a strategy is optimal for escaping the demon<br />B) no strategy can do better than such-and-such<br /><br />Moral: These toy problems, though simple and well-defined, can be tougher than they look when you actually start to bite into them. I was originally just attracted to the little trick of realizing you could force the demon away from you by making small circles about radii close to the center of the pond. But after I got that I naturally wanted to go a step further. Turned out the next step was biting off more than I could chew.Markkimarkkonnenhttp://www.blogger.com/profile/11122368349695914223noreply@blogger.com0tag:blogger.com,1999:blog-5497138109765858383.post-22826436715921276702008-09-12T23:26:00.000-07:002008-09-12T23:40:22.948-07:00Lake and DemonThis is a slightly-modified form of one of the most interesting problems from a book I just finished, called <a href="http://www.amazon.com/How-Would-Move-Mount-Fuji/dp/0316778494/ref=pd_bbs_sr_1?ie=UTF8&s=books&qid=1221287509&sr=8-1%20">How Would You Move Mount Fuji?</a> The book is about using such problems in job interviews, but the context is irrelevant to how clever the problem is. So without further ado:<br /><br />You're on a boat in the middle of a circular lake. On the outside edge of the lake is a demon who wants to kill you to death. He is bigger than you and has pointy teeth. With poison on them. Death poison. He can't go in the water, so he runs around the edge of the lake, waiting to intercept you when you land.<br /><br />If you can make it to shore without the demon being right where you land, you'll be okay. You're fast enough over land to run away and escape. But if the demon is right exactly there when you hit shore, you lose. Your wife and kids grieve. Your life insurer goes broke.<br /><br />Assume the demon and your boat both have a maximum speed, which they can maintain indefinitely. What is the minimum ratio of the speed of your boat to the speed of the demon for you to have an infallible strategy for escape? What is that strategy, and how long does it take you to get to the edge of the lake evading the demon (as a function of the ratio of your speeds, and the ratio of your speed to the size of the lake)?Markkimarkkonnenhttp://www.blogger.com/profile/11122368349695914223noreply@blogger.com0tag:blogger.com,1999:blog-5497138109765858383.post-89067229061273949022008-09-12T17:47:00.000-07:002008-09-12T17:54:38.080-07:00Interview with Brian GreeneBrian Greene gave me <a style="font-weight: bold;" href="https://ideotrope.org/index.pl?node_id=80382">a half hour interview</a> earlier this week. I asked him some obligatory questions about his new book, <span style="font-style: italic;">Icarus at the Edge of Time</span>. After reading the book (twice. It's very short.), I think Greene is a bit out of his league in writing fiction. Nonetheless, he gave some good insight on the relationship between thinking like a scientist and thinking like a writer, and how one could do both.<br /><br />I was nervous enough during the interview to have a hard time afterwards remembering exactly what we had talked about (I recorded it, of course), but not so nervous as to be completely incapable of creating follow-up questions on the spot after getting his answers to my prepared ones.<br /><br />What's more - Brian Greene appeared a little bit nervous, too. His tapped his foot throughout the interview and began every single response with the word "well" (which I deleted from the transcription, along with about two hundred of my own "um"s, "uhh"s, and "so"s.)<br /><br />This should come out in the school paper, in some shorter form, once the new school year starts, but for the time being check out the link to Ideotrope at the top of the post.Markkimarkkonnenhttp://www.blogger.com/profile/11122368349695914223noreply@blogger.com0tag:blogger.com,1999:blog-5497138109765858383.post-76351122414066277282008-09-08T10:32:00.000-07:002008-09-08T10:57:43.554-07:00Conjuring Trees From Thin Air?At <a style="font-weight: bold;" href="http://www.ted.com/">TED.COM</a>, Jonathan Drori <a style="font-weight: bold;" href="http://www.ted.com/index.php/talks/jonathan_drori_on_what_we_think_we_know.html">gave a talk</a> claiming that graduates from MIT could not correctly answer the following four questions:<br /><ol><li>An apple seed is small, but an apple tree is large. Where did all that extra mass come from?</li><li>It's easy to light a flashlight bulb with a 1.5V battery and two wires, but can it be done with just one wire? (Presumably, no "lateral thinking" solutions, such as cutting one long wire to make two wires, are allowed.)</li><li>Why is it hotter in summer than in winter?</li><li>Draw a diagram of the solar system.</li></ol>I believe that most students from Caltech would get all four mostly correct, although of course no one would be able to draw a perfect diagram of the solar system off the top of their heads. (However, I did recently have a conversation with one Caltech graduate who, although he could answer question one correctly, tripped up on its inverse. That is, he thought that when a human loses weight, that weight comes out in their poo.)<br /><br />Question 1 is the one I want to focus on, because I think that the guy who was giving a talk about how people get it wrong, was himself wrong. In his talk, Drori claims that 99% of the mass of a piece of wood comes from the air.<br /><br />Eighth grade biology gives us the tools to figure this out. Trees are made of cells, and cells bags holding organic molecules floating around in water. So the mass of a tree is mostly organic molecules and water. The organic molecules come in different varieties, but lots of them are carbohydrates having the empirical formula CH<sub>2</sub>O. The series of reactions that makes these carbohydrates is photosynthesis, summarized by:<br /><br />CO<sub>2</sub> + H<sub>2</sub>O -> CH<sub>2</sub>O + O<sub>2</sub><br /><br />So the mass of the tree comes primarily from carbon dioxide and water. There are other things sucked up by the roots - the tree needs plenty of nitrogen to build amino acids, various other chemicals such as phosphorus and iron, and these along probably add up to more than 1%, but let's just think about the carbon dioxide and water.<br /><br />The carbon dioxide comes from the air - that was Drori's point. The most important element in life is carbon -sometimes people refer to terrestrial beings as carbon-based life forms. Trees get their carbon from the air. Fine. But that's not the whole story. Trees need lots of water, too. And they get it from the ground.<br /><br />How do I know they get their water from the ground? Because in my new apartment, there are a bunch of trees on the back patio. They're all dead, because the previous tenants believed what Drori said, and thought the trees would suck up water from the air around them. Nope. Empirical evidence in the form of very dead trees now suggests that you have to pour water on the ground around the trees, and then they can suck up that water through the ground.<br /><br />PS - when you lose weight, you're going the other way from photosynthesis, turning carbohydrate and oxygen into water and carbon dioxide. So the weight you lose comes out partially as water, which has lots of ways to exit the body, and partially as CO<sub>2</sub>, which is exhaled. Every time you breathe out, you're losing weight - at rate of about a pound every 50,000 breaths (based on breathing 15 times a minute and burning 2000 calories a day).Markkimarkkonnenhttp://www.blogger.com/profile/11122368349695914223noreply@blogger.com7tag:blogger.com,1999:blog-5497138109765858383.post-26475846849541469812008-08-12T01:14:00.000-07:002008-08-12T01:21:43.306-07:00End of the Problem-of-the-Day Contest, On to New AdventuresThe problem of the day contest ended one week ago. Our winner at camp was Lucas Brown, followed closely by Jonathan Lilley. Tyler Fitzgerald and Ben Zinberg were third and fourth. Fantastic prizes were awarded to all.<br /><br />Now that I'm going on to new things and new ways of life, the problem of the day will probably become the "occasional problem when I see an interesting one, supplemented by other various things that are on my mind". That's sort of what the blog was intended for anyway. It covers various things I'm thinking about, but only has a one-arcsecond window on the breadth of human experience.<br /><br />My next adventure is to travel to the <a href="http://www.dhamma.org/en/">California Vipassana Center </a>in North Fork, to learn a certain form of meditation. It's a ten-day course that involves a rather extreme schedule of 10 hours of meditation a day, no talking, reading, writing, exercising, doing physics, looking at females, contacting the outside world, or in general doing anything that would distract from the task at hand (you wouldn't want to be distracted from doing nothing, would you?) So I'll be completely off the radar until Aug 25 or so.Markkimarkkonnenhttp://www.blogger.com/profile/11122368349695914223noreply@blogger.com0tag:blogger.com,1999:blog-5497138109765858383.post-65626144541177446092008-08-08T17:39:00.000-07:002008-08-08T18:13:58.511-07:00Answer: Square Root Calculator<img src="http://3.bp.blogspot.com/_2y3Souq-IUU/SJjKhMSE24I/AAAAAAAABA4/nM0m-iYsSDc/s1600/rootcalc.jpg" width="300" /><br /><br />Pick a number. Set the ball at that number on the ramp and let it go. Watch where it falls. The number where it falls is the square root of the number on the ramp.<br /><br />This works because the velocity of the ball as it leaves the ramp is proportional to the square root of its kinetic energy. Its kinetic energy comes from the potential energy it originally had while sitting on the ramp. That's linear in the height.<br /><br />Incidentally, not all the ball's initial potential energy gets converted into kinetic energy of translation. Some is also converted into kinetic energy of rotation. However, the fraction of kinetic energy that goes into rotation is constant, so this doesn't affect our ability to build a calculator. It just means the marks on the landing zone have to be closer together than they would for a frictionless, sliding ball.<br /><br />The thing about the square root calculator is that it uses two linear scales. That's why it's a great machine - it doesn't take any difficult work to decide where the marks go. The laws of physics do the actual computation.<br /><br />There were a variety of answers to the challenge to create your own analog calculator. Most of them missed the mark of what I was shooting for.<br /><br />For example, an abacus can do multiplication. It's a physical system, as well. So I suppose it's a physical system that does a useful computation. But it's not interesting from a physics perspective. It's interesting only from a logical perspective. The abacus is probably made of wood, but doesn't depend essentially on the properties of wood to work. I could make the abacus out of stones sitting in from of me, or fancy metal rings, or little holes that I dig or fill in with sand at the beach. The abacus only cares about the positional relationship between its parts, not their physical interactions. I could even, if my powers of imagination were strong enough, picture an abacus with all its beads in my head, move them around mentally, and do abacus-style multiplication completely inside my head. If you imagine a different universe, in which the laws of physics work differently, the abacus wouldn't care. Multiplication would still be the same, and you could still do it with any device whose pieces could move in relation to each other the same way they can in an abacus.<br /><br />So slide rules are out, too. Electronic computers can certainly compute useful quantities and solve differential equations numerically, and they are physical systems. But again, the magic of the computer is in its logical gates rather than its physics. Some other answers, like tree rings and sundials counting the passage of time, were not really doing a mathematical computation.<br /><br />If you're good at circuit analysis, you could build an analog computer to calculate all kinds of interesting things. In fact, people used to do this all the time before digital computers got so good. So I won't talk about circuit analysis.<br /><br />Instead, here's a fairly simple system to do multiplication. It's a tube of dilute gas, say argon. There's a piston allowing you to change the volume to whatever you want. The volume is marked off on the sides like it is in graduated cylinders or Nalgene bottles. You adjust the piston up or down with a stack of weights. If you want the cylinder to have less volume, stack on more weight. By counting the weights you have on there, you can get the pressure. There's a thermometer inside the tube positioned so you can easily read it. Finally, there's a bunsen burner and some cold packs lying around, so you can add or remove heat at will.<br /><br />To a good approximation, the gas obeys the equation of state PV = nRT. So choose the amount of gas in your cylinder, along with your units pressure, volume, and temperature, so that nR = 1. Then PV = T. You can now do multiplication. Start by setting the weight stack so that the pressure inside is the first number you want to multiply. If the volume is lower than the second number you want to multiply, add heat. If it's greater, suck heat out. Keep adding or subtracting heat until the volume is equal to the second number you want to multiply. Now read off the temperature from the thermometer, and that is the multiple you sought.<br /><br />The above machine is also a division machine, because you could equally well set the pressure to the denominator, add/subtract heat until the temperature is the numerator, and read off the volume using V = T/P. Similarly the square root calculator is also a square calculator. Take a guess at the square of a number, drop the ball from there and see where it lands. If it lands in front of the number you're squaring, drop it from higher. Repeat until you find the number on the ramp such that when you drop the ball from there, it hits your target number on the landing zone. That's the square.<br /><br />Any system that satisfies the differential equation df/dx = a*f is a potential logarithm/exponential calculator, since its solution is f = e^ax = (e^a)^x. An example is a system whose friction is proportional to its amplitude. This occurs in the viscous flow of a fluid, for example. A runaway chemical reaction might obey this law, at least over some regime. Population growth of organism does this. The expansion of the universe due to a nonzero vacuum energy does this, since the rate of growth is proportional to the amount of vacuum, but growth is itself the change in vacuum.<br /><br />Trig functions come from any physical system that obeys d<sup>2</sup>f/dx<sup>2</sup> = -a*f, where "a" is positive. A mass on a spring is the canonical example, all sorts of things will exhibit this same behavior. The reason is that particles will minimize their potential energy. When you're near a potential energy minimum, you can expand the potential as a Taylor series, but the constant term is not physically relevant, and the linear term will be zero because you're at a minimum. Then the quadratic term dominates over a certain regime, and in that regime the system obeys the differential equation for trig functions.Markkimarkkonnenhttp://www.blogger.com/profile/11122368349695914223noreply@blogger.com2tag:blogger.com,1999:blog-5497138109765858383.post-40189948754755694782008-08-05T13:44:00.000-07:002008-08-05T14:51:36.787-07:00Answer: Lifeguard New Problem: Square Root Calculator<span style="font-weight: bold;">Lifeguard</span><br />Let the guard be a distance "b" from the edge of the beach, a horizontal distance "h" from where the drowning guy is, and let the drowning guy be a distance "d" from the shoreline.<br /><br />Finally, let the lifeguard run to a point "x" on the shore before entering the water. See picture:<br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://2.bp.blogspot.com/_2y3Souq-IUU/SJjKYIl35gI/AAAAAAAABAw/8InIaJsiGmM/s1600-h/beach.jpg"><img style="margin: 0pt 10px 10px 0pt; float: left; cursor: pointer;" src="http://2.bp.blogspot.com/_2y3Souq-IUU/SJjKYIl35gI/AAAAAAAABAw/8InIaJsiGmM/s320/beach.jpg" alt="" id="BLOGGER_PHOTO_ID_5231153483162838530" border="0" /></a><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br />Then the time it takes him to get out there is<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://2.bp.blogspot.com/_2y3Souq-IUU/SJjKrUNKZeI/AAAAAAAABBA/4bxTEg_wCpI/s1600-h/lifeguard_1.gif"><img style="margin: 0pt 10px 10px 0pt; float: left; cursor: pointer;" src="http://2.bp.blogspot.com/_2y3Souq-IUU/SJjKrUNKZeI/AAAAAAAABBA/4bxTEg_wCpI/s320/lifeguard_1.gif" alt="" id="BLOGGER_PHOTO_ID_5231153812697933282" border="0" /></a><br /><br /><br /><br /><br />We want to find "x" that minimizes this time, so take a derivative and set it equal to zero.<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://1.bp.blogspot.com/_2y3Souq-IUU/SJjKrsEUNBI/AAAAAAAABBI/Uchnq5e1C-U/s1600-h/lifeguard_2.gif"><img style="margin: 0pt 10px 10px 0pt; float: left; cursor: pointer;" src="http://1.bp.blogspot.com/_2y3Souq-IUU/SJjKrsEUNBI/AAAAAAAABBI/Uchnq5e1C-U/s320/lifeguard_2.gif" alt="" id="BLOGGER_PHOTO_ID_5231153819103278098" border="0" /></a><br /><br /><br /><br /><br /><br /><br /><br />Now take another look at the geometry of the problem. The expression above can be rewritten as<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://1.bp.blogspot.com/_2y3Souq-IUU/SJjKrmzsopI/AAAAAAAABBQ/QJbZDs6csIQ/s1600-h/lifeguard_3.gif"><img style="margin: 0pt 10px 10px 0pt; float: left; cursor: pointer;" src="http://1.bp.blogspot.com/_2y3Souq-IUU/SJjKrmzsopI/AAAAAAAABBQ/QJbZDs6csIQ/s320/lifeguard_3.gif" alt="" id="BLOGGER_PHOTO_ID_5231153817691398802" border="0" /></a><br /><br /><br /><br />which is our answer: Snell's Law. Notice that all the actual distances fall out - only two of their proportions matter.<br /><br /><span style="font-weight: bold;">Square Root Calculator<br /></span>The Exploratorium in San Francisco has a device they call a "square root calculator". It's just a ramp, a ball, a jump, and a landing zone, like this:<br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://3.bp.blogspot.com/_2y3Souq-IUU/SJjKhMSE24I/AAAAAAAABA4/nM0m-iYsSDc/s1600-h/rootcalc.jpg"><img style="margin: 0pt 10px 10px 0pt; float: left; cursor: pointer;" src="http://3.bp.blogspot.com/_2y3Souq-IUU/SJjKhMSE24I/AAAAAAAABA4/nM0m-iYsSDc/s320/rootcalc.jpg" alt="" id="BLOGGER_PHOTO_ID_5231153638772366210" border="0" /></a><br />How is this a square root calculator? How does it work? Can you suggest another simple physical system that would operate as an multiplication calculator, logarithm calculator, trigonometric calculator, etc?<br /><span style="font-weight: bold;"></span>Markkimarkkonnenhttp://www.blogger.com/profile/11122368349695914223noreply@blogger.com5tag:blogger.com,1999:blog-5497138109765858383.post-30924909371206536782008-08-04T01:11:00.000-07:002008-08-04T01:43:08.008-07:00Answer: To Swim or Not to Swim New Problem: Lifeguard<span style="font-weight: bold;">To Swim or Not to Swim</span><br />The lake is a hyperbola.<br />Actually, it's a section of a hyperbola on one side of the axis of symmetry, then the mirror image of that section on the other side.<br /><br />This is good knowledge to have if you're building optical equipment. Light follows paths of stationary time, so light passing through a lens is very similar to this problem. Most lenses are built with edges that are sections of sphere, not hyperbolas. This is just because it's easier to manufacture them that way. Then if you're doing some fancy optics, you'll have to correct for the "spherical aberration" of your lens.<br /><br />Proving the the hyperbola is straightforward, but algebra-y. Just set up a coordinate system where the middle of the lake is (0,0) and the edge of the lake is given by x(y). Then the edge of the make must be long enough to satisfy the condition<br />((d-x)<sup>2</sup> + y<sup>2</sup>)<sup>.5</sup> + 2x = c<br />where c is some constant, equal to (d<sup>2</sup> + (w/2)<sup>2</sup>)<sup>.5</sup>.<br /><br />I got this by letting "y" be the height at which you cross the lake. Then I counted the distance to the point where you enter the lake, and the twice the distance to the axis of symmetry (since you go twice as slow in water). This must be equal to the same number for all "y" by the specification of the problem. If it is, the rest of the problem (you're going from A to B, not A to the axis of symmetry) follows because the second half is the same as the first half backwards, and also takes time "c".<br /><br />The equation you get is a hyperbola. This makes some sense. Consider the limit as the width of the lake is much greater than your initial distance from it. Then in that limit, the additional path length added by crossing the lake one unit higher is one unit (for a trip to the axis of symmetry). So in that same limit, the width of the lake must be directly proportional to height of the lake, so that it slows you down the right amount. That's what a hyperbola does - goes asymptotically to a line. Ah well, I just realized that random babbling probably doesn't get the idea across very well. But I also don't care that much because if you're reading this I'm sure you can figure it out for yourself anyway. I just thought it was neat.<br /><br />Also, note that you can quickly get into situations where the lake extends back behind your starting point, so if the lake is too wide, it's impossible. It'll be faster to swim across. This happens when (w/2)<sup>2</sup> + d<sup>2</sup>>4d<sup>2</sup>. If that happens, you'll need the start farther away from the lake, or make it out of something that's harder to swim through.<br /><br /><span style="font-weight: bold;">New Problem: Lifeguard<br /></span>Today's and yesterday's problems were inspired by Feynman's popular book "QED", which in turn is based on <a style="font-weight: bold;" href="http://vega.org.uk/video/subseries/8">these lectures</a>.<br /><br />You're a lifeguard who sees a swimmer drowning somewhere off shore. You need to get out to the swimmer by a combination of running on the beach and swimming in the water. See the picture:<br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://3.bp.blogspot.com/_2y3Souq-IUU/SJbA-u-0cCI/AAAAAAAABAo/0FHIgWmZrnE/s1600-h/lifeguard.jpg"><img style="margin: 0px auto 10px; display: block; text-align: center; cursor: pointer;" src="http://3.bp.blogspot.com/_2y3Souq-IUU/SJbA-u-0cCI/AAAAAAAABAo/0FHIgWmZrnE/s320/lifeguard.jpg" alt="" id="BLOGGER_PHOTO_ID_5230580201233346594" border="0" /></a><br />One particular path is fastest. Assume you run at speed n<sub>1</sub> and swim at speed n<sub>2</sub>. You take the fastest path. What is the relationship between the angle of your path on the beach relative to the normal, and your angle into the water relative to the normal?<br /><span style="font-weight: bold;"></span>Markkimarkkonnenhttp://www.blogger.com/profile/11122368349695914223noreply@blogger.com1tag:blogger.com,1999:blog-5497138109765858383.post-9557896238047755902008-08-02T22:30:00.000-07:002008-08-03T00:58:55.102-07:00Answer: Square Wheel New Problem: To Swim or Not To Swim<span style="font-weight: bold;">Square Wheel</span><br />Sorry to be lame, but I'm not going to prove this one. <a href="http://www.jstor.org/stable/2691240?seq=1">It's a catenary.</a><br /><br /><span style="font-weight: bold;">To Swim or Not to Swim</span><br />There's a lake in the middle of the field, as shown. You want to go from point A to point B as quickly as possible. You can go over land or over water, but you're twice as slow in the water as one the land.<br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://3.bp.blogspot.com/_2y3Souq-IUU/SJVigEhXnBI/AAAAAAAABAg/8W2muq8C-F0/s1600-h/lens.jpg"><img style="margin: 0px auto 10px; display: block; text-align: center; cursor: pointer;" src="http://3.bp.blogspot.com/_2y3Souq-IUU/SJVigEhXnBI/AAAAAAAABAg/8W2muq8C-F0/s320/lens.jpg" alt="" id="BLOGGER_PHOTO_ID_5230194845369736210" border="0" /></a><br />Assume (or prove) the best strategy is to head straight to some spot on the shore, swim straight across, and then head on from the far shore to point B by land. The shape of the lake is such that it doesn't matter where you choose to enter - all paths take the same amount of time. What is the shape of the lake?<br /><br />Assume: points A and B are equidistant from the center of the lake. The lake is symmetric about a vertical line through its center.Markkimarkkonnenhttp://www.blogger.com/profile/11122368349695914223noreply@blogger.com4tag:blogger.com,1999:blog-5497138109765858383.post-59247114411955153392008-08-01T13:18:00.000-07:002008-08-01T13:53:57.161-07:00Answer: Hourglass New Problem: Square Wheel<span style="font-weight: bold;">Hourglass<span style="font-weight: bold;"><br /></span></span><span>It's hard to say exactly what the plot would look like, if you try to account for things like the varying pressure on the sand, the varying height of the pile in the bottom of the hourglass, etc. We can say generally what the weight-vs-time graph looks like by finding an important property. Start with<br /><br />F = dp/dt<br />Fdt = dp<br /><br />Integrating, we find that the total force-seconds is equal to the change in momentum. But the hourglass has zero momentum when the sand is all up top, and again zero when the sand is all in the bottom. So the integral of force over time for the whole experiment must be zero.<br /><br />The force on the hourglass is from gravity and from the scale. Because the force of gravity isn't changing, that means the force read on the scale must, on time-average, be equal to the weight of the hourglass. Any "bumps" in the scale's readings must be canceled out by "valleys".<br /><br />That tells us about the entire outline of the plot. As for the details, it depends on the acceleration of the center of mass of the hourglass (F=ma). Just as the sand starts falling, the center of mass is certainly accelerating down. The force of gravity is stronger than the force exerted by the scale, and so the reading goes down for a time. When the sand hits the bottom, it accelerates up. If the rate of sand flowing were constant, and all the sand fell the same distance, and the sand in the top weren't sinking, then the total acceleration of the hourglass would be zero. There are complications so that while the sand is falling, the center of mass might have some nonzero acceleration, but it would be small.<br /><br />As the last bits of sand fall, the acceleration of the center of mass must be upwards - it's going towards a state where it's no longer falling. So the basic idea of the outline should be something like this:<br /><br /></span><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://1.bp.blogspot.com/_2y3Souq-IUU/SJN3PHUDdsI/AAAAAAAABAY/bQrppNS6Ycg/s1600-h/hourglass.jpg"><img style="margin: 0pt 10px 10px 0pt; float: left; cursor: pointer;" src="http://1.bp.blogspot.com/_2y3Souq-IUU/SJN3PHUDdsI/AAAAAAAABAY/bQrppNS6Ycg/s320/hourglass.jpg" alt="" id="BLOGGER_PHOTO_ID_5229654693852182210" border="0" /></a><br /><span><br /><span style="font-weight: bold;"><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br />Square Wheel<br /></span>A square wheel rolls on a bumpy road smoothly, and without slipping. What is the shape of the road?<br /><object height="344" width="425"><param name="movie" value="http://www.youtube.com/v/aCSGdowQlEo&hl=en&fs=1"><param name="wmode" value="transparent"><param name="allowFullScreen" value="true"><embed src="http://www.youtube.com/v/aCSGdowQlEo&hl=en&fs=1" type="application/x-shockwave-flash" allowfullscreen="true" wmode="transparent" height="344" width="425"></embed></object><span style="font-weight: bold;"><span style="font-weight: bold;"><br /></span></span></span><span style="font-weight: bold;"><span style="font-weight: bold;"></span></span>Markkimarkkonnenhttp://www.blogger.com/profile/11122368349695914223noreply@blogger.com0tag:blogger.com,1999:blog-5497138109765858383.post-7519268394128151022008-07-31T12:19:00.000-07:002008-07-31T12:23:41.179-07:00Answer: Writhing Chain New Problem: Hourglass<span style="font-weight:bold;">Writhing Chain</span><br />The speed of sound in the chain depends on its tension, the same way it does in a violin string. You can raise the pitch of a violin string by tightening it. Similarly, the speed of traveling waves increases where the chain has increasing tension. The tension in the chain is high near the top because it's supporting the weight of all the chain below it. Near the bottom, the speed is low. The chain is set to rotate at just the speed of traveling waves near the bottom of the chain. That way, patterns can persist there even as the chain cycles around.<br /><br /><span style="font-weight:bold;">Hourglass</span><br />You sit an hourglass on a scale with the sand in the top, but blocked from falling. Then you remove the block and let the sand fall down the hourglass. Make a plot of the reading on the scale as a function of time.Markkimarkkonnenhttp://www.blogger.com/profile/11122368349695914223noreply@blogger.com0