Worked Problems
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enHow often does the sun emit 1 TeV photons?
https://scienceblogs.com/builtonfacts/2013/11/27/how-often-does-the-sun-emit-1-tev-photons
<span>How often does the sun emit 1 TeV photons?</span>
<div class="field field--name-body field--type-text-with-summary field--label-hidden field--item"><p>I had an interesting question posed to me recently: how frequently does the sun emit photons with an energy greater than 1 TeV?</p>
<p>All of you know about the experiments going on at the LHC, where particles are accelerated to an energy which is equivalent to an electron being accelerated through a potential difference of trillions of volts (which is what a "trillion electron volts" - a TeV is). During the ensuing collisions between particles, high-energy TeV photons are produced. Of course everything is emitting light in the form of blackbody radiation all the time. Human beings emit mostly long-wavelength infrared, hot stoves emit shorter-wavelength infrared and red light, hotter objects like the sun emit across a broad range of wavelengths which include the entire visible spectrum. Here, from Wikipedia, is the spectrum of the sun:</p>
<p> </p>
<div style="width: 570px;display:block;margin:0 auto;"><a href="https://en.wikipedia.org/wiki/File:Solar_Spectrum.png"><img class=" wp-image-1972 " alt="Solar spectrum." src="https://scienceblogs.com/files/builtonfacts/files/2013/11/Solar_Spectrum.png" width="560" height="417" /></a> Solar spectrum.
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<p>This graph is given in terms of wavelength. For light, energy corresponds to frequency, and frequency is inversely proportional to wavelength. Longer wavelength, lower frequency. A TeV is a gigantic amount of energy, which corresponds to a gigantically high frequency and thus a wavelength that would be pegged way the heck off the left end of this chart pegged almost but not quite exactly at 0 on the x axis. Let me reproduce the same blackbody as the Wikipedia diagram, but cast in terms of frequency:</p>
<div style="width: 370px;display:block;margin:0 auto;"><a href="https://scienceblogs.com/files/builtonfacts/files/2013/11/spectrum.png"><img class="size-full wp-image-1975" alt="spectrum" src="https://scienceblogs.com/files/builtonfacts/files/2013/11/spectrum.png" width="360" height="208" /></a> Same spectrum, in terms of frequency
</div>
<p>Here the x-axis is in hertz, and the y-axis is spectral irradiance in terms of watts per square meter <em>per hertz</em>. (That <a href="https://scienceblogs.com/node/121045">makes a difference</a> - it's not just the Wikipedia graph with the x-axis relabeled although it gives the same watts-per-square-meter value when integrated over the same bandwidth region.)</p>
<p>Ok, so what's the frequency of a 1 TeV photon? Well, photon energy is given by E = hf, where h is Planck's constant and f is the frequency. Plugging in, a 1 TeV photon has an frequency of about 2.4 x 10<sup>26</sup> Hz. That's way off the right end of the graph. Thus you might think the answer is zero - the sun never emits such high-energy photons. But then again that tail never quite reaches zero, and there's a lot of TeVs per watt, and there's a lot of square meters on the sun...</p>
<p>So to find out more exactly, let's take a look at the actual equation which gave us that chart: Planck's law for blackbody radiation:</p>
<p style="text-align: center;">$latex \displaystyle B(f) = \frac{ 2 h f^3}{c^2} \frac{1}{e^\frac{h f}{kT} - 1}&s=2$</p>
<p style="text-align: left;">So you'd integrate that from 2.4 x 10<sup>26</sup> Hz to infinity if you wanted to find how many watts per square meter the sun emits at those huge frequencies. (Here k is Boltzmann's constant, which is effectively the scale factor that converts from temperature to energy.) That's kind of an ugly integral though, but we can simplify it. That $latex e^\frac{h f}{kT}$ term? It's indescribably big. The hf term is 1 TeV, and the kT is about 0.45 eV (which is a "typical" photon energy emitted by the sun), so the exponential is on the order of e<sup>2200000000000</sup>. (The number of particles in the observable universe is maybe 10<sup>80</sup> or so, for comparison.) Subtracting 1 from that gigantic number is absolutely meaningless, so we can drop it and end up with:</p>
<p style="text-align: center;">$latex \displaystyle B(f) = \frac{ 2 h f^3}{c^2} e^{-\frac{h f}{kT}}&s=2$</p>
<p style="text-align: left;">which means the answer in watts per square meter is</p>
<p style="text-align: center;">$latex \displaystyle I = \int_{a}^{\infty}\frac{ 2 h f^3}{c^2} e^{-\frac{h f}{kT}} \, df&s=2$</p>
<p style="text-align: left;">where "a" is the 1 TeV lower cutoff (in Hz). That exponential term now has a negative sign, so it's on the order of e<sup>-2200000000000</sup>. I'd say this is a safe place to stop and say "The answer is zero, the sun has never and will never emit photons of that energy through blackbody processes." But let's press on just to be safe.</p>
<p style="text-align: left;">That expression above can be integrated pretty straightforwardly. I let Mathematica do it for me:</p>
<p style="text-align: center;">$latex \displaystyle I = e^{-\frac{a h}{k T}}\frac{2 k T (a^3 h^3+3 a^2 h^2 k T+6 a h k^2 T^2+6 k^3 T^3 )}{c^2 h^3}&s=2$</p>
<p style="text-align: left;">So that's an exponential term multiplying a bunch of stuff. That bunch of stuff is a big number, because "a" is a big number and h is a tiny number in the denominator. I plug in the numbers and get that the stuff term is about 10<sup>93</sup> watts per square meter, and you have to multiply that by the 10<sup>18</sup> or so square meters on the surface of the sun. That's a very big number, but it's not even in the same sport as that e<sup>-2200000000000</sup> term. Multiplying those terms together doesn't even dent the e<sup>-2200000000000</sup> term. It's still zero for all practical purposes</p>
<p style="text-align: left;">Which is a lot of work to say that our initial intuition was correct. 1 TeV from blackbody processes in the sun? Forget it.</p>
<p style="text-align: left;">Now blackbody processes aren't the only things going on in the sun. I don't think there are too many TeV scale processes of other types, but stars can be weird things sometimes. I'd be curious to know if astrophysicists would know of other processes which might bump the TeV rate to something higher.</p>
<p style="text-align: left;">[<em>Personal note: I've been absent on ScienceBlogs since April, I think. Why? Writing my dissertation, defending, and summer interning. The upshot of all that is those things are done and I'm now Dr. Springer, and I have a potentially permanent position lined up next year. And now I might even have time to write some more!</em>]</p>
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<span><a title="View user profile." href="https://scienceblogs.com/author/mspringer" lang="" about="https://scienceblogs.com/author/mspringer" typeof="schema:Person" property="schema:name" datatype="" xml:lang="">mspringer</a></span>
<span>Wed, 11/27/2013 - 09:46</span>
Wed, 27 Nov 2013 14:46:14 +0000mspringer121051 at https://scienceblogs.comQuick, hit the brakes!
https://scienceblogs.com/builtonfacts/2013/02/27/quick-hit-the-brakes
<span>Quick, hit the brakes!</span>
<div class="field field--name-body field--type-text-with-summary field--label-hidden field--item"><p>A reader emailed me a fun question from a physics exam he took, along these lines:</p>
<blockquote><p>A car driver going at some speed v suddenly finds a wide wall at a distance r. Should he apply brakes or turn the car in a circle of radius r to avoid hitting the wall?</p></blockquote>
<p>My first thought was that surely the question wasn't doable without more information, but it turns out that we do have enough to give a straightforward answer. Let's take the "turns in a circle" and "slams on brakes" scenarios one at a time.</p>
<h4>Turns in a circle:</h4>
<p>Velocity is a vector whose magnitude is the speed and whose direction is the direction of travel. If you turn, your speed remains the same but your direction of travel changes. So the velocity is changing even if the speed isn't. A changing velocity is by definition an acceleration, and one of the key equations of first semester physics is the acceleration required to produce uniform circular motion. It turns out to be a function of the speed and the radius of the circle:</p>
<p style="text-align: center;">$latex \displaystyle a = \frac{v^2}{r} &s=1$</p>
<p>Since we don't have any numbers to plug in or really anywhere else to go with this, we're done with this part. The required acceleration to avoid the wall is equal to the square of the speed divided by the radius of the circle, which is just the initial distance to the wall.</p>
<h4>Slams on brakes:</h4>
<p>This one is a little more involved. The direction of the velocity is not changing, but the speed is. Another of the key equations of freshman physics is the formula for position in uniformly accelerated motion. It's:</p>
<p style="text-align: center;">$latex \displaystyle x = \frac{1}{2}a t^2 + v_0 t + x_0 &s=1$</p>
<p>where <em>a</em> is the acceleration, <em>v0</em> is the initial velocity, <em>x0</em> is the initial position, and <em>t</em> is the elapsed time. In this case we'd like to solve for a at the point where x = r (we define our coordinates such that x0 = 0). But we don't know how much time has elapsed by the time the car reaches the wall, so we need the formula for velocity in uniformly accelerated motion, which we might write from memory or find by differentiating the position equation if we know calculus:</p>
<p style="text-align: center;">$latex \displaystyle v = at + v_0 &s=1$</p>
<p>Now I'll start subscripting the letter f on the specific time when the car reaches the wall. We know that we've come to a stop at at that time, so we have:</p>
<p style="text-align: center;">$latex \displaystyle 0 = at_f + v_0 &s=1$</p>
<p>Which means</p>
<p style="text-align: center;">$latex \displaystyle t_f = -\frac{v_0}{a} &s=1$</p>
<p>Don't worry about the negative sign. <em>a</em> is itself negative (we're decelerating), so <em>tf</em> will be positive as well. Now that we know how much time has elapsed when the motion is complete, we can plug that into our position formula:</p>
<p style="text-align: center;">$latex \displaystyle r = \frac{1}{2}a(-\frac{v_0}{a})^2 + v_0 (-\frac{v_0}{a}) &s=1$</p>
<p>Remembering that at the wall, <em>x</em> = <em>r</em> and that we defined <em>x0</em> = 0. You can do the algebra to solve for <em>a</em>, and you'll find that</p>
<p style="text-align: center;">$latex \displaystyle a = -\frac{v_{0}^{2}}{2r} &s=1$</p>
<p>Which is (ignoring the minus sign that just tells us which way the acceleration is pointed) just half the acceleration we found for the turning scenario. So purely from a standpoint of the acceleration car tires can produce, braking works better than swerving.</p>
<div style="width: 441px;display:block;margin:0 auto;"><a href="http://www.imcdb.org/vehicle_99079-Chevrolet-4100-1947.html"><img class=" wp-image-1906 " alt="From Back to the Future - Biff shoulda braked..." src="https://scienceblogs.com/files/builtonfacts/files/2013/02/manure.jpg" width="431" height="234" /></a> From Back to the Future - Biff shoulda braked...
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<p>After writing this post, I came across a <a href="http://scienceblogs.com/dotphysics/2010/08/05/turn-or-go-straight-quick/">ScienceBlogs post on Dot Physics</a> a few years ago on the same subject. He approaches the problem in a different way, and I think it's well worth reading both solution methods.</p>
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<span><a title="View user profile." href="https://scienceblogs.com/author/mspringer" lang="" about="https://scienceblogs.com/author/mspringer" typeof="schema:Person" property="schema:name" datatype="" xml:lang="">mspringer</a></span>
<span>Wed, 02/27/2013 - 04:47</span>
Wed, 27 Feb 2013 09:47:50 +0000mspringer121047 at https://scienceblogs.comSolving the Schrodinger equation with a linear potential
https://scienceblogs.com/builtonfacts/2010/11/05/so-consider-the-one-dimensiona
<span>Solving the Schrodinger equation with a linear potential</span>
<div class="field field--name-body field--type-text-with-summary field--label-hidden field--item"><p>So consider the one-dimensional time-independent Schrodinger equation:</p>
<p><img src="http://scienceblogs.com/builtonfacts/wp-content/blogs.dir/477/files/2012/04/i-d4bb9fe09d6b78eb832985821d8a307f-1.png" alt="i-d4bb9fe09d6b78eb832985821d8a307f-1.png" /></p>
<p>In some ways it's not really an equation as such, because you have to plug in some function V(x) that describes the potential in the problem you're solving. When you first learn quantum mechanics you'll learn the big ones: V(x) = 0, V(x) = V > E, V(x) = 1/2 kx^2, V(x) = k/x, etc. One that's conspicuously absent is usually one of the simplest looking potentials: V(x) = -Fx. That's just the potential of a constant force, say a particle in a uniform gravitational field. Rarely do textbook authors bother to work out a solution - they'll just say "This can be solved with the Airy function", and maybe give the solution as an accomplished fact.</p>
<p>Which isn't too bad - in most cases if you're the kind of person who knows what an Airy function is you can probably work it out yourself. But there's something to be said for a walkthrough for those who're just getting to the point of understanding how to solve differential equations in terms of special functions. So let's do it:</p>
<p>For the Schrodinger equation with a linear potential, we have to solve:</p>
<p><img src="http://scienceblogs.com/builtonfacts/wp-content/blogs.dir/477/files/2012/04/i-88516ca505d84f969b7ab15a965c3fb3-2.png" alt="i-88516ca505d84f969b7ab15a965c3fb3-2.png" /></p>
<p>Rearrange and cancel the negative signs:</p>
<p><img src="http://scienceblogs.com/builtonfacts/wp-content/blogs.dir/477/files/2012/04/i-d18063683dcc0d42b9be45451a84d1e3-3.png" alt="i-d18063683dcc0d42b9be45451a84d1e3-3.png" /></p>
<p>This looks kinda sorta familiar. The Airy function - by defintion - is the function solving the differential equation:</p>
<p><img src="http://scienceblogs.com/builtonfacts/wp-content/blogs.dir/477/files/2012/04/i-bb5c2c6b0452df43a61e3974bd9b473f-5.png" alt="i-bb5c2c6b0452df43a61e3974bd9b473f-5.png" /></p>
<p>Our equation looks kinda sorta like that. What we'd like to do is to finagle our Schrodinger equation into that form, and then we'd know that the Airy function of some argument would be our solution. We might suspect the substitution u = Fx + E would be a good start. But that can't quite be it, because we'd still have that hbar/2m term floating out in front of the derivative. There may be some coefficient that can simultaneously get rid of the stuff in front of the derivative and turn Fx + E into just u. Let's call that coefficient (if it exists) A. We then have:</p>
<p><img src="http://scienceblogs.com/builtonfacts/wp-content/blogs.dir/477/files/2012/04/i-80f9713f6d11461837a9f9b540684e36-6.png" alt="i-80f9713f6d11461837a9f9b540684e36-6.png" /></p>
<p>By the chain rule,</p>
<p><img src="http://scienceblogs.com/builtonfacts/wp-content/blogs.dir/477/files/2012/04/i-9ef27a331268eac0961c3fd4d1a55446-7.png" alt="i-9ef27a331268eac0961c3fd4d1a55446-7.png" /></p>
<p>If the u-x relationship weren't linear we'd have to be a little careful here, but it is and so it's clear that:</p>
<p><img src="http://scienceblogs.com/builtonfacts/wp-content/blogs.dir/477/files/2012/04/i-2a911bc4888cdd6dfd35cad351fab7d5-8.png" alt="i-2a911bc4888cdd6dfd35cad351fab7d5-8.png" /></p>
<p>So we need the coefficient in front of the u to be -1, so go ahead and multiply through by -A:</p>
<p><img src="http://scienceblogs.com/builtonfacts/wp-content/blogs.dir/477/files/2012/04/i-44ec8cae73c5984c6c575d0dba7bf900-9.png" alt="i-44ec8cae73c5984c6c575d0dba7bf900-9.png" /></p>
<p>Plug that sucker in and we have:</p>
<p><img src="http://scienceblogs.com/builtonfacts/wp-content/blogs.dir/477/files/2012/04/i-c824f1ecbc6c0640a24a0c6253be158a-10.png" alt="i-c824f1ecbc6c0640a24a0c6253be158a-10.png" /></p>
<p>The solution to which is, by definition:</p>
<p><img src="http://scienceblogs.com/builtonfacts/wp-content/blogs.dir/477/files/2012/04/i-1d3223abadf16fa3303c041bf9fa4404-11.png" alt="i-1d3223abadf16fa3303c041bf9fa4404-11.png" /></p>
<p>Well, there's also the Airy function of the second kind Bi(x), but that diverges for positive arguments and usually isn't relevant. In occasional circumstances it is, and you should be aware of it.</p>
<p>I hereby declare us done. Now in practice you have to apply some boundary conditions. For instance, in solving the problem of a quantum ball bouncing on a hard surface we have the boundary condition psi(x = 0) = 0, and only particular values of E will satisfy those conditions. Those are the eigenvalues that make up the energy spectrum. That's just details though.</p>
<p>Details which you'll have to know if you're studying for exams - which are rapidly approaching if you're a student. Good luck! It's also interesting to compare this quantum result to the classical one, which we'll do shortly.</p>
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<span><a title="View user profile." href="https://scienceblogs.com/author/mspringer" lang="" about="https://scienceblogs.com/author/mspringer" typeof="schema:Person" property="schema:name" datatype="" xml:lang="">mspringer</a></span>
<span>Fri, 11/05/2010 - 06:48</span>
Fri, 05 Nov 2010 10:48:32 +0000mspringer120993 at https://scienceblogs.comA train leaves Cleveland...
https://scienceblogs.com/builtonfacts/2010/08/16/a-train-leaves-cleveland
<span>A train leaves Cleveland...</span>
<div class="field field--name-body field--type-text-with-summary field--label-hidden field--item"><p>The new academic year is starting, and if there's one thing students love it's a good word problem. <em>If Sue is four times as old as John will be when Sue is one year than John...</em> So in that spirit I was amused to find basically this kind of problem in a college physics textbook I was perusing for post ideas as I get back into the swing of blogging. It runs thusly:</p>
<blockquote><p>A father racing his son has half the kinetic energy of the son, who has half the mass of the father. The father speeds up by 1.0 m/s and then has the same kinetic energy as the son. What are the original speeds of the father and the son?</p></blockquote>
<p>You don't really need to know any physics to do this, other than just the equation for kinetic energy. The kinetic energy of an object of mass m and velocity v is:</p>
<p><img src="http://scienceblogs.com/builtonfacts/wp-content/blogs.dir/477/files/2012/04/i-d4bb9fe09d6b78eb832985821d8a307f-1.png" alt="i-d4bb9fe09d6b78eb832985821d8a307f-1.png" /></p>
<p>All right, give it a shot. Click below for the solution! (Seriously, if you have math or intro physics classes this semester it's a good warm-up)</p>
<!--more--><p>My guess is that the reason people hate these so much is that if you either do the guess/check method or just think really hard you can often get the solution. But really you don't have to do either, the point is to write the problem algebraically and then just turn the crank. That first step is what people have trouble with sometimes.</p>
<p>So initially the father has half the kinetic energy of the son. Using notation that I trust is self-explanatory, this is just:</p>
<p><img src="http://scienceblogs.com/builtonfacts/wp-content/blogs.dir/477/files/2012/04/i-88516ca505d84f969b7ab15a965c3fb3-2.png" alt="i-88516ca505d84f969b7ab15a965c3fb3-2.png" /></p>
<p>And the son has half the mass of the father:</p>
<p><img src="http://scienceblogs.com/builtonfacts/wp-content/blogs.dir/477/files/2012/04/i-d18063683dcc0d42b9be45451a84d1e3-3.png" alt="i-d18063683dcc0d42b9be45451a84d1e3-3.png" /></p>
<p>So we can write down the kinetic energy equations explicitlyl:</p>
<p><img src="http://scienceblogs.com/builtonfacts/wp-content/blogs.dir/477/files/2012/04/i-8d7070ce28f0af28047c7a5fb8246d53-4.png" alt="i-8d7070ce28f0af28047c7a5fb8246d53-4.png" /></p>
<p>Where on the right side the son has half the mass of the father (one of the 1/2s) and the father has one half the kinetic energy of the son (the other 1/2). Let this rat's nest of parentheses cancel and take the square root:</p>
<p><img src="http://scienceblogs.com/builtonfacts/wp-content/blogs.dir/477/files/2012/04/i-bb5c2c6b0452df43a61e3974bd9b473f-5.png" alt="i-bb5c2c6b0452df43a61e3974bd9b473f-5.png" /></p>
<p>So initially the son is running at twice the speed of the father. That's a good thing to know, but it doesn't quite answer the problem yet. Now we also know that when the father speeds up by 1 m/s, the kinetic energies are equal:</p>
<p><img src="http://scienceblogs.com/builtonfacts/wp-content/blogs.dir/477/files/2012/04/i-80f9713f6d11461837a9f9b540684e36-6.png" alt="i-80f9713f6d11461837a9f9b540684e36-6.png" /></p>
<p>Plugging in the stuff we already figured out:</p>
<p><img src="http://scienceblogs.com/builtonfacts/wp-content/blogs.dir/477/files/2012/04/i-1a518f2655b02f9d30ddb0c8226ad8eb-1a.png" alt="i-1a518f2655b02f9d30ddb0c8226ad8eb-1a.png" /></p>
<p>Solve that with the quadratic formula, you get that the initial velocity for the father is 2.4 m/s. The son's velocity was (as we saw) twice that, so he was going 4.8 m/s. Yay for math!</p>
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<span><a title="View user profile." href="https://scienceblogs.com/author/mspringer" lang="" about="https://scienceblogs.com/author/mspringer" typeof="schema:Person" property="schema:name" datatype="" xml:lang="">mspringer</a></span>
<span>Mon, 08/16/2010 - 04:57</span>
Mon, 16 Aug 2010 08:57:03 +0000mspringer120981 at https://scienceblogs.comAstrology and Gravity
https://scienceblogs.com/builtonfacts/2010/04/02/theres-a-classic-problem-in
<span>Astrology and Gravity</span>
<div class="field field--name-body field--type-text-with-summary field--label-hidden field--item"><p>There's a classic problem in physics textbooks which asks you about astrology. It's sometimes said - the problem will tell you - that the gravitational pull of the doctor delivering you is stronger than that of Jupiter, therefore it's unlikely that the planets are exerting a whole lot of influence on your life. The problem asks you to check this.</p>
<p>Of course gravity is not generally purported to be the conduit of the supposed influence of the zodiac, but it's an interesting problem. And in fact it will turn out that Jupiter generally has the doctor beat, but not by that much. It's a neat little order-of-magnitude demonstration of the how gravity scales.</p>
<p>Still, I'm not a huge fan of the problem for different reasons. The problem almost always just uses Newton's equation for the gravitational force due to a point object. A doctor isn't - his head is farther from the baby than his hands, and so are his feet, and so on and so forth. As such it's pretty hard to figure out what the "distance" is from the doctor to the baby. Using the center of mass would be a very good approximation if the distance was much longer than the doctor's height, but it's not.</p>
<p>So I propose we use this classic problem to learn a little more about approximations. What if we assume that instead of being a point mass, we assume the doctor's mass is evenly distributed along a line? Sure it's still a "spherical cow" type of situation, but it's a lot better than the previous point approximation. The situation will look something like this, with the blue dot representing the baby and the black line representing the doctor. The distances are labeled:</p>
<p><img src="http://scienceblogs.com/builtonfacts/wp-content/blogs.dir/477/files/2012/04/i-3378c9501b9495957ca861b600f6583a-doctor.png" alt="i-3378c9501b9495957ca861b600f6583a-doctor.png" /></p>
<p>Each little bit of mass on the line will contribute a bit of gravitational force. I've picked out one representative bit and labeled its distances specifically, and the sum of all the forces from all the bits will be the force on the baby. From Newton's law of gravitation for a point mass, the force from one bit will be equal to the gravitational constant, times the mass of the first object, times the mass of the second object, divided by the square of the distance between those objects. So calling the baby's mass M, the mass of that bit of the doctor dm, and plugging in the square of the distance:</p>
<p><img src="http://scienceblogs.com/builtonfacts/wp-content/blogs.dir/477/files/2012/04/i-86dffb9df7de0ecdc5e32a033c16f6e3-1c.png" alt="i-86dffb9df7de0ecdc5e32a033c16f6e3-1c.png" /></p>
<p>However, this is the force toward that mass dm. Because I've drawn the situation symmetrically with the baby even with the doc's center of mass, the upward component of the force from dm will be exactly canceled by the downward component of the force from the bit of mass located at -h. So only the horizontal forces contribute since they're all pulling in the same way. To get that force, we multiply by the cosine of the angle between the line r and the hypotenuse. Cosine is equal to the length of the line opposite the angle divided by the length of the hypotenuse, so plugging that in we see that the total horizontal force is:</p>
<p><img src="http://scienceblogs.com/builtonfacts/wp-content/blogs.dir/477/files/2012/04/i-d18063683dcc0d42b9be45451a84d1e3-3.png" alt="i-d18063683dcc0d42b9be45451a84d1e3-3.png" /></p>
<p>It is in general pretty difficult to add up this quantity mass by mass. We'd prefer to do it length by length, so let's express dm in terms of the mass per length - ie, the density λ = m/L so that dm = λ dh, where dh is the tiny height of that bit of the mass. Doing that and collecting terms:</p>
<p><img src="http://scienceblogs.com/builtonfacts/wp-content/blogs.dir/477/files/2012/04/i-8d7070ce28f0af28047c7a5fb8246d53-4.png" alt="i-8d7070ce28f0af28047c7a5fb8246d53-4.png" /></p>
<p>Now add up all the little lengths dh from the bottom to the top by integrating:</p>
<p><img src="http://scienceblogs.com/builtonfacts/wp-content/blogs.dir/477/files/2012/04/i-bb5c2c6b0452df43a61e3974bd9b473f-5.png" alt="i-bb5c2c6b0452df43a61e3974bd9b473f-5.png" /></p>
<p>It would be a tremendous pain to typeset the solution to this integral, as it's somewhat involved. I'll just present the solution, but for those interested in trying it yourself the best method is trig substitution. Let r = h tan(u) and go from there. After the fireworks are done, you'll get:</p>
<p><img src="http://scienceblogs.com/builtonfacts/wp-content/blogs.dir/477/files/2012/04/i-80f9713f6d11461837a9f9b540684e36-6.png" alt="i-80f9713f6d11461837a9f9b540684e36-6.png" /></p>
<p>So assuming an 8 pound baby and a 180 pound doctor with a height of 6 feet standing 1 foot from the baby, the total gravitational force is about 3.36 x 10<sup>-8</sup> newtons. Way too small to detect, but it's a real number. At its average distance from earth, Jupiter would produce a force of about 7.6 x 10<sup>-7</sup> newtons, or a little more than 20 times more force. Twenty times pretty much nothing is still pretty much nothing, but still - Jupiter wins.</p>
<p>I still can't say I'd recommend going to an astrologer though...</p>
</div>
<span><a title="View user profile." href="https://scienceblogs.com/author/mspringer" lang="" about="https://scienceblogs.com/author/mspringer" typeof="schema:Person" property="schema:name" datatype="" xml:lang="">mspringer</a></span>
<span>Fri, 04/02/2010 - 10:11</span>
Fri, 02 Apr 2010 14:11:22 +0000mspringer120950 at https://scienceblogs.comBrilliant Pebbles & A Physics Festival
https://scienceblogs.com/builtonfacts/2010/03/26/brilliant-pebbles-a-physics-fe
<span>Brilliant Pebbles & A Physics Festival</span>
<div class="field field--name-body field--type-text-with-summary field--label-hidden field--item"><p>If you happen to be in the Bryan/College Station area tomorrow, you might consider checking out the <a href="http://physicsfestival.tamu.edu/">Texas A&M Physics Festival</a>. It's sort of an open house with a ridiculous number of top-notch physics demonstrations as well as some very interesting talks. It's free! I'll be there helping out with some of the optics demos.</p>
<p>While we're here, we may as well do some physics. This problem is from Halliday, Resnick, and Walker 14.62. It asks (fitting well with the space stealth theme from earlier):</p>
<blockquote><p>One way to attack a satellite in Earth orbit is to launch a swarm of pellets in the same orbit as the satellite but in the opposite direction. Suppose a satellite in a circular orbit 500km above the Earth's surface collides with a pellet having a mass of 4g. a) What is the kinetic energy of the pellet in the reference frame of the satellite just before the collision? b) What is the ratio os this kinetic energy to the kinetic energy of a 4g rifle bullet from a modern army rifle with a muzzle speed of 950m/s?</p></blockquote>
<p>Well, we could "cheat" and use the virial theorem, but since at the intro level the students don't know that yet we'll do it the long (but still not difficult) way.</p>
<p>To find the kinetic energy we need to know the velocity of the pellet. We don't know the velocity, but we do know that the pellet is in a circular orbit. We know that for any object to be in uniform circular motion, the force on that object has to be equal to m v^2 / r, where r is the radius of the circle.</p>
<p>But we also know the equation describing the force of gravity, as given by Newton. That force has to be providing the requisite pull to keep the object in uniform circular motion:</p>
<p><img src="http://scienceblogs.com/builtonfacts/wp-content/blogs.dir/477/files/2012/04/i-d4bb9fe09d6b78eb832985821d8a307f-1.png" alt="i-d4bb9fe09d6b78eb832985821d8a307f-1.png" /></p>
<p>With M the mass of the Earth, and m the mass of the pellet. Solve for v:</p>
<p><img src="http://scienceblogs.com/builtonfacts/wp-content/blogs.dir/477/files/2012/04/i-88516ca505d84f969b7ab15a965c3fb3-2.png" alt="i-88516ca505d84f969b7ab15a965c3fb3-2.png" /></p>
<p>Now r is not 500 km. r is the radius of the orbit, which is the radius of the Earth plus 500 km. It works out to about 7.6 km/s. But we can't just get the kinetic energy from that either - we're asked for the energy in the reference frame of the satellite, which is <em>also</em> moving at 7.6 km/s in the opposite direction. The total closing speed is 15.2 km/s. The satellite "sees" itself as at rest with a projectile incoming at that closing speed. Kinetic energy is 1/2 times the mass of the projectile times the square of its closing speed, for a total of some 462 kJ. This is a pretty hefty quantity of energy. The rifle bullet postulated in the problem only has 1805 J of kinetic energy, so the orbital pellet hits with something like 256 times more energy.</p>
<p><a href="http://en.wikipedia.org/wiki/File:SDIO_KEW_Lexan_projectile.jpg"><br /><img src="http://scienceblogs.com/builtonfacts/wp-content/blogs.dir/477/files/2012/04/i-6df153fe8013b5feb3a9c8b54d5d119f-ouch.png" alt="i-6df153fe8013b5feb3a9c8b54d5d119f-ouch.png" /></a><em><br /><div style="text-align: center;">Fig 1: Aluminum block hit by 7g projectile at ~7km/s</div>
<p></p></em></p>
<p>This is why proposals for space warfare rarely involve the "explosives on rockets" paradigm which is so dominant in terrestrial warfare. At the speeds involved, adding an explosive is just an extra complication that doesn't add anything of consequence to the destructive energy available just by virtue of the orbital velocity.</p>
</div>
<span><a title="View user profile." href="https://scienceblogs.com/author/mspringer" lang="" about="https://scienceblogs.com/author/mspringer" typeof="schema:Person" property="schema:name" datatype="" xml:lang="">mspringer</a></span>
<span>Fri, 03/26/2010 - 05:32</span>
Fri, 26 Mar 2010 09:32:45 +0000mspringer120948 at https://scienceblogs.comSimple Harmonic Oscillator #1 - Differential Equation
https://scienceblogs.com/builtonfacts/2009/11/26/simple-harmonic-oscillator-1
<span>Simple Harmonic Oscillator #1 - Differential Equation</span>
<div class="field field--name-body field--type-text-with-summary field--label-hidden field--item"><p>First of all, happy Thanksgiving everyone! I hope you spend the day happily with the people you care about, and remember to spend a moment or two reflecting on the things for which you're thankful this year. Now on with the show:</p>
<p>Back when I first started writing this blog, I focused mostly on problem solving. The goal was to bridge the gap between popularization and textbook. I was always doubtful there was much of a market for this, but of course there are at least some interested people and especially since writing is so fun I was and am I'm more than happy to fill that gap. Over the last few months though, general grad student busyness has greatly reduced the time available for those kinds of posts. And so there's been more soft-physics kind of posts around here. Still interesting, I hope, but there's really no shortage of that sort of thing elsewhere. As such I'm going to try to improve the ratio of more in-depth fare a bit. I can't promise I'll be super consistent about it, but here's hoping y'all will bear with me!</p>
<p>Let's kick it off with perhaps the most important model in physics: the simple harmonic oscillator. It's ubiquitous in everything from solid state physics to quantum field theory, but when it comes right down to it, the harmonic oscillator is a spring. Its defining property is that the force acting on the spring is proportional to the displacement of the mass from equilibrium. Move the mass farther from its resting point, and the restoring force is proportionally stronger. Wikipedia has a nice image:</p>
<form mt:asset-id="22835" class="mt-enclosure mt-enclosure-image" style="display: inline;"><a href="http://scienceblogs.com/builtonfacts/2009/11/26/spring.gif"><img src="http://scienceblogs.com/builtonfacts/wp-content/blogs.dir/477/files/2012/04/i-9f57fc1fc9127f630fb3cc2cdff834f5-spring-thumb-400x90-22835.gif" alt="i-9f57fc1fc9127f630fb3cc2cdff834f5-spring-thumb-400x90-22835.gif" /></a></form>
<p>We can write "the force is proportional to the stretch" mathematically in the following way:</p>
<form mt:asset-id="22837" class="mt-enclosure mt-enclosure-image" style="display: inline;"><img src="http://scienceblogs.com/builtonfacts/wp-content/blogs.dir/477/files/2012/04/i-d4bb9fe09d6b78eb832985821d8a307f-1.png" alt="i-d4bb9fe09d6b78eb832985821d8a307f-1.png" /></form>
<p>The variable x is the position of the mass on the spring, and it's a function of time. Dots denote differentiation with respect to time, so x-dot-dot is the rate of change in the rate of change of position. Sounds bad, but that's just another name for the acceleration. The spring constant (the force produced by the spring per unit of stretch beyong equilibrium) is k, the mass of the object is m. Now if you know about solving differential equations, we can actually find the particular function x(t) that satisfies that equation. Physicists usually solve this kind of equation by the method of recognition - we've seen it so much we just know what the solution is. For those who haven't seen it so much, it's the pretty much the first thing you'll learn in differential equations class, and if you don't want to take the class it's ok because the problem is not difficult and either way I'm just going to tell you the solution. ;)</p>
<p>The solution is thus:</p>
<form mt:asset-id="22838" class="mt-enclosure mt-enclosure-image" style="display: inline;"><img src="http://scienceblogs.com/builtonfacts/wp-content/blogs.dir/477/files/2012/04/i-88516ca505d84f969b7ab15a965c3fb3-2.png" alt="i-88516ca505d84f969b7ab15a965c3fb3-2.png" /></form>
<p>Actually the solution contains k and m, but to make the equation <em>look</em> simpler, I've just substituted in omega, where it's an abbreviation for a slightly clumsier expression:</p>
<form mt:asset-id="22839" class="mt-enclosure mt-enclosure-image" style="display: inline;"><img src="http://scienceblogs.com/builtonfacts/wp-content/blogs.dir/477/files/2012/04/i-d18063683dcc0d42b9be45451a84d1e3-3.png" alt="i-d18063683dcc0d42b9be45451a84d1e3-3.png" /></form>
<p>And that's the general solution, for arbitrary constants A and B. Now really that's not good enough. We might have started this oscillator off close or far from equilibrium, or we might have just let it go or given it a good shove. These initial conditions - the initial position and initial velocity - determine what A and B are for our specific physical situation. Let's go ahead and nail the situation down. Call the initial position x0. That's the value of the position at t = 0 when the clock started and the system started oscillating. At t = 0, the value being plugged into sin and cos is 0. But sin(0) = 0 and cos(0) = 1, so we see that x0 = A. Nice, that pegs one of our constants. Now do the other. To find the velocity from the position equation, we differentiate with respect to time. Doing this with the knowledge of A, we see that:</p>
<form mt:asset-id="22840" class="mt-enclosure mt-enclosure-image" style="display: inline;"><img src="http://scienceblogs.com/builtonfacts/wp-content/blogs.dir/477/files/2012/04/i-8d7070ce28f0af28047c7a5fb8246d53-4.png" alt="i-8d7070ce28f0af28047c7a5fb8246d53-4.png" /></form>
<p>Now we're working with initial conditions, so set t = 0 again. The sin term goes to zero, the cos term goes to 1, and therefore the initial velocity v0 = B*omega. Solve for B, substitute into the general solution:</p>
<form mt:asset-id="22841" class="mt-enclosure mt-enclosure-image" style="display: inline;"><img src="http://scienceblogs.com/builtonfacts/wp-content/blogs.dir/477/files/2012/04/i-bb5c2c6b0452df43a61e3974bd9b473f-5.png" alt="i-bb5c2c6b0452df43a61e3974bd9b473f-5.png" /></form>
<p>Ok, that's great but what does it mean? For starters, it means that the simple harminic oscillator <em>oscillates</em>. It repeats its motion over and over with an angular frequency of omega. We might have guessed that it would oscillate, but thanks to the math we know it does so in an exactly sinusoidal way. Further, we now know the timing of this oscillation in terms of other physical constants, and we can relate amplitudes and velocities to the initial conditions. It's quite a bit of an improvement over the qualitative "back and forth" description we might have managed without math.</p>
<p>Why is this post given a #1 in the title, by the way? It's because there's about a zillion different and important ways to so physics with the harmonic oscillator. Even from a purely classical perspective, there's this, the Lagrangian formulation, the Hamiltonian formulation, the Poisson bracket formulation, action-angle variables, you name it. Plenty of these are graduate level, but I think I can make the interesting in a guided-tour way for those who aren't fluent in math-speak. I plan to tackle many of these methods over the next weeks.</p>
</div>
<span><a title="View user profile." href="https://scienceblogs.com/author/mspringer" lang="" about="https://scienceblogs.com/author/mspringer" typeof="schema:Person" property="schema:name" datatype="" xml:lang="">mspringer</a></span>
<span>Thu, 11/26/2009 - 04:00</span>
Thu, 26 Nov 2009 09:00:59 +0000mspringer120896 at https://scienceblogs.comLongfellow, Wellington, and Coriolis.
https://scienceblogs.com/builtonfacts/2009/10/07/longfellow-wellington-and-cori
<span>Longfellow, Wellington, and Coriolis.</span>
<div class="field field--name-body field--type-text-with-summary field--label-hidden field--item"><p><em>I shot an arrow into the air,<br />
It fell to earth, I knew not where;<br />
For, so swiftly it flew, the sight<br />
Could not follow it in its flight.<br /></em><br />
- Henry Wadsworth Longfellow</p>
<p>This is what happens when you don't pay attention in freshman physics. You don't need to see where the arrow goes to find out where it lands, you just need the initial speed and angle. Two simple accelerated motion equations later, you have the range. The method isn't perfect since it fails to account for air resistance and other factors, but for finding an arrow it's a lot better than shrugging and giving up once you lose sight of it.</p>
<p>In the early days of physics the question of refining the predictions of projectile motion was not merely one of academic or poetic interest. If you're commander Wellington of the noble British army wishing to lob cannon shells at the nefarious French (or vice versa), these other more delicate physics corrections to the projectile equations can mean the difference between life and death - or at least which side experiences which.</p>
<p>One of these corrections is due to the fact that the earth is not standing still. When a projectile leaves the ground, the earth keeps on rotating right under it. While this is a tiny effect for (say) a line drive in a baseball stadium, it can be substantial for artillery firing over longer ranges. The effect is called the Coriolis force, with the usual caveat that it's not a force as such but rather a "fictitious" force due to the fact that the earth is an accelerated coordinate system.</p>
<p>The acceleration of the projectile from the path it would have taken on a non-rotating earth is to first order:</p>
<form mt:asset-id="20364" class="mt-enclosure mt-enclosure-image" style="display: inline;"><img src="http://scienceblogs.com/builtonfacts/wp-content/blogs.dir/477/files/2012/04/i-d4bb9fe09d6b78eb832985821d8a307f-1.png" alt="i-d4bb9fe09d6b78eb832985821d8a307f-1.png" /></form>
<p>Where v is the velocity in the inertial frame. For simplicity, let's assume we're just throwing the projectile straight up into the air. On a stationary earth, it would just land exactly where it started. On the actual moving earth, it'll have that additional acceleration deflecting it to the side as a function of its velocity. Plug in the velocity in the above equation:</p>
<form mt:asset-id="20367" class="mt-enclosure mt-enclosure-image" style="display: inline;"><img src="http://scienceblogs.com/builtonfacts/wp-content/blogs.dir/477/files/2012/04/i-88516ca505d84f969b7ab15a965c3fb3-2.png" alt="i-88516ca505d84f969b7ab15a965c3fb3-2.png" /></form>
<p>Now omega is along the earth's axis, and z is the upward direction at the point of the throw. Thus the cross product will be in the y direction, defining y as east:</p>
<form mt:asset-id="20368" class="mt-enclosure mt-enclosure-image" style="display: inline;"><img src="http://scienceblogs.com/builtonfacts/wp-content/blogs.dir/477/files/2012/04/i-d18063683dcc0d42b9be45451a84d1e3-3.png" alt="i-d18063683dcc0d42b9be45451a84d1e3-3.png" /></form>
<p>Integrate that twice to get the position:</p>
<form mt:asset-id="20369" class="mt-enclosure mt-enclosure-image" style="display: inline;"><img src="http://scienceblogs.com/builtonfacts/wp-content/blogs.dir/477/files/2012/04/i-8d7070ce28f0af28047c7a5fb8246d53-4.png" alt="i-8d7070ce28f0af28047c7a5fb8246d53-4.png" /></form>
<p>Though there's not much to be gained by going through the algebra directly, it's not so difficult to find t and the initial velocity in terms of h. Doing this, we get a total deflection of:</p>
<form mt:asset-id="20370" class="mt-enclosure mt-enclosure-image" style="display: inline;"><img src="http://scienceblogs.com/builtonfacts/wp-content/blogs.dir/477/files/2012/04/i-bb5c2c6b0452df43a61e3974bd9b473f-5.png" alt="i-bb5c2c6b0452df43a61e3974bd9b473f-5.png" /></form>
<p>in the westerly direction. Theta is the angle between omega and your local z - in other words, your <a href="http://en.wikipedia.org/wiki/Colatitude">colatitude</a>. Plugging in a value of h = 100 meters, the earth's rotational angular frequency, and my colatitude of roughly 60 degrees, we get a total deflection of 7.6 centimeters. Very small considering the total distance of travel, but not microscopic either. It would be especially difficult to test due to the difficulty of throwing a projectile exactly straight upward.</p>
<p>The experiment has been done with the much more simple method of dropping a projectile off of a tall building, therefore automatically causing it to fall exactly straight. Bizarrely, if you work out the calculation the total deflection is exactly 1/8th of what we get for the up-and-down case - in the opposite direction (east). Nobody said being a modern major general would be easy.</p>
<p>(This problem more-or-less happens to be <a href="http://www.amazon.com/gp/product/0486432610?ie=UTF8&tag=buionfac-20&linkCode=as2&camp=1789&creative=9325&creativeASIN=0486432610">Fetter and Walecka</a><img src="http://www.assoc-amazon.com/e/ir?t=buionfac-20&l=as2&o=1&a=0486432610" width="1" height="1" border="0" alt="" style="border:none !important; margin:0px !important;" /> 2.5. Great book, btw.)</p>
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<span><a title="View user profile." href="https://scienceblogs.com/author/mspringer" lang="" about="https://scienceblogs.com/author/mspringer" typeof="schema:Person" property="schema:name" datatype="" xml:lang="">mspringer</a></span>
<span>Wed, 10/07/2009 - 06:50</span>
Wed, 07 Oct 2009 10:50:02 +0000mspringer120870 at https://scienceblogs.comA quick contour integral.
https://scienceblogs.com/builtonfacts/2009/09/21/a-quick-contour-integral
<span>A quick contour integral.</span>
<div class="field field--name-body field--type-text-with-summary field--label-hidden field--item"><p><em>Note to the reader: this post is relatively stiff mathematically. For those not mathematically inclined, I think you might enjoy reading it anyway and enjoying it as you would a tour of a widget factory; even if you're not worried about the details of the nuts and bolts, it's fun to see it done.</em></p>
<p>While watching football yesterday, I got a phone call from a great friend whom I first met in our halcyon high school days. He's a fellow physics student a few years younger than me, calling to discuss a problem in contour integration. Most people like watching football more than they like doing contour integrals, but not me. And I like watching football*, so that's saying something.</p>
<p>This one is Arfken and Weber 6.3.4. It effectively asks us to compute this integral:</p>
<form mt:asset-id="19369" class="mt-enclosure mt-enclosure-image" style="display: inline;"><img src="http://scienceblogs.com/builtonfacts/wp-content/blogs.dir/477/files/2012/04/i-d18063683dcc0d42b9be45451a84d1e3-3.png" alt="i-d18063683dcc0d42b9be45451a84d1e3-3.png" /></form>
<p>Where C is a circle in the complex plane, centered on the origin, with a radius strictly greater than 1. (Note: the problem is given before the section on the residue theorem, so we're going to evaluate the integral directly rather than by the more direct method of residues.) </p>
<p>A person just encountering this integral might think "Ah! It's a closed curve, so Cauchy's integral theorem guarantees the answer is 0." Unfortunately the theorem only holds for a function which is analytic within the closed curve, and this one isn't. The denominator is zero for z = 0 and z = -1. Both of those are inside the curve, so the function isn't analytic in our region and the theorem doesn't apply. We can fix this problem by finagling our region in the following way: instead of just that one circle, we can dent the circle such that we dodge around those bad points and have a region inside which our function is analytic. The integral for our modified contour is thus 0 by Cauchy's theorem. I propose some dents that look like this:</p>
<form mt:asset-id="19370" class="mt-enclosure mt-enclosure-image" style="display: inline;"><img src="http://scienceblogs.com/builtonfacts/wp-content/blogs.dir/477/files/2012/04/i-d3fdbda776403268fadccbd8598f89a3-graph3.png" alt="i-d3fdbda776403268fadccbd8598f89a3-graph3.png" /></form>
<p>This modified contour is composed of three sub-contour circles labeled C1, C2, and C3. They're connected by straight lines. The complete contour fails to enclose any of the non-analytic points, so the integral over this modified contour is certainly 0 no matter how weird it looks.</p>
<p>Now the paths connecting the three circle regions have an interesting and important property. In the limit that the lines come very close together leaving the C1, C2, and C3 circles gap-free, the paths will overlap each other going in opposite directions. This means whatever the value of the integral over the ingoing line happens to be, it will be exactly the negative of its corresponding outgoing line integral. Thus the total contribution to the integral from the lines is zero. As such we can scrap the lines completely, as our single weird path is this exactly the same as the sum of these three sub-paths:</p>
<form mt:asset-id="19384" class="mt-enclosure mt-enclosure-image" style="display: inline;"><img src="http://scienceblogs.com/builtonfacts/wp-content/blogs.dir/477/files/2012/04/i-d8107f2fa66bfd8e5e6f522cd5b441bc-graph4.png" alt="i-d8107f2fa66bfd8e5e6f522cd5b441bc-graph4.png" /></form>
<p>As a quick summary of what's important so far: the total integral of the weird path is zero, and the straight lines contribute nothing at all. Thus the sum of the the integrals over the circles is zero. C1 + C2 + C3 = 0,if we take each term to be the value of the integral over that path. Important: notice that to be consistent with the weird path, the interior cicles are integrated in the opposite direction as the exterior circle. This is easy, it's just an overall minus sign.</p>
<p>So let's evaluate the interior circle integrals. First, C2. Let's make our lives easier by breaking up the fraction in the integral by means of partial fraction expansion. (You probably already know how to do this; if not, don't worry. It's in any calculus book and it's super easy.)</p>
<form mt:asset-id="19385" class="mt-enclosure mt-enclosure-image" style="display: inline;"><img src="http://scienceblogs.com/builtonfacts/wp-content/blogs.dir/477/files/2012/04/i-bb5c2c6b0452df43a61e3974bd9b473f-5.png" alt="i-bb5c2c6b0452df43a61e3974bd9b473f-5.png" /></form>
<p>The second integral on the right is zero over the contour C2. Why? It's analytic in that region: C2 doesn't contain z = -1, which is where that integral fails to be analytic. Thus we only have to evaluate the first integral on the right.</p>
<p>It's a fundamental fact in complex analysis that a complex number z can be written in a polar form, where:</p>
<form mt:asset-id="19386" class="mt-enclosure mt-enclosure-image" style="display: inline;"><img src="http://scienceblogs.com/builtonfacts/wp-content/blogs.dir/477/files/2012/04/i-1a518f2655b02f9d30ddb0c8226ad8eb-1a.png" alt="i-1a518f2655b02f9d30ddb0c8226ad8eb-1a.png" /></form>
<p>Where r is the distance from the origin and theta is the angle with respect to the positive x axis, as usual. We're going to want to make a substitution, so we also differentiate this to find:</p>
<form mt:asset-id="19387" class="mt-enclosure mt-enclosure-image" style="display: inline;"><img src="http://scienceblogs.com/builtonfacts/wp-content/blogs.dir/477/files/2012/04/i-ec6170d6ae0330319bd539b5264503e8-2a.png" alt="i-ec6170d6ae0330319bd539b5264503e8-2a.png" /></form>
<p>We're integrating over a circle, so theta runs from 0 to 2 pi. The radius r of the tiny circle we're free to specify, but it won't matter. Substitute into the integral:</p>
<form mt:asset-id="19397" class="mt-enclosure mt-enclosure-image" style="display: inline;"><img src="http://scienceblogs.com/builtonfacts/wp-content/blogs.dir/477/files/2012/04/i-5e62ab89294132669fadf517e98dfbb8-3a.png" alt="i-5e62ab89294132669fadf517e98dfbb8-3a.png" /></form>
<p>The negative sign is because we're integrating clockwise, remember? Now do the same thing for C3:</p>
<form mt:asset-id="19389" class="mt-enclosure mt-enclosure-image" style="display: inline;"><img src="http://scienceblogs.com/builtonfacts/wp-content/blogs.dir/477/files/2012/04/i-31a7f984f6bf3bd3a24b94b10ad187cf-4a.png" alt="i-31a7f984f6bf3bd3a24b94b10ad187cf-4a.png" /></form>
<p>As with the similar argument for C2, the first integral on the right is zero - because C3 doesn't enclose the bad point z = 0. Therefore we have only to evaluate that second term on the right. To do it, make a substitution. Let u = z + 1, which will also give us du = dz. This turns this integral into the exact same one as the integral we just did, modulo the extra minus sign from the partial fraction. Minus a minus is plus, so:</p>
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<p>All right. We're almost done. We know C1 + C2 + C3 = 0. We know C2 = -2πi. We know C3 = 2πi. Therefore:</p>
<p>C1 -2πi + 2πi = 0</p>
<p>And thus C1 = 0. C1 is the same contour as our original un-dented C, so the answer to our original problem is 0. This looks difficult, but that's just because I've explained it in laborious detail. As far as contour integral problems go this one is pretty elementary. People write dissertations on the hard ones. Seriously - a person in my group is about to get a Ph.D. largely from work on a particularly hair-raising contour integral in the theory of classical electromagnetism.</p>
<p>*All three of my favorite teams - LSU, Texas A&M, and the New Orleans Saints - are undefeated at the moment. I don't expect this to last, but I'll take it while I've got it.</p>
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<span><a title="View user profile." href="https://scienceblogs.com/author/mspringer" lang="" about="https://scienceblogs.com/author/mspringer" typeof="schema:Person" property="schema:name" datatype="" xml:lang="">mspringer</a></span>
<span>Mon, 09/21/2009 - 03:00</span>
Mon, 21 Sep 2009 07:00:59 +0000mspringer120862 at https://scienceblogs.comDiavolo!
https://scienceblogs.com/builtonfacts/2009/08/31/diavolo
<span>Diavolo!</span>
<div class="field field--name-body field--type-text-with-summary field--label-hidden field--item"><p>This guy is Allo Diavolo:</p>
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<p>He was a circus daredevil. At the dawn of the 20th century he worked on a number of stunts dressed in his ominous horned outfit. These days a lot of people, including me, have heard of him as an example in the pages of physics textbooks. In my case it was <a href="http://www.amazon.com/gp/product/0471758019?ie=UTF8&tag=buionfac-20&linkCode=as2&camp=1789&creative=9325&creativeASIN=0471758019">Halliday and Resnick</a><img src="http://www.assoc-amazon.com/e/ir?t=buionfac-20&l=as2&o=1&a=0471758019" width="1" height="1" border="0" alt="" style="border:none !important; margin:0px !important;" />, a standard (and good) freshman physics text. Diavolo did a trick where he looped the loop riding a bicycle. It's pretty impressive:</p>
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<p>The breathless circus ad copy claims that this stunt is more or less certain death. It wasn't, of course. I wouldn't want to try it, but it's certainly a physical possibility. How?</p>
<p>As it happens, it's a basic fact of classical mechanics at an object in instantaneous uniform circular motion will undergo an acceleration. After all, an acceleration is just change in velocity, and velocity is not just a speed. It's a speed in a particular direction. If the direction is changing, the velocity is changing. If the velocity is changing, there's an acceleration. Newton was the guy who figured out acceleration hundreds of years ago. One of his famous laws of motion is that force equals mass times acceleration. Therefore if we know something about the forces and accelerations acting on Diavolo, we can say something about the requirements to do this trick without going splat.</p>
<p>An object in uniform circular motion is undergoing an acceleration equal to:</p>
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<p>The velocity is v, the radius of the circle is r. The bicycle's motion isn't quite uniform; the speed is changing along with the direction. But this is ok, our equation remains valid for circular motion as long as we're interested in the component of the acceleration that's perpendicular to the surface of the track. And that's what we're after. We want to see whether he stays on it or not. To do so, he needs that acceleration we wrote above.</p>
<p>There's two forces acting on him. One is the force of gravity. It's equal to mg, where m is his mass and g is the acceleration due to gravity at the earth's surface. The other force is the so-called normal force, which is just the force the track exerts on him as it prevents him from ghosting through the surface. If his speed is just barely enough to prevent a fall, the normal force will be zero. Therefore the only force acting is the mg gravity, which will as Newton said be equal to the mass times acceleration:</p>
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<p>Do a little algebra jazz and find v:</p>
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<p>Halliday and Resnick estimate the radius of the loop at 2.7 meters, which plugged into the above equation give a minimum speed of about 5 m/s, or around 11.2 miles per hour at the top of the loop.</p>
<p>What makes it especially tough to do in practice is that this translates to a considerably higher speed at the bottom of the loop. There the normal force is going to be quite high, and it's the normal force that we think of and feel as the "g-force". Keeping balanced and steering while experiencing these powerful and rapidly changing forces is quite the trick.</p>
<p>How about we make this a quick little Monday exercise? Using the equations for <a href="http://en.wikipedia.org/wiki/Kinetic_energy#Newtonian_kinetic_energy">kinetic energy</a> and <a href="http://en.wikipedia.org/wiki/Potential_energy#Gravitational_potential_energy">gravitational potential energy</a>, estimate his speed at the bottom of the loop. From there, repeat the procedure we used above to find the g-force. (Hint: g-force is N/m). Good luck!</p>
<p>(Slight personal ad: It's the beginning of a new semester, and that means it's time to be violated by horrible textbook prices! As always, one good way to wave some money is to buy online. If do do so via Amazon.com, consider getting there via the link on the left sidebar. At zero extra cost to you, your purchase will help support me financially. For science!)</p>
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<span><a title="View user profile." href="https://scienceblogs.com/author/mspringer" lang="" about="https://scienceblogs.com/author/mspringer" typeof="schema:Person" property="schema:name" datatype="" xml:lang="">mspringer</a></span>
<span>Mon, 08/31/2009 - 04:00</span>
Mon, 31 Aug 2009 08:00:59 +0000mspringer120848 at https://scienceblogs.com