mimoSite - PhysicsProgramming and some other things
http://mimosite.com/blog/
http://www.rssboard.org/rss-specificationBlogEngine.NET 2.7.0.0en-UShttp://mimosite.com/blog/opml.axdhttp://www.dotnetblogengine.net/syndication.axdMy namemimoSite0.0000000.000000Learning quantum physics<p>I just finished Coursera’s <a href="https://www.coursera.org/course/eqp" target="_blank">Exploring Quantum Physics</a>. It was pretty good: the Coursera Web sites works well, both lectures and homework exercises were good and the discussion forums were helpful. </p> <p>The course starts from the basics, but it is not an introductory course: it requires a decent mathematical background (linear algebra and calculus, Fourier transforms) and some previous knowledge of quantum physics is useful, although not strictly necessary.</p> <p>The course is taught by two teachers with very different styles covering quite different material, this makes it a bit disjointed. I would have preferred a single teacher thorough the whole course. </p> <p>The program covers various advanced subjects like the Feynman path integral, quantum localization, superconductivity, the Dirac equation, the time-dependent Schrodinger equation and so on. The problem is that having to start more or less from the basics it is impossible to cover any of these advanced subjects in any detail: they are just introduced but not explained enough to get a real understanding. I think it would be better to limit the course to the more ‘standard’ introductory stuff, with maybe just one or two advanced topics covered in more details.</p> <p>Through a post in one of the Coursera’s forum I discovered the <a href="http://theoreticalminimum.com/" target="_blank">‘Theoretical minimum’</a> Stanford lectures. It is not a course but just a collection of live lectures by the same professor. I watched a couple of them on <a href="http://theoreticalminimum.com/courses/advanced-quantum-mechanics/2013/fall" target="_blank">quantum field theory</a>, the first <a href="http://theoreticalminimum.com/courses/particle-physics-1-basic-concepts/2009/fall" target="_blank">particle physics series</a> and now I am watching <a href="http://theoreticalminimum.com/courses/quantum-entanglement/2006/fall" target="_blank">the quantum entanglement ones</a>. </p> <p>The lectures are for a general audience, so they usually start from the very basics and then build the whole theory step by step. The teacher – professor Susskind – manages to introduce very complex stuff in an approachable way.</p> <p>Being live lectures, there tend to be quite a lot of question from the audience, backtracking and interruptions – these makes them much longer that the recorded Coursera lectures, but I found that this makes the material more easily absorbed (by me at least).</p> <p>They are very good introduction to modern physics for anyone with some scientific background and enough time to watch an entire series: watching individual lectures would probably not work.</p>
http://mimosite.com/blog/post/2013/12/07/Learning-quantum-physics
http://mimosite.com/blog/post/2013/12/07/Learning-quantum-physics#commenthttp://mimosite.com/blog/post.aspx?id=e72e1a0b-c653-4dc4-9915-5c01f4f7cbc9Sat, 07 Dec 2013 14:51:13 -0700OpinionPhysicsadminhttp://mimosite.com/blog/pingback.axdhttp://mimosite.com/blog/post.aspx?id=e72e1a0b-c653-4dc4-9915-5c01f4f7cbc91http://mimosite.com/blog/trackback.axd?id=e72e1a0b-c653-4dc4-9915-5c01f4f7cbc9http://mimosite.com/blog/post/2013/12/07/Learning-quantum-physics#commenthttp://mimosite.com/blog/syndication.axd?post=e72e1a0b-c653-4dc4-9915-5c01f4f7cbc9Relativistic billiard problems<p>I did not find anything ‘pre cooked’ to convert my billiard simulator to use special relativity, so I started to work on the necessary math from scratch – that turned out to be problematic (to say the least).</p> <p>For example, the collision of a ball with the side of the table become something like this:</p> <p><a href="http://mimosite.com/blog/image.axd?picture=ballside.png"><img title="ballside" style="border-left-width: 0px; border-right-width: 0px; background-image: none; border-bottom-width: 0px; padding-top: 0px; padding-left: 0px; display: inline; padding-right: 0px; border-top-width: 0px" border="0" alt="ballside" src="http://mimosite.com/blog/image.axd?picture=ballside_thumb.png" width="237" height="244" /></a></p> <p>because – in the frame of reference of the table – the ball contracts along its direction of motion. This deformation affects the computation of the collision time and changes the dynamic of the collision. The same collision seen in the frame of reference of the ball becomes instead:</p> <p><a href="http://mimosite.com/blog/image.axd?picture=ballside2.png"><img title="ballside2" style="border-top: 0px; border-right: 0px; background-image: none; border-bottom: 0px; padding-top: 0px; padding-left: 0px; border-left: 0px; display: inline; padding-right: 0px" border="0" alt="ballside2" src="http://mimosite.com/blog/image.axd?picture=ballside2_thumb.png" width="244" height="225" /></a></p> <p>because now it is the distance from the ball to the side that contracts, changing the collision angle. </p> <p>A ball-ball collision is even worse, being something like this (in the table frame of reference):</p> <p><a href="http://mimosite.com/blog/image.axd?picture=ballball.png"><img title="ballball" style="border-top: 0px; border-right: 0px; background-image: none; border-bottom: 0px; padding-top: 0px; padding-left: 0px; border-left: 0px; display: inline; padding-right: 0px" border="0" alt="ballball" src="http://mimosite.com/blog/image.axd?picture=ballball_thumb.png" width="244" height="155" /></a></p> <p>I found some pretty <a href="http://www.oberlin.edu/physics/dstyer/Modern/RelativisticDynamics.pdf" target="_blank">good notes on relativistic dynamics</a> by an Oberlin college professor. The last chapter has a section about ‘hard sphere forces’ – e.g. billiard balls – that concludes that the whole idea of handling them within special relativity is ‘ludicrous’. At the same  time I found <a href="http://arxiv.org/abs/1105.3899" target="_blank">one article</a> about rigid bodies in special relativity. I think that some kind of solution should be possible, but I have to study more. </p> <p>For the moment being I am going through the relativistic dynamics notes (including the exercises!) and then I’ll have a look at the rigid body article.</p>
http://mimosite.com/blog/post/2013/09/18/Relativistic-billiard-problems
http://mimosite.com/blog/post/2013/09/18/Relativistic-billiard-problems#commenthttp://mimosite.com/blog/post.aspx?id=98447960-5f63-4299-b1e0-8d64cf2a86eaWed, 18 Sep 2013 07:57:08 -0700Physicsadminhttp://mimosite.com/blog/pingback.axdhttp://mimosite.com/blog/post.aspx?id=98447960-5f63-4299-b1e0-8d64cf2a86ea1http://mimosite.com/blog/trackback.axd?id=98447960-5f63-4299-b1e0-8d64cf2a86eahttp://mimosite.com/blog/post/2013/09/18/Relativistic-billiard-problems#commenthttp://mimosite.com/blog/syndication.axd?post=98447960-5f63-4299-b1e0-8d64cf2a86eaRelativistic collision in one dimension<p>I want to modify my <a href="http://mimosite.com/blog/rpool/" target="_blank">billiard simulator</a> to use special relativity instead than classical mechanics. The idea is to add the speed of light as a parameter, so that it is possible to see the relativistic effects increase as the speed of light decreases.</p> <p>I need to derive the relativistic formulas for ball-side and ball-ball collisions, as well as adjusting the collision detection to account for relativistic time dilation and length contraction. I did not find anything much ‘ready made’ in various books and papers found on the Internet, so I’ll start working out the simple cases and (hopefully) build from that.</p> <p>The simplest collision is ball-ball head-on:</p> <p><a href="http://mimosite.com/blog/image.axd?picture=onedimension_1.png"><img title="onedimension" style="border-top: 0px; border-right: 0px; background-image: none; border-bottom: 0px; padding-top: 0px; padding-left: 0px; border-left: 0px; display: inline; padding-right: 0px" border="0" alt="onedimension" src="http://mimosite.com/blog/image.axd?picture=onedimension_thumb_1.png" width="301" height="128" /></a></p> <p>It can be treated as one-dimensional, ignoring the vertical coordinate. The energy-momentum conservation equations are:</p> <p>$$ \frac{E}{c^2} = m_0\gamma_0 + m_1\gamma_1 = m’_0\gamma’_0 + m’_1\gamma’_1 = \frac{E’}{c^2} $$</p> <p>$$ p = m_0 \gamma_0 v_0 + m_1 \gamma_1 v_1 = m’_0 \gamma’_0 v’_0 + m’_1 \gamma’_1 v’_1 = p’$$</p> <p>where \(m_0\) and \(m_1\) are the rest masses of the two balls and </p> <p>$$ \gamma_0 = \frac{1}{\sqrt{1 - v_0^2 / c^2}},  \gamma_1 = \frac{1}{\sqrt{1 - v_1^2 / c^2}}$$</p> <p>with \(c\) the speed of light. The primed symbols indicate the same quantities after the collision. </p> <p>The collision can be not perfectly elastic, and so the rest masses of the balls can change, i.e. some of the kinetic energy gets converted into internal energy of the balls – hence \(m’_0\) and \(m’_1\) instead than just \(m_0\), \(m_1\) in the right-hand sides of the conservation equations.</p> <p>These equations are difficult to solve for \( v’_0 \) and \( v’_1 \). The standard way to simplify them is to switch to the rest frame of the two-balls system, i.e. the frame of reference where the total momentum is zero. Indicating with \(\bar v_0\) and \(\bar v_1\) the ball velocities in this frame of reference, the momentum conservation equation becomes:</p> <p>$$ m_0 \bar\gamma_0 \bar v_0 + m_1 \bar \gamma_1 \bar v_1 = 0 = m’_0 \bar\gamma’_0 \bar v’_0 + m’_1 \bar\gamma’_1 \bar v’_1 $$</p> <p>(note that the rest masses are Loretz-invariant, so they stay the same). This equation has solutions</p> <p>$$ m’_0 \bar\gamma’_0 \bar v’_0 = -\epsilon m_0 \bar\gamma_0 \bar v_0 $$</p> <p>$$ m’_1 \bar\gamma’_1 \bar v’_1 = -\epsilon m_1 \bar\gamma_1 \bar v_1 $$</p> <p>where \(\epsilon\) is a positive constant, and the minus sign indicates that the velocities change direction after the collision. Even with this simplification, the single conservation of energy equation is not enough to determine both the rest masses and the velocities after the collision, so we need to make some further assumption.  In the classical, perfectly elastic case the energy of each ball in the zero momentum frame of reference is conserved after the collision. Assuming that this is still the case in a relativistic collision gives the equations:</p> <p>$$ m_0 \bar\gamma_0 = m’_0 \bar\gamma’_0 $$</p> <p>$$ m_1 \bar\gamma_1 = m’_1 \bar\gamma’_1 $$</p> <p>Solving for \( \bar v’_0 \), \(m’_0\) and \( \bar v’_1 \), \(m’_1\) gives:</p> <p>$$ \bar v’_0 = –\epsilon \bar v_0 $$</p> <p>$$ \bar v’_1 = -\epsilon \bar v_1 $$</p> <p>$$ m’_0 = m_0 \frac{\bar\gamma_0}{\bar\gamma’_0} = m_0 \sqrt\frac{1-\epsilon^2 \bar v_0^2 / c^2}{1- \bar v_0^2 / c^2} $$</p> <p>$$ m’_1 = m_1 \frac{\bar\gamma_1}{\bar\gamma’_1} = m_1 \sqrt\frac{1-\epsilon^2 \bar v_1^2 / c^2}{1- \bar v_1^2 / c^2} $$</p> <p>The first two equations are exactly as in the classical case, with \(\epsilon\) being the coefficient of restitution.</p> <p>To compute the final velocities in the frame of reference of the billiard table it is necessary to convert the initial velocities to the zero-momentum frame of reference, apply the formulas above and then convert back the results to the original frame of reference. </p> <p>The Lorentz transformation of the momentum \(p\) of a body with energy \(E\) to a frame of reference with velocity \(u\) is:</p> <p>$$ p’ = \gamma_u (p – \frac{u}{c^2}E) $$</p> <p>so the velocity of the zero-momentum frame of reference is </p> <p>$$ \begin{equation}\tag{U} u = \frac{p}{E/c^2} = \frac{m_0\gamma_0 v_0 + m_1 \gamma_1 v_1}{m_0 \gamma_0 + m_1 \gamma_1} \end{equation}$$</p> <p>and the velocity transformations are:</p> <p>$$ \bar v = \frac{v-u}{1-vu / c^2} $$</p> <p>$$ v’ = \frac{\bar v’+u}{1+\bar v’u / c^2} $$</p> <p>hence:</p> <p>$$ \begin{equation}\tag{V} v’ = \frac{u(1-u v / c^2) + \epsilon (u – v)}{1- u v / c^2+\epsilon u / c^2 (u-v)} \end{equation} $$</p> <p>$$ \begin{equation}\tag{M} m’ = m \sqrt{\frac{1 – 2(1-\epsilon^2)vu/c^2 – \epsilon^2 (v^2 + u^2) / c^2 + v^2u^2/c^4}{1- (v^2+u^2)/c^2 + v^2u^2/c^4}} \end{equation} $$</p> <h4><font style="font-weight: bold">Classical limit</font></h4> <p>In the classical limit (U) becomes:</p> <p>$$ u = \frac{m_0 v_0 + m_1 v_1}{m_0 + m_1} $$</p> <p>and (V) becomes:</p> <p>$$ v’ = (1 + \epsilon) u – \epsilon v$$</p> <p>combining the two:</p> <p>$$ v’ = \frac {1+\epsilon} {m_0+m_1} [m_0 (v_0 – v) + m_1 (v_1 – v)] + v$$</p> <p>that gives:</p> <p>$$ v’_0 = –m_1 \frac {1+\epsilon} {m_0 + m_1} (v_0 – v_1) + v_0 $$</p> <p>$$ v’_0 = m_0 \frac {1+\epsilon} {m_0 + m_1} (v_0 – v_1) + v_1 $$</p> <p>that is the same solution as in the <a href="http://mimosite.com/blog/post/2013/05/13/Billiard-simulation-part-3-collision-between-two-balls">two dimensional classical case</a> when \(\alpha = 0\).</p> <h4><font style="font-weight: bold">Inelastic collision</font></h4> <p>For a perfectly inelastic collision \(\epsilon = 0\) and so (V) becomes simply \( v’ = u \): the two balls have always the same velocity after the collision, i.e. they ‘stick together’ as expected.</p>
http://mimosite.com/blog/post/2013/08/22/Relativistic-collision-in-one-dimension
http://mimosite.com/blog/post/2013/08/22/Relativistic-collision-in-one-dimension#commenthttp://mimosite.com/blog/post.aspx?id=2cd606e2-e75f-43e5-a8a2-77dbcfc38edeThu, 22 Aug 2013 09:05:35 -0700Physicsadminhttp://mimosite.com/blog/pingback.axdhttp://mimosite.com/blog/post.aspx?id=2cd606e2-e75f-43e5-a8a2-77dbcfc38ede1http://mimosite.com/blog/trackback.axd?id=2cd606e2-e75f-43e5-a8a2-77dbcfc38edehttp://mimosite.com/blog/post/2013/08/22/Relativistic-collision-in-one-dimension#commenthttp://mimosite.com/blog/syndication.axd?post=2cd606e2-e75f-43e5-a8a2-77dbcfc38ede