Geometric Algebra

Another teacher and I have been exploring Geometric Algebra in application to geometry and physics together on most weekends using the online collaboration suite Elluminate (through LearnCentral, which allows one up to three collaborators for free).  I love David Hestenes’ Oersted Medal Lecture, but it was a bit deep for someone not steeped in the culture of physics.  So, I was delighted when I found a series of more introductory lectures by Chris Doran and Anthony Lasenby.  Occasionally they get into deep physics, but the bulk of the treatment is easy to follow for the mathematician, and I enjoy their style of content presentation.

Using the tools of geometric algebra I was able to prove (easily) a theorem that I have never before seen in physics.  Concerning a constantly accelerating model

\begin{array}{rcl} a &=& \text{constant} \\ v &=& v_0 + a t \\ x &=& x_0 + v_0 t + \frac{1}{2} a t^2 \end{array}

one can (and usually does) prove the following intermediate steps

\begin{array}{rcl} v-v_0 &=& a t \\ v+v_0 &=& \frac{2}{t}(x-x_0) \end{array}

on the way to proving

v^2-v_0^2 = 2 a\cdot(x-x_0)

but, by taking the wedge product instead of the dot product, one gets

v\wedge v_0=a\wedge(x-x_0).

In other words, the parallelogram formed by final and initial velocity has the same area as the parallelogram formed by acceleration and displacement. Not only this, but on the way to proving it, one finds that (x-x_0)=\frac{v+v_0}{2}t, so that the displacement lies in the same direction as the average velocity, which we proved intermediately. It makes for a great GeoGebra applet, but I can’t embed it with WordPress. It’s also the quickest way to prove (also using the first relation that v^2=v_0^2) that range is \frac{v_0^2\sin(2\theta_0)}{|a|}.

Published in: on 2011.01.09 at 11:44  Leave a Comment  
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