Geometric Algebra applied to Physics

Geometric Algebra can be applied to Physics, and many of the introductions to GA online cover this, but they immediately jump to electromagnetic fields or quantum mechanics, which is unfortunate since GA can also greatly simplify 2D kinematics. One such example is uniform circular motion.

You should be familiar with all the concepts presented in An Introduction to Geometric Algebra over R^2 before proceeding.

If we have a vector p that moves at a constant rate of ω rad/s and has a starting position p0, then we can describe the vector p very easily:

\bm{p} = \bm{p_0} e^{\omega t \bm{I}}

Let's figure out what the derivative of a Rotor looks like, by first recalling its definition:

 e^{\theta \bm{I}} := \cos(\theta) + \sin(\theta)\bm{I}

We take the derivative with respect to θ:

          \frac{d}{d \theta} e^{\theta \bm{I}} &=  \frac{d}{d \theta} (\cos(\theta) + \sin(\theta)\bm{I}) \\
            &=  -\sin(\theta) + \cos(\theta)\bm{I} \\

At this point observe that cos and sin just changed places, along with a sign change, but we know of another operation that does the same thing, which is multiplication by I, so we get:

          \frac{d}{d \theta} e^{\theta \bm{I}} &= \frac{d}{d \theta} (\cos(\theta) + \sin(\theta)\bm{I}) \\
            &= -\sin(\theta) + \cos(\theta)\bm{I}          \\
            &= \bm{I} (\cos(\theta) + \sin(\theta)\bm{I})  \\
            &= \bm{I} e^{\theta \bm{I}}                    \\

Not only does the derivative have a nice neat expression, we can read off from the formula what is happening, which is that the derivative is a vector that is rotated 90 degrees from the original vector. Also note that normally the geometric product ins't commutative, but in this case both parts are rotors, so the order doesn't matter.

We can go through the same process to show what happens if θ has a constant multiplier k:

          \frac{d}{d \theta} e^{k \theta \bm{I}} &= \frac{d}{d \theta} (\cos(k \theta) + \sin(k \theta)\bm{I}) \\
            &= k \bm{I} e^{k \theta \bm{I}} \\

With our new derivative in hand we can now find the velocity vector for our position vector p, since velocity is just the derivative of position with respect to time.

        \bm{v}  &= \frac{d}{dt} \bm{p} \\
                &= \frac{d}{dt} \bm{p_0} e^{\omega t \bm{I}} \\
                &= \bm{p_0} \omega \bm{I}  e^{\omega t \bm{I}} \\
                &= \omega \bm{p_0} \bm{I} e^{\omega t \bm{I}} \\

Again, because we using Geometric Algebra, we can read off what is going on geometrically from the formula, that is, the derivative is a vector orthogonal to the position vector that is scaled by ω.

Note that we've drawn the vector as starting from the position, but that's not required.

We get the acceleration vector in the same manner, by taking the derivative of the velocity vector with respect to time.

        \bm{a}  &= \frac{d}{dt} \bm{v}                                      \\
                &= \frac{d}{dt} \omega \bm{p_0} \bm{I} e^{\omega t \bm{I}}  \\
                &= \omega \bm{p_0} \bm{I} \omega \bm{I} e^{\omega t \bm{I}} \\
                &= \omega^2 \bm{p_0} \bm{I} \bm{I} e^{\omega t \bm{I}}      \\
                &= - \omega^2 \bm{p_0} e^{\omega t \bm{I}}                  \\

And again we can just read off from the formula what is going on geometrically, which is that we end up with a vector that is rotated 180 degrees from the position vector, and scaled by ω2.

We can place the acceleration and velocity vectors as starting from the positition vector, and that looks like:

Note how simple this was to derive and that the geometric interpretation could be read off of the resulting formulas. We didn't need to leave the 2D plane, that is, all of these calculations took place in 𝔾2. The more classical derivations for uniform circular motion rely on the cross-product which takes you out of ℝ2 into ℝ3 and which doesn't work in higher level dimensions.