Eigenmumbles

In order to figure out some gyroscope simulations recently, I had to go back and revisit the derivation of fictitious forces in rotating frames. I always got confused by this the first time I was exposed to it, partly because I didn’t really understand the relation between cross products, rotations, and skew symmetric matrices back then. I worked it out for myself, and my notes have been sitting quietly in a corner of my drive since then. So this is my excuse to see how well MathJax works.

Suppose that $x(t) \in \mathbb{R}^3$ represents the position of a material point in a rotating reference frame, and $y(t) \in \mathbb{R}^3$ represents the point in an inertial frame. The two coordinate vectors are connected by an orthogonal matrix, $y(t) = Q(t) x(t).$ Acceleration is the second derivative in the inertial frame, i.e. $a = \ddot{y} = \ddot{Q} x + 2\dot{Q} \dot{x} + Q \ddot{x},$ and the inertial force is $ma(t)$ in the inertial frame.

In the rotating frame, the inertial force is $m Q^T a = m ( Q^T \ddot{Q} x + 2Q^T \dot{Q} \dot{x} + \dot{x} ).$ The $m \ddot{x}$ term is straightforward enough, but clearly we need to understand $Q^T \dot{Q}$ and $Q^T \ddot{Q}$. To do this, it’s useful to write $\dot{Q}$ as $\dot{Q} = Q S$. Remember that $Q$ satisfies the identity $Q^T Q = I,$ so $\dot{Q}^T Q + Q^T \dot{Q} = S^T + S = 0$; that is, $S$ is a skew-symmetric matrix. Multiplying a vector by a skew symmetric matrix is exactly equivalent to taking a cross product with a vector, so we might also write $S = \Omega \times$. If we differentiate the relation $\dot{Q} = Q S$, we have $\ddot{Q} = \dot{Q} S + Q \dot{S} = Q (S^2 + \dot{S}).$ Therefore, we can write the inertial force in the rotating frame as $m Q^T a = m ( S^2 x + \dot{S} x + 2 S \dot{x} + \ddot{x} ).$ We now give names to each of the pieces of this expression. The centrifigul force is $-m S^2 x = -m \Omega \times (\Omega \times x),$ the Coriolis force is $-2m S \dot{x} = 2 m \Omega \times \dot{x},$ and the Euler force (which vanishes if the rate of rotation $S$ is constant) is $-\rho \dot{S} x = -\rho \dot{\Omega} \times x.$

What do I take away from all this? Three things, I think:

1. “Fictitious forces” is just a fancy way of refering to the product rule.
2. Textile intrudes just enough into the raw input text that it can screw up MathJax. Better be careful mixing the two systems.
3. Debugging problems with MathJax – like debugging problems with LaTeX, I suppose – is not altogether easy.

UPDATE (2012-01-31): Actually, I’m now using Markdown via kramdown, and it works just fine with MathJax.