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## Classical Mechanics Euler Angles (lecture for phy354 taught by Prof E. Poppitz)

Posted by peeterjoot on March 23, 2012

[Click here for a PDF of this post with nicer formatting and figures if the post had any (especially if my latex to wordpress script has left FORMULA DOES NOT PARSE errors.)]

# Pictorially.

We want to look at some of the trig behind expressing general rotations. We can perform a general rotation by a sequence of successive rotations. One such sequence is a rotation around the $z,x,z$ axes in sequence. Application of a rotation of angle $\phi$ takes us from our original (\ref{fig:classicalMechanicsEulerAngles:classicalMechanicsEulerAnglesFig1}) frame to that of (\ref{fig:classicalMechanicsEulerAngles:classicalMechanicsEulerAnglesFig2}). A second rotation around the (new) $x$ axis by angle $\theta$ takes us to (\ref{fig:classicalMechanicsEulerAngles:classicalMechanicsEulerAnglesFig3}), and finally a rotation of $\psi$ around the (new) $z$ axis, takes us to (\ref{fig:classicalMechanicsEulerAngles:classicalMechanicsEulerAnglesFig4}).

A composite image of all of these rotations taken together can be found in figure (\ref{fig:classicalMechanicsEulerAngles:classicalMechanicsEulerAnglesFig5}).

# Relating the two pairs of coordinate systems.

Let’s look at this algebraically instead, using figure (\ref{fig:classicalMechanicsEulerAngles:classicalMechanicsEulerAnglesFig6}) as a guide.

Step 1. Rotation of $\phi$ around $z$

\begin{aligned}\begin{bmatrix}x' \\ y' \\ z'\end{bmatrix}=\begin{bmatrix}\cos\phi & \sin\phi & 0 \\ -\sin\phi & \cos\phi & 0 \\ 0 & 0 & 1\end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix}\end{aligned} \hspace{\stretch{1}}(2.1)

Step 2. Rotation around $x'$.

\begin{aligned}\begin{bmatrix}x'' \\ y'' \\ z''\end{bmatrix}=\begin{bmatrix}1 & 0 & 0 \\ 0 & \cos\theta & \sin\theta & 0 \\ 0 & -\sin\theta & \cos\theta & 0 \end{bmatrix}\begin{bmatrix}x' \\ y' \\ z'\end{bmatrix}\end{aligned} \hspace{\stretch{1}}(2.2)

Step 3. Rotation around $z''$.

\begin{aligned}\begin{bmatrix}x''' \\ y''' \\ z'''\end{bmatrix}=\begin{bmatrix}\cos\psi & \sin\psi & 0 \\ -\sin\psi & \cos\psi & 0 \\ 0 & 0 & 1\end{bmatrix}\begin{bmatrix}x'' \\ y'' \\ z''\end{bmatrix}\end{aligned} \hspace{\stretch{1}}(2.3)

So, our full rotation is the composition of the rotation matrices

\begin{aligned}\begin{bmatrix}x''' \\ y''' \\ z'''\end{bmatrix}=\begin{bmatrix}\cos\psi & \sin\psi & 0 \\ -\sin\psi & \cos\psi & 0 \\ 0 & 0 & 1\end{bmatrix}\begin{bmatrix}1 & 0 & 0 \\ 0 & \cos\theta & \sin\theta & 0 \\ 0 & -\sin\theta & \cos\theta & 0 \end{bmatrix}\begin{bmatrix}\cos\phi & \sin\phi & 0 \\ -\sin\phi & \cos\phi & 0 \\ 0 & 0 & 1\end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix}\end{aligned} \hspace{\stretch{1}}(2.4)

Let’s introduce some notation and write this as

\begin{aligned}B_z(\alpha)=\begin{bmatrix}\cos\alpha & \sin\alpha & 0 \\ -\sin\alpha & \cos\alpha & 0 \\ 0 & 0 & 1\end{bmatrix}\end{aligned} \hspace{\stretch{1}}(2.5)

\begin{aligned}B_x(\theta) =\begin{bmatrix}1 & 0 & 0 \\ 0 & \cos\theta & \sin\theta & 0 \\ 0 & -\sin\theta & \cos\theta & 0 \end{bmatrix}\end{aligned} \hspace{\stretch{1}}(2.6)

so that we have the mapping

\begin{aligned}\mathbf{r} \rightarrow B_z(\psi) B_x(\theta) B_z(\phi) \mathbf{r}\end{aligned} \hspace{\stretch{1}}(2.7)

Now let’s write

\begin{aligned}\mathbf{r} = \begin{bmatrix}x_1 \\ x_2 \\ x_3\end{bmatrix}\end{aligned} \hspace{\stretch{1}}(2.8)

We’ll call

\begin{aligned}A(\psi, \theta, \phi) = B_z(\psi) B_x(\theta) B_z(\phi) \end{aligned} \hspace{\stretch{1}}(2.9)

so that

\begin{aligned}x_i' = \sum_{j = 1}^3 A_{ij} x_j.\end{aligned} \hspace{\stretch{1}}(2.10)

We’ll drop the explicit summation sign, so that the summation over repeated indexes are implied

\begin{aligned}x_i' = A_{ij} x_j.\end{aligned} \hspace{\stretch{1}}(2.11)

This matrix $A(\psi, \theta, \phi)$ is in fact a general parameterization of the $3 \times 3$ special orthogonal matrices. The set of three angles $\theta$, $\phi$, $\psi$ parameterizes all rotations in 3dd space. Transformations that preserve $\mathbf{a} \cdot \mathbf{b}$ and have unit determinant.

In symbols we must have

\begin{aligned}A^\text{T} A = 1\end{aligned} \hspace{\stretch{1}}(2.12)

\begin{aligned}\det A = +1.\end{aligned} \hspace{\stretch{1}}(2.13)

Having solved this auxiliary problem, we now want to compute the angular velocity.

We want to know how to express the coordinates of a point that is fixed in the body. i.e. We are fixing $x_i'$ and now looking for $x_i$.

The coordinates of a point that has $x'$, $y'$ and $z'$ in a body-fixed frame, in the fixed frame are $x$, $y$, $z$. That is given by just inverting the matrix

\begin{aligned}\begin{bmatrix}x_1 \\ x_2 \\ x_3\end{bmatrix}&=A^{-1}(\psi, \theta, \phi)\begin{bmatrix}x_1' \\ x_2' \\ x_3'\end{bmatrix} \\ &=B_z^{-1}(\phi)B_x^{-1}(\theta)B_z^{-1}(\psi)\begin{bmatrix}x_1' \\ x_2' \\ x_3'\end{bmatrix} \\ &=B_z^{\text{T}}(\phi)B_x^{\text{T}}(\theta)B_z^{\text{T}}(\psi)\begin{bmatrix}x_1' \\ x_2' \\ x_3'\end{bmatrix}\end{aligned}

Here we’ve used the fact that $B_x$ and $B_z$ are orthogonal, so that their inverses are just their transposes.

We have finally

\begin{aligned}\begin{bmatrix}x_1 \\ x_2 \\ x_3\end{bmatrix}=B_z(-\phi)B_x(-\theta)B_z(-\psi)\begin{bmatrix}x_1' \\ x_2' \\ x_3'\end{bmatrix} \end{aligned} \hspace{\stretch{1}}(2.14)

If we assume that $\psi$, $\theta$ and $\phi$ are functions of time, and compute $d \mathbf{r}/dt$. Starting with

\begin{aligned}x_i = [A^{-1}(\phi, \theta, \psi)]_{ij} x_j',\end{aligned} \hspace{\stretch{1}}(2.15)

\begin{aligned}\Delta x_i = \left([A^{-1}(\phi + \Delta \phi, \theta + \Delta \theta, \psi + \Delta \psi)]_{ij} -[A^{-1}(\phi, \theta, \psi)]_{ij} \right)x_j'.\end{aligned} \hspace{\stretch{1}}(2.16)

For small changes, we can Taylor expand and retain only the first order terms. Doing that and dividing by $\Delta t$ we have

\begin{aligned}\frac{d x_i}{dt} =\left(\frac{\partial {}}{\partial {\psi}} A^{-1}_{ij} \dot{\psi}+\frac{\partial {}}{\partial {\theta}} A^{-1}_{ij} \dot{\theta}+\frac{\partial {}}{\partial {\phi}} A^{-1}_{ij} \dot{\phi}\right) x_j'\end{aligned} \hspace{\stretch{1}}(2.17)

Now, we use

\begin{aligned}x_j' = A_{jl} x_l,\end{aligned} \hspace{\stretch{1}}(2.18)

so that we have

\begin{aligned}\frac{d x_i}{dt} =\left(\left(\frac{\partial {}}{\partial {\psi}} A^{-1}_{ij} \right) A_{jl} \dot{\psi}+\left(\frac{\partial {}}{\partial {\theta}} A^{-1}_{ij} \right) A_{jl} \dot{\theta}+\left(\frac{\partial {}}{\partial {\phi}} A^{-1}_{ij} \right) A_{jl} \dot{\phi}\right) x_l.\end{aligned} \hspace{\stretch{1}}(2.19)

We are looking for a relation of the form

\begin{aligned}\frac{d\mathbf{r}}{dt} = \boldsymbol{\Omega} \times \mathbf{r}.\end{aligned} \hspace{\stretch{1}}(2.20)

We can write this as

\begin{aligned}\begin{bmatrix}v_x \\ v_y \\ v_z\end{bmatrix} =\left(\dot{\theta} \frac{\partial {A^{-1}}}{\partial {\theta}} A+\dot{\phi} \frac{\partial {A^{-1}}}{\partial {\phi}} A+\dot{\psi} \frac{\partial {A^{-1}}}{\partial {\psi}} A\right)\begin{bmatrix}x \\ y \\ z\end{bmatrix} \end{aligned} \hspace{\stretch{1}}(2.21)

Actually doing this calculation is asked of us in HW6. The final answer is

\begin{aligned}\frac{d x_i}{dt} = \left(\dot{\phi} \epsilon_{ijk} j_n^\phi+\dot{\theta} \epsilon_{ijk} j_n^\theta+\dot{\psi} \epsilon_{ijk} j_n^\psi\right) x_j.\end{aligned} \hspace{\stretch{1}}(2.22)

Here $\epsilon_{ijk}$ is the usual fully antisymmetric tensor with properties

\begin{aligned}\epsilon_{ijk} =\left\{\begin{array}{l l}0 & \quad \mbox{when any of the indexes are equal.} \\ 1 & \quad \mbox{for any oflatex ijk = 123, 231, 312(cyclic permutations of $123$.} \\ -1 & \quad \mbox{for any of $ijk = 213, 132, 321$.} \\ \end{array}\right.\end{aligned} \hspace{\stretch{1}}(2.23)

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