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For $\rm SO(3)$ rotations, the group elements are given by the standard Euler matrices $R_x(\theta_x)$, $R_y(\theta_y)$ and $R_z(\theta_z)$ for rotations in 3D space: $$R_x(\theta_x)=\begin{bmatrix} 1 &0 &0 \\ 0& \cos\theta_x &\sin\theta_x \\0 &-\sin\theta_x & \cos\theta_x \end{bmatrix} $$ $$R_y(\theta_y)=\begin{bmatrix} \cos\theta_y & 0 & -\sin\theta_y \\ 0& 1 &0 \\ \sin\theta_y& 0& \cos\theta_y \end{bmatrix} $$ $$R_z(\theta)=\begin{bmatrix} \cos\theta_z &\sin\theta_z & 0\\ -\sin\theta_z& \cos\theta_z &0 \\0 & 0& 1 \end{bmatrix} $$ The corresponding generators $J_x$, $J_y$ and $J_z$ are defined such that $$R_x(\theta_x)=e^{i\theta_xJ_x} , R_y(\theta_y)=e^{i\theta_yJ_y},R_z(\theta_z)=e^{i\theta_zJ_z}.$$

I then read that the general rotation transformation is given by $$R(\vec{\theta})=e^{i\vec{\theta}\cdot\vec{J}},$$ where $\vec{J}=(J_x, J_y, J_z)$.

I am confused about what the components for $\vec{\theta}$ will be.

Consider a rotation in space where we first rotate by $\theta_z$, then by $\theta_y$ and lastly $\theta_x$, the rotation matrix should be $$R(\theta_x,\theta_y,\theta_z)=R_x(\theta_x)R_y(\theta_y)R_z(\theta_z)=e^{i\theta_xJ_x}e^{i\theta_yJ_y}e^{i\theta_zJ_z}.$$ Intuitively I would think that using the $R(\vec{\theta})=e^{i\vec{\theta}\cdot\vec{J}}$ formula means using $\vec{\theta}=(\theta_x, \theta_y, \theta_z)$: $$R(\vec{\theta})=e^{i\theta_xJ_x+i\theta_yJ_y+i\theta_zJ_z}$$

However, as the generators $J_x, J_y, J_z$ don't commute, $$e^{i\theta_xJ_x}e^{i\theta_yJ_y}e^{i\theta_zJ_z} \neq e^{i\theta_xJ_x+i\theta_yJ_y+i\theta_zJ_z}.$$ So it is wrong to say that $\vec{\theta}=(\theta_x, \theta_y, \theta_z)$. What then should the components for $\vec{\theta}$ be?

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Your reasoning is correct. The Euler angles are not the components of $\vec{\theta}$. Here is what $\vec{\theta}$ is.

Let $\vec{\theta}=(\theta_1,\theta_2,\theta_3)=\theta \hat{n}$. Let's derive the 3x3 matrix (ie: the group elements $R(\vec{\theta})$) for rotating an object by $\theta$ radians about an arbitrary direction specified by the unit vector $\hat{n}$. This means put your right hand thumb along the unit vector $\hat{n}$ and rotate the object by pushing with your right hand fingers through the angle $\theta$. For me, this is a much easier way to parameterize and visualize an arbitrary rotation than Euler angles. Notice $\theta=\sqrt{\theta_1^2+\theta_2^2+\theta_3^2}$ .

As you say in your question, the group element is $R(\vec{\theta})=e^{i\vec{\theta}\cdot \vec{J}}$. This rotates any object which includes vectors with any number of components (for example, an arrow, a rock, a tensor, or particles with different spins). We want to rotate a 3-vector so we put in the 3x3 matrix representation of each of the 3 generators $\vec{J}=(J_1,J_2,J_3)$. $$ \begin{align} \Theta & =i\vec{\theta}\cdot \vec{J} \\ & = i\theta_1\begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & i \\ 0 & -i & 0 \\ \end{bmatrix} +i\theta_2\begin{bmatrix} 0 & 0 & -i \\ 0 & 0 & 0 \\ i & 0 & 0 \\ \end{bmatrix} +i\theta_3\begin{bmatrix} 0 & i & 0 \\ -i & 0 & 0 \\ 0 & 0 & 0 \\ \end{bmatrix} \\ & = \begin{bmatrix} 0 & -\theta_3 & \theta_2 \\ \theta_3 & 0 & -\theta_1 \\ -\theta_2 & \theta_1 & 0 \\ \end{bmatrix} \end{align} $$ Notice that $[J_1,J_2]=iJ_3$ which is correct for rotation generators (=angular momentum).

Finally we expand $e^{\Theta}$ in a power series and matrix multiply $\Theta$ 's together to calculate each term. You will find $\Theta^3=-\theta^2\Theta$. $$ \begin{align} R(\Theta) & =e^{\Theta} \\ & =I+\Theta+\dfrac{\Theta^2}{2!}+ \dfrac{\Theta^3}{3!}+ \dfrac{\Theta^4}{4!} +… \\ & =I+\Theta(1-\frac{\theta^2}{3!}+\frac{\theta^4}{5!}-...)+\Theta^2(\frac{1}{2!}-\frac{\theta^2}{4!}+\frac{\theta^4}{6!}-...) \\ \\ R(\Theta) & =I+ \frac{\Theta}{\theta}sin(\theta)+\frac{\Theta^2}{\theta^2}(1-cos(\theta)) \end{align} $$ This $R(\Theta)$ is the matrix for rotating any 3-vector about an arbitrary unit vector $\hat{n}$ by angle $\theta$. As an example, suppose $\hat{n}=(0,0,1)$, which is a rotation about the z-axis by theta. Then the final equation for $R$ yields the familiar rotation matrix $$ R(\Theta) = \begin{bmatrix} cos(\theta) & -sin(\theta) & 0 \\ sin(\theta) & cos(\theta) & 0 \\ 0 & 0 & 1 \\ \end{bmatrix} $$ Notice that my $sin(\theta)$ is the opposite sign as yours because I am doing an active transformation on the object, whereas your formula is for a passive transformation on the coordinate axis (ie: $\vec{\theta}_{passive}=-\vec{\theta}_{active}$) .

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  • $\begingroup$ Thanks for this very clear answer! $\endgroup$
    – TaeNyFan
    Commented Nov 30, 2020 at 6:06
  • $\begingroup$ Can you recommend where I can read how the formula for rotation about an arbitrary axis is derived? $\endgroup$
    – TaeNyFan
    Commented Nov 30, 2020 at 6:11
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    $\begingroup$ I'm sorry but I don't have a book/paper reference. The final $R(\Theta)$ is known as the Rodrigues formula. The Wikipedia derivation looks complicated to me. The above expansion of the exponential is a simpler derivation. You may verify the thumb/fingers interpretation of what $\theta$ does by choosing $\hat{n}=(0,0,1)$ and seeing that it rotates say $(1,0,0)$ correctly, The $\vec{\theta}$ and $\vec{J}$ transform as vectors if you view the rotation from a rotated coordinate system. $\endgroup$ Commented Nov 30, 2020 at 22:58
  • $\begingroup$ I referred back to this answer quite a few times this week. It helped me a lot. Thanks and hope you like the bounty! $\endgroup$
    – TaeNyFan
    Commented Dec 31, 2020 at 10:02
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take the Taylor series for this rotation matrix:

$$R_x(\theta_1)=\left[ \begin {array}{ccc} 1&0&0\\ 0&\cos \left( \theta_{{1}} \right) &-\sin \left( \theta_{{1}} \right) \\ 0&\sin \left( \theta_{{1}} \right) &\cos \left( \theta_{{1}} \right) \end {array} \right] $$

you obtain

$$R_x(\theta_1)=\left[ \begin {array}{ccc} (1)&0&0\\ 0&(1-{\frac {1 }{2}}{\theta_{{1}}}^{2}+{\frac {1}{24}}{\theta_{{1}}}^{4}+O \left( { \theta_{{1}}}^{6} \right) )&(-\theta_{{1}}+{\frac {1}{6}}{\theta_{{1}} }^{3}-{\frac {1}{120}}{\theta_{{1}}}^{5}+O \left( {\theta_{{1}}}^{6} \right) )\\ 0&(\theta_{{1}}-{\frac {1}{6}}{\theta_{ {1}}}^{3}+{\frac {1}{120}}{\theta_{{1}}}^{5}+O \left( {\theta_{{1}}}^{ 6} \right) )&(1-{\frac {1}{2}}{\theta_{{1}}}^{2}+{\frac {1}{24}}{ \theta_{{1}}}^{4}+O \left( {\theta_{{1}}}^{6} \right) )\end {array} \right] \tag 1$$

the rotation matrix $R_x$ is also

$$R_x=\text{exp}(i\,\theta_1\tau_1)$$

take the Taylor series

$$R_x=I_3+x+\frac 12 x\,x+\frac 16 x\,x\,x+\ldots$$

with $x=i\,\theta_1\,\tau_1$

and: $$\tau_1=-i\,\left[ \begin {array}{ccc} 0&0&0\\ 0&0&1 \\ 0&-1&0\end {array} \right] $$

thus:

$$R_x(\theta_1)=\left[ \begin {array}{ccc} (1)&0&0\\ 0&(1-{\frac {1 }{2}}{\theta_{{1}}}^{2}+{\frac {1}{24}}{\theta_{{1}}}^{4}+O \left( { \theta_{{1}}}^{6} \right) )&(-\theta_{{1}}+{\frac {1}{6}}{\theta_{{1}} }^{3}-{\frac {1}{120}}{\theta_{{1}}}^{5}+O \left( {\theta_{{1}}}^{6} \right) )\\ 0&(\theta_{{1}}-{\frac {1}{6}}{\theta_{ {1}}}^{3}+{\frac {1}{120}}{\theta_{{1}}}^{5}+O \left( {\theta_{{1}}}^{ 6} \right) )&(1-{\frac {1}{2}}{\theta_{{1}}}^{2}+{\frac {1}{24}}{ \theta_{{1}}}^{4}+O \left( {\theta_{{1}}}^{6} \right) )\end {array} \right] \tag 2$$

analog for the rotation matrix $R_y~$ and $R_z$

the rigid body rotation matrix is now:

$$R_x(\theta_1)\,R_y(\theta_2)\,R_z(\theta_3)\mapsto \text{e}^{(i\,\theta_1\,\tau_1)}\,\text{e}^{(i\,\theta_2\,\tau_2)}\,\text{e}^{(i\,\theta_3\,\tau_3)}$$

with:

$$\tau_2= -i\,\begin{bmatrix} 0 & 0 & -1 \\ 0 & 0 & 0 \\ 1 & 0 & 0 \\ \end{bmatrix}~,\tau_3=-i\,\begin{bmatrix} 0 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 0 \\ \end{bmatrix}$$

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