# Different formulas for $\rm SO(3)$ rotations

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?

• Commented Nov 29, 2020 at 11:24
• @Qmechanic So components for $\vec{\theta}$ are determined using the Baker-Campbell-Hausdorff relation. Is that correct? Commented Nov 29, 2020 at 11:37

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: $$\vec{\theta}_{passive}=-\vec{\theta}_{active}$$) .

• Thanks for this very clear answer! Commented Nov 30, 2020 at 6:06
• Can you recommend where I can read how the formula for rotation about an arbitrary axis is derived? Commented Nov 30, 2020 at 6:11
• 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. Commented Nov 30, 2020 at 22:58
• I referred back to this answer quite a few times this week. It helped me a lot. Thanks and hope you like the bounty! Commented Dec 31, 2020 at 10:02

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:

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