Rotation

A rotation is a circular movement of an object around a central line, known as the axis of rotation. A rotation $R(\alpha)$ in $\mathbb{R}^{N}$ is a linear operator parameterized by a real value, typically interpreted as the angle of rotation and denoted $\theta$ or $\alpha$. It is also a transformation.

In the context of a rigid body, it is a movement that maintains at least one point fixed. A rotation where the axis passes through an object's center of mass is called a spin rotation, whereas a rotation around an axis completely outside of the body is called a revolution (or orbit). In this case, the axis is called the axis of revolution.

Rotations form a group with interesting properties. Since rotations are, geometrically speaking, planar operations, the most fundamental group is the one in $\mathbb{R}^{2}$. Here they form a group known as

where $M_{2}(\mathbb{R})$ is the set of all real $2\times 2$ matrices. It is called the (two-dimensional) special orthogonal group¹. It can also be written as $SO(2)=\{ R(\alpha)\ |\ \alpha \in[0,2\pi[\ \}$ and is the group of all antisymmetric matrices in $\mathbb{R}^{2}$. For rotations in $\mathbb{R}^{N}$, they form the group

Properties#

• They are linear operators.
• They are antisymmetric: $R(\alpha)^{T}R(\alpha)=R(\alpha)R^{T}(\alpha)=\mathrm{I}$, where $\mathrm{I}$ is the Identity matrix.
• They have unit determinant: $\det R(\alpha)=1$.
• Given a vector $\mathbf{v}\in \mathbb{R}^{N}$, the application of a rotation $R(\alpha)$ rotates it by an angle $\alpha$ with respect to its previous orientation.

Matrix representation#

Being linear operators, rotations can be represented through square matrices. This is a particularly convenient form in more theoretical math, especially when operator theory can be used. Given a vector $(x,y)$ in a plane, a rotation about the origin by an angle $\theta>0$ is the vector $(x',y')$ given by

which can be derived using plain trigonometry. The rotation is counterclockwise (by convention). These equations can be written in matrix form as

In three dimensions, a rotation about the $z$ axis can be trivially defined as

The 3D rotation matrix columns can be represented in the standard $\mathbb{R}^{3}$ Basis $\{\mathbf{i},\mathbf{j},\mathbf{k}\}$:

where $c=\cos\theta$ and $s=\sin\theta$. Consider a right-handed orthonormal set $\{\mathbf{a},\mathbf{b},\mathbf{d}\}$. $R_{0}$ is a rotation in this basis. The matrix $R_{1}$ that represents a rotation in the standard basis will transform $\mathbf{a}$, $\mathbf{b}$ and $\mathbf{d}$ with

and in matrix form

Calling $P=(\begin{array}{c|c}\mathbf{a} & \mathbf{b} & \mathbf{d}\end{array})$, we can see that the rotation matrices are similar:

where $P^{-1}=P^{T}$ comes from antisymmetry. We can also solve for

which gives us

where all the $\mathbf{x}\mathbf{x}^{T}$ are $3\times3$ matrices. This allows us to calculate $R_{1}$, though at the cost of having to define $\mathbf{a}$ and $\mathbf{b}$. Still, there is a more convenient form that can be derived to remove this dependency as, intuitively, a rotation is independent of which vectors on the plane you refer it to.

The vector $\mathbf{v}$ that's being rotated is represented in $\{\mathbf{a},\mathbf{b},\mathbf{d}\}$ as a Linear combination of the basis vectors:

A couple of useful vectors are

The first one can be written as a matrix multiplied by a vector

The matrix $D$ is antisymmetric. We can then also say $\mathbf{d}\times(\mathbf{d}\times\mathbf{v})=D^{2}\mathbf{v}$. This is (one of) the matrix representation(s) of the cross product.

Using $(2)$ and the unitary matrix $I$ we can find

which means

Similarly, we have

and so

Finally, we have

and so

We can combine all these relations to find $R_{1}$ from $(1)$:

In this form, the rotation matrix is independent from the choice of $\mathbf{a}$ and $\mathbf{b}$ when applied to a vector. In fact

Rotation vector spaces#

It is possible to take the exponential of a matrix by using the series definition

This has some interesting effects in the context of rotations. Consider the most general antisymmetric $2\times 2$ matrix:

The exponential is

where we used that $\mathrm{A}^{2m}=(-1)^{m}\alpha^{2m}\mathrm{I}$ and $\mathrm{A}^{2m+1}=\mathrm{A}^{2m}\mathrm{A}=(-1)^{m}\alpha^{2m+1}\mathrm{E}$ and the Sine and cosine series. As such, in the most general case, the exponential of an antisymmetric matrix is

Evidently, all (finite) rotations are exponentials of the antisymmetric matrix $\mathrm{E}$, weighed by some factor $\alpha$. We can therefore represent them as

Since all rotations are exponentials of $\mathrm{E}$, we say that $\mathrm{E}$ is the generator of $SO(2)$. In fact, antisymmetric matrices form a Vector space (sum and scalar multiplication are both defined and closed) and $\mathrm{E}$ forms a Basis of this space. Thus, any rotation is also a Linear combination of basis elements. In 2D the basis is just $\{ \mathrm{E} \}$, but in $N$-dimensional spaces this provides an easy extension of the definition of rotation: just add more elements to the basis so that it generates $SO(N)$.

The number of generators, that is elements in the basis, is found to be $N(N-1)/2$. In $N=3$ we have $3$ generators and these are

A generic infinitesimal rotation $\Omega$ is given by

For example, the rotation $R_{3}$ is given by the exponential of $\mathrm{E}_{3}$

But this is just a 2D rotation around an axis, specifically around the $z$ axis. The same can be found for $R_{1}$ and $R_{2}$ too, which end up being rotations around the $x$ and $y$ axes. Evidently, then, $R_{i}$ is the rotation around the $i$-th axis. A generic rotation can then be described as a Linear combination of basis rotations, just like how a vector is a linear combination of basis vectors (again, rotations constitute a vector space).

Intuitively, this makes sense: rotating something on a "diagonal" is like rotating it "horizontally" and then "vertically".

Look for a moment at the $\mathrm{E}$ basis matrices. They all have the same blueprint: antisymmetric, all zeros on the diagonal and exactly a single pair of $1$ and $-1$, with one matrix for every entry outside of the diagonal. For one, this gives proof on why the number of basis elements is what it is: the number of nondiagonal entries in an $N\times N$ matrix is precisely $N(N-1)/2$. But more than that, it gives us a way to formalize how this basis is built, which is through the Levi-Civita tensor $\epsilon_{ijk}$. For instance, in 3D each basis matrix $i$ is defined by its elements $(\mathrm{E}_{i})_{jk}=-\epsilon_{ijk}$. A generic rotation basis then is just the application of the Levi-Civita tensor in whatever dimension you need.

Infinitesimal rotations are hence defined in a similar manner. In 3D:

This permits a more refined analysis of what infinitesimal rotations even do. The variation of a vector $v$ subject to such a rotation must then be

using $\epsilon_{ijk}=-\epsilon_{ikj}$ and the tensor definition of the vector product. Putting components together we can say that, in general, $\boldsymbol{\omega}\times \mathbf{v}$ is the variation of $\mathbf{v}$ under an infinitesimal rotation of angle $\lvert \boldsymbol{\omega} \rvert$ around an axis parallel to $\boldsymbol{\omega}$. The norm of this variation is

To further analyze $\mathrm{E}_{i}$, we can calculate the Commutator between the $\mathrm{E}_{i}$ and $\mathrm{E}_{j}$:

We need to figure out the product between these matrices:

using the Dirac delta. The commutator components then are

Simplify:

You can verify that each component $[\mathrm{E}_i, \mathrm{E}_j]_{mn}$ is actually an entry of $\mathrm{E}_k$ weighted by $\epsilon_{ijk}$. In particular, the commutator simplifies to:

This gives a compact formula for the commutator of the generators of rotation. This might seem insignificant, but the fact is that we now have a well-behaved commutator in the vector space of rotations. This means that it's not any old vector space, but specifically it's a Lie algebra. This provides access to all of the machinery specific to Lie algebras and helps to study the special orthogonal groups such as $SO(3)$.

Conservation of angular momentum#

Rotations have a connection of fundamental importance to the conservation of angular momentum. We'll use the Lagrangian formalism for this proof, but this fact is true in general regardless of formalism.

Consider a 3D rotation $R$ around an axis parallel to $\boldsymbol{\omega}$ and write it as the exponential of a linear combination of generators as above:

where $\hat{\boldsymbol{\omega}}=\boldsymbol{\omega}/\lvert \boldsymbol{\omega} \rvert$. We define the Coordinate transformation

We can take the derivative of this in $\alpha$ to find

The sum in parenthesis is an infinitesimal rotation. When evaluated in $\alpha=0$ we get

Now, assuming the Lagrangian $\mathcal{L}(\mathbf{q}, \dot{\mathbf{q}})$ is invariant under this rotation, we can invoke Nöther's theorem to guarantee the existence of a constant of motion:

Substituting $\frac{ \partial \varphi }{ \partial \alpha } (q,0)$ gives

This can be rearranged as

Here, $p_j = \frac{\partial \mathcal{L}}{\partial \dot{q}_j}$ is the conjugate momentum and $(\mathrm{E}_i)_{jk}$ is the $j, k$-th component of the $i$-th rotation generator.

In vector form, recognize that

where $\mathbf{L}$ is the angular momentum. The conserved quantity becomes:

which is the projection of the total angular momentum onto the rotation axis $\hat{\boldsymbol{\omega}}$.

This is an universally important fact: if the Lagrangian (or Hamiltonian for that matter) is invariant under rotation on an axis, then angular moment is conserved on that axis. This is true always as long as rotation invariance persists and in fact, for isolated systems, there is no restriction on what the axis is, which means that all angular momentum is conserved in such a system. This is one of the several important manifestations of Nöther's theorem.

Specific cases#

Circular motion#

If the axis $\mathbf{D}$ is fixed and the distance is a constant $r_{0}$, the coordinate system can be chosen to be Polar coordinates or Cylindrical coordinates depending on dimensions. If cylindrical, the plane of rotation should be the plane of reference, with the axis of rotation being the $z$-axis.

A set of orthonormal axes can be chosen with the Cartesian tangent-normal Moving frame. Call $\mathbf{\xi}$ and $\mathbf{\eta}$ these two axes, potentially with $\mathbf{D}$ as the third if in 3D. Therefore, the position unit vector of the rotating object is $\mathbf{R}=\mathbf{\xi}\sin\theta+\mathbf{\eta}\sin\theta$. In this system, angular speed is $\sigma(t)=\dot{\theta}(t)$, angular velocity is $\mathbf{w}(t)=\sigma(t)\mathbf{D}=\dot{\theta}(t)\mathbf{D}$ and angular acceleration is $\mathbf{\alpha}(t)=\dot{\sigma}(t)\mathbf{D}=\ddot{\theta}(t)\mathbf{D}$.

The position therefore changes like

where $h_{0}$ is the height of the movement above the plane $\mathbf{D}\cdot\mathbf{r}=0$ if in 3D. The velocity therefore is $\mathbf{v}(t)=r_{0}\sigma\mathbf{P}=r_{0}\sigma\mathbf{D}\times\mathbf{R}=\mathbf{w}\times\mathbf{r}$, so

so the tangential velocity can be derived from the angular velocity if the position is known. The acceleration then is $\mathbf{a}(t)=-r_{0}\sigma^{2}\mathbf{R}+r_{0}\ddot{\theta}\mathbf{P}=-r_{0}\sigma^{2}\mathbf{R}+r_{0}\dot{\sigma}\mathbf{D}\times\mathbf{R}=-r_{0}\sigma^{2}\mathbf{R}+\alpha\times\mathbf{r}$, so

where $-r_{0}\sigma^{2}\mathbf{R}$ is the centripetal acceleration and $\mathbf{\alpha}\times\mathbf{r}$ is the tangential acceleration.

Moving axis#

If the axis $\mathbf{D}(t)$ is moving, let's consider the static axis case in a bit more detail before. We can define an antisymmetric matrix for any vector $\mathbf{u}$ as

This is the matrix representation of the cross product, $C(\mathbf{u})\mathbf{r}=\mathbf{u}\times\mathbf{r}$ thanks to $(3)$, and thanks to $(4)$ we can say²

Linear velocity is worked out as

If we differentiate $\mathbf{r}$ directly, we get

Therefore

or

$(5)$ and $(6)$ equations tell us the rate of change for the rotation in terms of the current rotation and angular velocity.

Equipped with these, let's consider an actually moving axis $\mathbf{D}(t)$. Using $(4)$, the rotation matrix corresponding to this vector for angle of rotation $\theta(t)$ is

which acts like

The linear velocity is given by $(5)$.

For a rotation matrix, $I=RR^{T}$, so taking the time derivative

which means $S(t)=\dot{R}(t)R^{T}(t)$ is antisymmetric and because of that we can write $S(t)=A(\mathbf{w}(t))$.

The angular velocity can be calculated using $R(t)$ from $(7)$ and then calculating $\dot{R}R^{T}$. Using $C(\mathbf{D})^{3}=-C(\mathbf{D})$ we can prove

As $\mathbf{D}$ is a unit vector, the triad $\{\mathbf{D},\dot{\mathbf{D}},\dot{\mathbf{D}}\times\mathbf{D}\}$ forms an orthonormal basis. The angular velocity is expressed in this basis.