Anti symmetric matrix and rotations

Question

Studying the lagrangian formulation of Noether's theorem and came upon how the invariance under rotations gives conservation of angular momentum.

Whilst setting up the problem the notes state that if a potential only depends on the distance between 2 points, namely $V(|r_i-r_j|)$, then you can apply the transformation:

$$\textbf{r}\rightarrow \textbf{r}+\epsilon T\textbf{r}$$

where $\epsilon$ is a small variation, $\textbf{r}$ is just a vector and $T$ is a rotation matrix. I'm confused about the fact that the notes state that $T$ is an anti-symmetric matrix, I thought rotation matrices where orthogonal.

Answer 1

If you consider the set of $n$-by-$n$ rotation matrices $SO(n)$ as a Lie group, then the corresponding Lie algebra is the set of antisymmetric or skew symmetric $n$-by-$n$ matrices. I.e., in the limit of $\epsilon \to 0$, the any rotation matrix $U$ is equal to $I + \epsilon T$ up to first-order. This is known as an "infinitessimal rotation". See this Wiki article for more details and references:

https://en.wikipedia.org/wiki/Skew-symmetric_matrix#Infinitesimal_rotations

Answer 2

Just to give a down-to-earth explanation without developing Lie theory.

Suppose the rotation afte time $t$ is given by $R(t)$, in particular $R(0)=I$ (that is we shall use the coordinate system at the initial time as our standard reference frame), then we have $R(t)R(t)^T=I$. To differentiate both sides, we get $$\dot{R}(t)R(t)^T+R(t)\dot{R}(t)^T=0$$

Let $t=0$, we get $\dot{R}(0)=-\dot{R}(0)^T$, hence $$R(\epsilon)\textbf{r}\approx(R(0)+\dot{R}(0)\epsilon)\textbf{r}=\textbf{r}+\epsilon\dot{R}(0)\textbf{r}$$ where $\dot{R}(0)$ is skew-symmetric as shown above.