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.


You are absolutely right, a rotation matrix should be orthogonal. However, when we investigate symmetry transformations, we often like to consider infinitesimal variations, as they are often much easier to investigate.

Consider then a orthogonal matrix $R$, the rotation matrix. The rotation transformation is written as follows :

$$\vec{r}\rightarrow R\vec{r}$$

Now, let us consider an infinitesimal rotation. Since it is infinitesimal, it must be very close to the "$0$" rotation, or in other words, the identity $I$. Thus, we must write it in some way as :

$$R = I+\epsilon T$$

Where $\epsilon$ is a small parameter. Now, we want to find which $T$ we should use. Obviously, a random matrix $T$ wouldn't work in general. We must enforce the fact that $R$ is a rotation, i.e. that $R^T R = I$ :

$$I = R^T R = (I+\epsilon T^T)(I+\epsilon T) = I+\epsilon(T+T^T)+O(\epsilon^2)$$

Here, we don't care about terms in $\epsilon^2$, as they are much smaller than terms in $\epsilon$. So, to enforce $R^T R = I$, we must have simply $T+T^T = 0 \Leftrightarrow T = -T^T$. This is exactly the condition of anti-symmetricity you were looking for !

Thus, for an infinitesimal rotation, we can conveniently write $R = I +\epsilon T$ with T antisymmetric. Note that we often use an even more compact notation, $R = e^{\epsilon T}$, which you can check is equivalent to the one you gave for small $\epsilon$.

P.S. : Physicist use yet another notation, where they write $R = I -i\epsilon (iT) = I -i\epsilon T'$, so we would have purely imaginary antisymmetric matrices. This is mainly because in this way, when you consider unitary matrices $U^\dagger U =I$, then the associated $T'$ is hermitian $T'^\dagger = T'$

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