# Expressing general rotation in terms of tensors

1. Feb 15, 2014

### CAF123

1. The problem statement, all variables and given/known data
A general rotation through angle $a$ about the axis $\underline{n}$, where $\underline{n}^2 = 1$ is given by $$R(a,\underline{n}) = \exp(-ia\underline{n} \cdot \underline{T}),$$ where $(T_k)_{ij} = -i\epsilon_{ijk}$. By expanding the exponential as a power series in $a$, and explicitly summing the resulting series, show that $$R_{ij}(a,\underline{n}) = \delta_{ij}\cos a + n_i n_j (1-\cos a) - \epsilon_{ijk} n_k \sin a$$ using $(\underline{n} \cdot \underline{T})_{ij}^2 = \delta_{ij} - n_i n_j$ and $(\underline{n} \cdot \underline{T})_{ij}^3 = (\underline{n} \cdot \underline{T})_{ij}$, (which I derived earlier)

2. Relevant equations
$\exp(ix) = \cos x + i\sin x$ and cos and sin power series.

3. The attempt at a solution
$$R(a,\underline{n})_{ij} = [\cos(a (\underline{n} \cdot \underline{T})) - i\sin(a (\underline{n} \cdot \underline{T}))]_{ij}$$ Reexpressing in terms of the Taylor series for sin and cos: $$\left[\sum_{k=0}^{\infty} (-1)^k \frac{(a (\underline{n} \cdot \underline{T}))^{2k}}{(2k)!} - i \sum_{k=0}^{\infty} (-1)^k \frac{(a (\underline{n} \cdot \underline{T}))^{2k+1}}{(2k+1)!}\right]_{ij}$$ Reorganising the even and odd terms gives $$\left[\sum_{k=0}^{\infty} (-1)^k \frac{(\delta_{ij} - n_i n_j)^k}{(2k)!} - i \sum_{k=0}^{\infty} (-1)^k\frac{[(\delta_{ij} - n_i n_j)^{2k} (-i\epsilon_{ijk}n_k)]}{(2k+1)!}\right]$$

I am not really sure how to progress. Thanks for any help.

2. Feb 15, 2014

### tiny-tim

Hi CAF123!

(btw, you missed out all the aks
sooo you should have no powers of $(\delta_{ij} - n_i n_j)$ or of $(\underline{n} \cdot \underline{T})_{ij}$ above 1 !

3. Feb 16, 2014

### CAF123

Hi tiny-tim,

I was able to progress and obtained for $k \geq 1$, $(\underline{n} \cdot \underline{T})_{ij}^{2k} = \delta_{ij} - n_in_j$ and for $k \geq 0$, $(\underline{n} \cdot \underline{T})_{ij}^{2k+1} = (\underline{n} \cdot \underline{T})_{ij}$.

For the case $k=0$ in $(\underline{n} \cdot \underline{T})_{ij}^{2k}$, I was wondering if this should be $\delta_{ij}$. If so, then I have at the end $$\delta_{ij} + (\delta_{ij} - n_i n_j) (\cos a - 1) - \epsilon_{ijl} n_l \sin a = \text{result}$$ My reasoning being that for all other powers of $k=0$ the result is a matrix, so for $k=0$, the result should also be a matrix, (= the identity, which is naturally $\delta_{ij}$.) Thanks.

4. Feb 16, 2014

### tiny-tim

Hi CAF123!
Yes, A0 for any matrix is the identity, I,

so (A0)ij = δij