# Rotation matrix - levi-civita symbol

I'm trying to solve the following problem:

Given a rotation matrix $$R_{ij}$$, show that
$$n_k=\frac{-R_{ij}\epsilon_{ijk}}{\sqrt{(3-tr(R))(1+tr(R))}}$$
and that $$\sin(\phi)=-\frac{\epsilon_{ijk}R_{ij}n_k}{2}.$$
The trace of the matrix was obtained on a previous item and it is equal to
$$tr(R)=1+2\cos(\phi)$$ the rotation formula for the problem is
$$\vec{r}'=\vec{r}\cos(\phi)+\hat{n}(\hat{n}\cdot\vec{r})(1-\cos(\phi))-(\vec{r}\times\hat{n})\sin(\phi)$$
where $$\hat{n}$$ is the unitary vector parallel to the axis of rotation and $$\phi$$ is the rotation angle taken anti-clockwise to the orientation of $$\phi$$.

What I've tried (using the rotation formula):
$$\begin{gather*} x'_i=x_i\cos(\phi)+n_in_jx_j(1-\cos(\phi))-\epsilon_{ijk}x_jn_k\sin(\phi) \\ \Rightarrow\epsilon_{lkm}(x'_i-x_i\cos(\phi)-n_in_jx_j(1-\cos(\phi)))=-\epsilon_{lkm}\epsilon_{ijk}x_jn_k\sin(\phi) \\ \Rightarrow\epsilon_{ijk}(x'_i-x_i\cos(\phi)-n_in_jx_j(1-\cos(\phi)))=-2\delta_{mk}x_jn_k\sin(\phi). \end{gather*}$$ - I've used the relation: $$\epsilon_{ijk}\cdot\epsilon_{lkm}=2\delta_{mk}$$ on the right side.
- I'm trying to isolate the $$n_k$$ from the right side of the equation, but I don't think it should be zero for $$m\ne k$$ when evaluating the $$\delta$$. I mean, $$x_j$$=0?
That does not seem right. Any help is aprecciated.

EDIT: After some time banging my head against the wall and with the help from @LonelyProf, the answer to the problem is below: $$\begin{gather*} R_{ij}=\delta_{ij}cos(\phi)+n_in_j(1-cos(\phi))-\epsilon_{ijk}n_ksin(\phi) \\ \epsilon_{ijm}R_{ij}=\epsilon_{ijm}\delta_{ij}cos(\phi)+\epsilon_{ijm}n_in_j(1-cos(\phi))-2\delta_{km}n_ksin(\phi) \\ \epsilon_{ijm}n_in_j(1-cos(\phi))=0 \\ \epsilon_{ijm}\delta_{ij}cos(\phi)=0 \\ \Rightarrow n_k=\frac{-\epsilon_{ijk}R_{ij}}{2sin(\phi)} \\ tr(R)^2=1+2cos(\phi)+2cos(\phi)+4cos^2(\phi) \\ tr(R)^2=2tr(R)+3-4sin^2(\phi) \\ \Rightarrow 2sin(\phi)=\sqrt{(3-Tr(R))(1+Tr(R))} \\ \therefore n_k=\frac{-\epsilon_{ijk}R_{ij}}{\sqrt{(3-Tr(R))(1+Tr(R))}} \\ sin(\phi)=\frac{-\epsilon_{ijk}R_{ij}n_k}{2} \end{gather*}$$

• I assume $n_k$ is the axis of rotation and $\phi$ the angle of rotation. You should write it in the question so that potential answerers have all the data. – MannyC Apr 18 '19 at 14:06
• Still unclear, I'm afraid. Are you trying to use Einstein's repeated index convention? Your indices look all wrong, e.g. $k$ which occurs once on the left and three times on the right. Presumably you are trying to prove the equations linking the rotation matrix with the axis-angle form, given at en.m.wikipedia.org/wiki/Rotation_formalisms_in_three_dimensions. – user197851 Apr 18 '19 at 14:33
• What I've tried is possible wrong, this is the central issue of my question. I'm trying to isolate the $n_k$ from the right side and so I separated the rotation formula in its components thinking that would make my life easier. The question is: how do I isolate the $n_k$ from the rotation formula? – Lincon Ribeiro Apr 18 '19 at 14:43
• Tks. I've added more info regarding my attempt to solve the problem. Not sure if it is more clear now. – Lincon Ribeiro Apr 18 '19 at 14:54

To tackle this problem, it is probably best to write out the form of the rotation matrix $$R_{ij} = \cos\phi\, \delta_{ij} + (1-\cos\phi)\, n_i n_j -\sin\phi \,\epsilon_{ijk} n_k$$ where the repeated index $$k$$ is implicitly summed over. This is equivalent to your starting point.
Now you can multiply both sides by $$\epsilon_{ijm}$$, summing over $$i$$ and $$j$$. The identity you were trying to use is $$\epsilon_{ijk}\epsilon_{ijm}=2\delta_{km}$$ (you wrote it down wrongly) and it is also useful to note that $$\epsilon_{ijm}n_in_j$$ will vanish. From this it is easy to prove your first relation, and then the second relation follows by multiplying both sides by $$n_k$$ and summing over $$k$$ (it is a unit vector).
• I'm moving forward now. Tks. New doubt: When I have the following: $\delta_{ij}\cdot\epsilon_{ijm}$ for $i=j$ then $\delta_{ij}=1$ but $\epsilon_{iim}=0$. Is this correct? From the equation you wrote I got $\epsilon_{ijm}R_{ij}=\delta_{ij}\cdot\epsilon_{ijm}cos(\phi)-2\delta_{km}n_ksin(\phi)$ and I'm thinking the first term on the right should not be zero. – Lincon Ribeiro Apr 18 '19 at 18:09
• The first thing you said is correct: $\epsilon_{iim}=0$. Why, then, would you have any doubt about the $\delta_{ij}\epsilon_{ijm}$ term? I think I've said enough now. If you judge that the answer, as it stands, is acceptable, then please consider accepting it by clicking the check mark to the left. Otherwise, of course, you are free to wait for more detailed answers. – user197851 Apr 18 '19 at 18:55