User talk:Paul Wormer/scratchbook

Given are two unnormalized, non-parallel vectors, the rotation axis n and the vector r to be rotated. Decompose r into two orthogonal components:

$\mathbf {r} =\mathbf {n} {\frac {\mathbf {n} \cdot \mathbf {r} }{n^{2}}}+\underbrace {(\mathbf {r} -\mathbf {n} {\frac {\mathbf {n} \cdot \mathbf {r} }{n^{2}}})} _{\equiv \;\mathbf {x} }\quad {\hbox{with}}\quad n^{2}\equiv \;\mathbf {n} \cdot \mathbf {n}$

Clearly, n and x are orthogonal. Define further y as a cross product, a vector orthogonal to the plane containing n, r, and x,

$\mathbf {y} =\mathbf {n} \times \mathbf {r} .$

As is well-known the cross product can be written as a matrix-vector product

$\mathbf {y} =\mathbf {n} \times \mathbf {r} ={\begin{pmatrix}n_{y}r_{z}-n_{z}r_{y}\\n_{z}r_{x}-n_{x}r_{z}\\n_{x}r_{y}-n_{y}r_{x}\end{pmatrix}}=\underbrace {\begin{pmatrix}0&-n_{z}&n_{y}\\n_{z}&0&-n_{x}\\-n_{y}&n_{x}&0\end{pmatrix}} _{\mathbf {N} }{\begin{pmatrix}r_{x}\\r_{y}\\r_{z}\end{pmatrix}}$

The matrix N has as general element

$N_{\alpha \beta }=-\epsilon _{\alpha \beta \gamma }n_{\gamma }\,$

where ε_αβγ is the antisymmetric Levi-Civita tensor.

For further use we compute normalization constants of x and y,

$x^{2}=\mathbf {x} \cdot \mathbf {x} =\left(\mathbf {r} -\mathbf {n} {\frac {\mathbf {n} \cdot \mathbf {r} }{n^{2}}}\right)\cdot \left(\mathbf {r} -\mathbf {n} {\frac {\mathbf {n} \cdot \mathbf {r} }{n^{2}}}\right)=r^{2}-{\frac {(\mathbf {n} \cdot \mathbf {r} )^{2}}{n^{2}}}$

$y^{2}=(\mathbf {n} \times \mathbf {r} )\cdot (\mathbf {n} \times \mathbf {r} )=n^{2}\,r^{2}-(\mathbf {n} \cdot \mathbf {r} )^{2},$

and divide the two

${\frac {x^{2}}{y^{2}}}={\frac {r^{2}-{\frac {(\mathbf {n} \cdot \mathbf {r} )^{2}}{n^{2}}}}{n^{2}\,r^{2}-(\mathbf {n} \cdot \mathbf {r} )^{2}}}={\frac {1}{n^{2}}}.$

When we rotate r over an angle φ around n, the component of r along n is unchanged, while the component x of r perpendicular to n becomes x′

$\mathbf {x} '=\cos \phi \;\mathbf {x} +{\frac {x}{y}}\sin \phi \;\mathbf {y} =\cos \phi \;\mathbf {x} +{\frac {1}{n}}\sin \phi \;\mathbf {y}$

Hence the rotated vector r′ is

$\mathbf {r} '=\mathbf {n} {\frac {\mathbf {n} \cdot \mathbf {r} }{n^{2}}}+\cos \phi (\mathbf {r} -\mathbf {n} {\frac {\mathbf {n} \cdot \mathbf {r} }{n^{2}}})+{\frac {1}{n}}\sin \phi \;\mathbf {N} \;\mathbf {r}$

We may introduce the dyadic product of the vector n with itself, which has the form of a 3 × 3 symmetric matrix, and write

$\mathbf {n} {\frac {\mathbf {n} \cdot \mathbf {r} }{n^{2}}}={\frac {1}{n^{2}}}\;{\big (}\mathbf {n} \otimes \mathbf {n} {\big )}\;\mathbf {r} \quad {\hbox{with}}\quad {\big (}\mathbf {n} \otimes \mathbf {n} {\big )}_{\alpha \beta }\equiv n_{\alpha }\;n_{\beta }.$

Now,

$\mathbf {r} '=\left[\cos \phi \;\mathbf {E} +{\frac {(1-\cos \phi )}{n^{2}}}\;{\big (}\mathbf {n} \otimes \mathbf {n} {\big )}+{\frac {1}{n}}\sin \phi \;\mathbf {N} \right]\mathbf {r} ,$

where E is the identity matrix. The quantity between square brackets is the matrix R that rotates r around n over an angle φ. This equation is very well-known and was first derived by Leonhard Euler [check]. A general element of R is

$R_{\alpha \beta }=\cos \phi \;\delta _{\alpha \beta }+{\frac {(1-\cos \phi )}{n^{2}}}n_{\alpha }n_{\beta }-{\frac {\sin \phi }{n}}\epsilon _{\alpha \beta \gamma }n_{\gamma }=\cos \phi \;\delta _{\alpha \beta }+(1-\cos \phi ){\hat {n}}_{\alpha }{\hat {n}}_{\beta }-\sin \phi \;\epsilon _{\alpha \beta \gamma }\;{\hat {n}}_{\gamma },$

where the unit vector is

${\hat {\mathbf {n} }}\equiv {\frac {\mathbf {n} }{n}}.$

Consider now two non-parallel unit vectors f and t making an angle φ

$\mathbf {f} \cdot \mathbf {t} =\cos \phi \quad {\hbox{and}}\quad |\mathbf {f} \times \mathbf {t} |=\sin \phi$

We want to find the matrix that rotates f (the "from" vector) to t (the "to" vector). An obvious way is using the cross product f×t as a rotation axis and rotating f over φ. We can use the equation just derived by substituting

$\mathbf {n} \rightarrow \mathbf {f} \times \mathbf {t} \quad {\hbox{and}}\quad n\rightarrow \sin \phi .$

Hence

${\frac {(1-\cos \phi )}{n^{2}}}={\frac {(1-\cos \phi )}{1-\cos ^{2}\phi }}={\frac {1}{1+\cos \phi }}$

so that an element of the rotation matrix takes the form

$R_{\alpha \beta }=\cos \phi \;\delta _{\alpha \beta }+{\frac {1}{1+\cos \phi }}n_{\alpha }n_{\beta }-\epsilon _{\alpha \beta \gamma }n_{\gamma },$

which is Eq. (1) of Möller and Hughes (1999).