# Rotor Identity $\frac{1+ba}{|a+b|} = e^{-B\theta /2}$

To prove:the identity given above where $a, b$ are vectors, $B$ is the unit bivector in the $a\wedge b$ plane and $\theta$ is the angle between $a$ and $b$. (From "Geometric Algebra for Physicists" by Doran and Lasenby).
Expanding the L.H.S i get $$\frac{1+b.a}{|a+b|} - \frac{|a\wedge b|}{|a+b|}B$$ The R.H.S gives me by definition, $$\cos(\theta/2) - \sin(\theta/2)B$$ Using grade projection, we should have $$\frac{1+b.a}{|a+b|} = \cos(\theta/2)$$ and $$\frac{|b\wedge a|}{|a+b|} = \sin(\theta/2)$$ But i can't think of an easy way to prove either. I am trying to prove it using geometry and the rules of GA, rather than trigonometry.

• Please check theta=0 and a=b. can this be true for any a, b pair? – ahala Jul 15 '13 at 4:02
• For theta = 0 the plane of rotation is not defined anyway – user997712 Jul 18 '13 at 7:35

Remember that a rotation can be performed through a composition of two reflections. Let $c$ be the vector in the $a \wedge b$ plane that has an angle $\theta/2$ with both $a$ and $b$. Then a rotation that would rotate $a$ to $b$ can be seen as

$$\underline R(s) = \hat c \hat a s \hat a \hat c$$

for any vector $s$. (Here, we're choosing trivially to reflect over $a$ and then reflect over the angle bisector.)

The question then becomes how we can compute the vector $c$ that bisects that angle. In fact, $c = \hat a + \hat b$ does this nicely (not well defined if $\hat b = - \hat a$, but then the plane is not well-defined anyway). Then $\hat c = (\hat a + \hat b)/|\hat a + \hat b|$ and we get the resulting rotor to be

$$R = \hat c \hat a = \frac{\hat a \hat a + \hat b \hat a}{|\hat a + \hat b|}$$

which takes the form you want. It is, however, not immediately clear to me how to generalize the problem to using non-unit vectors. This argument that $c$ is the right vector to reflect over doesn't work if you don't use unit vectors.

• Upon a careful reading of the text, i think $a$ and $b$ are meant as unit vectors, although it is not explicitly stated in the question. Thanks for laying out the geometric picture. – user997712 Jul 18 '13 at 7:33