4
$\begingroup$

Im trying to express the following quaternion in polar form (axis-angle)

$a=1+i-2j+k$

Would this be the resultant ?

$$\cos \frac{θ}{2} +\sin \frac{θ}{2} (i-2j+k)$$

$\endgroup$
3
$\begingroup$

To find the polar (exponential) form of a quaternion we have to start from the definition of the exponential of a quaternion. ( see :Exponential Function of Quaternion - Derivation)

For a quaternion $z=a+b\mathbf{i}+c\mathbf{j}+d\mathbf{k} = a+\mathbf{v}$ this is given by: $$ e^z = e^{a+\mathbf{v}}=e^a \left( \cos |\mathbf{v}| +\dfrac{\mathbf{v}}{|\mathbf{v}|} \,\sin |\mathbf{v}| \right) $$

Now note that we have obviously: $$ z=a+\mathbf{v}= |z| \left( \dfrac{a}{|z|}+\dfrac{|\mathbf{v}|}{|z|} \dfrac{\mathbf{v}}{|\mathbf{v}|} \right) $$

end, since: $$ a^2+|\mathbf{v}|^2=|z|^2 $$ we have: $$ \left(\dfrac{a}{|z|}\right)^2+ \left( \dfrac{|\mathbf{v}|}{|z|}\right)^2 =1 $$ so there exists $0\le\theta<2\pi$ such that: $$ \dfrac{a}{|z|}=\cos \theta \qquad \dfrac{|\mathbf{v}|}{|z|}=\sin \theta $$ and every quaternion $z=a+ib+jc+kd=a+\mathbf{v}$ can be expressed in exponential (polar) form as: $$ z = |z|\left(\cos \theta +\mathbf{n}\sin \theta \right)=|z| e^ {\mathbf{n} \theta}= e^{\ln |z| + \mathbf{n} \theta} $$ with: $$ \begin{cases} & |z|= \sqrt{a^2+b^2+c^2+d^2}\\ & \cos \theta=\dfrac{a}{|z|} \qquad \sin \theta=\dfrac{|\mathbf{v}|}{|z|}\\ & \mathbf{n}=\dfrac{\mathbf{v}}{|z|\sin \theta}=\dfrac{\mathbf{v}}{|\mathbf v|} \end{cases} $$

Now you can apply this general result to your case (note that your answer is wrong and I don't see where it come from).

$\endgroup$
2
$\begingroup$

In general if you have a number $z=a+b_1i+b_2j+b_3k$, you can write it as $$ z = A\cos\theta + A\sin\theta(\cos\varphi_1 i+\cos\varphi_2 j+\cos\varphi_3 k) $$ where $A = \sqrt{a^2+b_1^2+b_2^2+b_3^2}$, and $\cos\theta = a/A$. And if you define $B=\sqrt{A^2-a^2}$, then $\cos\varphi_i = b_i/B$. Clearly $\cos^2\varphi_1+\cos^2\varphi_2+\cos^2\varphi_3=1$. A quick calculation shows $$(\cos\varphi_1 i+\cos\varphi_2 j+\cos\varphi_3 k)^2 = -1$$ This means that in fact you can also write $$z = A\exp\Big[(\cos\varphi_1 i+\cos\varphi_2 j+\cos\varphi_3 k)\theta\Big]$$ This is one possibility of polar coordinates for quarternions. One can do other things too.

$\endgroup$
  • $\begingroup$ So is my answer from above an acceptable answer. Or should i say the correct answer ? $\endgroup$ – JekasG Oct 25 '15 at 7:29
  • $\begingroup$ I doubt it is correct but then again depends on what's you definition of $\theta$? $\endgroup$ – Hamed Oct 25 '15 at 7:34
  • $\begingroup$ So how would i work out the answer ? And what is the correct answer ? $\endgroup$ – JekasG Oct 25 '15 at 8:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.