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I was reading Geometric Algebra for Physicists, by Doran and Lasenby, and, in section 5.5.2, they calculate the Thomas Precession. However, at a certain point, they have the exponential of an exponential and lower it, can someone explain how is that done in detail?

I show the calculations here:

$$ n=\mathrm{e}^{-\omega t I \sigma_{3}} \boldsymbol{\sigma}_{2}=R_{\omega} \boldsymbol{\sigma}_{2} \tilde{R}_{\omega} \tag{5.150} $$ where $R_{\omega}=\exp \left(-\omega t I \sigma_{3} / 2\right) .$ We now have $$ \mathrm{e}^{\alpha n / 2}=\exp \left(\alpha R_{\omega} \sigma_{2} \tilde{R}_{\omega} / 2\right) \overset{??}{=} R_{\omega} R_{\alpha} \tilde{R}_{\omega} \tag{5.151} $$ where $$ R_{\alpha}=\exp \left(\alpha \sigma_{2} / 2\right) $$

How does one perform the last step in (5.151)?

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1 Answer 1

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The third term in Eq (5.151) stems from two definitions:

  1. The definition of the exponential operator on the multivector $A$ is, $$ \exp\left(A\right)\triangleq\sum_{j=0}^\infty\frac{A^j}{j!}=1+A+\frac12A^2+\frac16A^3+\cdots\tag{1}. $$
  2. Rotors obey $R_\omega \tilde R_\omega=1$.

Thus, if you use Equation (1) with $A=R_\omega\boldsymbol\sigma_2\tilde R_\omega$ and point #2, you will find, $$ \exp\left(R_\omega\boldsymbol\sigma_2\tilde R_\omega\right)=R_\omega \tilde R_\omega+R_\omega\boldsymbol\sigma_2\tilde R_\omega+\frac{1}{2}\left(R_\omega\boldsymbol\sigma_2\tilde R_\omega\right)^2+\frac{1}{6}\left(R_\omega\boldsymbol\sigma_2\tilde R_\omega\right)^3+\cdots\tag{2} $$ For the higher order terms, you can expand out the products to find that each of them will obey, $$ \left(R_\omega\boldsymbol\sigma_2\tilde R_\omega\right)^j=\left(R_\omega\boldsymbol\sigma_2\tilde R_\omega\right)\left(R_\omega\boldsymbol\sigma_2\tilde R_\omega\right)\cdots\left(R_\omega\boldsymbol\sigma_2\tilde R_\omega\right)=R_\omega\boldsymbol\sigma_2^j\tilde R_\omega\tag{3} $$ due to repeated use of point #2.

Hence, you can combine Equations (2) and (3) here along with the associative property of the geometric product to write Eq (5.151) as, \begin{align} \exp\left(\alpha R_\omega\boldsymbol\sigma_2\tilde R_\omega/2\right)&=R_\omega\left(1+\alpha\boldsymbol\sigma_2/2+\frac{1}{2}\left(\alpha\boldsymbol\sigma_2/2\right)^2+\frac{1}{6}\left(\alpha\boldsymbol\sigma_2/2\right)^3+\cdots\right)\tilde R_\omega \\ &=R_\omega\mathrm{e}^{\alpha\boldsymbol\sigma_2/2}\tilde R_\omega \\ &= R_\omega R_\alpha \tilde R_\omega \end{align}

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