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.

I got lost there. Can someone explain how is that done in detail?

I show the calculations here:

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

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

Thanks.

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)?