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--- advanced_notions:thomas_precession [2019/06/10 10:13]
jakobadmin [Abstract]
+++ advanced_notions:thomas_precession [2019/06/10 10:19]
jakobadmin [Abstract]
@@ Line 12: / Line 12: @@
 <tabbox Abstract>
+Abstractly, the Thomas precession is due to an additional rotation the spin of the particle pics up as a result of of the orbit movement. In this sense, the phenomenon is due to a [[advanced_tools:geometric_phase|geometric phase]]/Berry phase.
+From a group theoretic perspective, Thomas precession occurs because Lorentz boost do not commute. For example, we have
+$$[K_y,\,K_x] = i\,J_z \, .$$
+---
 <blockquote>{{ :advanced_notions:thomas_precession.jpg?nolink&200|}}[T]he revolution of the particle around the nucleus can be interpreted as a sequence of Lorentz boosts.[...] Abstractly, the effect is due to Wigner rotations [3] that appear in representations of the Poincaré group, so it is purely kinematical—it follows solely from space-time symmetries. <cite>https://arxiv.org/abs/1711.05753</cite></blockquote>
+<blockquote>To get the Lorentz transformation mapping from frame 1 to 3 directly, we compose boosts:
+$$\exp\left(\frac{a_y}{c} \,\delta t\, K_y + \frac{a_x}{c} \,\delta t\, K_x\right) \exp\left(\frac{v_x}{c}\,K_x\right) = \exp\left(\frac{v_x}{c}\,K_x + \frac{a_x}{c}\,\delta t\,K_x + \frac{a_y}{c}\,\delta t\, K_y + \frac{a_y\,v_x}{2\,c^2}\,[K_y,\,K_x]\,\delta t + \cdots\right)\tag{1}$$
+by the Dynkin formula version of the Baker-Campbell-Hausdorff theorem. From the commutation relationships for the Lorentz group we get $[K_y,\,K_x] = i\,J_z$, so that $\frac{a_y\,v_x}{2\,c^2}\,[K_y,\,K_x]\,\delta t$ corresponds to a rotation about the $z$ axis of $a_y\,v_x\,\delta t / (2\,c^2)$ radians (here, somewhat obviously $J_z,\,J_y,\,J_z$ are "generators of rotations", *i.e.* tangents to rotations about the $x,\,y,\,z$ axes at the identity, respectively, whilst $J_z,\,J_y,\,J_z$ are "generators" of boosts).<cite>https://physics.stackexchange.com/a/99459/37286</cite></blockquote>
 <tabbox Why is it interesting?>

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