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advanced_notions:thomas_precession [2019/06/10 10:13]
jakobadmin [Abstract]
advanced_notions:thomas_precession [2019/06/10 10:20]
jakobadmin [Abstract]
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 <tabbox Abstract> ​ <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 
 +$$[K_y,​\,​K_x] = i\,J_z \, .$$
 +The rotation that we get by combining boosts is known as a Wigner rotation.
 +
 +
 +
 +
 +---
  
 <​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>​{{ :​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?> ​   <tabbox Why is it interesting?> ​  
  
advanced_notions/thomas_precession.txt · Last modified: 2019/06/10 10:20 by jakobadmin