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--- advanced_notions:thomas_precession [2019/06/10 10:13]
jakobadmin [Abstract]
+++ advanced_notions:thomas_precession [2019/06/10 10:15]
jakobadmin [Abstract]
@@ Line 16: / Line 16: @@
+<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?>