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Do I need to use the angular velocity vector in the rotating or inertial reference frame for this? Yes. You can do it either way. I start with the expression that relates the time derivative of a vector quantity $\boldsymbol u$ in the inertial and rotating frames: $$\left(\frac {d\boldsymbol u}{dt}\right)_I = \left(\frac {d\boldsymbol u}{dt}\right)_R ...


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To add to David Hammen's answer on the question: When numerically integrating this, together with Euler's equation of rotation, is there a way to ensure that the determinant of $R$ remains equal to one (otherwise $\vec{x}(t)$ will also be scaled)? Method 1 Dumb But Effective Naïve Multiplication Whilst you are getting up to speed with more ...


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The Lorentz group is the group of matrices that conserve the quadratic form: $$\mathscr{Q}(X,\,Y) = X^T\,\eta\,Y\tag{1}$$ where here $X$ and $Y$ are $1\times 4$ column vectors, the $4\times 4$ group member matrices act on these from the left and $\eta$ is the Minkowski (pseuso) metric. Therefore, $\Lambda\in O(1,\,3)$ if and only if: ...


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The three generators of right-handed spinor rotations are given by $\left\{- i\sigma_x,-i\sigma_y,-i\sigma_z\right\}$, see for instance Peskin & Schroeder page 44, and the rotation matrix for a spinor rotation over an angle $\phi$ around a unit vector $\hat{s}$ is given by: $R~=~ \exp\left(-i\frac{\phi}{2}~\hat{s}\cdot\vec{\sigma}\right) ~=~ ...


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This question inspired me to try to write a conceptual introduction at the wikipedia article. To save you the trouble of clicking, I copied it below. (It's slightly inspired by what @Kostia wrote here) Motivating example: Position operator matrix elements for 4d→2s transition Let's say we want to calculate transition dipole moments for an electron to ...


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The defining property of the fundamental representation of the Lorentz group $\mathrm{SO}(1,3)$ $$ M^T\eta M = \eta \quad \forall M\in\mathrm{SO}(1,3)$$ and hence the defining property of the Lorentz group itself does not make sense in representations other than the fundamental, because those are not naturally equipped with a metric "$\eta$" from a physics ...


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[I somewhat haphazardly pieced this answer together, so I'm not absolutely certain the conclusion is correct.] Cayley's theorem is useless here, because the group isomorphism it produces is not required to preserve any kind of topology on the groups, in particular not notions of continuity or differentiability. On the infinite symmetric group $S_\infty$ on ...



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