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Recall that the Faraday tensor in this form is a linear mapping that maps a charged particle's contravariant four-velocity to the latter's rate of change, wrt proper time (modulo scaling by invariant rest mass $m$ and invariant charge $q$): $$m\,\frac{\mathrm{d} v^\mu}{\mathrm{d}\tau} = q\, F^\mu{}_\nu\,v^\nu\tag{1}$$ Now let's think of a particle's ...
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) ~=~ ... 4 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: ... 1 [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 ... 1 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 ... 0$SO(1,\,2)$, with two spatial dimensions, only has rotations in one plane - that of all space. So if our co-ordinates are$(t,\,x,\,y)\$ (spatial co-ords last), its unique (up to a real scale factor of course) rotational Lie algebra generator must be: $$R=\left(\begin{array}{ccc}0&0&0\\0&0&-1\\0&1&0\end{array}\right)$$ where we've ...