Let $V^{\mu}$ be a vector field defined in a Minkowski spacetime and suppose it transforms under a Lorentz transformation $V'^{\mu} = \Lambda^{\mu}_{\,\,\,\nu}V^{\nu}$. We can write this like $V'^{\mu} = (e^{i\omega})^{\mu}_{\,\,\,\nu}V^{\nu}$ I think where $\omega$ denotes a rotation in some plane spanned by indices $\left\{\rho \sigma\right\}$, say. In 2D Euclidean space time, we can write the matrix representation of $\Lambda$ as $$\begin{pmatrix} \cos \omega & \sin \omega\\-\sin \omega&\cos \omega\end{pmatrix}$$ and in Minkowski space this changes to the 'hyperbolic' rotation. Linearising the above yields $$\begin{pmatrix}1&\omega\\-\omega&1\end{pmatrix} = \text{Id} + \begin{pmatrix} 0&\omega\\-\omega&0\end{pmatrix} = \text{Id} + \omega \begin{pmatrix} 0&1\\-1&0\end{pmatrix}$$

Now compare with the more general treatment:
$V'^{\mu} = \Lambda^{\mu}_{\,\,\,\nu}V^{\nu} \approx (\delta^{\mu}_{\nu} + \omega^{\mu}_{\,\,\,\nu})V^{\nu}$, where $\omega^{\mu}_{\,\,\,\nu} \equiv (\omega^{\rho \sigma} S_{\rho \sigma})^{\mu}_{\,\,\,\nu}$ In 2D, the spin matrix $S$ when acting on vectors in 2D Euclidean space time is therefore the matrix multiplying $\omega$ above, which agrees with the single generator of the SO(2) group.

If we continue with the general analysis, we obtain $V'^{\mu} - V^{\mu} = \omega^{\mu}_{\,\,\,\nu}V^{\nu} = \eta^{\mu \rho}\omega_{\rho \nu} V^{\nu} = \omega^{\rho \sigma} \delta^{\mu}_{\rho} \eta_{\sigma \nu} V^{\nu}$ Now use the antisymmetry of $\omega_{\rho \sigma}$ gives $$2(V'^{\mu} - V^{\mu}) = \omega^{\rho \sigma}(\delta^{\mu}_{\rho} \eta_{\sigma \nu} - \delta^{\mu}_{\sigma} \eta_{\rho \nu})$$ from which we can identify $S$ to be $\delta^{\mu}_{\rho} \eta_{\sigma \nu} - \delta^{\mu}_{\sigma} \eta_{\rho \nu}$. I am wondering how this agrees with the matrix I obtained above.

Many thanks.

This post imported from StackExchange Physics at 2014-06-17 07:53 (UCT), posted by SE-user CAF