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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.

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  • $\begingroup$ Isn't it the same in both cases? Remember your $\eta_{\mu \nu}= \delta_{\mu \nu}$ as you are considering the 2D Euclidean case. $\endgroup$ – user7757 Jun 16 '14 at 17:50
  • $\begingroup$ @ramanujan_dirac: Thanks for your comment. Yes, I just verified it. However, it only seems to work provided I take $\rho = 1, \sigma = 2$. (So rotation in 1-2 plane). It does not yield correct results for a rotation in 2-3 or 1-3 plane, which is peculiar because there is nothing special about a rotation around the 3 axis. $\endgroup$ – CAF Jun 16 '14 at 18:29
  • $\begingroup$ In 2D there is just one parameter, the angle of rotation, the group is abelian. There is no 2-3 and 1-3 plane! The other S matrices would be zero. $\endgroup$ – user7757 Jun 16 '14 at 19:31
  • $\begingroup$ Ok, but what is special about the 1-2 plane? Why doesn't the S matrices vanish for say, the 1-3 and 1-2 plane? After all, they are just numbers denoting a plane of rotation. Thanks! $\endgroup$ – CAF Jun 17 '14 at 6:27
  • $\begingroup$ The other matrices are 0 because even if one of the $\rho$ or $\sigma$ takes the value 3, one of the kronecker delta's in each term will always be zero, as the matrix indices $\mu \nu$ can take only values 1 or 2. You shouldn't even begin with a two indexed object $\omega_{\mu \nu}$ , because you only need 2 basis vectors in 2 dimensions, the other being the identity matrix, you just need one other generator. $\endgroup$ – user7757 Jun 17 '14 at 6:47

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