- #1
PhyAmateur
- 105
- 2
If we have:
$$F_{\mu\nu} \rightarrow \cos\alpha F_{\mu\nu} +\sin\alpha \star G_{\mu\nu}$$
$$G_{\mu\nu} \rightarrow \cos\alpha G_{\mu\nu} +\sin\alpha \star F_{\mu\nu}$$
for rotation $\alpha$.
If infinitesimal transformation for small alpha one gets
$$\delta F_{\mu\nu} = \delta\alpha~\star G_{\mu\nu}$$
$$\delta G_{\mu\nu} = \delta\alpha~\star F_{\mu\nu}.$$
How do we get the infinitesimal transformation? I didn't understand the procedure. I know that $\cos\alpha \sim1$ and $\sin\alpha \sim\alpha$ but when I am substituting back I am not getting the same $\delta F_{\mu\nu}$ as above.
$$F_{\mu\nu} \rightarrow \cos\alpha F_{\mu\nu} +\sin\alpha \star G_{\mu\nu}$$
$$G_{\mu\nu} \rightarrow \cos\alpha G_{\mu\nu} +\sin\alpha \star F_{\mu\nu}$$
for rotation $\alpha$.
If infinitesimal transformation for small alpha one gets
$$\delta F_{\mu\nu} = \delta\alpha~\star G_{\mu\nu}$$
$$\delta G_{\mu\nu} = \delta\alpha~\star F_{\mu\nu}.$$
How do we get the infinitesimal transformation? I didn't understand the procedure. I know that $\cos\alpha \sim1$ and $\sin\alpha \sim\alpha$ but when I am substituting back I am not getting the same $\delta F_{\mu\nu}$ as above.