# Infinitesimal transformation

1. Dec 10, 2014

### PhyAmateur

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.

2. Dec 10, 2014

### Einj

Your new tensors are going to be:
$$\begin{cases} F_{\mu\nu}^\prime=\cos\alpha F_{\mu\nu}+\sin\alpha\star G_{\mu\nu}\simeq F_{\mu\nu}+\alpha\star G_{\mu\nu} \\ G_{\mu\nu}^\prime=\cos\alpha G_{\mu\nu}+\sin\alpha\star F_{\mu\nu}\simeq G_{\mu\nu}+\alpha\star F_{\mu\nu} \end{cases}.$$
The variations of the tensors themselves are defined as the new tensors minus the old ones: $\delta F_{\mu\nu}=F_{\mu\nu}^\prime-F_{\mu\nu}$ and $\delta G_{\mu\nu}=G_{\mu\nu}^\prime-G_{\mu\nu}$. And thus you obtain what you are looking for.

3. Dec 11, 2014

### PhyAmateur

Thank youuu a lot!!