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Infinitesimal transformation

  1. Dec 10, 2014 #1
    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. jcsd
  3. Dec 10, 2014 #2
    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: [itex]\delta F_{\mu\nu}=F_{\mu\nu}^\prime-F_{\mu\nu}[/itex] and [itex]\delta G_{\mu\nu}=G_{\mu\nu}^\prime-G_{\mu\nu}[/itex]. And thus you obtain what you are looking for.
     
  4. Dec 11, 2014 #3
    Thank youuu a lot!!
     
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