What is the exact form of local Lorentz transformations (from the point of view of the metric) in a curved spacetime background like in general relativity? It should deviate substantially from ordinary Lorentz transformations in Minkowski space.
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$\begingroup$ Related: physics.stackexchange.com/q/190243/2451 and links therein. $\endgroup$– Qmechanic ♦Commented Oct 19, 2020 at 13:57
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$\begingroup$ @Qmechanic Thank you.i was thinking if one can do this directly?suppose for static observer we got a form of metric and from that we directly go to another inertial frame and then get lorentz transform directly.can this be done? $\endgroup$– RoyCommented Oct 19, 2020 at 14:11
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$\begingroup$ The metric is INVARIANT under local Lorentz transformations! In addition to local vs. global, there is a fine distinction (often not pointed out in text books) between the local Lorentz transformations in general relativity and global Lorentz transformations in special relativity. See details here: physics.stackexchange.com/questions/502982/… $\endgroup$– MadMaxCommented Oct 19, 2020 at 19:31
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Yes, there is such a thing in GR. These are the rotations of tetrads.
At each point of a space define an orthonormal basis: $$ (\vec e_{(a)}, \vec e_{(b)}) = \eta_{ab} $$ Where $a, b$ - denote the indices, corresponding to the local frame, in constract to the Greek spacetime indices $\mu, \nu$. The coordinate basis is related to the local basis, by some invertible $4 \times 4$ matrix : $$ \vec e_{\mu} = e_{\mu}^{a} \vec e_a $$ The metric in the coordinate space is expressed therefore, as: $$ g_{\mu \nu}= e_{\mu}^{a} e_{\nu}^{b} \eta_{ab} $$ Local Lorentz transformations can be made at any point: $$ \vec e_a = \Lambda_a^{b} (x) \vec e_b $$ Where the matrix of transformation depends on the point $x$.
For a good reference I recommend - https://arxiv.org/abs/1106.2037.
answered Oct 19, 2020 at 13:41
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$\begingroup$ Thank you.of course this can be done.i was thinking if one can do this directly?suppose for static observer we got a form of metric and from that we directly go to another inertial frame and then get lorentz transform directly.can this be done? $\endgroup$– RoyCommented Oct 19, 2020 at 13:52
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$\begingroup$ "from the point of view of the metric": The metric is INVARIANT under local Lorentz transformations! If you use eq(27) from the link (arxiv.org/abs/1106.2037) provided in the answer, you can clearly see that $g_{\mu \nu}= e_{\mu}^{a} e_{\nu}^{b}\eta_{ab}$ is invariant. $\endgroup$– MadMaxCommented Oct 19, 2020 at 19:01