# What are the local Lorentz transformations in general relativity?

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.

• Related: physics.stackexchange.com/q/190243/2451 and links therein. – Qmechanic Oct 19 '20 at 13:57
• @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? – Roy Oct 19 '20 at 14:11
• 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/… – MadMax Oct 19 '20 at 19:31

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 $$x$$.
• "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. – MadMax Oct 19 '20 at 19:01