# Interpretation of Ricci rotation coefficients in tetrad formalism

Given an orthonormal frame (tetrad in 4 dimensions, vielbein more generally) $$\{(e_{\mu})^{a}\}$$ with $$g(e_{\mu}, e_{\nu}) = \eta_{\mu\nu}$$, the Ricci rotation coefficients are defined as $$\omega_{\lambda\mu\nu} = (e_{\lambda})^{a}(e_{\mu})^{b}\nabla_{a}(e_{\nu})_{b}$$ and you can extract information about the curvature of the spacetime in question if you have all the relevant Ricci rotation coefficients. My question is: what is, exactly, the geometrical/physical interpretation of the Ricci rotation coefficients? In a vague sense, I can see that they account for "non-inertial" effects that a local Lorentz observer would detect if them chose this respective vielbein as their frame, which in turn would give information about the local curvature, but that is very hand-wavy; I wanted something more illuminating. So is there any intuitive sense I can give to these coefficients?

• What is your interpretation of the Christoffel symbols?
– MBN
Oct 11 '18 at 10:45
• @MBN Somewhat in analogy to what happens in a noninertial frame of reference in newtonian mechanics, and so summarize it in a few words, the Christoffel symbols would carry the information on how the coordinate basis is varying thoughout spacetime, or how the coordinate basis fails to be parallel transported in the directions of its respective basis vectors. So that would be precisely the same as the Ricci rotation coefficients, right (now that you've made me explain that to myself, it already seems a bit more clear)? Oct 11 '18 at 12:39

I think what you said in the comments is right. I don't know if what follows give you more intuition. In your notation, $$\big\{e_{\nu} \, : \, \nu = 1...n \big\}$$ is a frame of $$n$$ linearly independent vector fields $$e_{\nu}(x) = \big(\,e_{\nu}(x)\,\big)^b \,\frac{\partial}{\partial x^b}$$ that determine the frame. The part $$(e_{\lambda})^a \nabla_a \, (e_{\nu})_b$$ should more accurately be written as $$\Big((e_{\lambda})^a \nabla_a \, e_{\nu} \Big)_b$$ and so $$\Big((e_{\lambda})^a \nabla_a \, e_{\nu} \Big)^b \frac{\partial}{\partial x^b}= \nabla_{e_{\lambda}} \, e_{\nu}$$ is the covariant derivative of the frame vector field $$e_{\nu}$$ in the direction of the frame vector $$e_{\lambda}$$. The derivative $$\nabla_{e_{\lambda}} \, e_{\nu}$$ is a vector that describes how the frame vector $$e_{\nu}$$ deviates from its parallel transport when moved slightly in the direction of the frame vector $$e_{\lambda}$$. In other words, it measures the "rotation" of vector $$e_{\nu}$$ when moved along vector $$e_{\lambda}$$. Then you take the "projections" of these deviation vectors $$\nabla_{e_{\lambda}} \, e_{\nu}$$ onto the axes of the frame $$\Big(e_{\mu}\circ \big(\,\nabla_{e_{\lambda}} \, e_{\nu}\,\big)\Big) = (e_{\mu})^b \, g_{bs} \, \Big((e_{\lambda})^a \nabla_a \, e_{\nu} \Big)^s = (e_{\mu})^b \Big((e_{\lambda})^a \nabla_a \, e_{\nu} \Big)_b = \omega_{\lambda \mu \nu}$$ where $$g_{bs}$$ is the Riemannian (or Lorentz) metric.