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JamesJ Group TitleBest ResponseYou've already chosen the best response.1
Ricci curvature a slightly sophisticated concept and it's hard to know what simple might be here. Mathematically, maybe one simple way to understand is as a function of the metric tensor, g, in a local coordinate system is \[R_{ij} = (3/2) \Delta g_{ij}\] But even that can be obscure. The best thing to do at first to begin to get a sense of R_ij is is to work some examples for simple geometries: e.g., flat space, a sphere and a cylinder.
 3 years ago

amazingopen45 Group TitleBest ResponseYou've already chosen the best response.0
Thanks so much! But what is delta g_ij? I only know that it is used in finding distance in 4d where c^2=a^2+b^2 doesn't work. So the ds^2=g_ij*dx_i*dx_j=dt^2dx^2dy^2dz^2. Also what is the first, second, and third componets of momentum?
 3 years ago

JamesJ Group TitleBest ResponseYou've already chosen the best response.1
I recommend you go to the library and find a good textbook that works some examples; that is the best way to get a feel for this. In the expression I wrote, g_ij is the metric tensor and the delta is the Laplacian operator.
 3 years ago
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