I recently calculated the Einstein tensor for the exterior Schwarzchild solution. Here it is:(adsbygoogle = window.adsbygoogle || []).push({});

G_{00}= 0

G_{11}= [-2GM/(r^{3}c^{2}- 2GMr^{2})] - G^{2}M^{2}/[r^{2}(rc^{2}- 2GM)^{2}] - [-G^{2}M^{2}/(r^{4}c^{4}- 2GMr^{3}c^{2})] - (-2GM)/(r^{3}c^{2}) - (-2G^{3}M^{3})/[r^{3}c^{2}(rc^{2}- 2GM)^{2}]

G_{22}= (2G^{2}M^{2}rc^{2}- 2G^{3}M^{3}) / (r^{3}c^{6}- 2GMr^{2}c^{4})

G_{33}= G_{22}sin^{2}(θ)

Every other element is 0. Just so you know, I used the (+ - - -) signature, so here is the line element:

ds^{2}= [1 - 2GM/(rc^{2})]c^{2}dt^{2}- dr^{2}/[1 - 2GM/(rc^{2})] - r^{2}(dθ^{2}+ sin^{2}(θ)d∅^{2})

Now what exactly does this Einstein tensor (or any Einstein tensor for that matter) tell you about the curvature of space-time? I have some speculation about this question based on some dimensional analysis that I did.

After calculating this Einstein tensor (while keeping in all factors of c and G) I did some dimensional analysis on this Einstein tensor using SI units, and I found that in SI units, the radial element G_{11}of this tensor has the units:

1/m^{2}(or just simply m^{-2}). I noticed that this is a common unit for things involving curvature. On an interesting note however, I found the angular elements G_{22}and G_{33}to be dimensionless. I wonder if this is because the Schwarzchild metric is isometric, and doesn't have a function in front of the angular section of the line element.

Any way, based on the dimensions that I found for the radial coordinate, I speculated that G_{11}gives you the amount of space-time curvature in the radial direction. Would this speculation be correct? Also, what would the angular components tell you, since they are dimensionless in this case? If my speculation is not correct then what exactly would this tensor tell me?

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# I What does the Einstein tensor actually tell you?

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