In Misner, Thorne & Wheeler, they say, in their delightful 'word equations' that

$$\left(\frac{\mathrm{radius\,\, of \,\,curvature}}{\mathrm{of\,\, spacetime}}\right) = \left(\frac{\mathrm{typical\,\, component\,\, of\,\, Riemann\,\, tensor}}{\mathrm{as\,\, measured\,\, in \,\,a \,\,local \,\,Lorentz\,\, frame}}\right)^{-\frac{1}{2}}.$$

My question is: does this definition of radius of curvature (and others like it - where tensor are described in words) depend on the valence of the Riemann curvature tensor?


You link to a definition of curvature radius that is appropriate for "extrinsic" curvature – the curvature of a line/submanifold embedded into a higher-dimensional space. The Riemann tensor measures all the components of the intrinsic curvature so they're not exactly the same. However, they're of the same order and the MTW equation you mention is meant to be only an order-of-magnitude estimate, too. Of course, by the typical component, they mean roughly speaking the largest components. If some of them are zero, they're not typical.

To see that the estimate is right, just calculate the Riemann tensor for a sphere of radius $a$. You will get the Ricci scalar equal to something like $2/a^2$. The only thing you need to fix in your equation is the power. The typical component of the Riemann tensor goes like $a^{-2}$ where $a$ is the curvature radius. The power may be calculated by dimensional analysis. I guess that you – or they? – just forgot the power.

  • $\begingroup$ Thanks for the reply. Your comment on the difference between in/extrinsic curvature is helpful (note that the links were added by Qmechanic). Also, very insightful is the comment on the word 'typical', +1 for those. Yes, I forgot the power, I'll edit that. Although, one thing I'm still not clear on is how the valence of the Riemann tensor affects its physical meaning, or does it not? E.g. does it matter if in the above equation we use $R^{a}_{bcd}$ or $R_{abcd}$? $\endgroup$ – user12345 Feb 16 '13 at 14:04
  • 1
    $\begingroup$ Thanks for your understanding. It shouldn't matter whether you raise/lower indices because you should be expressing the components in a local frame where the metric tensor is essentially the identity matrix, perhaps up to the signs (signatures), so the raising/lowering of indices just changes some signs. This form of the coordinates in which the metric tensor is this simple is imposed by the condition "local Lorentz frame" on the right hand side. Note that the signs of the Riemann tensor components must be ignored although there's a big difference between positive/negative curvature. $\endgroup$ – Luboš Motl Feb 16 '13 at 14:14
  • $\begingroup$ Oh, that's an interesting conclusion! $\endgroup$ – user12345 Feb 16 '13 at 14:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.