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May 7, 2017 at 8:18 comment added OTH Nevermind, I got it now! For clarification to others it may be helpful to note how I repeated the above derivation: We have $d\xi^m=e^m_{\ \ \mu} dx^\mu$, and if we set $dx^0=0$, we get the above result for $d^3 \xi=\sqrt{det\gamma}^{-1} dx^1 dx^2dx^3$
Apr 7, 2017 at 3:25 vote accept OTH
Mar 7, 2017 at 9:12 comment added OTH Ah, I see, so in this case the proper volume would be: $d\Sigma = det (e_m^\mu)^-1 dx dy dz=r^2 \sin (\theta ) \sqrt{\frac{r}{r-2 m}} dx dy dz$? Just to confirm I understood it properly.
Mar 7, 2017 at 8:22 history answered Bence Racskó CC BY-SA 3.0