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Oct 27, 2022 at 17:24 comment added OTH AHH! FINALLY! I thought I was going crazy. I was trying to compute $e^\mu_{\,\,\,\,\nu}$ inverse and then multiplying it with the 4-vector, and it wasn't working. Turns out I was missing the transpose from my matrix calculation. Scratch the last part. However, I think it's still missing a part of the answer, in particular how the lorenz boost is related to the freely falling observer.
Oct 27, 2022 at 17:14 comment added OTH I think there's something missing here. How do you get the "remember that gamma = (..)"? Shouldn't we specifically figure out what the magnitude of the boost is for us to go from the stationary observer frame to the moving observer frame? Furthermore, this basis does not seem to satisfy $u^{\prime\, m}=e^{\prime\,m}_{\,\,\mu}u^\mu=(1,0,0,0)$ when the $u^\mu$ is the 4-velocity of a freely falling observer, i.e., $u^\mu=(g^{-1},\pm (1-r_s/r),0,0)$. What gives? It does not seem like it's the correct solution.
Feb 26, 2018 at 8:25 history edited Michele Grosso CC BY-SA 3.0
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Feb 26, 2018 at 0:41 history edited Michele Grosso CC BY-SA 3.0
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Feb 26, 2018 at 0:22 history answered Michele Grosso CC BY-SA 3.0