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Dec 31, 2020 at 6:47 history bounty ended Faber Bosch
Dec 31, 2020 at 6:45 vote accept Faber Bosch
Dec 30, 2020 at 4:08 history edited user142288 CC BY-SA 4.0
add discussion on correction
Dec 28, 2020 at 8:45 vote accept Faber Bosch
Dec 28, 2020 at 15:44
Dec 28, 2020 at 8:02 vote accept Faber Bosch
Dec 28, 2020 at 8:02
Dec 28, 2020 at 7:14 comment added user142288 1, $dA=\sqrt{\gamma} dr d\phi$; 2, taking the limit, such that the sun can be regarded as point, 3, the $1/2$ comes from the integrating out the $dr$
Dec 28, 2020 at 7:11 comment added Faber Bosch I don't see where the factor of half comes from. Can you please elaborate? $dA=\sqrt{\gamma}rd\phi=r^2/\sqrt{1-r_s/r}d\phi$ ?. We are not allowed to use the formula $\frac{1}{2}(\text{base})(\text{height})$ in curved spaces, right?
Dec 28, 2020 at 7:04 comment added user142288 The mesure is $\sqrt{\gamma}=r/\sqrt{1-1/r}$, then take the limit, where the sun can be regarded as point.
Dec 28, 2020 at 7:01 comment added Faber Bosch Okay, I see. I didn't refresh the page earlier. Please write down the precise formula for $dA$ before taking the limit. Because I want to know the order of the error.
Dec 28, 2020 at 6:56 comment added user142288 @Faber Bosch see the renewed answer.
Dec 28, 2020 at 6:55 comment added Faber Bosch State the reason why you think @JerrySchirmer 's comment is irrelevant? How do you calculate infinitesimal area on the $r$-$\phi$ plane? and how do you adjust for the $\sqrt{-g_{00}}$ factor for considering coordinate time instead of proper time?
Dec 28, 2020 at 6:49 comment added user142288 @Faber Bosch, I think that is sufficient.
Dec 28, 2020 at 6:48 history edited user142288 CC BY-SA 4.0
improve
Dec 28, 2020 at 6:21 comment added Faber Bosch Showing $r^2\frac{d\phi}{d\tau}$ is not sufficient. See @JerrySchirmer 's comment.
Dec 28, 2020 at 5:43 history answered user142288 CC BY-SA 4.0