Timeline for Can we measure gravitational constant at high accuracy using a black hole?
Current License: CC BY-SA 4.0
14 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 2 at 17:10 | vote | accept | Tentacles3587 | ||
| Apr 2 at 17:10 | vote | accept | Tentacles3587 | ||
| Apr 2 at 17:10 | |||||
| Sep 7, 2022 at 20:15 | comment | added | PM 2Ring | @Jerry Mea culpa, although I did mention "that critical impact parameter calculation ignores the spacetime curvature caused by the photon itself" in astronomy.stackexchange.com/questions/25552/… OTOH, I expect that the error in the measurement of the potential energy of the chunk of matter that we drop into the BH is larger than the errors due to the photon energy. | |
| Sep 7, 2022 at 19:29 | comment | added | Zo the Relativist | @PM2Ring: to be hyper-pedantic (because I know it doesn't really matter, you're neglecting the back-reaction of the photon onto the spacetime, so you're still taking the same geometric approximation you use when you treat schwarzschild as a background for computing planetary orbits. | |
| Sep 6, 2022 at 16:51 | comment | added | PM 2Ring | @safesphere You don't actually need to do a 2 body calculation. Photon trajectories around a Schwarzschild black hole can be computed using $\phi=\int\frac{du}{\sqrt{b^{-2}-u^2+u^3}}$, where $u=r_s/r$ and $b\cdot r_s$ is the impact parameter. That's an elliptic integral of the 1st kind, which can be rapidly calculated to high precision using the arithmetic-geometric mean. See Carlson (1994) for details. | |
| Sep 6, 2022 at 16:15 | comment | added | TimRias | @safesphere How much accuracy can be achieved (and through what method) depends the orbital separation of the binary, the mass-ratio, compactness of the object orbitting the black hole. For example, of the object is compact and small enough, its internal composition would not be relevant for high accuracy calculation. All this is moot though, since no matter how accurate the simulation data, you would still not be able to measure G. | |
| Sep 6, 2022 at 15:29 | comment | added | John Doty | @safesphere Perhaps you're right for close orbits, but gravity has long range. See ssd.jpl.nasa.gov/orbits.html for extremely accurate orbital simulations. | |
| Sep 6, 2022 at 15:18 | comment | added | safesphere | @TimRias Are you saying that numerical gravity can produce a high accuracy result for a realistic two body problem? Even if this were the case, you would not know the density distribution of the real planet and therefore the actual stress-energy tensor. Plus a real black hole would be spinning with an accretion disk that would is described by the vacuum Kerr solution. So calculating a real binary system with numerical gravity to a high precision is a just pipe dream. | |
| Sep 6, 2022 at 9:58 | comment | added | Tentacles3587 | I agree with @TimRias (en.wikipedia.org/wiki/…) but how to find that data? | |
| Sep 6, 2022 at 9:56 | comment | added | TimRias | @safesphere No analytic solution, does not imply no highly accurate solution. Some aspects of orbital dynamics around a black hole have been calculated with thousands (!) of digits of accuracy. | |
| Sep 5, 2022 at 20:45 | comment | added | safesphere | @mmesser314 No, because there is no analytic solution for a two-body system in GR. So the result would not be of high accuracy. | |
| Sep 5, 2022 at 16:57 | comment | added | John Doty | @mmesser314 How, then, would you measure the mass of the planet? | |
| Sep 5, 2022 at 16:02 | comment | added | mmesser314 | +1. Could you have a planet and the black hole orbit their mutual center of mass, and infer the black hole mass from the radius of its orbit? | |
| Sep 5, 2022 at 15:13 | history | answered | John Doty | CC BY-SA 4.0 |