| ▲ | lupsasca 6 days ago |
| Yes, but one issue is that the amount of redshift depends on the motion of the emitter, so we would have to artificially assign some four-velocity to your surroundings in order to give them some redshift. There doesn't seem to be a "natural" choice for how to do this. TLDR: redshift depends not only on the position of the source, but also its velocity. |
|
| ▲ | floxy 6 days ago | parent [-] |
| Since you don't notice any red-shift with your eyes in daily life, why is zero velocity relative to the camera not a natural choice? Or maybe I'm not following you? |
| |
| ▲ | lupsasca 6 days ago | parent [-] | | If you put the source infinitely far away (as we are doing here) and at zero velocity relative to the camera, then there is no redshift effect at all, so you could say that this is the choice of redshift that we made :) If you want the details, they're too long to put in this comment but essentially what I mean is that the r->infty limit of the redshift factor in Eq. (B22) of this paper is unity: https://arxiv.org/abs/2211.07469 | | |
| ▲ | floxy 6 days ago | parent [-] | | Maybe my original question wasn't specific enough? So you are saying there is no "extra" red-shifting due to the gravitational effects of the black hole for light coming from behind the black hole? Some photons skim "close" to the event horizon, and have to climb out of the gravitational well, but they don't end up red-shifted for the camera (because maybe they were blue-shifted on the way in from infinity?). Some photons sufficiently close to the event horizon will even do an orbit around the black hole (one or more times) before reaching the camera. And those photons that made the orbit maybe do have "extra" red-shift, because they are essentially no longer coming from infinity? Maybe? ??? |
|
|
