9
$\begingroup$

In the recent releases of images of Sgr A*, simulated versions of what they expected were included along side the actual images they were able to get. What confuses me about these simulated images (and I believe this was the case for M87 as well) is that the shadow of the black hole seems to be off center from the ring that is visible on the image.

So my question is why is that the case? My initial thoughts are that it might be related to the anisotropic nature of accretion disks and the fact that we are looking at it at an unknown inclination, but even then it still seems a bit odd. It reminds me of an Einstein ring in the sense that gravitational lensing is playing a role here for light found around the black hole, but the fact that it’s off center confuses me and I’m not sure if that’s the right term given this isn’t lensing in the sense that it’s normally used.

Included is an image of what I’m referencing, taken from one of the videos from the press release from the other day.

Screenshot from one of the press release videos of Sgr A*

I got this image from a drive folder shared as ‘Additional visuals’ at the bottom of this website

Improve this question
$\endgroup$
2
  • $\begingroup$ @planetmaker Unfortunately I got this off the Google drive folder they put out there so I’m not sure where I could find this in an easy to include reference, but I’ll put up the link to the page with the Google drive folder $\endgroup$
    – Justin T
    May 13, 2022 at 18:58
  • $\begingroup$ Thanks for adding a link :) $\endgroup$
    – planetmaker
    May 13, 2022 at 19:01

2 Answers 2

Reset to default
11
$\begingroup$

The photon ring around a non-spinning Schwarzschild black hole is perfectly circular and centered on the black hole.

The photon ring around a spinning Kerr black hole is almost circular (except for very high spins) but is displaced from the centre of the black hole (e.g. see Takahashi 2005; Johanssen 2015). The amount of displacement is related to the spin and the inclination of the black hole spin to the line of sight (if the rotation axis points towards you with $i=0$, then the ring is undisplaced). The plot below (from Johanssen 2015) shows a calculation of the displacement (in units of $M$, where the Schwarzschild radius is $2M$) as a function of inclination for values of the spin parameter ($GJ/cM$) varing from $a=0.0$ (no displacement) to 0.9 in steps of 0.1 and then 0.998.

Displacement of photon ring

Improve this answer
$\endgroup$
1
  • 2
    $\begingroup$ Very interesting!! I have a follow up question based on your answer, but I’ll create a separate question for that and I’ll link it here once I do! $\endgroup$
    – Justin T
    May 13, 2022 at 23:22
6
$\begingroup$

The folder with the video this image comes from has a google docs document with captions and credits. The caption for this video reads "A video demonstrating the effect of inclination on the appearance of Sagittarius A* using a simulation."

An inclination of a thin disk with a hole in it and a light absorbing sphere in the center would look exactly like this with the disk part towards us unobscured and the part behind the black hole partially invisible due to being blocked-out by the black hole itself. (Yes, this simplistic view ignores many intricacies but might serve as mental picture nonetheless).

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .