NASA published a computer generated black hole image.

In the image you can clearly see the event horizon and the light of background stars graduating from "smeared" to normal.

However, between the event horizon and the smeared stars there is an area which looks rather normal. Why is this? If the smear is due to gravitational force, it should start right from the event horizon across the image, and not after some seemingly random distance.

  • $\begingroup$ en.wikipedia.org/wiki/Einstein_ring $\endgroup$
    – lemon
    Commented Jun 10, 2016 at 21:08
  • $\begingroup$ See also physics.stackexchange.com/q/148567/109928 for a rotating black hole with an accretion disk. $\endgroup$ Commented Jun 10, 2016 at 21:19
  • 2
    $\begingroup$ Why the down vote? It's a perfectly good question unless it's a duplicate? I am still looking for a simple derivation of the optical imaging properties of a black hole (or a general mass distribution), but I have not found anything that I find to be a satisfactory and easy to understand derivation. I think the OP made a good observation and the question is very reasonable. Given the importance of lensing in cosmology, we should both understand and explain it well. $\endgroup$
    – CuriousOne
    Commented Jun 10, 2016 at 22:29
  • $\begingroup$ @CuriousOne Agree, this is a good question and I also don't know any good answer short of 'numerically, this is what comes out'. $\endgroup$
    – user107153
    Commented Jun 10, 2016 at 22:57
  • 1
    $\begingroup$ Look carefully at the "normal" stars, trace a line from them past the BH and eventually, voila, there's that same star again. The smeared region is where light from directly behind the black hole is lensed to (lemon's link). Inside of that region light is being lensed by even larger angles, and is effectively looping under the BH. The smeared region looks like that because a very small area of space is being projected across a large area and so is "grainy", much like zooming too much on an image, but all the light within the smeared region (and all without to a lesser extent) is being lensed. $\endgroup$
    – zephyr
    Commented Jun 10, 2016 at 23:00

2 Answers 2


Fleshing out my answer in the comment above:

The light from all the stars in the picture is lensed (i.e. it's path bent) by the black hole (or BH). The closer it passes to the BH, the larger the angle it's path is bent by.

enter image description here

Stars far (in projection) from the BH, like A, are hardly lensed at all (in fact I significantly over-egged it in my diagram, should be a much straighter line from A to the observer!)

A star directly behind the BH (B) will be lensed in a ring (https://en.wikipedia.org/wiki/Einstein_ring). This is because multiple light rays, all bent by the same amount, make their way to the observer. In 2D only two of these rays are visible, but in 3D it forms a ring. This is the smear you're seeing, a very small area of space projected across a large area, giving a nasty smudge effect. Think of zooming too far on any picture and it becoming grainy.

But stars slightly off-center, like C, will also have light emitted in two different directions that reaches the observer. The first, the bottom ray in the picture, does not pass very close to the black hole, is not bent by a huge amount, and appears outside of the "smear" ring. The second must pass very close the BH, be diverted by a large angle, and will appear within the ring.

Unlike the area directly behind the black hole, this area of sky is projected over a similar (or even smaller) area than it came from, and so does not appear to be a grainy smudge.

So the stars within the ring can all be seen twice, once on the outside of the ring, then again, on the opposite side of the BH, within the ring.

See this excellent image that Previous has made and given me permission to reproduce (as well as his great, more optics based explanation below) to trace which stars are being seen twice.

enter image description here


A magnifying glass will produce an upright, enlarged (virtual) image when the distance to the object is smaller than the focal length, and an inversed image when the distance is larger than the focal length (in which case the size of the image decreases rapidly with increasing distance).

image size for different distances

(image by cmglee, wikimedia commons)

For a black hole, the gravitational lensing is stronger closer to the black hole: the focal length is variable, and shortens the closer you come to the black hole. Objects will have two images, an enlarged upright image further from the center, and an inversed (for the most part reduced) image closer to the center. In between you get a region where (angular) magnification becomes maximal, resulting in a "smeared" image.

The "smeared" part is the region where the image switches from upright to inverse (the equivalent of the region where the distance is close to the focal length, like the letters h, i, j in the picture).

  • $\begingroup$ Both yours and zephyr's answers helped me a lot but unfortunately I may select only one accepted answer. I really wish you could have merged it somehow, at least your edited image. $\endgroup$
    – Bharel
    Commented Jun 11, 2016 at 10:24
  • $\begingroup$ @Bharel zephyr can use the image if he wants to (and remove it from my answer if he does, no use in showing it twice). $\endgroup$
    – Previous
    Commented Jun 11, 2016 at 12:09
  • $\begingroup$ That's very generous, thank you! I think I will, but also try and lead poeple to read your explanation as well. It's a great discussion of the optics of the problem! $\endgroup$
    – zephyr
    Commented Jun 12, 2016 at 17:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.