3
$\begingroup$

Is there an upper limit to how far out an Einstein ring can be visible? For black holes, is it a fixed multiple of their Schwarzschild radius?

$\endgroup$
2
  • $\begingroup$ What do you mean by "how far out"? $\endgroup$
    – ProfRob
    Mar 1, 2022 at 21:48
  • 1
    $\begingroup$ @ProfRob Based on the second sentence, one would guess it's angular distance seen by an observer. $\endgroup$
    – uhoh
    Mar 1, 2022 at 22:04

1 Answer 1

5
$\begingroup$

The angular diameter (in radians) of an Einstein ring is $$\theta = \sqrt{\frac{4GM}{c^2}\;\frac{D_{LS}}{D_S D_L}}\ ,$$ where $M$ is the lens mass, $D_S$ and $D_L$ are the angular diameter distance to the source and lens respectively and $D_{LS}$ is the angular diameter distance between source and lens.

There isn't therefore any mathematical maximum, it depends on the geometry and lens mass.

$\endgroup$
0

You must log in to answer this question.

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