In Schwarzschild coordinates the photon sphere is located at an "r-parameter" of $r=3GM/c^2$. If we are watching the photon sphere from infinity using a telescope such as the Event Horizon Telescope there will be a gravitational lensing effect making the photon sphere look larger. The expression for this is $r_{obs}=r(1-\frac{2GM}{rc^2})^{-0.5}$. I do not know if this formula for gravitational lensing enlargement is specific to Schwarzschild coordinates or if you get the very same expression using isotropic coordinates. Anyhow, using Schwarzschild coordinates the expression for the apparent radius of the photon sphere for an observer watching from infinity as a function of the central mass and the "r-parameter" is $r_{obs= \sqrt{27}GM/c^2}$.

Question: What is the expression for the apparent radius of the photon sphere for an observer watching from infinity in isotropic coordinates as a function of the central mass and the "isotropic r-parameter"?

The Schwarzschild isotropic metric looks like: $$ -c^2d\tau^2 = -\left(\frac{1-GM/2rc^2}{1+GM/2rc^2}\right)^2c^2dt^2 +(1+GM/(2rc^2))^4[dr^2 +r^2( d\theta^2 +\sin^2\theta d\phi^2)]$$

From this I am able to get that the time dilation formula must be:


and the velocity of light:


I guess that you can somehow decide the orbital velocity of a circular orbit in isotropic coordinate and when this velocity becomes equal to the speed of light, that is where the radius of the photon sphere is. I really have no clue to what the expression for gravitational lensing enlargment will be using isotropic coordinates.

In solar system dynamics relativistic effects are accounted for by using the post-newtonian expansion which is a low order expansion of the isotropic metric and, as I understand it, treat the "r-parameter" as a real radial distance.


1 Answer 1


The isotropic coordinate $r'$ is obtained from the Schwarzschild radial coordinate $r$ using the coordinate transformation:

$$ r = r'\left(1 + \frac{M}{2r'}\right)^2 $$

The derivation of this is described in my answer to Schwarzschild metric in Isotropic coordinates. So the position of the photon sphere in the isotropic coordinates is given by:

$$ \frac{3GM}{c^2} = r'\left(1 + \frac{M}{2r'}\right)^2 $$

  • $\begingroup$ It is really that simple? I mean if you define the photon sphere radius to be $r=3GM/c^2$ in Schwarzschild coordinates and use the coordinate transformation that is what you end up with. If you do it the other way around and first find the radius in isotropic coordinates where the orbital velocity of a body in circular motion calculated from the metric expressed in isotropic coordinates is equal to the speed of light in isotropic coordinates and then "transform back" to Schwarzschild coordinates it is obvious that you will end up with $r=3GM/c^2$? $\endgroup$
    – Agerhell
    May 2, 2019 at 14:57
  • $\begingroup$ John Rennie's answer is not well analyzed, because he interchanges $M$ with $GM/c^2$ in his last equation. $\endgroup$
    – omivela17
    Feb 20 at 3:00
  • $\begingroup$ What's more, there is no further discussion about the implication of solving this equation for $r'$ and we can notice it has two positive possible roots, which one is the correct? And if both roots rise up as the single Schwarzschild photon radius, what is the meaning or the physics behind this having two values for $r'$? $\endgroup$
    – omivela17
    Feb 20 at 3:03

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