We are currently studying general relativity in school and also had a brief look at gravitational lensing. My teacher gave us the angle of deflection between the apparent location of a star and its actual location (as seen on picture) to be

theta = 1.75" R/d where R is the radius of the object and d is the distance at which the light beam passes the massive object.

Apparent and real locations of a star who's light has been deflected by a massive object

Our teacher was not very clear on where this is applicable and showed no derivation.

The equation does not contain mass or density as a variable which irritates me. Does this mean that the formula only works for a specific stellar object (e.g. Is it tailored to the sun's lensing or does it work for a different object as well?)

Could it also be that R and d are different variables than what I have mentioned?


1 Answer 1


The derivation of the gravitational lensing equation requires more maths than you'll learn at school, and indeed more than you'll learn in your first year or two as an undergraduate. General relativity isn't taught in any detail until your postgraduate studies, or possibly in the final year of your degree.

The full equation is:

$$ \theta = \frac{4GM}{c^2b} $$

where $G$ is Newton's constant, $M$ is the mass of the object, $b$ is the distance of closest approach and $c$ is the speed of light. Strictly speaking $b$ is a quantity called the impact parameter, but it's approximately the distance of closest approach.

The radius of the object doesn't appear in the equation because the spacetime geometry outside a spherically symmetric object depends only on the object mass and not on its radius.

This equation is derived in an approximation called the weak field limit, which basically means the spacetime curvature is small. It works as long as $\theta$ is small but would fail very close to a black hole where the bending of the light becomes large. In practice the gravitational lensing we've observed is all in the weak field limit so the equation works fine.

  • $\begingroup$ Where does the 1.75 arcseconds come from? $\endgroup$
    – Jaywalker
    Jan 27, 2016 at 18:04
  • $\begingroup$ I understand also from what I have googled that $2GMc^-2$ is the Schwarzschildradius which is probably the R in my original equation right? $\endgroup$
    – Jaywalker
    Jan 27, 2016 at 18:07
  • $\begingroup$ @Jaywalker: $1.75$ arcseconds is the value of $\theta$ when light just grazes the surface of the Sun. So the equation you give only applies to the Sun and $R$ is the radius of the Sun. At grazing $d=R$ so $R/d=1$ and we get $\theta = 1.75$ arc seconds. $\endgroup$ Jan 27, 2016 at 18:18

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