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Can a photon be made to orbit a known (or undiscovered theoretical) body?

How massive would a black hole have to be for light to orbit it at 1km away from the singularity? Given that Gravitational lensing occurs, I assume that it should be possible to bend light in a full circle. How would you calculate, using minimal advanced general relativity, show the minimum mass of that black hole? What would be the orbital period according to different reference frames?

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marked as duplicate by Qmechanic, dmckee Oct 5 '12 at 14:32

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

related:… – DJBunk Oct 5 '12 at 14:27
up vote 1 down vote accepted

It's not hideously difficult to calculate the trajectory of a light beam in the Schwarzschild metric. You'll find the calculation in any introductory book on General Relativity. The result is that light will orbit the mass (whether it's a black hole or not) at a distance of $r = 3M$, where $M$ is the geometrical mass $Gm/c^2$. So:

$$ r_{orbit} = \frac{3Gm}{c^2} $$


$$ m = \frac{r_{orbit} c^2}{3G} $$

For a 1km orbit I get the mass to be $4.5 \times 10^{29}$ kg.

Later: Since the duplicate question doesn't give you the period of the orbit I'll go ahead and calculate it.

The radius is 1000m, and this is the radius in Schwarzschild co-ordinates, so by definition the circumference of the orbit is just $2 \pi r$ or $2000\pi$. But, if we are Schwarzschild observers time is dilated at a radius $r$, and the dilation factor is $\sqrt{1 - 2M/r}$. At $r = 3M$ the factor is $\sqrt{1/3}$ so we measure the light to be moving at $c/\sqrt{3}$. That means that we observe the period to be:

$$ \tau = \frac{2000\pi}{c/\sqrt{3}} $$

which is about 36$\mu$s.

Response to comment:

I don't think this is a good place to go through the derivation of the photon orbit radius because any GR textbook will have it. My preferred introductory book "A first course in general relativity" by Bernard F. Schutz derives this at the beginning of chapter 11. The derivation takes three pages and even then relies on results from earlier in the book. So instead I'll just outline the derivation.

The symmetry of the metric can be used to obtain an expression for the 4-momentum. Using this it's relatively easy to calculate an expression for the orbit, and from this we extract an effective potential, which for a photon is:

$$ V^2(r) = \left( 1 - \frac{2M}{r}\right) \frac{L^2}{r^2} $$

where $L$ is the angular momentum. The reason the effective potential is so useful is that we get a circular orbit only when $V(r)$ is at a minimum or maximum. A minimum give a stable orbit and a maximum gives an unstable orbit. So we look for a minimum or maximum by differentiating $V(r)$ with respect to $r$ and setting this to zero. Actually we'll make life easier by differentiating $V^2(r)$ and setting this to zero to get:

$$ \frac{-2L^2}{r^3} - \frac{-6ML^2}{r^4} = 0 $$

and this has just one solution at $r = 3M$. This turns out to be a maximum, so the orbit at $r = 3M$ is unstable i.e. if we perturb the orbit the tiniest distance away from a perfect circle the photon will either fly away from or fly into the black hole.

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Thank you, much clearer than the answer of the duplicate. – Meow Oct 5 '12 at 16:11
Why did you unaccept the answer? Not that I'm complaining, but I'm curious to know the reason. – John Rennie Oct 17 '12 at 15:48
On review, the 3 in the r=3Gm/c^2 seemed plucked out of thin air (undoubtedly because of my inexpertise in the subject). The first time I accepted it in a rush. – Meow Oct 17 '12 at 17:25
OK I've tried to explain where the $r = 3M$ comes from, but I think that to go any deeper than this you'll need to gird up your loins and dive into a GR textbook. – John Rennie Oct 18 '12 at 6:38
Thank you, I suppose I was overly harsh where the fault was my lethargy. – Meow Oct 18 '12 at 16:48

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