Photon sphere

From Wikipedia, the free encyclopedia

A photon sphere is a spherical region of space where gravity is strong enough that photons of light are forced to travel in orbits. The formula to find the radius for a circular photon orbit is: r=3GM/C2. Because of this equation photon spheres can only exist in the space surrounding an extremely compact object, such as a black hole.

As photons travel near the event horizon of a black hole they can escape being pulled in by the gravity of a black hole by traveling at a nearly vertical direction known as an exit cone. A photon on the boundary of this cone will not completely escape the gravity of the black hole. Instead it orbits the black hole. These orbits are not stable.

The photon sphere is located further from the center of a black hole than the event horizon and ergosphere. Within a photon sphere it is possible to imagine a photon that starts at the back of your head and orbits around a black hole only then be seen by your eyes. For non-rotating black holes, the photon sphere is a sphere of radius 3/2 Rs, where Rs denotes the Schwarzschild radius (the radius of the event horizon) - see below for a derivation of this result. No unaccelerated orbit with a semi-major axis less than this distance is possible, but within the photon sphere, a constant acceleration will allow a spacecraft or probe to hover above the event horizon.

A rotating black hole has two photon spheres. As a black hole rotates it drags space with it. The photon sphere that is closer to the black hole is moving in the same direction as the rotation, whereas the photon sphere further away is moving against it. The greater the angular velocity of the rotation of a black hole the greater distance between the two photon spheres. Because the black hole has an axis of rotation this only holds true if approaching the black hole in the direction of the equator. If approaching at a different angle, such as one from the poles of the black hole, to the equator there is only one photon sphere. This is because approaching at this angle the possibility of traveling with or against the rotation does not exist.

[edit] Derivation of the Photon Sphere Radius

This derivation involves using the Schwarzschild metric, given by:

ds^{2} = (1 - \frac{2GM}{rc^{2}})c^{2}dt^{2} - (1 - \frac{2GM}{rc^{2}})^{-1}dr^{2} - r^{2}(\textrm{sin}^{2}\theta d\phi^{2} + d\theta^{2})

For a photon travelling at a constant radius r (ie. in the Φ-coordinate direction), ds, dr and dθ all must equal zero (the consequence of ds = 0 is a "light-like interval").

Setting ds, dr and dθ to zero, we have:

(1 - \frac{2GM}{rc^{2}})c^{2}dt^{2} = r^{2}\textrm{sin}^{2}\theta d\phi^{2}

Re-arranging gives:

\frac{d\phi}{dt} = \frac{c}{r\textrm{sin}\theta}\sqrt{1 - \frac{R_s}{r}}

where Rs is the Schwarzschild radius

Now, the speed of light in the Φ-coordinate direction is given by r\frac{d\phi}{dt}, so we can write the speed of light along the orbital path as:

v = \frac{c}{\textrm{sin}\theta}\sqrt{1 - \frac{R_s}{r}}

But, as a direct result of Newton's Law of Universal Gravitation for an object travelling in a circular orbit at a radius r about a central body, M, the speed of light may also be given by:

v = \sqrt{\frac{GM}{r}}

We can write this in terms of the Schwarzschild radius:

v = c\sqrt{\frac{R_s}{2r}}

Putting our two expressions for the speed of light, we have:

c\sqrt{\frac{R_s}{2r}} = c\sqrt{1 - \frac{R_s}{r}}

where we have inserted \theta = \frac{\pi}{2} radians (imagine that the central mass, about which the photon is orbitting, is located at the centre of the coordinate axes. Then, as the photon is travelling along the Φ-coordinate line, for the mass to be located directly in the centre of the photon's orbit, we must have \theta = \frac{\pi}{2} radians).

Hence, rearranging this final expression gives:

r = \frac{3}{2}R_s

which is the result we set out to prove.

[edit] External links

[edit] References

General Relativity: An Introduction for Physicists

This relativity-related article is a stub. You can help Wikipedia by expanding it.
Languages