Photon sphere

This article is about the physics of photons under the influence of gravity. For the part of a star that emits light, seephotosphere.For thepseudoscientificspiritual belief, seePhoton belt.

An animation of how light rays can be gravitationally bent to form a photon sphere

Aphoton sphereorphoton circlearises in a neighbourhood of the event horizon of a black hole where gravity is so strong that emitted photons will not just bend around the black hole but also return to the point where they were emitted from and consequently display boomerang-like properties.As the source emitting photons falls into the gravitational field towards the event horizon the shape of the trajectory of each boomerang photon changes, tending to a more circular form. At a critical value of the radial distance from the singularity the trajectory of a boomerang photon will take the form of a non-stable circular orbit, thus forming a photon circle and hence in aggregation a photon sphere. The circular photon orbit is said to be thelast photon orbit.The radius of the photon sphere, which is also thelower bound for any stable orbit,is, for aSchwarzschild black hole,

$r={\frac {3GM}{c^{2}}}={\frac {3r_{\text{s}}}{2}},$

whereGis thegravitational constant,Mis the mass of theblack hole,cis thespeed of lightin vacuum, andr_sis theSchwarzschild radius(the radius of theevent horizon); see below for a derivation of this result.

This equation entails that photon spheres can only exist in the space surrounding an extremely compact object (ablack holeor possibly an "ultracompact"neutron star).

The photon sphere is located farther from the center of a black hole than the event horizon. Within a photon sphere, it is possible to imagine aphotonthat is emitted (or reflected) from the back of one's head and, following an orbit of the black hole, is then intercepted by the person's eye, allowing one to see the back of the head, see e.g. For non-rotating black holes, the photon sphere is a sphere of radius 3/2r_s.There are no stable free-fall orbits that exist within or cross the photon sphere. Any free-fall orbit that crosses it from the outside spirals into the black hole. Any orbit that crosses it from the inside escapes to infinity or falls back in and spirals into the black hole. No unaccelerated orbit with asemi-major axisless 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.

Another property of the photon sphere iscentrifugal force(note: notcentripetal) reversal.Outside the photon sphere, the faster one orbits, the greater the outward force one feels. Centrifugal force falls to zero at the photon sphere, including non-freefall orbits at any speed, i.e. an object weighs the same no matter how fast it orbits, and becomes negative inside it. Inside the photon sphere, faster orbiting leads to greater weight or inward force. This has serious ramifications for the fluid dynamics of inward fluid flow.

Arotating black holehas two photon spheres. As a black hole rotates, itdragsspace 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 theangular velocityof the rotation of a black hole, the greater the distance between the two photon spheres. Since the black hole has an axis of rotation, this only holds true if approaching the black hole in the direction of the equator. In apolar orbit,there is only one photon sphere. This is because when approaching at this angle, the possibility of traveling with or against the rotation does not exist. The rotation will instead cause the orbit toprecess.

Further information:Photon surface

Derivation for a Schwarzschild black hole

Since a Schwarzschild black hole has spherical symmetry, all possible axes for a circular photon orbit are equivalent, and all circular orbits have the same radius.

This derivation involves using theSchwarzschild metric,given by

$ds^{2}=\left(1-{\frac {r_{\text{s}}}{r}}\right)c^{2}\,dt^{2}-\left(1-{\frac {r_{\text{s}}}{r}}\right)^{-1}\,dr^{2}-r^{2}(\sin ^{2}\theta \,d\phi ^{2}+d\theta ^{2}).$

For a photon traveling at a constant radiusr(i.e. in theφ-coordinate direction),$dr=0$.Since it is a photon,$ds=0$(a "light-like interval" ). We can always rotate the coordinate system such that$\theta$is constant,$d\theta =0$(e.g.,$\theta =\pi /2$).

Settingds,dranddθto zero, we have

$\left(1-{\frac {r_{\text{s}}}{r}}\right)c^{2}\,dt^{2}=r^{2}\sin ^{2}\theta \,d\phi ^{2}.$

Re-arranging gives

${\frac {d\phi }{dt}}={\frac {c}{r\sin \theta }}{\sqrt {1-{\frac {r_{\text{s}}}{r}}}}.$

To proceed, we need the relation${\frac {d\phi }{dt}}$.To find it, we use the radialgeodesic equation

${\frac {d^{2}r}{d\tau ^{2}}}+\Gamma _{\mu \nu }^{r}u^{\mu }u^{\nu }=0.$

Non vanishing$\Gamma$-connection coefficients are

$\Gamma _{tt}^{r}={\frac {c^{2}BB'}{2}},\quad \Gamma _{rr}^{r}=-{\frac {B^{-1}B'}{2}},\quad \Gamma _{\theta \theta }^{r}=-rB,\quad \Gamma _{\phi \phi }^{r}=-Br\sin ^{2}\theta,$

where$B'={\frac {dB}{dr}},\ B=1-{\frac {r_{\text{s}}}{r}}$.

We treat photon radial geodesics with constantrand$\theta$,therefore

${\frac {dr}{d\tau }}={\frac {d^{2}r}{d\tau ^{2}}}={\frac {d\theta }{d\tau }}=0.$

Substituting it all into the radial geodesic equation (the geodesic equation with the radial coordinate as the dependent variable), we obtain

$\left({\frac {d\phi }{dt}}\right)^{2}={\frac {c^{2}r_{\text{s}}}{2r^{3}\sin ^{2}\theta }}.$

Comparing it with what was obtained previously, we have

$c{\sqrt {\frac {r_{\text{s}}}{2r}}}=c{\sqrt {1-{\frac {r_{\text{s}}}{r}}}},$

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

Hence, rearranging this final expression gives

$r={\frac {3}{2}}r_{\text{s}},$

which is the result we set out to prove.

Photon orbits around a Kerr black hole

Views from the side (l) and from above a pole (r). A rotating black hole has 9 radii between which light can orbit on a constantrcoordinate. In this animation, all photon orbits fora=Mare shown.

In contrast to a Schwarzschild black hole, aKerr (spinning) black holedoes not have spherical symmetry, but only an axis of symmetry, which has profound consequences for the photon orbits, see e.g. Cramerfor details and simulations of photon orbits and photon circles. There are two circular photon orbits in the equatorial plane (prograde and retrograde), with differentBoyer–Lindquistradii:

$r_{\pm }^{\circ }=r_{\text{s}}\left[1+\cos \left({\frac {2}{3}}\arccos {\frac {\pm |a|}{M}}\right)\right],$

where$a=J/M$is the angular momentum per unit mass of the black hole. There exist other constant-radius orbits, but they have more complicated paths which oscillate in latitude about the equator.

External links

• Step by Step into a Black Hole
• Virtual Trips to Black Holes and Neutron Stars
• Guide to Black Holes
• Spherical Photon Orbits Around a Kerr Black Hole