# Photons emitted at the event horizon?

While looking through the questions, a came across a section about black holes. I immediately though; what would happen if an atom is orbiting a black hole and emitted a photon perpendicular to the event horizon, going away from the black hole.

How would light going away from a black hole react to the gravity?

Photon "stuck" on the event horizon of a black hole actually talks about a photon stopping at the event horizon, but my question is about a photon just outside of the event horizon and if the photon is slowed down.

• It will lose energy as it is redshifted but it will escape at the speed of light. Commented Apr 22, 2014 at 20:21
• Possible duplicates: physics.stackexchange.com/q/98980/2451 and links therein. Commented May 1, 2014 at 13:42

The question Speed of light originating from a star with gravitational pull close to black-hole strength? is very nearly, but not quite, a duplicate. However my answer to that question applies to your question as well. When you ask:

How would light going away from a black hole react to the gravity?

you have to extend your question to indicate what observers you are asking about. For an observer hovering next to the atom emitting the light the photon will be travelling at $c$. For an observer far from the black hole the speed of the photon will be less than $c$ and given by:

$$v = c\left(1 - \frac{2GM}{c^2r}\right)$$

where $M$ is the mass of the black hole and $r$ is the distance from the centre of the black hole. It is common in general relativity that diffrent observers will observe different behaviour, which is one of the (many!) things that makes GR confusing for beginners.

In addition, as Brandon and Frederic have said, the light will lose energy and red shift as it moves away from the black hole. If $\nu_r$ is the original frequency when the light is emitted at a distance $r$ then the frequency at infinity is given by:

$$\nu_\infty = \nu_r\sqrt{1 - \frac{2GM}{c^2r}}$$

In the presence of strong gravitational sources, such as black holes, photons will experience redshift. This means that with respect to an observer in a region of a weaker gravitational field, they lose energy. This is equivalent to the statement that their frequency decreases and their wavelength increases. This is a consequence of time dilation. The velocity of the photon does not change: it is fixed at $c$.

The same holds true for a photon moving away from the black hole, starting just outside the horizon.

As stated in a previous answer, the photon emitted in a radial direction will move away from the black hole keeping the same direction with speed $$c$$ to infinity, with a decreasing frequency (measured by a stationary observer).

This can be completed with the effect of general relativity on the photon geodesic which will be different from that of classical mechanics.

Applying the Schwarzschild metric to an outgoing geodesic $$\frac{dr}{dt}>0$$, it is written as: $$ct=r+R_s\ln{(\frac{r}{R_s}-1)}+constant\ \ \ \ \ [A]$$ with $$r$$ radial coordinate of the photon, $$c$$ speed of the light in vacuum and $$R_s$$ Schwarzschild radius $$= \frac{2GM}{c^2}$$.

When $$r\gg R_s$$, the logarithm term is negligible and $$[A]$$ turns into: $$ct\simeq r+constant$$ which is the equation for radial light beams in a flat space-time (classical mechanics).

In the hope of having usefully completed the answer,

Best regards.