What does it exactly mean when we say that a photon or light is "frozen" at the event horizon?

Question

Objects on this site are said to "freeze" at the event horizon, but what about photons themselves?

I have read this question:

At the event horizon $v_{eff} = 0$ and the light beam is frozen at the horizon unable to move outwards.

If you shoot a light beam behind the event horizon of a black hole, what happens to the light?

And this one:

At large distances (large r) the velocity tends to 1 (i.e. c) but close to the black hole it decreases, and falls to zero at the event horizon. But, but, but, be absolutely clear what you're calculating. All you're calculating is the speed of light in your co-ordinate system i.e. the result you get applies only to you. Other observers in other places will calculate a different value, and every observer everywhere will find the local speed of light to have the same value of c.

Speed of light in a gravitational field?

Now two very important rules about photons/light that come to mind:

1. photons (or classically light) do not have a rest frame (special relativity)

2. photons (or light) can never be brought to rest (they move at speed c always)

The speed of photons is always c, when measured locally in vacuum. This question is about a non-local measurement.

However, when I measure the speed of light based on the second answer above, I get zero at the event horizon, which might be interpreted, as being at rest, that is having a rest frame (from an external point of view).

Question:

1. What does it exactly mean when we say that a photon or light is "frozen" at the event horizon?

Answer 1

It doesn't mean a lot.

Event horizons are outward-moving lightlike surfaces, but the curvature is such that usually their overall size doesn't increase (except during black hole formation, accretion of matter, and mergers).

If you cover a part of the event horizon with local inertial coordinates, the speed of the light on the horizon is $c$ with respect to those coordinates. But there isn't any inertial or approximately inertial coordinate system that covers the entire horizon (because of spacetime curvature). Some common coordinate systems that do cover the whole horizon use an $r$ coordinate that is constant on the horizon (when it isn't expanding). The $dr/dt$ speed of outward-moving light on the horizon is therefore zero. The same light at the same time has a speed of $c$ in the inertial coordinates. The light isn't at rest. It's moving at the speed of light, which just happens to be $0$ in certain coordinate systems.

Answer 2

As far as I understand, you won't see visible light as it is. Anything you see will be highly red-shifted. By frozen, we mean that we will never see it fall inside the black hole. From the perspective of a distant viewer, it will seem like it takes an infinite amount of time to cross the event horizon, because of time dilation. But if a person were, say, hypothetically riding on that photon, he would find it crossing the event horizon in a finite amount of time. And he would measure its speed to be c.

Answer 3

This is on the general expression "frozen light" .

In this article

A pulse of light has been stopped in its tracks with all its photons intact, reveal US physic

In a vacuum, light travels at the phenomenal speed of approximately 300,000,000 metres per second. Scientists can exploit the way that the electric and magnetic fields in light interact with matter to slow it down.

Over the last few years, scientists have become masters of the light beam. Speeds of a few metres per second are now reached routinely in laboratories around the world. It is rather harder, however, to stop light completely and previous attempts have halted light but lost its photons in the process.

Mikhail Lukin and colleagues at Harvard University in Cambridge, Massachusetts managed to stop light without this loss by firing a short burst of red laser light into a gas of hot rubidium atoms.

This is then “frozen” with the help of two control beams. The light in the control beams interacts with the rubidium atoms to create layers that alternately transmit and reflect the pulse.

As the signal tries to propagate through these layers, the photons bounce backwards and forwards between them. As a result, the pulse makes no forward progress – the light is “frozen” in place. The pulse is set free when the control beams are turned off.

italics mine.

In terms of photons, a frozen light beam means that the addition of the wavefunctions of the photons adds up to frozen light

The strong gravitational field of a black hole traps photons in trajectories around the hole at the horizon. It is the light that is frozen, not the photons. The photons are are in orbits, but the light is frozen. Their wavefunctions add up to a frozen electromagnetic wave, analogous to the frozen light in the experiment above.