# General Relativity (Level 2): Can you fall into a black hole?

These Level 2 videos analyze General Relativity topics that are a bit contentious.

In this video, we ask, “With the speed of light going to zero at the event horizon, can you actually cross it?” The standard answer is often formulated imperfectly. As an analogy we describe an artificial intelligence (AI) on a pendulum, with a clock rate that varies depending on angular speed.

Let me elaborate a bit.

There is a very common confusion, among us physicists, that conflates the proper time T of an infaller (finite time), with the Schwarzschild time t to reach the event horizon (infinite time). The latter, t, is not “just a coordinate” but physically measurable as the Shapiro delay (one of four classic tests of GR).

Thought experiment: shoot two photons, A and B, toward a black hole. Plot this in Schwarzschild coordinates, (r, t). Photon A bounces off a mirror a distance epsilon outside the event horizon, and returns after t = 1 billion years. (This is the Shapiro delay.) Photon B continues toward the horizon. But can we say that it did cross? There could be another mirror at 0.001*epsilon, so that B bounces and returns after 1 trillion years. Given the possibility of a bounce, you cannot say that B “has crossed” the event horizon.

For a non-evaporating black hole, we can map infinite-t to finite T: no problem, see for example Kruskal-Szekeres coordinates. For a finite-t, evaporating black hole you cannot do this.

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### 9 Responses to General Relativity (Level 2): Can you fall into a black hole?

1. Michael McGinnis says:

Because the Shapiro Delay to an event horizon is always infinite, no matter the starting location, it’s not possible for light to travel to an event horizon. Therefore, nothing can.

Einstein was right that black hole event horizons are impossible.

• bernander says:

Michael, you’re zeroing in on the key confusion addressed in the video.

If the lifetime of the black hole is also infinite (in faraway coordinate time t), then we have two infinities, and we need to be careful not to make an Achilles-and-tortoise argument that Achilles doesn’t catch up with the tortoise. Mapping to Kruskal-Szekeres coordinates, for example, and calculating null geodesics, we find that the photon indeed can reach the event horizon and cross it. This is the standard treatment of non-evaporating black holes: there is a region of the space-time manifold separated from us by a horizon.

But with an evaporating black hole, with a finite lifetime, the photon cannot catch up with the receding event horizon.

Many people conflate the two cases, which is why I made the video. Some insist that photons cross the horizon of evaporating black holes (they don’t). Some insist that photons cannot cross the horizon for eternal black holes (they can).

• Michael McGinnis says:

I disagree that light can cross the event horizon of even an eternal black hole. Anything that cannot happen in finite time cannot happen. If it happened, that event would define a definite time.

2. Michael McGinnis says:

The speed of light doesn’t drop to zero at the event horizon. The speed of light in vacuum is the same, everywhere. That’s the foundation of Relativity. Any interpretation that has light slowing down due to gravity is a non-relativitstic interpretation.

Spacetime is dilated in a gravity well. The radial distance about a mass is stretched.

If you’re going to use Relativity then you are required to adhere to the assumptions of Relativity.

• bernander says:

Michael, the speed of light is always measured locally to be the same everywhere — this is indeed the foundation of Relativity, as you say.

However, the Shapiro delay can be interpreted as “slow light.” Again, it’s measured, but over a distance. “Slow light” is the interpretation used by Taylor and Wheeler in their textbook “Exploring Black Holes” which has an entire chapter with the title “Light Slowed Near Sun.”

I elaborate on the two viewpoints in this video.

Writers are sloppy saying the speed of light is c, while they really should write is measured locally to be c. (That’s a key message of this video)

• Michael McGinnis says:

In Science, the final arbiter of what is reality is observation. To make a distinction between the way the universe is measured to be and the way the universe “really” is, is unjustifiable.

3. Michael McGinnis says:

If event horizons cannot form then this should have observable consequences for merging black holes. If two Schwarzschild black holes were to spiral into each other and merge, there would be an abrupt cutoff for the “ringdown” of the gravitational waves. But, if nothing can fall to an event horizon in finite time, the ringdown should not have an abrupt cutoff but should fade to undetectability while increasing in frequency.

4. Anwar Shiekh says:

Nice video, but I have only one slight disagreement

You have the faller ‘stuck’ at the event Horizon as the black hole Hawking evaporates;
I think it important to realize that the faller is now part of the black hole and so will
also evaporate and so not follow the Horizon down.

http://m-hikari.com/astp/astp2012/astp21-24-2012/shiekhASTP21-24-2012.pdf

• bernander says:

Thanks. Nice paper yourself.

And yes, you may be correct that a finite falling mass would evaporate too. It might depends on the exact mechanism of evaporation.

I intended for the falling astronaut to represent either a time-like geodesic or an infinitesimal test mass in the absence of any black-hole growth.

Either way, it sounds like we agree that the astronaut would not hit the center before evaporation.