# Maximum velocity of a falling object at the event horizon?

1. Dec 30, 2015

### Bernie G

If an object is an infinite distance from a black hole and falls directly to the black hole without being affected by any other force, what is its velocity at the event horizon?

2. Dec 30, 2015

### jambaugh

Velocity is a frame dependent quantity and in GR a "frame" can be quite varied as you can pick all kinds of different coordinate systems, even with the same asymptotic behavior of a stationary (relative to the BH) distant observer. Keep in mind also that in the 4-dim formulation all 4-velocities are either null (for light like paths) or unit length (for massive objects). The magnitude means nothing beyond this. (your speed is always 1 second per second). In short...

Your question as it stands is ill posed.

This is in part because the way GR and SR formulates velocity is as a direction of motion within the unified space-time manifold. Acceleration then becomes a (pseudo) rotation leaving the magnitude invariant but changing the direction. You of course want the "inward" component but then the relativity creeps in because that depends on the orientation of the observer... and keep in mind we are talking here space-time orientation so "how fast the observer frame is moving" too. Complicating things further is that within the black hole "inward" is no longer a spatial direction but rather time-like.

One way to answer your question, and one that may fit well with ones visualization of black holes as static objects is to look at an embedding of the Black hole's Schwarzschild geometry into a larger dimensional flat space-time. This is not unique and you may find thus the answer I get here is also --again-- not unique but here it is.

The simplest embedding has as its spatial component a (half) parabola shape with the vertex of the cross section occuring at the event horizon. (see: http://demonstrations.wolfram.com/SchwarzschildSpaceTimeEmbeddingDiagram/) Thus as an in-falling object achieves the event horizon his spatial direction no longer has an "inward" spatial component as defined by the embedding. He is traveling along an infinite "parallel" spatial direction as he "rounds the turn" at the parabolic vertex on the event horizon. However this direction is the axis of a "hyper-tube" which is itself shrinking circumferentially and will in finite time become singular. What's basically happened is as you cross the event horizon the external non-compact coordinate t, becomes space-like, and the circumferential radius r, has become time-like (and decreasing in the future direction).

So the answer to your question, formulated this way, is his inward speed is zero at the event horizon. That's the best I can do in giving you a definitive answer.

3. Dec 30, 2015

### bcrowell

Staff Emeritus
The other natural frame of reference to consider is that of a static observer hovering at some infinitesimal distance above the horizon. Then the velocity of the infalling matter turns out to be $c$. So the answer can be zero, as in jambaugh's answer, or c, or anything in between. That covers all possible velocities, which shows that the question is ill-posed.

4. Jan 7, 2016

### Bernie G

In the same way the event horizon escape velocity is considered c, shouldn't the undisturbed infalling velocity be considered c? I can't put my finger on it but this seems disturbing.

5. Jan 7, 2016

### bcrowell

Staff Emeritus
You're missing the basic message, which is that none of these velocities are uniquely well defined.