# Can someone adjudicate the conflicting answers to "what is the velocity when falling at the event horizon of a black hole"?

I personally like Sarat Kant's answer at Why can an object not reach the speed of light by falling in a gravitational field with constant acceleration?. It makes sense to me. I would just add that plugging in the numbers, Sarat's math leads to the interesting result that the fall velocity across the event horizon from a very far away distance is 0.75c.

But others say that the speed cannot be defined at all becasue there is no meaningful reference frame to use, which seems inconsistent to me. After all, we have a physical length for the event horizon radius. Does not the specification of a length imply a reference frame?

And yet others say that the fall velocity is the same as the escape velocity at If I throw a ball at a black hole, will the ball exceed the speed of light when it reaches the event horizon?.

Is there some sort of officiating body at Physics Stack Exchange who can pick one of these three answers as the "conventionally accepted" answer? That is, to that question about throwing a ball at a black hole, is the velocity as it crossed the event horizon (a) slower than lightspeed, (b) undefinable, or (c) lightspeed? Or is this a question with no conventionally accepted answer?

To clarify: at least at wikipedia, I can see a description of the black hole at the center of the Milky Way as being at a distance of 25,000 lightyears and having a radius of 118 million meters. So the reference frame I am using is the one defined by those lengths (which, in turn, implies that time is passing as measured by my wrist watch here on Earth, since length can always be measured as ct, so the 25,000 lightyears implies time passes as I measure it). So I am postulating a falling ball that is currently at some definable position, say 118 million meters + 300 meters from the black hole center, at time = noon. My definition of velocity will be the (distance 300 meters)/(time to reach the event horizon). So if the object fell at light speed of 300 m per microsecond, it would arrive at 1 microsecond past noon, as I would calculate it (not as I would see it, I understand that I cannot see it ever). I think that I would calculate it arriving at noon + 1/0.745 microseconds by my watch based on the final equation at Paul T's answer at Why can an object not reach the speed of light by falling in a gravitational field with constant acceleration? plugging in an initial distance of very large and a final distance of the black hole radius.

• Dec 5, 2020 at 17:54
• The velocity” is not meaningful but various radial-infall velocities like $dr/d\tau$ and $dr/dt$ in Schwarzschild coordinates or other coordinates are well-defined and can be calculated. Dec 5, 2020 at 18:03
• There is no officiating body at PSE. Dec 5, 2020 at 18:05
• If an answer isn’t using General Relativity, and doesn’t explain how it is defining velocity, it can’t properly answer a question about a black hole. Dec 5, 2020 at 18:19
• Thank you, G. Smith! I believe that you answered my question as option (a), but I wonder if it can be expressed as for a simpleton like myself. I live on planet Earth. There is a black hole 25,000 light years away from me with radius 118 million meters per wikipedia. I want to know the velocity of a ball falling into that black hole in terms of dr/dt - my measure of r and my measure of t. That is, it was 118 million plus 300 meters away from the center by my reckoning at noon. Where will it be at noon plus 1 microsecond by my watch? I think the answer will be 118 million plus 75 m. Dec 5, 2020 at 19:01

That is, to that question about throwing a ball at a black hole, is the velocity as it crossed the event horizon (a) slower than lightspeed, (b) undefinable, or (c) lightspeed? Or is this a question with no conventionally accepted answer?

The question is not meaningful. It would be if it is related to shell observers close to the horizon. They would record the speed of the ball close to the speed of light. This is a limiting case though because such an observer can't exist at r=2M, the horizon.

But you can relate your question to a photon emitted radially outwards at the horizon. This photon "stays" at r=2M and moves with the speed of light in the ball's restframe at r=2m. The other way round makes no sense because said photon doesn't have a restframe.

All the answers are correct. It all depends on your frame of reference (where the observer stands) - i.e. what you measure your velocity relative to.

Assume the falling object was dropped (zero initial velocity) at infinity from the black hole

If you ride with the falling object, then you would not, initially at least, detect any acceleration. You would feel weightless. From this perspective, you have not accelerated at all. Your velocity relative to yourself at the event horizon is zero. Of course, eventually the difference in gravity between your feet and your head would rip you apart.

If you are far away from the black hole, you would see the in-falling object slow down, go dark and freeze at the event horizon. That's because the light is struggling to get out, so you could argue the velocity of the in-falling object at the event horizon is zero.

If you were able to maintain a constant distance from the black hole, outside the event horizon (you would need to be accelerating away from it), you would see the in-falling object accelerate past you and then slow, fade and freeze as before. The velocity as it past would be the same as the escape velocity at that point, which would be less than the speed of light - by definition the event horizon is where the escape velocity is the speed of light.

Velocity is always relative so without a reference frame, it is a meaningless term. At no point would the object exceed the speed of light for any observer. There are some pretty weird effect around black holes - frame dragging, space and time dilation so if you really want to understand this you need to go to the maths.

• I understand what I would see, but what I would see is often different than what I would calculate. For example, say an astronaut is traveling to a point one lightyear distant at 0.5c. Two years after he left the Earth, I could look through a telescope and see him enroute, but I would calcuate that he is arriving at his destination at that time. I want to know is what value would I calculate for dr/dt when the object is at r = the black hole radius (say, 10 km) and t is time as recorded on my wrist watch. Dec 5, 2020 at 21:20
• @Ralph exactly as the answerer said, what you calculate depends on your frame of reference. It is manifestly a coordinate dependent calculation. An observer close to the horizon measures the velocity to be greater than the observer who measures it at infinity. Dec 5, 2020 at 21:43
• @ralphberger: Your $t$ is a local time coordinate defined near you/now. It has no meaning at an event far away from you/now. Hence neither does $dr/dt$ Dec 6, 2020 at 13:06

It's absolutely impossible for a massive object to move at the speed of light or higher in general relativity. At every location in spacetime, there's a light cone. A beam of light has a worldline that lies on the light cone, and any massive object has a worldline that lies strictly inside of the light cone. If you throw a massive object and a moment later emit a pulse of light, the light will always catch up to the object.

It's common in general relativity to use coordinates in which the speed (defined as something like $$|dx/dt|$$) of massive objects can be larger than $$c$$. In these coordinates, the speed of a beam of light is not $$c$$; it is (at least in some directions) larger than $$c$$, and in all directions it's larger than the maximum speed of any massive object. Unfortunately, the constant $$c$$ is often called "the speed of light", and many people don't understand the difference, so they may think that a speed larger than $$c$$ implies motion that's faster than light in some sense. It doesn't.

Another source of confusion in the case of black holes is that you can (in principle) conduct the following experiment. You station traffic cops at various heights above the event horizon, armed with radar guns, and toss an object past all of them into the hole. As the object passes each cop, they report the reading on their radar gun. These readings approach the speed of light as the object approaches the horizon. This does not mean that the object's speed equals the speed of light at the horizon. The divergence of the relative speed of the object and the cops is actually due to a coordinate singularity in the placement of the cops (their acceleration approaches infinity at the horizon). It has nothing to do with the motion of the infalling object.

The motion of a massive object can only be sensibly defined relative to the motion of some other massive object at the same location. Outside the event horizon, you can argue that objects hovering at a fixed height make reasonable reference objects. At the horizon and inside it, it's impossible to hover at a fixed height, and there's no obviously sensible family of reference objects that you can use. The infalling object's speed relative to any other nearby object is still well defined, but it can be any sublight speed, depending on the motion of the nearby object, and none of those values is more correct than any other.

Sarat Kant's answer was (in part)

When a object nears the speed of light, its mass starts increasing in relation to its speed. So the acceleration produced due to the gravitational force on that body will decrease due to its increasing mass. This acceleration keeps decreasing at such a rate that the particle will never reach the speed of light.

As Paul T. pointed out in a followup comment, the gravitational force is also proportional to mass, so wouldn't it increase to compensate? Given the freedom to choose any coordinates, you could probably invent some interpretation of the terms (mass, speed, etc.) such that the answer is technically correct, but I think it's at least misleading and unhelpful.

Sarat's math leads to the interesting result that the fall velocity across the event horizon from a very far away distance is 0.75c.

How did you calculate that? Note that the equation $$mgh=\frac12mv^2$$ in that question is completely wrong.

There is a black hole 25,000 light years away from me with radius 118 million meters per wikipedia. I want to know the velocity of a ball falling into that black hole in terms of dr/dt - my measure of r and my measure of t.

There's no such thing as "your" $$r$$ or $$t$$. You can use whatever coordinates you want.

In special relativity it's common to work in an inertial frame in which you (the person making the observation) are at rest. This is only a convention, not a physical law. Its only value is that it often makes problems easier to solve (but sometimes it doesn't).

In general relativity there are no inertial frames, so "the inertial frame in which you're at rest" is not well defined and you can't use it. That's fine, because you were never obliged to use it even in special relativity.

• I got to the 75% from Paul T's response. I don't have whatever software is used to write equations (sorry) so I get dinged when I try, but I see in that link that when the initial distance is infinite, v = cSQRT(1 - 1/(rs/2r +1)^2). So when r = rs, the equation reduces to v = cSQRT(1-1/2.25) = 0.745c. Dec 5, 2020 at 21:13
• I don't have whatever software is used to write equations (sorry) You don’t need any; it is built into this site. Here is a MathJax tutorial. You can learn what you need in a few minutes. Dec 5, 2020 at 22:07