How Long Does it Take to Fall into a Black Hole?

[Jackzim]

TL;DR Summary

This forum has been asked many times about whether an infalling mass can cross the Schwarzschild Radius. The answer is YES, it will & I agree. However, what I don’t know is the amount of “gravity-free” time it will take to reach the Event Horizon; is it hours or trillions of years?

Gratings
The following is an example of a question that I have been battling to get my mind around for some time:
How long, in gravity-free time (Tᴓ-time), would it take for a stationary, one kilogram, point-mass to fall directly towards a non-spinning, accretion-disk-free, singularity in a Black Hole, weighing ten solar masses, from ten times the Schwarzschild Radius (Rs) to the Rs?
I would really like to know the exact Tᴓ-time taken together with a brief outline of the calculation.
I don’t have the mathematical skills to prove/disprove it but I suspect that due to extreme distortions of time & space, there may be a point just outside Rs where the point mass will be moving at near the speed of light and only move one micrometer in one billion years in Tᴓ-time?
To those that may reply, I wish to thank you in advance for your kindness, time & effort to enlighten me. And especial thanks to those who may come up with an actual time.

 

[PeroK]

This is problem 5 in chapter 12 of Gravity: An Introduction to GR by James B. Hartle.

I got the proper time from $R = 10M$ to the singularity to be $\Delta \tau = 5\sqrt 5 \pi M$.

And for $M = 10M_s$, I get that $M \equiv 5 \times 10^{-5} s$.

That means that $\Delta \tau \approx 2 \times 10^{-3} s$. Not long!

Corrected, as out by a factor of 10.

 

[Ibix]

What do you mean by "gravity free time"?

 

[phyzguy]

This is problem 5 in chapter 12 of Gravity: An Introduction to GR by James B. Hartle.

I got the proper time from $R = 10M$ to the singularity to be $\Delta \tau = 5\sqrt 5 \pi M$.

And for $M = 10M_s$, I get that $M \equiv 5 \times 10^{-5} s$.

That means that $\Delta \tau \approx 2 \times 10^{-4} s$. Not long!

But for a super-massive black hole (SMBH) with a mass of, say, 10^10 Msun (these exist), the time is ~2E5 seconds = about 2-1/2 days. For these objects, an observer could fall in at a reasonable rate, and without experiencing excessive tidal forces. Of course, after crossing the event horizon, the observer is still doomed, it just takes longer.

 

[Jackzim]

This is problem 5 in chapter 12 of Gravity: An Introduction to GR by James B. Hartle.

I got the proper time from $R = 10M$ to the singularity to be $\Delta \tau = 5\sqrt 5 \pi M$.

And for $M = 10M_s$, I get that $M \equiv 5 \times 10^{-5} s$.

That means that $\Delta \tau \approx 2 \times 10^{-4} s$. Not long!

 

[Jackzim]

Thank you for your prompt and precise reply - much appreciated

 

[Jackzim]

What do you mean by "gravity free time"?

My apologies for being unclear.
What I meant was the time measured by an observer in an ideal position where space is undistorted by gravity - to avoid any time distortions on the observer's cloc; and who can ideally measure the time, instantaneously, that it takes the mass to travel from 10xRs to1xRs.
I should also have said that the distance between the observer and singularity is fixed - to avoid any time dilation due to relative speeds of the two.

 

[phinds]

My apologies for being unclear.
What I meant was the time measured by an observer in an ideal position where space is undistorted by gravity - to avoid any time distortions on the observer's cloc; and who can ideally measure the time, instantaneously, that it takes the mass to travel from 10xRs to1xRs.
I should also have said that the distance between the observer and singularity is fixed - to avoid any time dilation due to relative speeds of the two.

In that case ("measured by an observer"), the answer is "the life time of the BH". If you mean CALCULATED by an observer then it's a function of the in-falling distance and the size of the BH as already noted.

 

[Ibix]

What I meant was the time measured by an observer in an ideal position where space is undistorted by gravity - to avoid any time distortions on the observer's cloc;

It's spacetime that's curved - space may or may not be curved whether or not there is spacetime curvature. The only way to avoid spacetime curvature is to be infinitely far away from any mass, although in this case that may be taken to mean just very, very far away.

and who can ideally measure the time, instantaneously, that it takes the mass to travel from 10xRs to1xRs.

However, then "measure instantaneously" is a problem. The only way to measure instantaneously a time when something crosses the event horizon is to be there at that same event (or to fall in very near by). And this isn't an area of negligible curvature.

I should also have said that the distance between the observer and singularity is fixed - to avoid any time dilation due to relative speeds of the two.

This isn't well-defined, since the singularity is more like a moment in time than a place in space. I take it you mean hovering at constant altitude and not orbiting.

The upshot of this is that your question doesn't really have an answer. The number @PeroK calculated is, I believe, the time on the infalling astronaut's watch when they cross the horizon free falling from rest at $10R_S$. That doesn't seem to me to be what you are asking. A distant observer cannot measure a time when anything crosses the horizon because they can never see it happen (until, in principle, the black hole evaporates trillions of years from now, as @phinds notes). They can calculate the astronaut's wristwatch time, or they may impose an arbitrary simultaneity convention and calculate an arbitrary answer for the coordinate time.

 

[PeterDonis]

What I meant was the time measured by an observer in an ideal position where space is undistorted by gravity

Such an observer is infinitely far away from the hole, and can only measure time where he is, not anywhere else. And there is no invariant way of relating measurements of time made by spatially separated observers in a curved spacetime. So there is no way to answer your question; the question itself is not well defined.

 

[PeroK]

The number @PeroK calculated is, I believe, the time on the infalling astronaut's watch when they cross the horizon free falling from rest at $10R_S$.

It was the proper time from $10M$ (initially at rest) to the singularity. Although I guess it should have been from $20M$ to $2M$. Which is about $6ms$. I hope that's correct.

Correction: $5ms$ if you do the calculation properly.

 

[pervect]

My apologies for being unclear.
What I meant was the time measured by an observer in an ideal position where space is undistorted by gravity - to avoid any time distortions on the observer's cloc; and who can ideally measure the time, instantaneously, that it takes the mass to travel from 10xRs to1xRs.
I should also have said that the distance between the observer and singularity is fixed - to avoid any time dilation due to relative speeds of the two.

Any such answer will be ill-defined because of the problem of specifying an observer, also the issue of a hypothetical "undistorted space" that doesn't exist in the real world.

The answers you have gotten are for "proper time", sometimes called wristwatch time. It's the amount of time that elapses for the infalling object, i.e. the time they'd measure on a "wristwatch", a clock they carried with them. No external observer needs to be specified.

 

[meekerdb]

It's spacetime that's curved - space may or may not be curved whether or not there is spacetime curvature. The only way to avoid spacetime curvature is to be infinitely far away from any mass, although in this case that may be taken to mean just very, very far away.

However, then "measure instantaneously" is a problem. The only way to measure instantaneously a time when something crosses the event horizon is to be there at that same event (or to fall in very near by). And this isn't an area of negligible curvature.

This isn't well-defined, since the singularity is more like a moment in time than a place in space. I take it you mean hovering at constant altitude and not orbiting.

The upshot of this is that your question doesn't really have an answer. The number @PeroK calculated is, I believe, the time on the infalling astronaut's watch when they cross the horizon free falling from rest at $10R_S$. That doesn't seem to me to be what you are asking. A distant observer cannot measure a time when anything crosses the horizon because they can never see it happen (until, in principle, the black hole evaporates trillions of years from now, as @phinds notes). They can calculate the astronaut's wristwatch time, or they may impose an arbitrary simultaneity convention and calculate an arbitrary answer for the coordinate time.

There's a myth that an object falling into a black hole can never be seen by a distant observer to reach the event horizon. This is only true of a "test particle", something with infinitesimal mass. Real objects have their own gravitational field and this causes the event horizon to expand out an engulf them. Of course this is a small effect in a sense for small objects falling in, but it's also the difference between infinite time and finite time. And it's most easily appreciated in the events detected by LIGO in which two black holes merge. It obviously makes no sense to picture one black stuck just outside the event horizon of another black hole. And in fact they merge, as seen by the distant LIGO observers, in a fraction of a second.

 

[PeterDonis]

There's a myth that an object falling into a black hole can never be seen by a distant observer to reach the event horizon.

It's not a myth, it's a fact, and not just for test particles. See below.

Real objects have their own gravitational field and this causes the event horizon to expand out an engulf them.

And that means no outside observer will ever see the object cross the expanded horizon, which is the event horizon--there is only one event horizon.

it's also the difference between infinite time and finite time.

No, it isn't. An event horizon is an event horizon, whether it expands from more matter falling in or not.

it's most easily appreciated in the events detected by LIGO in which two black holes merge

Observing a black hole merger does not involve seeing anything cross any event horizon. The gravitational waves we observe with LIGO are emitted just outside the event horizon of the merged hole.

 

[PeterDonis]

It obviously makes no sense to picture one black stuck just outside the event horizon of another black hole.

In a spacetime containing a black hole merger, there are not multiple event horizons; there is only one. If visualized in a spacetime diagram with time going upwards, the horizon just looks like a pair of trousers instead of a cylinder: the legs of the trousers are the two holes that merge, and the top of the trousers is the merged hole. (The legs are also twisted around each other because the holes orbit each other before merging.) So it is indeed never the case that one hole is "stuck" just outside the horizon of the other.

 

[Ibix]

There's a myth that an object falling into a black hole can never be seen by a distant observer to reach the event horizon.

You might need to review the definition of an event horizon if you think this is a myth.

And it's most easily appreciated in the events detected by LIGO in which two black holes merge. It obviously makes no sense to picture one black stuck just outside the event horizon of another black hole. And in fact they merge, as seen by the distant LIGO observers, in a fraction of a second.

All gravitational waves we detect come from events outside the event horizon - again, see the definition of an event horizon. Furthermore, merging event horizons is not at all the same thing as an object made of matter crossing a horizon (as Peter has just posted).

 

[Nugatory]

There's a myth that an object falling into a black hole can never be seen by a distant observer to reach the event horizon. This is only true of a "test particle", something with infinitesimal mass. Real objects have their own gravitational field and this causes the event horizon to expand out an engulf them.

There's a subtlety here. The outside observer will observe the increase in the size of the black hole and this happens in a finite time, but that's not the same thing as seeing the infalling object cross the horizon - light from that event is still trapped at the horizon.

There's an older thread where this was discussed at more length - I'll see if I can find it.

(And "their own gravitational field" isn't right, it's their mass/energy that matters. That's a bit of a digression in this context)

 

[Guillermo Navas]

This is problem 5 in chapter 12 of Gravity: An Introduction to GR by James B. Hartle.

I got the proper time from $R = 10M$ to the singularity to be $\Delta \tau = 5\sqrt 5 \pi M$.

And for $M = 10M_s$, I get that $M \equiv 5 \times 10^{-5} s$.

That means that $\Delta \tau \approx 2 \times 10^{-4} s$. Not long!

The falling time between $r_1 = 5 R_s$ and $r_2 = 0$ is not $\approx 2 \times 10^{-4} s$ but
$\Delta \tau=1.729 \times 10^{-3} s$

The falling time between $r_1 = 10 R_s$ and $r_2 = 1$ is
$\Delta \tau=4.823 \times 10^{-3} s$

 

[PeroK]

The falling time between $r_1 = 5 R_s$ and $r_2 = 0$ is not $\approx 2 \times 10^{-4} s$ but
$\Delta \tau=1.729 \times 10^{-3} s$

The falling time between $r_1 = 10 R_s$ and $r_2 = 1$ is
$\Delta \tau=4.823 \times 10^{-3} s$

Is that for $M$ equal to 10 solar masses?

 

[Guillermo Navas]

Summary:: This forum has been asked many times about whether an infalling mass can cross the Schwarzschild Radius. The answer is YES, it will & I agree. However, what I don’t know is the amount of “gravity-free” time it will take to reach the Event Horizon; is it hours or trillions of years?

Gratings
The following is an example of a question that I have been battling to get my mind around for some time:
How long, in gravity-free time (Tᴓ-time), would it take for a stationary, one kilogram, point-mass to fall directly towards a non-spinning, accretion-disk-free, singularity in a Black Hole, weighing ten solar masses, from ten times the Schwarzschild Radius (Rs) to the Rs?
I would really like to know the exact Tᴓ-time taken together with a brief outline of the calculation.
I don’t have the mathematical skills to prove/disprove it but I suspect that due to extreme distortions of time & space, there may be a point just outside Rs where the point mass will be moving at near the speed of light and only move one micrometer in one billion years in Tᴓ-time?
To those that may reply, I wish to thank you in advance for your kindness, time & effort to enlighten me. And especial thanks to those who may come up with an actual time.

I follow Taylor & Wheeler's "Exploring Black Holes" book, Chapter 3.
From the conservation of energy equation:
$$\frac E m=(1-\frac {2M} r)\frac {dt} {d\tau}=(1-\frac {2M} {r_1})^{1/2}$$
where $(1-\frac {2M} {r_1})^{1/2}$ is the energy at position $r_1$ when it starts to fall.
$$(1-\frac {2M} r)^2dt^2=(1-\frac {2M} {r_1})d\tau^2 ~~~(1)$$
From the Schwarzschild metric
$$d\tau^2=(1-\frac {2M} r)dt^2-\frac {dr^2} {(1-\frac {2M} r)}$$
$$\Rightarrow (1-\frac {2M} r)^2dt^2=(1-\frac {2M} r)d\tau^2+dr^2 ~~~ (2)$$
Substituting (2) in (1):
$$dr^2=[\frac {2M} r-\frac {2M} {r_1}]d\tau^2$$
$$d\tau=-\frac {dr} {\sqrt {\frac {2M} r-\frac {2M} {r_1}}}$$
$$\tau=-\int_{r_1}^{r_2}{\frac 1 {\sqrt {\frac {2M} r-\frac {2M} {r_1}}} \, dr}$$
We make the variable change $r=2Mx$
$$\tau=2M\int_{x_2}^{x_1}{\frac 1 {\sqrt {\frac 1 x-\frac 1 {x_1}}} \, dr}$$
For $x_1=10$ and $x_2=1$
$$\int_{1}^{10}{\frac 1 {\sqrt {\frac 1 x-\frac 1 {10}}} \, dx}=48.985$$
And $\tau=2*10*M_s*48.985\frac 1 c=4.823 \times 10^{-3}~s$
where $M_s=1,477 ~m$

Falling to the singularity, for example, from $x_1=5$ to $x_2=0$
$$\int_{0}^{5}{\frac 1 {\sqrt {\frac 1 x-\frac 1 {5}}} \, dx}=17.562$$
And $\tau=2*10*M_s*17.562\frac 1 c=1.729 \times 10^{-3}~s$

 

[Guillermo Navas]

Is that for $M$ equal to 10 solar masses?

Yes, 10 Msun.
##M_s=1

Is that for $M$ equal to 10 solar masses?

Yes, 10 solar mass.
With your formula and data: $1.756\times 10^{-3} ~s$

 

[PeroK]

Yes, 10 solar mass.
With your formula and data: $1.756\times 10^{-3} ~s$

Yes, you're right. I was out by a factor of 10 on the first one.

 

[PeroK]

If you do the integral you get:
$$\Delta \tau = \frac{r_1}{\sqrt{2M}}\bigg (\sqrt{r_2}\sqrt{1 - \frac{r_2}{r_1}} + \sqrt{r_1}\tan^{-1}\big ( \sqrt{\frac{r_1}{r_2} -1} \big ) \bigg )$$ And, when $r_2 = 0$ we get:
$$\Delta \tau = \frac{\pi r_1}{2}\sqrt{\frac{r_1}{2M}} = \pi M x_1\sqrt{x_1} \ \ \ \ \ \ \ \ (r_1 = 2Mx_1)$$

 

[Guillermo Navas]

Thank you.

 

[hilbert2]

Even this Newtonian problem of two gravitating masses falling together is quite difficult to solve for a beginner.

https://physics.stackexchange.com/q...-masses-will-collide-due-to-Newtonian-gravity

 

[PeroK]

Even this Newtonian problem of two gravitating masses falling together is quite difficult to solve for a beginner.

https://physics.stackexchange.com/q...-masses-will-collide-due-to-Newtonian-gravity

Or, closer to home:

https://www.physicsforums.com/threa...ith-varying-acceleration.866218/#post-5437928

 

[JLowe]

There's a subtlety here. The outside observer will observe the increase in the size of the black hole and this happens in a finite time, but that's not the same thing as seeing the infalling object cross the horizon - light from that event is still trapped at the horizon.

So is it right to say there is a moment where an outside observer can definitively say an object crossed the horizon? Assuming the object is large enough to cause a detectable increase in size of the event horizon?

 

[PeterDonis]

So is it right to say there is a moment where an outside observer can definitively say an object crossed the horizon?

There are at least two ways of defining such a moment. The first is to detect the mass increase; more on that below.

The second way is to find the time by the outside observer's clock when he can no longer send a light signal to the object that will reach it before it crosses the horizon. In more technical language, we find the point where the past light cone of the event where the object crosses the horizon intersects the observer's worldline, and look at the reading on the observer's clock at that point on his worldline. This is a straightforward computation and makes sense as a criterion: the object has crossed the horizon when you can no longer stop it from doing so by any means whatever (since any such means would involve sending some influence down to the object that could travel, at most, at the speed of light).

Assuming the object is large enough to cause a detectable increase in size of the event horizon?

If it is large enough to do this, it is large enough to cause a detectable increase in the mass of the hole, and the outside observer will detect that increase (by, for example, detecting a change in his orbital parameters, or a change in the proper acceleration he has to maintain to keep the same altitude) at a time that depends on where the object fell in, relative to where he is. If the object falls right past him, for example, he will actually detect the mass increase of the hole right away; in other words, before he would even say that the object has crossed the horizon (using the criterion I gave above). If the object falls in on the other side of the hole, it would take longer for him to detect the mass increase (heuristically, because the information about the object has to propagate to him at the speed of light).

These observations illustrate a key fact: that what we call the "mass of the black hole" is actually a global property of the entire spacetime geometry; it is not something that we can say is "contained inside the hole", since, as above, an increase in that mass due to an object falling in can be detected, in some cases, before the object crosses the horizon.

 

[JLowe]

These observations illustrate a key fact: that what we call the "mass of the black hole" is actually a global property of the entire spacetime geometry; it is not something that we can say is "contained inside the hole", since, as above, an increase in that mass due to an object falling in can be detected, in some cases, before the object crosses the horizon.

This is something I was thinking about. How would outside observer be able to say an object crossed the horizon by detecting a change in mass alone? An object just hugging the horizon would still have added it's mass to some region of spacetime much closer to the hole than the outside observer.

However, a change in the radius of the horizon itself by some significant mass falling in would be detectable no?

 

[Nugatory]

An object just hugging the horizon would still have added it's mass to some region of spacetime much closer to the hole than the outside observer.

The most easily analyzed case is a uniform spherical shell of negligible thickness that surrounds the black hole and is collapsing into it. Everywhere outside the shell, the physics is the same as if we had a black hole whose mass included the shell.

However, a change in the radius of the horizon itself by some significant mass falling in would be detectable no?

Yes, but this still doesn’t allow us to identify “a moment where an outside observer can definitively say an object crossed the horizon” (quoting from your post 27 above). We still have the problems that @PeterDonis identifies in #28

 

[PeterDonis]

An object just hugging the horizon would still have added it's mass to some region of spacetime much closer to the hole than the outside observer.

The mass isn't a property of "some region of spacetime". It's a global property of the spacetime geometry as a whole.

An object can't "just hug the horizon" unless it has some way of "hovering" at a constant altitude; below $r = 6M$ (3 times the horizon radius), there are no stable free-fall orbits, and below $r = 3M$ (3/2 the horizon radius) there are no free-fall orbits, period. And any way of "hovering" (such as, for example, firing a rocket) will require energy, and that energy will have been part of the object when it fell past you, so it will get counted in the increased mass of the hole that you detect; and since that energy will have to be finite, at some point it will run out and the object will fall into the hole anyway. It is not possible for an object containing finite energy to "hug the horizon" indefinitely.

a change in the radius of the horizon itself by some significant mass falling in would be detectable no?

You might be able to detect it by observing a change in the behavior of light rays coming from behind the hole, yes.

 

[JLowe]

An object can't "just hug the horizon" unless it has some way of "hovering" at a constant altitude; below r=6M (3 times the horizon radius), there are no stable free-fall orbits, and below r=3M (3/2 the horizon radius) there are no free-fall orbits, period. And any way of "hovering" (such as, for example, firing a rocket) will require energy, and that energy will have been part of the object when it fell past you, so it will get counted in the increased mass of the hole that you detect; and since that energy will have to be finite, at some point it will run out and the object will fall into the hole anyway. It is not possible for an object containing finite energy to "hug the horizon" indefinitely.

Yes, I understand that. I wasn't necessarily talking about an object doing so indefinitely. Just if outside observer would be able to tell the difference between object being outside or inside the horizon from calculation of mass alone.

In any case, thanks for the replies, I don't really have the math chops for GR.

 

[Vanadium 50]

If it is large enough to do this, it is large enough to cause a detectable increase in the mass of the hole, and the outside observer will detect that increase (by, for example, detecting a change in his orbital parameters, or a change in the proper acceleration he has to maintain to keep the same altitude) at a time that depends on where the object fell in, relative to where he is.

Can you elaborate?

If I am in orbit far from both the BH and the approaching object, I am in orbit around both of them - in orbit around their center of mass which has whatever momentum it has. That's true both before and after the object crosses the horizon.

 

[PeterDonis]

If I am in orbit far from both the BH and the approaching object,

Then the approaching object has already fallen past you (more precisely, past the distance from the hole that you are), long enough ago that the effects on the spacetime geometry will have had time to propagate to you, so yes, you will count the mass of the object as part of the overall mass of the thing you are orbiting.

 

[Vanadium 50]

Let's call it a javelin. I'm in orbit around the BH + javelin system. To me, nothing special happens when the javelin crosses the horizon (or, to be a little more accurate, gets close enough to the BH for the horizon to form around the javelin).

I don't think I can do better than calculating when the javelin must have crossed the horizon.