B Possibility of Black Hole research

[Marshall_Mathers]

Hey,

If we take a hypothetical black hole with the mass of thirty billion suns which has the schwarzschild radius of roughly the orbit of the venus and let a spaceship fly into it, would it be possible for the spaceship to do some reseach on the physical nature of the singularity and then somehow fly out of it again or is the space around it while it is in the black hole always too contorted so that there is no possibility of escape? Tidal power lessens the farther you are from the singularity (is that right?), so if we throw all possibility overboard: Is there a hypothetical mass limit when a stellar body is able to (by the means of external energy) escape the event horizon again if he crosses it once? A black hole with the mass of trillions, maybe zillions of stars?
Have a lovely day,
Marshall

 

[Nugatory]

Once through the event horizon, there is no escape.

 

[Marshall_Mathers]

Excuse my question, but why? Wait, it's not the tidal spaces that we are battling, it's space itself, right?
Then my followup question: Is there a hipothetical black hole with enough mass that a spaceship could just chill out at the borders of the event horizon without being ripped apart by tidal forces?

 

[weirdoguy]

Excuse my question, but why?

Well, one might consider it as a definition of an event horizon of a black hole. You can't escape horizon because it's horizon.

 

[Orodruin]

Well, one might consider it as a definition of an event horizon of a black hole.

I would go as far as saying that it is the definition of the event horizon.

 

[Dale]

Is there a hipothetical black hole with enough mass that a spaceship could just chill out at the borders of the event horizon without being ripped apart by tidal forces?

Yes. Super massive black holes can have little tidal gravity.

 

[PeterDonis]

it's not the tidal spaces that we are battling, it's space itself, right?

No, it's spacetime. Once you are inside the black hole, moving into the future is the same thing as moving towards the singularity. And there's no way to avoid moving into the future.

 

[Heikki Tuuri]

The Schwarzschild radius of the Sun is 3 kilometers. The Schwarzschild radius of 30 billion suns is 90 billion kilometers, which is 1,000 times the orbital radius of Venus.

https://arxiv.org/abs/0705.1029

90 billion kilometers is 9 * 10^13 meters. Divide that by the speed of light 3 * 10^8 m. It is 300,000 light seconds, or over 3 light days.

The formula given in the reference gives you a maximum time of 3.14 / 2 * 3 days = 5 days to reach the singularity.

 

[Ibix]

The formula given in the reference gives you a maximum time of 3.14 / 2 * 3 days = 5 days to reach the singularity.

15$\mu$s per solar mass is my rule of thumb (derived from the paper you linked). Note that this is a maximum survival time - no path has more proper time from horizon to singularity, and a lot have less.

 

[PAllen]

Upshot: you can do all the research you want, you just can't publish.

 

[alantheastronomer]

Hey,

If we take a hypothetical black hole with the mass of thirty billion suns which has the schwarzschild radius of roughly the orbit of the venus and let a spaceship fly into it, would it be possible for the spaceship to do some reseach on the physical nature of the singularity and then somehow fly out of it again or is the space around it while it is in the black hole always too contorted so that there is no possibility of escape? Tidal power lessens the farther you are from the singularity (is that right?), so if we throw all possibility overboard: Is there a hypothetical mass limit when a stellar body is able to (by the means of external energy) escape the event horizon again if he crosses it once? A black hole with the mass of trillions, maybe zillions of stars?
Have a lovely day,
Marshall

The reason your spaceship can't escape once inside the event horizon is because it would need to be moving faster than the speed of light in order to cross back out. So while you're right that the tidal forces are less for a black hole that is massive enough, the spacetime within the event horizon is what prevents you from leaving once you've crossed over. What's more, inside the event horizon, no signal from any other body can ever reach you( a condition known as "spacelike light cones" among physicists) so there's absolutely no possibility of being able to do any research whatsoever anyway!

 

[PeterDonis]

inside the event horizon, no signal from any other body can ever reach you( a condition known as "spacelike light cones" among physicists)

I have no idea where you are getting this from, but it's wrong. It is perfectly possible for objects inside the horizon to exchange light signals. There are limitations on it due to the spacetime geometry, but that's true everywhere in any spacetime, and such limitations have nothing whatever to do with "spacelike light cones" (light cones by definition are null surfaces, not spacelike).

 

[Justin Hunt]

What if two black holes are spirally in towards each other? Could you hypothetically fly between them in such a way that the gravity on both sides net to zero? Or dip into into one BH event horizon and then allow the presence of the second BH to allow u to escape? Or possibly waiting right before they collide when a portion of their combined mass is transformed into gravitational wave energy does the event horizon shrink?

 

[PeterDonis]

Could you hypothetically fly between them in such a way that the gravity on both sides net to zero?

In the sense that you would stay equidistant between them, yes. But you would still most likely be trapped inside the new hole that the two spiraling in holes merged into.

Or dip into into one BH event horizon and then allow the presence of the second BH to allow u to escape?

No. Once you're inside the horizon, you can't escape. There actually is only one horizon in spacetime terms; it just looks like a pair of trousers instead of a cylinder, heuristically speaking.

Or possibly waiting right before they collide when a portion of their combined mass is transformed into gravitational wave energy does the event horizon shrink?

Again, once you're inside the horizon, you can't escape. Also, gravitational wave emission doesn't shrink the horizon; the GWs from the merger get emitted from outside the horizon.

 

[Ibix]

Could you hypothetically fly between them in such a way that the gravity on both sides net to zero?

Depends what you mean. You can (at least sometimes) find places where your path is not deviating towards one hole or another, simply from symmetry. But you will always be able to see light from distant stars Doppler shifted due to the holes' gravity.

Or dip into into one BH event horizon and then allow the presence of the second BH to allow u to escape?

No. Once you cross an event horizon, that's it.

Or possibly waiting right before they collide when a portion of their combined mass is transformed into gravitational wave energy does the event horizon shrink?

The event horizon(s) of a combining black hole are not spherically (or even cylindrically) symmetric, but settle down to it quite quickly. They change shape, rather than shrinking. And the key point is that once you are inside the horizon, you're inside. Whatever changes happen to the horizon, you can't escape it.

 

[Justin Hunt]

What about gravity waves themselves? Is it theoretically possible to create them? And if so encode information in them if u do? Because apparently they are the only thing that could escape from a BH

 

[PeterDonis]

What about gravity waves themselves? Is it theoretically possible to create them?

Sure, just smash some neutron stars or black holes into each other.

If you mean, create them with technology we are going to be able to develop any time soon, no. At least not gravitational waves of an intensity anywhere close to what we can detect.

apparently they are the only thing that could escape from a BH

No. Gravitational waves cannot escape from inside a black hole's event horizon, any more than anything else can. As I've already said, the GWs we observe from black hole mergers come from outside the horizon.

 

[Ibix]

Is it theoretically possible to create them? And if so encode information in them if u do?

You mean artificially? Sure, you just wave a couple of neutron stars around and modulate the motion. Actually doing that is left as an exercise for the reader.

Because apparently they are the only thing that could escape from a BH

They don't escape from a black hole. Any gravitational wave you receive must come from within your past lightcone, and this cannot include anything inside an event horizon.

 

[Justin Hunt]

So my understanding of gravity in relativity is that it is the curvature of spacetime rather than a force in Newtonian physics. My understanding of gravitational waves are a disturbance of the spacetime curvature. Isn’t this fundamentally different from electromagnetic radiation? Does spacetime not exist within the event horizon? I understand why light can’t escape, but please explain why GW can’t as well

Also, I get that for far away objects LIGO can only detect things like BH merges, but u could have a second ship orbiting the BH that could detect weaker gravitational waves and convert the encoded information into another form of communication.

Possibly something like a particle accelerator could make tiny detectable gravity waves.

 

[Ibix]

Possibly something like a particle accelerator could make tiny detectable gravity waves.

The Earth going round the Sun produces around 100W of gravitational radiation. Nothing less than a couple of stars is going to produce anything detectable.

Also - note that gravity waves are a kind of water wave. We are talking about gravitational waves.

Isn’t this fundamentally different from electromagnetic radiation? Does spacetime not exist within the event horizon?

Yes, and yes it does.

I understand why light can’t escape, but please explain why GW can’t as well

Because they propagate at the speed of light. If they could escape, so could light. Or, to put it simply, nothing can escape the event horizon.

 

[PeterDonis]

Isn’t this fundamentally different from electromagnetic radiation?

There are some differences, but there are also many similarities.

please explain why GW can’t as well

Because GWs have to move on the light cones, just like light does. They can't escape from anywhere that light can't escape from.

Possibly something like a particle accelerator could make tiny detectable gravity waves.

Just accelerating particles isn't enough. You need a nonzero third time derivative of the quadrupole moment; basically that means you need two massive objects orbiting each other, and their masses need to be not exactly the same. If the detector is close to the source, you might not need neutron star masses, but you would still need objects much more massive than anything we can manipulate now or in the foreseeable future.

 

[Justin Hunt]

It seems odd to me that on a BH merge that a significant portion of their total mass can be converted to Gravitational waves that are released outside the BH event horizon. Intuitively it would make sense if they were created from the impact and originated from the impact. Is this related to Hawking radiation in any way?

 

[PeterDonis]

Intuitively it would make sense if they were created from the impact and originated from the impact.

There is no impact; black holes are vacuum. Their "mass" is not because they contain matter; it's a property of the spacetime geometry.

Is this related to Hawking radiation in any way?

No. Hawking radiation is a quantum process. Gravitational wave production from a black hole merger is a classical process.

 

[Ibix]

Intuitively

Unfortunately, intuition is a very poor guide to fundamental physics. You can develop and train an intuitive feel for how the maths goes, but that's a rather different thing.

If relativity were intuitive, we would not have all the threads on the twin paradox that we do. And SR is the friendly and straightforward bit of GR.

 

[Orodruin]

Put in another way, intuition is something you get from experience. The typical person has very little experience with situations where relativity becomes noticeably different from Newtonian physics and therefore should expect to have their intuition lead them wrong and full of misconceptions from every-day intuition that no longer applies. After working with teaching SR and GR for years, I feel that I finally have some actual intuition that is applicable to those subjects as a result of being familiar with the theories and the math they contain.

 

[alantheastronomer]

It is perfectly possible for objects inside the horizon to exchange light signals.

How so?

 

[Ibix]

How so?

You can see it's possible from a Kruskal diagram. It obeys the same rules as a Minkowski diagram - light travels on 45° lines and timelike paths are above 45°. There's a fairly narrow window for communication between horizon and singularity, but it can happen. And one way communication is always possible, at least in principle.

It's also obvious from the locally Minkowski nature of spacetime. If we are close enough together then we see spacetime between us as flat, so obviously we can communicate. That can't cut off dead outside "local" distances unless something catastrophic (like the singularity) happens to one partner.

 

[pervect]

A practical example. Suppose you are on the stern of a somewhat long spaceship, and you're looking at an object in the bow of the spaceship. While you are looking, the front of the space-ship crosses the event horizon of a very large black hole. You snap snapshots just before, at the same time, and just after the front of the spaceship crosses the event horizon of a very large black hole. The black hole is so large that the curvature at the event horizon is negligible.

What do the photos show?

In all three cases, the photos show the object as normal, as if the black hole were not present. The assumptions we made that the black hole is large enough that curvature / tidal effects can be ignored is important for this result to be correct.

For completeness, I should add an explanation of what I mean by "at the same time". Due to the fact that space-time curvature can be neglected, there is essentially a local inertial frame of reference for the space-ship. It is in this local inertial frame that we judge the idea of "at the same time", via the usual convetions we apply to an inertial frame in Special relativity. You might be familiar with the Schwarzschild time coordinate "t". We are NOT using that time coordinate to judge "at the same time", we are using the notion of time associated with the local inertial frame of the ship. Due to the relativity of simultaneity, we need to specify a particular inertial frame to define the particular notion of simultaneity we are using.

To be specific, consider the photograph taken after the bow of the ship falls through the horizon. The r-coordinate of the light is decreasing with time, but slowly. The r-coordiate of the back of the ship is increasing more quickly. At some point, the light ray from the front of the ship hits the back of the ship. This will only happen after the stern of the ship has crossed the event horizon. But it is an example of "objects inside the horizon to exchange light signals."

Technically, we've only analyzed one signal of the two-way exchange. But the analysis is similar for the case of an exchange of signals in the other direction.,

 

[timmdeeg]

The r-coordiate of the back of the ship is increasing more quickly.

You seem to mean 'is decreasing more quickly', right?

At some point, the light ray from the front of the ship hits the back of the ship. This will only happen after the stern of the ship has crossed the event horizon.

And is it correct that this happens in the moment the stern is crossing the event horizon as this light ray is "frozen" at $r=2m$ because it was emitted in the moment the bow was crossing the event horizon?

 

[pervect]

You seem to mean 'is decreasing more quickly', right?

Oops, yes.

And is it correct that this happens in the moment the stern is crossing the event horizon as this light ray is "frozen" at $r=2m$ because it was emitted in the moment the bow was crossing the event horizon?

I had three cases in the original - just before, just at, and just after the crossing. For the middle case, yes, the r-coordinate of the light signal emitted from the bow as it crosses the stern hangs there, and it is actually photographed when the camera reaches the horizon.

 

[timmdeeg]

I had three cases in the original - just before, just at, and just after the crossing.

To follow this up I think after the crossing the stern will receive light from the bow for a certain time only depending on the length of the ship. Light emitted later at the bow will no more hit the stern but the singularity.

 

[PAllen]

To follow this up I think after the crossing the stern will receive light from the bow for a certain time only depending on the length of the ship. Light emitted later at the bow will no more hit the stern but the singularity.

Yes, but the stern will continue to “see” the bow until it reaches the singularity. That is, at any point at all in the stern’s world line, there will be a bow emission event just becoming visible.

 

[timmdeeg]

Technically, we've only analyzed one signal of the two-way exchange. But the analysis is similar for the case of an exchange of signals in the other direction.,

Could you please elaborate a bit on the "other direction". By means of an Eddington-Finkelstein diagram I understand that the stern receives light from the bow but I fail to see the other way round. From the diagram it seems that the bow is in the past of the stern.

 

[timmdeeg]

Yes, but the stern will continue to “see” the bow until it reaches the singularity.

Yes, until the stern "reaches the singularity". Just to avoid any misunderstanding.

 

[PAllen]

Could you please elaborate a bit on the "other direction". By means of an Eddington-Finkelstein diagram I understand that the stern receives light from the bow but I fail to see the other way round. From the diagram it seems that the bow is in the past of the stern.

Just think light cones. There are stern world line events in the bow past light cone until the bow reaches the singularity. Thus there is a window of time in which two way signals can be exchanged between the bow and stern, with all relevant emission events occurring inside the horizon. And, of course, the bow therefore sees the stern until it hits the singularity.

It really becomes obvious in the Kruskal chart.

 

[timmdeeg]

Just think light cones.

Ah, that clarifies the issue, thanks.

 

[pervect]

Could you please elaborate a bit on the "other direction". By means of an Eddington-Finkelstein diagram I understand that the stern receives light from the bow but I fail to see the other way round. From the diagram it seems that the bow is in the past of the stern.

As I recall, the ingoing Eddington-Finkelstein diagram is different than the outgoing one. I'd recommend Kruskal-Szerkeres coordinates and/or a Penrose diagram, where light always travels at a 45 degree angle.

To quote Wiki

Kruskal–Szekeres coordinates have a number of useful features which make them helpful for building intuitions about the Schwarzschild spacetime. Chief among these is the fact that all radial light-like geodesics (the world lines of light rays moving in a radial direction) look like straight lines at a 45-degree angle when drawn in a Kruskal–Szekeres diagram (this can be derived from the metric equation given above. ... All timelike world lines of slower-than-light objects will at every point have a slope closer to the vertical time axis (the T coordinate) than 45 degrees. So, a light cone drawn in a Kruskal–Szekeres diagram will look just the same as a light cone in a Minkowski diagram in special relativity.

It shouldn't be too hard to see this that for a short enough space-ship, light from the stern can thus reach the bow, as well as the reverse, because locally the light is always traveling "faster" than any material object (the slope of the infalling object is more vertical).

This is only true in the appropriate limit of a "short enough" ship or "big enough" Schwarzschild radius.

Unfortunately, I didn't find any good diagrams of an infalling object in KS coordinates to illustrate this graphically.

The intuitive argument here is that if light from the bow couldn't reach the stern, or vica versa, an observer would obviously know that they crossed the event horizon when they lost sight of the other end of the ship. This just doesn't happen - in the limit of a short ship (which is the same as the limit of a large blac hole), one can't tell from inside the ship that one has crossed the horizon.

As one approaches the singularity , the space-time curvature keeps going, and it gets harder and harder to meet the requirements of a "short enough" ship.

 

[Ibix]

Unfortunately, I didn't find any good diagrams of an infalling object in KS coordinates to illustrate this graphically.

Your wish is my command:

This is the top right quadrant of a Kruskal diagram like the one (by @DrGreg, I believe) at Wikipedia. The fine grey line is the event horizon and the dark grey curve is the singularity (the shaded area doesn't correspond to anywhere in spacetime). The red lines show the final moments of three objects free-falling from rest at infinity. The two trailing objects emit blue light pulses as they cross the horizon - you can see that they catch up to the object in front, which has time to reply and receive a second reply (well, object one doesn't quite get to receive the second reply).

I actually set this up with a solar mass black hole ($R_S=3km$) and at about $1.5R_S$ the objects had Schwarzschild $r$ coordinates 100m apart.

 

[PeterDonis]

the ingoing Eddington-Finkelstein diagram is different than the outgoing one.

That's correct. However, the ingoing one is enough to analyze the experiment in question; it just isn't as easy to see the result as it is in a Kruskal diagram since only ingoing light rays go on 45 degree lines in the ingoing Eddington-Finkelstein diagram.

 

[pervect]

I did want to say a bit more about the "short space ship" condition. Material objects must follow a timelike worldline, and because of this no such object can ever go as fast as a light beam. So if a light beam and a material object have a race, if they both start at the same time and place, the lightbeam will always "be ahead of" the material object in a race.

However, if the accelerating object has a head start, even though it's slower than the lightbeam, it's possible (though a bit surprising) that the ligihtbeam will never be able to catch up. This is a feature of hyperbolic motion, for instance.

In the "short ship", limit, though, the space-time diagram will be linear, because terms of the second order will be negligible compared to terms of the first order. And for a linear diagram, the faster object will catch the slower one.

 

[pervect]

Your wish is my command:

Very nice! How did you calculalte and make the diagram?

 

[Ibix]

Very nice! How did you calculalte and make the diagram?

I took the geodesic equations in Schwarzschild coordinates and substituted $u$ and $v$ for $r$ and $t$. This works even at the horizon because the geodesic equation remains valid at and below the horizon - it's just its expression in Schwarzschild coordinates that goes wrong. I fed the result to one of scipy's numerical integrators and plotted results in pyplot. Confession: the light pulses were added by hand. My code can handle null geodesics, but I couldn't be bothered to do all the initialisation and intercept calculations in order to generate 45° straight lines.

I've had code lying around to do this for a while - I didn't do it all on Thursday night. It's not perfect, since you can't eliminate $r$ completely and you need to solve for it numerically. That generates numerical issues for long-lasting orbits since $u$ and $v$ grow while $u^2-v^2$ gets small, and $r$ depends on the latter (writing $(u+v)(u-v)$ postpones the problem, but ultimately doesn't solve it). And sometimes the integrator goes wrong near the singularity and objects skip merrily across its surface. But, those issues aside, the results are consistent with some analytical checking and a couple of things I found in the literature.

 

[timmdeeg]

Sorry for being a bit late and thanks for your response.

It shouldn't be too hard to see this that for a short enough space-ship, light from the stern can thus reach the bow, as well as the reverse, because locally the light is always traveling "faster" than any material object (the slope of the infalling object is more vertical).

Indeed, this is very convincing.

In the meantime I've found some Eddington-Finkelstein diagrams in Robert Geroch's nice book "General Relativity from A to B", which is written for laymen. One comes close to the space-ship scenario. Observer A falls into the black hole and after a while B follows him. Their wordlines are parallel and its easy to see that B receives light from A for some time beyond the horizon. In all the diagrams light goes up as time increases, so light emitted by B can't reach A. I'm not sure how to resolve this. In contrast in KS coordinates which @Ibix shows in #38 the situation is crystal clear.

 

[PeterDonis]

In all the diagrams light goes up as time increases, so light emitted by B can't reach A. I'm not sure how to resolve this.

If B falls in long enough after A, then light signals he emits inward won't reach A before A hits the singularity. To have A see any signals from B, you have to make sure the time between them falling in isn't too large.

 

[timmdeeg]

If B falls in long enough after A, then light signals he emits inward won't reach A before A hits the singularity. To have A see any signals from B, you have to make sure the time between them falling in isn't too large.

Heureka, I can see it now. https://www.researchgate.net/figure/Space-time-diagram-in-Eddington-Finkelstein-coordinates-showing-the-light-cones-close-to_fig4_260835665 is Fig. 93 in Geroch's book I've mentioned. It shows C freely falling and outgoing light, ingoing light according to the light cones. If I paint the worldline of B who jumped first together with the past light cone very close to that of C then it turns out that indeed B can receive light of C. It's a bit tricky though and as mentioned by others it's far better to be seen in a KS diagram.

 

[PAllen]

Heureka, I can see it now. https://www.researchgate.net/figure/Space-time-diagram-in-Eddington-Finkelstein-coordinates-showing-the-light-cones-close-to_fig4_260835665 is Fig. 93 in Geroch's book I've mentioned. It shows C freely falling and outgoing light, ingoing light according to the light cones. If I paint the worldline of B who jumped first together with the past light cone very close to that of C then it turns out that indeed B can receive light of C. It's a bit tricky though and as mentioned by others it's far better to be seen in a KS diagram.

There is something wrong with that diagram. It purports to show a timelike world line that does not remain inside its own future light cone. That is impossible. Your confusion is caused by a quantitatively inaccurate diagram. Consider especially the future light cone emanating from W, and the proposed world line beyond W. This cannot be right. It would make the free fall spacelike.

 

[timmdeeg]

Consider especially the future light cone emanating from W, and the proposed world line beyond W. This cannot be right. It would make the free fall spacelike.

Yes. The diagram which I mentioned in #43 doesn't show light cones. I was drawing the two parallel timelike geodesics close to each other and inside the respective light cones, with the ingoing null geodesics having an angle of 45° with the space axis.

 

[martinbn]

There is something wrong with that diagram. It purports to show a timelike world line that does not remain inside its own future light cone. That is impossible. Your confusion is caused by a quantitatively inaccurate diagram. Consider especially the future light cone emanating from W, and the proposed world line beyond W. This cannot be right. It would make the free fall spacelike.

It only looks like this. But in this diagram the light cones tilt as you go along the green/red world line. The closer to the singularity you are the more horizontal (open to the right) they are.

 

[PAllen]

It only looks like this. But in this diagram the light cones tilt as you go along the green/red world line. The closer to the singularity you are the more horizontal (open to the right) they are.

I know, but don’t think that affects my critique. The black dashed lines are paths of radially outgoing light, thus they form the left boundary of light cones. Along one of these dashes lines, the width of a light cone is shown constant. Following these rules of the diagram, the light cone at event W does not enclose the purported free fall path.

[edit: I should note that the issue is that this diagram is not remotely an accurate representation of Eddinton-Finkelstein coordinates. ]

 

[martinbn]

I know, but don’t think that affects my critique. The black dashed lines are paths of radially outgoing light, thus they form the left boundary of light cones. Along one of these dashes lines, the width of a light cone is shown constant. Following these rules of the diagram, the light cone at event W does not enclose the purported free fall path.

[edit: I should note that the issue is that this diagram is not remotely an accurate representation of Eddinton-Finkelstein coordinates. ]

I see, but I thought the diagram is only schematic.