Does Time Drive Black Hole Travelers to a Central Singularity?

[.Scott]

TL;DR Summary

I know its supposed to. But how sure are we of that?

Since my understanding of these geometries is wrong, I'll do this in numbered steps - the easier to correct my logic.
I think the big problem I have is with the time dimension. There seems to be a presumption that the time vector will drive a falling object into a central singularity. But how well attached is that notion to Einstein's model?

Step 1: With space-time curvature, the circumference of a circle around any gravitational body will be less than the diameter times $\pi$. This is because the diameter is following the space curvature and the circumference is avoiding some of that curvature.

Step 2: If an object is dropped from the circle and its progress is measured as a percent of the distance to the center, it will appear to travel somewhat slower than expected from Euclidean space - because it is following that curved path.

Step 3: Gravity follows the inverse square law. Intuitively, this makes sense. At any given radius from the center, the influence of the mass is spread out over a spherical area that varies in proportion to the square of that radius. So my presumption is that the force of gravity is proportional to the radius as computed from the circumference over $2 \pi$.

Step 4: In the case of a Black Hole, at some point (I'm guessing it's the event horizon, but perhaps it's closer in) the slope of the curve is so great that no further progress is made in decreasing the distance to the Euclidean center. This is where I have my greatest doubt. I'm thinking that gravity will cause acceleration along the space curvature ia a direct path towards the presumed singularity - not necessarily towards the center.

Step 5: Upon reaching that point described in step 4, there is no chance of "passing" any significant portion of the previously fallen mass, so the gravitational mass I am falling towards becomes constant. Likewise, my circumference is not changing, so acceleration due to gravity would become constant.

Step 6: As the conditions in step 5 are approached, there does not seem to be any reason for further change to the space geometry.

Step 7: Even if time did drive me closer to the center thus reducing the circumference, this would intensify gravity further - potentially bending space back to counter the effect - but still keeping the circumference from closing all the way to zero. If BH gravity can bend my timeline to point directly into the center, why can't it keep bending it until it has a component in the -t direction and starts to turn further away from the center?

 

[Dale]

Summary:: I know its supposed to. But how sure are we of that?

Step 1: With space-time curvature, the circumference of a circle around any gravitational body will be less than the diameter times π. This is because the diameter is following the space curvature and the circumference is avoiding some of that curvature.

Your logic starts failing immediately. Since the diameter includes the horizon it is not particularly well defined.

Please don’t speculate. Just ask questions about what the established understanding is.

There seems to be a presumption that the time vector will drive a falling object into a central singularity. But how well attached is that notion to Einstein's model?

Completely well attached. This is easiest to see in Kruskal coordinates:

https://en.m.wikipedia.org/wiki/Kruskal–Szekeres_coordinates

In these coordinates light travels at a 45 deg angle. So things with mass must go more vertically than 45 degrees. Once you cross the horizon there is no such path that avoids the singularity.

 

[PeterDonis]

There seems to be a presumption that the time vector will drive a falling object into a central singularity.

I'm not sure what you mean by "the time vector". What GR predicts is that any object that falls through the horizon of a black hole will eventually reach the singularity.

how well attached is that notion to Einstein's model?

It is an unequivocal prediction of the Schwarzschild solution of the Einstein Field Equation.

With space-time curvature, the circumference of a circle around any gravitational body will be less than the diameter times $\pi$.

No. In the complete Schwarzschild solution (i.e., where the gravitating body is a black hole--which is the only case where there is a singularity inside the horizon), there is no meaningful notion of the "diameter" of any such circle.

The rest of your reasoning is therefore meaningless, since it is based on a false premise.

 

[.Scott]

Your logic starts failing immediately. Since the diameter includes the horizon it is not particularly well defined.

Please don’t speculate. Just ask questions about what the established understanding is.

I am trying to get a better feel for it. So I'm for specific contradictions to my thinking.
I tend to think of gravity as light - but there's no Doppler effect - so they really are not alike.

https://en.m.wikipedia.org/wiki/Kruskal–Szekeres_coordinates

In these coordinates light travels at a 45 deg angle. So things with mass must go more vertically than 45 degrees. Once you cross the horizon there is no such path that avoids the singularity.

I've definitely seen those charts before, I'm looking more for how they work than what they say.

 

[.Scott]

I'm not sure what you mean by "the time vector". What GR predicts is that any object that falls through the horizon of a black hole will eventually reach the singularity.

Yes. I'm starting to have questions about my question. The fact that the "time vector" points in towards the mass is what makes gravity work.
By "time vector" I mean the direction along the space-time line that an object will follow when in free fall.

It is an unequivocal prediction of the Schwarzschild solution of the Einstein Field Equation.

No. In the complete Schwarzschild solution (i.e., where the gravitating body is a black hole--which is the only case where there is a singularity inside the horizon), there is no meaningful notion of the "diameter" of any such circle.

The rest of your reasoning is therefore meaningless, since it is based on a false premise.

So both you and @Dale don't like the notion of "diameter". Would radius be better? Even the Kruskal–Szekeres diagrams have an "r" value. But if it is ill-defined, would that eliminate the notion of the diameter going to zero? Or perhaps the problem with diameter (or radius) is only at the event horizon, but not within it.

If there really is a problem with diameter or radius, can there be a strong argument for ever reaching a "central singularity"?

What is more key is this:
My statement in step 1 was not intended to be applied immediately to black holes. Gravity is often visualized as a bulge in space - like the heavy ball sitting on the rubber sheet that we have all seen.
Certainly in the case of the rubber sheet the circumference is less that $2\pi r$ because "r" is stretched out more. So if we do the vertical trans-planetary tunnel with two satellites, one that falls back and forth through the tunnel and the other that orbits in a circle, does the one in a circular orbit get back to the starting point first? (assuming we have an appropriate planet of even density and no atmosphere)

If it does, I think my logic against the singularity may hold some weight (excuse the pun).
On the other hand, if the time to cross the diameter is not directly effected by the bulge, then I see nothing at all to avoid a BH singularity.

 

[PeterDonis]

The fact that the "time vector" points in towards the mass is what makes gravity work.

By "time vector" I mean the direction along the space-time line that an object will follow when in free fall.

The correct term for this is "tangent vector to the object's worldline" or "4-velocity of the object". And it has nothing whatever to do with "what makes gravity work". The fact that the 4-velocity of an object freely falling towards a massive points inward (towards the massive object) is an effect of gravity, not "what makes gravity work".

So both you and @Dale don't like the notion of "diameter".

It isn't a question of what anyone "likes". The concept of "diameter" is not well-defined for the circles you describe in the presence of a black hole. That is a simple mathematical fact about the spacetime geometry we are discussing. It is not a matter of opinion or preference.

Would radius be better?

That term does have a meaningful interpretation, but it is not the one you are intuitively thinking of. The "radius" $r$ is the "areal radius", i.e., it is $\sqrt{A / 4 \pi}$, where $A$ is the area of the 2-sphere in which the circle you describe lies. It has nothing whatever to do with "radius" meaning "half of the diameter".

if it is ill-defined, would that eliminate the notion of the diameter going to zero?

No such notion plays any role in the actual math of the Schwarzschild geometry. The areal radius $r$ goes to zero as the singularity is approached, but, as above, that has nothing whatever to do with any notion of "diameter".

perhaps the problem with diameter (or radius) is only at the event horizon, but not within it.

No. The notion of "diameter" is not well-defined anywhere in the spacetime geometry we are discussing.

If there really is a problem with diameter or radius, can there be a strong argument for ever reaching a "central singularity"?

There is more than a "strong argument"; as I have already said, it is an unequivocal prediction of the Schwarzschild solution of the Einstein Field Equation. There is no room for doubt about that at all.

My statement in step 1 was not intended to be applied immediately to black holes.

Since it can't be applied to black holes at all, it is immaterial whether you want to apply it to them "immediately" or not. A black hole is a fundamentally different kind of object from an ordinary gravitating body like a planet or star. You cannot treat them the same.

the heavy ball sitting on the rubber sheet

This analogy works (in a very limited sense--we have had plenty of other threads discussing the limitations) for an ordinary gravitating body like a planet or star. But it does not work for a black hole.

I think my logic against the singularity may hold some weight

No, it holds no weight at all. See above.

 

[PeterDonis]

@.Scott, I suggest that you read my series of Insights articles on the Schwarzschild geometry. This is the first of the series of four:

https://www.physicsforums.com/insights/schwarzschild-geometry-part-1/

Part 3, in particular, discusses some properties of the Kruskal diagram that are relevant to this discussion.

 

[Dale]

Even the Kruskal–Szekeres diagrams have an "r" value.

Indeed, there is a coordinate labeled r. However, recall that your original quote was

the circumference of a circle around any gravitational body will be less than the diameter times π.

but the circumference is exactly $2\pi$ times the radial coordinate, by definition. The radial coordinate does not measure the radius. It measures the circumference and scales it as though it were a radius.

I think my logic against the singularity may hold some weight

If you would like to continue posting here you need to recognize that we do not discuss personal theories. We can explain the geometry, but not discuss your speculation

 

[.Scott]

It isn't a question of what anyone "likes". The concept of "diameter" is not well-defined for the circles you describe in the presence of a black hole. That is a simple mathematical fact about the spacetime geometry we are discussing. It is not a matter of opinion or preference.

Even if only one of you "disliked" diameter, I would have presumed it to be problematic for good reasons.

That term does have a meaningful interpretation, but it is not the one you are intuitively thinking of. The "radius" $r$ is the "areal radius", i.e., it is $\sqrt{A / 4 \pi}$, where $A$ is the area of the 2-sphere in which the circle you describe lies. It has nothing whatever to do with "radius" meaning "half of the diameter".

Areal "radius" totally works for me. I was presuming the gravitational field was following the "inverse square" law based on area.

This analogy works (in a very limited sense--we have had plenty of other threads discussing the limitations) for an ordinary gravitating body like a planet or star. But it does not work for a black hole.

I found this:

Actually, to "explain" gravity, curved time only suffice. The curvature of space is very small in the solar system. It's almost flat, about 10^-8. That's why it took 100-year observation to find out something is wrong with the Mercury orbit. Since the perihelion shift of Mercury is just too small.
In his book, Gravity from the Ground Up, Schutz says, "All of Newtonian gravitation is simply the curvature of time".

(No, not Feynman)
And that's one of the more recent ones - links were included, but the better ones have expired.
But this does tell me that the space curvature is real enough to move Mercury. I had remembered Mercury, but didn't realize it was entirely the result of the space curvature.

But, on further searching, I found this:

Richard Feynman Lecture Notes
Using the definition of excess radius, we have Radius excess $= \frac{G}{3c^2}\cdot M$. ... This is Einstein’s law for the mean curvature of space".

(Yes, that Feynman - Actually, it's Einstein, but it's in Feynman's lecture notes)

What's key here is that you actually need mass within the region of space to curve it. Having a strong gravitational source nearby won't induce any curvature.

So, within the event horizon (unless this equation also comes with a limit warranty preventing BH use), where there is no significant mass until you reach the singularity, space would have a curvature of zero. So my notion that additional space curvature could somehow drive me away from the center is hopeless.

 

[.Scott]

@.Scott, I suggest that you read my series of Insights articles on the Schwarzschild geometry. This is the first of the series of four:

https://www.physicsforums.com/insights/schwarzschild-geometry-part-1/

Part 3, in particular, discusses some properties of the Kruskal diagram that are relevant to this discussion.

I will read this.

 

[.Scott]

Indeed, there is a coordinate labeled r. However, recall that your original quote was but the circumference is exactly $2\pi$ times the radial coordinate, by definition. The radial coordinate does not measure the radius. It measures the circumference and scales it as though it were a radius.

Only in Euclidean space. I was trying to determine exactly how consequential that space curvature was.
As it turns out, $2\pi r$ only works when there is no mass. So that satellite that falls back and forth through the tunnel will take longer to return to its starting position that the one in a circular orbit - because it will have about 6mm further to go than expected (1.5mm per Earth radius).

If you would like to continue posting here you need to recognize that we do not discuss personal theories. We can explain the geometry, but not discuss your speculation

I wasn't really speculating. I was trying to work out what I was missing.
By "holding weight", I meant if the curvature didn't cause a real change in "r", then I still hadn't found my misconception. So it held weight - but abruptly dropped it when I discovered that it only applied directly to the mass.
Perhaps I needed more caveats to that effect.

 

[PeterDonis]

I was presuming the gravitational field was following the "inverse square" law based on area.

This is only true in the Newtonian approximation, which breaks down as you get closer to the hole's horizon. Gravity is not a "force field" in GR and your Newtonian intuitions about such things will not work.

I found this

This is talking about the weak field approximation (which works fine in, for example, the solar system). You cannot apply intuitions from this to a black hole.

I had remembered Mercury, but didn't realize it was entirely the result of the space curvature.

"Space curvature" is a bad term in general, since "space" is not an invariant; it depends on your choice of coordinates. (For example, in Painleve coordinates, which are the natural coordinates for an observer free-falling into a black hole from rest at infinity, "space" is Euclidean--there is no "space curvature" at all.)

Notice that later on in the Feynman lectures chapter you linked to, he talks about spacetime curvature and drops the talk about "space" curvature. Spacetime curvature is independent of your choice of coordinates.

What's key here is that you actually need mass within the region of space to curve it. Having a strong gravitational source nearby won't induce any curvature.

This is wrong. A gravitating body like the Earth creates spacetime curvature in the vacuum region around it, not just within its own substance. And, with an appropriate choice of coordinates, some aspects of this spacetime curvature will appear as "space curvature" in the vacuum region (if you insist on viewing it that way, even though, as I said above, it's a bad way to view it).

within the event horizon (unless this equation also comes with a limit warranty preventing BH use), where there is no significant mass until you reach the singularity

This is not quite true for a real black hole that forms from the gravitational collapse of a massive objects. There is a portion of the spacetime inside the horizon that is not vacuum, but is occupied by the collapsing matter.

It is true that, unless you fall into the hole very soon after it forms, you won't be able to reach the collapsing matter region before you hit the singularity.

space would have a curvature of zero.

Wrong. The notion of "space" that is well-defined outside the horizon and in which "space curvature" exists, is not well-defined inside the horizon. So it doesn't even make sense to ask what the "space curvature" is inside the horizon.

So my notion that additional space curvature could somehow drive me away from the center is hopeless.

Your notion is hopeless, but not for this reason. It is hopeless because it is not even wrong: it is based on concepts that aren't even well-defined in the domain in which you are trying to apply them.

 

[Dale]

Only in Euclidean space.

No. That is how the coordinate $r$ is defined in the Schwarzschild spacetime. In the curved manifold of Schwarzschild spacetime the coordinate $r$ is defined as the circumference divided by $2\pi$.

As it turns out, $2\pi r$ only works when there is no mass.

You have to be careful when you say “works”. The circumference is always $2\pi r$. That works because it is how $r$ is defined. What does not work is that $r\ne \int_{0}^{r}ds$. In other words $r$ is not the radial distance from the center.

I wasn't really speculating. I was trying to work out what I was missing.
By "holding weight", I meant if the curvature didn't cause a real change in "r", then I still hadn't found my misconception. So it held weight - but abruptly dropped it when I discovered that it only applied directly to the mass.
Perhaps I needed more caveats to that effect.

Please just discard it entirely. It is difficult enough to convey these ideas without also having to fight through revisiting a fatally flawed list. It is not helpful to either us or to you.

“Let me state a bunch of wrong things and then argue about them” is not an effective way to learn. It is easier to build a house from scratch than to first demolish a previous house, pull out the old foundation, scavenge for any bits of useable material, and then rebuild the house.

Even now I am wasting time on the echoes of your OP instead of typing something useful for you to progress.

 

[PeterDonis]

As it turns out, only works when there is no mass. So that satellite that falls back and forth through the tunnel will take longer to return to its starting position that the one in a circular orbit - because it will have about 6mm further to go than expected (1.5mm per Earth radius).

This is wrong.

You really need to stop speculating based on your incorrect understanding, and learn the correct understanding instead.

This thread is closed.