Descending a Black Hole: Spaghettification or Compression?

[Jiyong Chung]

TL;DR Summary

Problems trying to visualize how one would fall into a black hole

I have heard from, many attractive looking physicists on Youtube, that if I fell into a black hole, as I fell, I'd be stretched out or be "spaghettified. Is this correct?

It seems to me that gravity compresses the free space, and that means, any object near a black hole should be compressed. If one fell into a black hole, with the feet first, wouldn't one get squashed toward the feet, rather than stretched out?

Also, does the compaction of space continue after one passes through the event horizon? Or does the density of space remain constant thereafter? It seems to me that there should be a limit to how dense a space can get, since a point singularity does not exist in the real world. BTW, by "density" or "compaction" of space, I'm being figurative, as I don't know the proper terms for these ideas.

 

[Ibix]

Is this correct?

Yes.

It seems to me that gravity compresses the free space,

No - GR models gravity as curvature, not compression.

If one fell into a black hole, with the feet first, wouldn't one get squashed toward the feet, rather than stretched out?

No. Basically the gravitational effect on the part of you closer to the black hole is stronger than on the part furthest away. Thus you are stretched out because your toes accelerate faster than your head.

BTW, by "density" or "compaction" of space, I'm being figurative, as I don't know the proper terms for these ideas.

The problem is that you don't seem to be describing anything I recognise as GR, which models gravity as spacetime curvature. That curvature (to theextent you can summarise it as a single number) tends to infinity as you approach the singularity. There is no upper bound to it.

 

[Jiyong Chung]

No - GR models gravity as curvature, not compression.

It seems to me that your "curved space" is the same thing as my "compressed space." If one maps a flat surface to a curved surface (e.g., say a plane on a sphere), the points on the curved surface would be closer to each other than on the flat surface, given a reference space. In the sense that the points on the curved surface are closer to one another, I'm saying that the curved space is "compressed."

Assuming that we use the term "curvature," it still seems to me that, in the falling body scenario, a space near one's head would map to a smaller space near the feet. This means that the feet would be squashed, not stretched.

Also, although there is a greater acceleration toward one's feet, there is also a greater time dilation. So, whether one gets stretched out - wouldn't that depend on whether the acceleration or the time dilation has the greater effect? How can you know whether the acceleration or the the time dilation wins out?

 

[Ibix]

It seems to me that your "curved space" is the same thing as my "compressed space." If one maps a flat surface to a curved surface (e.g., say a plane on a sphere), the points on the curved surface would be closer to each other than on the flat surface, given a reference space. In the sense that the points on the curved surface are closer to one another, I'm saying that the curved space is "compressed."

That would depend on the mapping you choose to use. The curvature tensor is an invariant and does not depend on your choices.

This means that the feet would be squashed, not stretched.

If it seems that way to you then your intuition is incorrect.

How can you know whether the acceleration or the the time dilation wins out?

If you state your argument formally, you may be able to answer that. I'm not sure what you are saying even makes sense.

The formally correct way to approach this is by considering geodesic deviation for two nearby infalling geodesics and showing that the lower one would move ahead of the upper one.

 

[PeterDonis]

If one maps a flat surface to a curved surface (e.g., say a plane on a sphere), the points on the curved surface would be closer to each other than on the flat surface, given a reference space.

This notion of "closer together" has nothing whatever to do with any actual physical distance, and I would not recommend using it.

although there is a greater acceleration toward one's feet, there is also a greater time dilation

Gravitational time dilation is not a well-defined concept inside a black hole. Also, even outside the hole, gravitational time dilation, strictly speaking, applies to objects that are stationary, i.e., that are maintaining a constant altitude above the hole's horizon. You can't apply it as you are doing here to objects free-falling towards the hole.

 

[Jiyong Chung]

The formally correct way to approach this is by considering geodesic deviation for two nearby infalling geodesics and showing that the lower one would move ahead of the upper one.

Thanks again, for being patient. At the start of the thread, I was seeking a simple layperson's understanding of some ideas. Based on your replies, it appears that understanding the math is required. I will look into the math.

 

[Ibix]

Based on your replies, it appears that understanding the math is required.

If you want to know facts about black holes, maths is not required. You seem to want to reason with those facts - for that you need maths. Sean Carroll's lecture notes are a good place to start, and Ben Crowell's free-to-download GR book (lightandmatter.com) is also good. There are significant prerequisites for understanding GR - notably special relativity. Taylor and Wheeler's Spacetime Physics is a good source and I have heard good things of Morin's Special Relativity for the Enthusiastic Beginner, although I haven't read the latter. First chapters of both are available online. And Ben Crowell has a free online text at the same link as above.

 

[pervect]

Summary:: Problems trying to visualize how one would fall into a black hole

I have heard from, many attractive looking physicists on Youtube, that if I fell into a black hole, as I fell, I'd be stretched out or be "spaghettified. Is this correct?

It seems to me that gravity compresses the free space, and that means, any object near a black hole should be compressed. If one fell into a black hole, with the feet first, wouldn't one get squashed toward the feet, rather than stretched out?

To the first part, yes, you'd be spaghettified.

To the second part, no. Even Newtonian gravity predicts that you'll be stretched apart radially by a large mass, and compressed in the other direction.

I think it would be most useful at this point for you to examine the Newtonian case, and understand that case first.

Then all you need to understand about GR to answer this specific question is that it's not any different from the Newtonian gravity that you're used to in regards to spaghetification. This should come as now surprise after all GR reduces to Newtonian theory in the weak field limit.

In Newtonian theory, if your feet are closer to a large mass than your head, the Newtonain gravity pulls harder on your feet than your head. The net result can be decomposed into two parts. An averaged force, that pulls everything equally downwards, and a "stretching" force that tries to pull you apart.

More can be said, even for the Newtonian case, but I think this is a good place to stop, to avoid causing unnecessary confusion. And it's probablyi best not to go into the curved space-time approach in this thread - it's interesting, but more advanced than understanding why you're worong.

 

[PAllen]

Summary:: Problems trying to visualize how one would fall into a black hole

I have heard from, many attractive looking physicists on Youtube, that if I fell into a black hole, as I fell, I'd be stretched out or be "spaghettified. Is this correct?

What is the gender of these attractive physicists?

Sorry, couldn’t resist. Mentors delete this if desired.

 

[Jiyong Chung]

If you want to know facts about black holes, maths is not required. You seem to want to reason with those facts - for that you need maths. Sean Carroll's lecture notes are a good place to start, and Ben Crowell's free-to-download GR book (lightandmatter.com) is also good. There are significant prerequisites for understanding GR - notably special relativity. Taylor and Wheeler's Spacetime Physics is a good source and I have heard good things of Morin's Special Relativity for the Enthusiastic Beginner, although I haven't read the latter. First chapters of both are available online. And Ben Crowell has a free online text at the same link as above.

Thank you for the references. I will begin with Sean Carroll's lectures, and see how I fare.

 

[Jiyong Chung]

In Newtonian theory, if your feet are closer to a large mass than your head, the Newtonain gravity pulls harder on your feet than your head. The net result can be decomposed into two parts. An averaged force, that pulls everything equally downwards, and a "stretching" force that tries to pull you apart.

Thank you. That explanation makes sense at one level.

What has been confounding me is that the force you refer to is siad to result from a spacetime curvature, which varies as 1/r near the black hole. The 1/r indicates that near a black hole, there are sharp changes in spacetime derivatives over a small volume (i.e., which indicates a "denser space" according to my layman's understanding). By that logic, the feet should be more squashed than the head, since the feet are closer to the black hole.

This appears to be in tension with the idea of the forces being stronger near the black hole (as 1/r) and stretching the feet more than the head. It seems I can't avoid the math to resolve the contradiction.

 

[PeterDonis]

What has been confounding me is that the force you refer to is siad to result from a spacetime curvature, which varies as 1/r near the black hole.

No, spacetime curvature varies as $M / r^3$. You are thinking of "gravitational potential", which is only well-defined in stationary spacetimes, and is not the same thing as spacetime curvature. Spacetime curvature is tidal gravity.

The 1/r indicates that near a black hole, there are sharp changes in spacetime derivatives over a small volume (i.e., which indicates a "denser space" according to my layman's understanding)

Your layman's understanding is incorrect. You should be able to see why even from a simple Newtonian picture. If you are free falling feet first towards a gravitating body, the Newtonian "gravitational acceleration" towards the body is slightly larger for your feet than for your head. This means your feet and your head will tend to move apart, because your feet are accelerating downward faster than your head is. This is tidal gravity.

This simple picture doesn't fully capture spacetime curvature in GR--for one thing, it uses the concept of Newtonian "gravitational acceleration", which is only well-defined in stationary spacetimes (roughly, as the gradient of the "gravitational potential", although there are complications here as well). In particular, it is not well-defined inside the horizon of a black hole. But the tidal gravity aspect of it turns out to work the same for a radially infalling object inside the horizon of a black hole.

 

[pervect]

Thank you. That explanation makes sense at one level.

What has been confounding me is that the force you refer to is siad to result from a spacetime curvature, which varies as 1/r near the black hole. The 1/r indicates that near a black hole, there are sharp changes in spacetime derivatives over a small volume (i.e., which indicates a "denser space" according to my layman's understanding). By that logic, the feet should be more squashed than the head, since the feet are closer to the black hole.

This appears to be in tension with the idea of the forces being stronger near the black hole (as 1/r) and stretching the feet more than the head. It seems I can't avoid the math to resolve the contradiction.

As others have noted, the curvature varies as 1/r^3, or, if you prefer, M/r^3, near a black hole. This still has the property of "blowing up" near the central singularity of the black hole (and not blowing up at the event horizon), though.

The above is a little misleading, because "the curvature" isn't a single number. It's a rank 4 tensor. As tensors are a graduate level subject, I'll assume that you (the OriginalPoster) is not familiar with them. So I'll give a very short and incomplete descrition of what tensor is. Rank 0 tensors are just numbers - scalars, and rank 1 tensors may be familiar as vectors, which are usually represented as little arrows that point in some direction, and require multiple numbers (3 or 4, depending on whether one is deal with space or space-time) to represent. Newtonian tidal gravity, which can be regarded as a map from one spatial vector (a displacement) to another vector (a force, the tidal force) is an example of a rank 2 tensor. The Riemann tensor is a rank 4 tensor, though, and doesn't have any easy non-mathematical description that I can come up with.

It does turns out that the rank 4 Riemann space-time tensor can be "decomposed" into 3 rank 2 spatial tensors by a process known as the Bel decomposition. Furthermore, one of these pieces of the decomposition is just the Newtonian tidal tensor I mentioned. It's a rank 2 tensor, because it can be regarded as a map from one vector (a displacement vector) to a force (the tidal force) on an extended body.

I'm not sure how much of this makes sense - but it is as simple as I know how to write it - it's probably even over-simplified.

When I think about the tidal forces around a black hole, though, I think of this one particular "piece" of the decomposition of the Riemann tensor, this piece being interpreted as the tidal tensor of Newtonian gravity. There are some other pieces too, but I won't get into them, this has already probably too technical and they're not so relevant anyway.

 

[PeterDonis]

The above is a little misleading, because "the curvature" isn't a single number.

Yes, in general it has 20 independent components. The particular component I was describing is the radial component of the piece of the Riemann tensor that corresponds to the Newtonian tidal tensor, in Schwarzschild spacetime, in Schwarzschild coordinates.