Explaining Spaghettification w/ Spacetime Curvature

[Karl Coryat]

TL;DR Summary

If you had to explain spaghettification to a lay person who's new to general relativity and spacetime curvature, how would you do it?

I enjoy explaining spacetime curvature to people with a rank-beginner understanding of GR. But someone asked about that favorite concept in pop-sci, spaghettification. I'm having a hard time with it.

If you fell into a black hole, there's no reference frame within which you could describe what's happening to your entire body. So how would you explain tidal forces emerging out of a spacetime curvature that has a gradient? I attempted a thought experiment involving a group of marbles descending radially toward a black hole, and then the same group connected by a string, but it didn't work out without appealing to a difference in Newtonian "pulling" forces. Doing a quick search in other forums, answers there also tend to mix in gravitational forces, which is equally unhelpful.

If possible, please keep technical terms to a minimum. Thanks!

 

[PeterDonis]

If you fell into a black hole, there's no reference frame within which you could describe what's happening to your entire body.

This is not correct. Why do you think this?

 

[PeterDonis]

how would you explain tidal forces emerging out of a spacetime curvature that has a gradient?

Tidal gravity is not the gradient of spacetime curvature. It is spacetime curvature itself.

 

[Karl Coryat]

This is not correct. Why do you think this?

I assumed that a reference frame cannot span two locations with different spacetime curvatures. For example, I have seen it said multiple times that no reference frame can span the entire earth. So I assumed the same principle applies here.

 

[PeterDonis]

I assumed that a reference frame cannot span two locations with different spacetime curvatures.

Why would you assume that? It's wrong. If it were true it would be impossible to have a coordinate chart that covered any finite region of a curved spacetime at all, which is absurd.

I suspect you are confusing "reference frame" in general with the much more specific and limited concept of a local inertial frame. See below.

I have seen it said multiple times that no reference frame can span the entire earth.

Please give a specific reference. I strongly suspect you are misinterpreting something.

A correct statement would be that no local inertial frame can span the entire earth. But that is a much, much weaker statement than the one you are making.

 

[Karl Coryat]

A correct statement would be that no local inertial frame can span the entire earth. But that is a much, much weaker statement than the one you are making.

Yes, that clears up some things. Thank you.

I'd still like to explain how tidal forces arise near a black hole without appealing to Newtonian forces of gravity pulling an object apart. Do you know a good online resource for that?

 

[Dale]

Summary:: If you had to explain spaghettification to a lay person who's new to general relativity and spacetime curvature, how would you do it?

So how would you explain tidal forces emerging out of a spacetime curvature that has a gradient?

You just take a geodesic congruence and calculate the expansion tensor.

Basically, if you have a spherical bunch of coffee grounds initially at rest wrt each other that are all free-falling near the event horizon you will see that the grounds do not remain at rest wrt each other and that the sphere elongates and thins into an ellipse with the same volume. In the extreme that is spaghettification.

 

[PeterDonis]

I'd still like to explain how tidal forces arise near a black hole without appealing to Newtonian forces of gravity pulling an object apart.

Tidal gravity/spacetime curvature in GR is geodesic deviation: the trajectories of nearby freely falling particles diverge (or converge). For example, if you drop two rocks above a massive body like the Earth, aligned radially but with slightly different altitudes, the distance between them will increase. Or if you drop them from the same altitude but separated tangentially, the distance between them will decrease.

Tidal gravity near or inside a black hole is the same thing, except that the divergence/convergence is more severe (and the severity increases as you get closer and closer to the singularity inside the hole).

Objects that are held together by internal forces will experience internal stresses in the presence of tidal gravity: these stresses are there because, for the object to hold together, its parts must not diverge or converge; so the internal forces inside the object pull (or push) the individual parts of the object away from the geodesic (freely falling) trajectories they would otherwise follow. However, any object has only a finite ability to withstand internal stresses; when that limit is exceeded, the object falls apart and its individual parts are no longer bound by internal forces, so they move on freely falling trajectories and will diverge or converge according to the spacetime curvature that is present. That is what spaghettification is: the parts of any object diverge in the radial direction and converge in the tangential directions as the singularity of a black hole is approached.

 

[hmmm27]

Shoulda bin called "lasagnafication" : barring itty bitty black holes, how does tangential convergence occur ?

 

[Dale]

In the vacuum region outside a black hole (or a regular star or planet), the volume of that ball of coffee grounds doesn't change. So if it stretches longitudinally then is must compress tangentially to maintain volume. Regardless of the size of the black hole or star or planet

 

[PeterDonis]

how does tangential convergence occur ?

Read the last sentence of the first paragraph of post #8 again.

 

[hmmm27]

Thanks guys, I get the basic geometry ; does the math actually work out for "constant volume" ?

 

[PeterDonis]

does the math actually work out for "constant volume" ?

If you mean, will a small ball of test particles in a curved spacetime maintain its volume as its shape changes, as long as the region it is moving through is vacuum, yes, that's a mathematical theorem in GR.

 

[hmmm27]

If you mean, will a small ball of test particles in a curved spacetime maintain its volume as its shape changes, as long as the region it is moving through is vacuum, yes, that's a mathematical theorem in GR.

Okay, I'm not sure that helps though it could be the same question. Mine concerned tidal differential vs radii convergence... strictly Newtonian, no ?

 

[PeterDonis]

Mine concerned tidal differential vs radii convergence

Convergence of objects falling from the same altitude and separated tangentially is tidal gravity. If that's not what you mean by "radii convergence", then I don't understand what you mean.

strictly Newtonian, no ?

Again, I don't know what you mean. Tidal gravity is not just present in Newtonian physics.

 

[pervect]

I assumed that a reference frame cannot span two locations with different spacetime curvatures. For example, I have seen it said multiple times that no reference frame can span the entire earth. So I assumed the same principle applies here.

Any flat map of the Earth's curved surface cannot be to scale over a large area. You can make a map, with various projection techniques, such as the common Mercator projection. But the resulting map won't be to scale, shapes and areas of countries get distorted for instance.

However, one can make a local map on a flat sheet of paper that's reasonable accurate.

For the problem of explaining spaghettification, the later is all one needs. One doesn't need to cover the entire black hole, just a small area of whatever is falling into it.

I don't know about the B level explanation, but at a higher level , one can talk about the tangent space of an observer. This is similar to the tangent space on the Earth, a flat plane tangent to the curved Earth's surface at any point. One can draw a local map of the Earth's surface in the flat tangent space, and the scale distortions won't be too severe as long as one doesn't try to cover too large a region.

From these beginnings, one can eventually grasp the idea of geodesic deviation, and it's relation to the Riemann curvature tensor. The geodesic equation is:

$$\frac{d^2 X}{df\tau^2} = R^{a}{}_{bcd} u^b u^d X^c$$

here X is the "separation vector" measured in the tangent space between two nearby geodesics, R is the Riemann curvature tensor representing the curvature of space-time, and u^a is the 4-velocity, which is a vector in the tangent space representing the 4-velocity of an object following the geodesic.

That may be a bit much to grasp, so let's go back to an analogy. You have two ships sailing on geodesics on the Earth's surface, regarded as spherical. The geodesics are great circles, curves of shortest distance on the sphere. Have the ships start at the south pole, and sail towards the north. THe ships will follow a great circle path, which means they'll separate at first, reaching maximum separation at the equator, then join togehter again at the north pole. There is no "force" involved, just the ships sailing on a curved surface.

To make this analogy work in space-time, the direction "north" should be taken as the direction of time on a space-time diagram, and east-west dispaclements woul be "spatial" displacements.

 

[A.T.]

So how would you explain tidal forces emerging out of a spacetime curvature that has a gradient?

The spacetime curvature doesn't need a gradient to produce tidal forces. Constant curvature would produce them as well. What you probably mean is that local gravity has a gradient (non-uniform gravitational field), which implies space-time curvature (not necessarily changing curvature).

In the diagram below cases A, B show flat space-time with no curvature and no tidial froces (geodesic free fall worldlines keep constant separation), while case C shows curved spacetime with tidal forces (geodesic free fall worldlines increase separation).

Credit to @DrGreg from this post with more explanations:
https://www.physicsforums.com/threa...-in-a-gravitational-field.673304/post-4281670

 

[italicus]

Tidal gravity/spacetime curvature in GR is geodesic deviation: the trajectories of nearby freely falling particles diverge (or converge). For example, if you drop two rocks above a massive body like the Earth, aligned radially but with slightly different altitudes, the distance between them will increase. Or if you drop them from the same altitude but separated tangentially, the distance between them will decrease.

It is possible to quantify the effects of tidal accelerations, due to non-uniformity of the gravitational field around a spherical body like the Earth, just using the Newtonian law for universal gravitation. Have a look at the two sheets at the end: they are written in Italian, but I think are not difficult to understand. Physics is made of concepts written in the language of mathematics, and formulae need not be translated. I wrote these pages many years ago, for an article to be published.
But for the sake of clarity I will clarify some terms :

The title is simply : “ Tidal accelerations, geodesic deviation in the Earth gravitational field”

Free explanation , not literally translated:

There are two material particles , P and Q, separated by a small vector $\vec d$, the distance of P from the center of the Earth is R . Let us assume a local reference frame (P,x,y,z) , with z vertical , x and y in the horizontal plane. As the field $vec g$ is not uniform , there are variations passing from P to Q , that is in all three directions (x,y,z).

The separation x (fig 2) causes a component of $\vec g(Q)$ toward P , because the two vertical directions from P and Q converge towards the center O of the Earth; we have equation (3) :

$\frac{d^2x}{dt^2} = -g \frac{x}{R} = - \frac{ {GM}}{R^3} x$

the same applies to y, equation (4) :

$\frac{d^2y}{dt^2} = -g \frac{y}{R} = - \frac{ {GM}}{R^3} y$

The “-“ sign means that the particles are approaching each other in the horizontal plane.
As for the separation along the z axis , taking into account that when the distance from the Earth’s center becomes (R+z) the magnitude of g decreases , because :

$g(R+z) = \frac{{GM}}{(R+z)^2}$

there is a difference between the radial components of g(P) and g(Q) , which can be expressed ( see (5) for details) as :

$\frac{d^2z}{dt^2} =+ 2 \frac{ {GM}}{R^3} z$

the “+” sign means that the two particles will increase their distance in the vertical direction.
These ones are called tidal accelerations, because the calculus is analog to that carried out when you have to consider Earth tides, due to the Moon ( and the Sun too).
Thank you for the attention.

(I have problems to insert images...)

 

[PeterDonis]

It is possible to quantify the effects of tidal accelerations, due to non-uniformity of the gravitational field around a spherical body like the Earth, just using the Newtonian law for universal gravitation.

Only to a good approximation, for the case of weak fields and velocities small compared to the speed of light. Outside that regime, the Newtonian approximation breaks down.

 

[phinds]

Tidal gravity near or inside a black hole is the same thing, except that the divergence/convergence is more severe (and the severity increases as you get closer and closer to the singularity inside the hole).

@PeterDonis I have a question about this. If my memory serves me correctly, I have seen numerous statements here on PF (and I think some of them were by you) that the singularity of a BH is not a point in space, it is a point in time. The statement I just quoted seems to imply that it is a point in space. What am I misinterpreting?

 

[italicus]

Only to a good approximation, for the case of weak fields and velocities small compared to the speed of light. Outside that regime, the Newtonian approximation breaks down

Yes, of course. That’s why I have clearly stated at the beginning that I’m concerned with Earth, not with a neutron star or a black hole, where one has to use the full GR equations, and determine the 20 components of the Riemann curvature tensor.
I remember that the gravitational potential around the Earth , compared to $c^2$ , is of the order of $10^{-9}$, while for the Sun it is of the $10^{-6}$.

 

[PeterDonis]

If my memory serves me correctly, I have seen numerous statements here on PF (and I think some of them were by you) that the singularity of a BH is not a point in space, it is a point in time.

Yes, that's correct.

The statement I just quoted seems to imply that it is a point in space.

No, it doesn't. You can get closer and closer to a moment in time.

 

[PeterDonis]

That’s why I have clearly stated at the beginning that I’m concerned with Earth

This thread is not about Earth, or about any ordinary gravitating object. It's specifically about a black hole, since near the singularity of a black hole is where "spaghettification" occurs. So your comments are irrelevant to this particular thread. Please do not hijack someone else's thread with irrelevant comments.

 

[italicus]

your comments are irrelevant to this particular thread. Please do not hijack someone else's thread with irrelevant comments.

I don’t think I have hijacked anybody. Anyway, if this is your opinion, I can only record it. I didn't want to irritate you.

 

[PeterDonis]

I don’t think I have hijacked anybody.

"Hijack" means that you, who did not start this thread (and you didn't), are making comments/posts which are not relevant to the topic of the thread. I have pointed out where you have done that.

I didn't want to irritate you.

I'm not irritated. I am simply trying to keep this thread on topic. That is one of my jobs as a moderator.

 

[italicus]

Ok , ok ! No problem.