Spaghettification & Black Holes: Classical & Einsteinian

[DaveC426913]

TL;DR

What is the source of the lateral "squashing" when falling into a black hole?

(Classical model)
The radial "stretching" is caused by differential gravity (tides), but what is the lateral squashing caused by? Is it because the "force" of gravity is not parallel, but instead comes from a point, forming an acute angle?

(Einsteinian model)
I guess it's pretty trivial to explain in curved spacetime - the curvature near a black hole can be measured both radially and circumferentially, yes? (That's just a little less intuitive.) And they curve in opposite "directions", so opposing "forces".

 

[PeterDonis]

The radial "stretching" is caused by differential gravity (tides), but what is the lateral squashing caused by?

Tides. For example, if you look at the tides on Earth caused by the Moon, there is stretching radially but squashing laterally. Low tide is not just the ocean being at the "normal" level it would be at without the Moon there; it's the ocean being squashed by lateral tidal effects. (Of course this is all at a very heuristic level, there are lots of complications to actual tides in the Earth's oceans, but in an idealized model the effects would be what I've described.)

Note, btw, that the basic effect is the same in both Newtonian gravity and relativity. The underlying conceptual basis is very different, of course, but the actual effect is basically the same.

the curvature near a black hole can be measured both radially and circumferentially, yes?

Yes. In units where $G = c = 1$, the radial tidal stretching at radial coordinate $r$ in Schwarzschild spacetime goes like $2M / r^3$, and the lateral squashing goes like $M / r^3$. In more technical terms, these are the relevant components of the Riemann curvature tensor in an orthonormal basis.

 

[DaveC426913]

the ocean being squashed by lateral tidal effects.

But is it because the Moon is effectively a point?

OK, I see. Even as I try to describe the counter-example - a laterally-uniform gravitational source - I realize I'm describing a massive body that is far away, meaning its influence is effectively parallel. Which is tides.

 

[PeterDonis]

is it because the Moon is effectively a point?

As far as tides on Earth are concerned, yes, the Moon can be considered a point mass. There are small theoretical corrections due to its finite size, but I think they're too small to be measurable.

a laterally-uniform gravitational source - I realize I'm describing a massive body that is far away, meaning its influence is effectively parallel. Which is tides.

I'm not sure what you mean here. Tides are not the same as the "acceleration due to gravity" vector. In the Newtonian approximation you can think of them as being due to spatial differences in the magnitude and direction of that vector. (In GR that's not quite correct, but it's still a reasonable approximation for cases like the Earth and the Moon.)

A "laterally uniform" source would be one in which the magnitude and direction of the vector does not change laterally. In such a case there would be no lateral tides. If you are very far away from a spherically symmetric mass (much farther than the Earth is from the Moon), the lateral change in the vector can become too small to measure--in which case the lateral tides would also become too small to measure. But in that limit, the radial tides would also become too small to measure, because both tides are of the same order of magnitude (look at the numbers I gave previously for Schwarzschild spacetime, for example). In other words, the radial change in the vector would also be too small to measure.

 

[DaveC426913]

A "laterally uniform" source would be one in which the magnitude and direction of the vector does not change laterally. In such a case there would be no lateral tides. If you are very far away from a spherically symmetric mass (much farther than the Earth is from the Moon), the lateral change in the vector can become too small to measure--in which case the lateral tides would also become too small to measure.

Yes, this is exactly what I mean.

But in that limit, the radial tides would also become too small to measure, because both tides are of the same order of magnitude (look at the numbers I gave previously for Schwarzschild spacetime, for example). In other words, the radial change in the vector would also be too small to measure.

Yes, exactly. That's the point where I realized they were one-and-the-same. Radial tides and lateral tides go hand-in-hand. I could visualize it but I didn't have the vocabulary to express it.

Thanks!

 

[PeterDonis]

Thanks!

You're welcome!