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I have a hard time wrapping my head around this "spaghettification" process that apparently takes places when getting close to a black hole.

Gravity is proportional to the distance of the object exerting gravity. Everything in space is unimaginably HUGE. This applies to, e.g., both distances and size of different bodies. Earth is big, but our sun could contain more than one million Earths in it. And a grain of sand is negligible compared to Earth.

The difference in gravity between the "front" and the "rear" of a grain of sand sent towards a black hole should/must be negligible, considering how weak gravity as a force is, compared to e.g., magnetism. So how can this weak force and miniscule gravity gradient cause spaghettification on something as small as a grain of sand?

(I believe I can understand why something as big as a star would become spaghetti when closing in towards a black hole.)

Second question, if grain of sand would spaghettify in these circumstances, would something much smaller like a hydrogen atom also spaghettify? A water molecule? A free electron? Neutrinos?

minseong
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d-b
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    Gravity is proportional to the distance of the object exerting gravity. No, it’s inversely proportional to the square of the distance. – Ghoster Feb 02 '23 at 02:02
  • Related: https://physics.stackexchange.com/q/631414/123208 – PM 2Ring Feb 02 '23 at 03:48
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    @Ghoster Proportional doesn't mean linearly. – d-b Feb 02 '23 at 08:18
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    “$A$ is proportional to $B$” means $A =cB$ for some constant $c$. That is indeed a linear relationship. Gravity is not proportional to distance in any sense. – Ghoster Feb 02 '23 at 08:23
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    would something much smaller like a hydrogen atom also spaghettify?” - There are no atoms in General Relativity, it is a classical theory. Atoms exist in Quantum mechanics, but there are no black holes there. It is a flat space theory. Atoms and black holes both exist in Semiclassical Gravity, but it is neither strict nor confirmed theory with potentially wrong conclusions. Atoms and black holes also both exist in Quantum Gravity, except thus theory has not yet been developed. Based on which of these theories would you like the answer to be? – safesphere Feb 02 '23 at 09:02
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    Note that the sand becomes vapor before it is stretched. – bandybabboon Feb 02 '23 at 10:07
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    @d-b you would be fine with that on [english.se], but here on [physics.se] "proportional" should only be used in its narrowest sense – AakashM Feb 02 '23 at 10:32
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    @d-b all proportional relationship are linear, but not all linear relationships are proportional. Ghoster gave a correct definition of proportional (it gets its name from the fact that the two variables have a constant ratio i.e. they grow in proportion to each other). A linear relationship adds the possibility of a constant offset – Tristan Feb 02 '23 at 15:08
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    @AakashM I'm not even sure that's true. The word "proportional" outside of STEM generally just means "not excessive (when compared to)" and can't really be used in the context given here because the two variables have different dimensions and so can't be directly compared – Tristan Feb 02 '23 at 15:12
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    "Everything in space is unimaginably HUGE" - not your typical stellar black hole. Massive but relatively small :) – Filip Milovanović Feb 02 '23 at 18:10
  • "proportional" has innumerable uses / senses in general everyday english. all of that has no connection at all to the meaning of proportional in maths and physics (ie, as stated by ghoster) – Fattie Feb 03 '23 at 15:38
  • @FilipMilovanović Also, I've got a pretty good imagination – Michael Feb 03 '23 at 15:43
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    @AakashM tristan fattie Linearly proportional is a subset of proportional. All proportional relationship ARE NOT linear. – d-b Feb 03 '23 at 23:38
  • @d-b in that case perhaps you should say "inversely proportional"! apologies to Ghoster, but this bears repeating IMO. In a sense, it it the "opposite" of proportional. – m4r35n357 Feb 04 '23 at 09:06
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    @Tristan "A linear relationship adds the possibility of a constant offset" That's affine, though unfortunately the term "linear" is sometimes misused to mean "affine". – user76284 Feb 06 '23 at 16:48
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    @Tristan : Since ws're being precise, no, a linear relationship does not allow for a constant offset. – WillO May 17 '23 at 12:34

6 Answers6

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Spagghetification occurs when the gravitational potential energy on the side of the grain of sand closer to the center of the black hole is much larger than the potential energy on the other side. The gradient in potential energy across the grain of sand leads to a force, and if this force is large enough it will be larger than the forces holding the grain of sand together.

For simplicity, let's assume a non-spinning black hole of mass $M$.

The potential energy on the "near side" of the grain of sand, at a distance $r$ from the black hole singularity, is $$ U_1 = \frac{GM}{r} $$ The potential energy on the "far side" is $$ U_2 = \frac{GM}{r + d} $$ where $d$ is the diameter of the grain of sand.

Now when $r \gg d$, it is true that $U_1 - U_2 \approx 0$. This is where your intuition likely comes in.

However, if we are imagining a grain of sand falling into a black hole and hitting the singularity, then we are ultimately imagining $r$ going to zero. So we cannot assume $r \gg d$.

In fact, we have $$ U_1 - U_2 = \frac{GM}{r}\left(1 - \frac{r}{r+d}\right) = \frac{GM}{r}\left(\frac{d}{r+d}\right) $$ As $r\rightarrow 0$, this energy difference will become arbitrarily large. So as $r$ becomes small enough, inevitably the tidal force ripping apart the grain of sand will become larger than whatever (electromagnetic) forces hold it together.

We expect spaghettification to happen to any finite size particle -- sand grains, hydrogen, neutrons, etc.

There are a few caveats, because most physicists don't expect general relativity to be a complete theory that tells us what happens near the singularity of a black hole. Probably we need a quantum theory of gravity to really answer these questions.

When you start to get to distances $r$ of order a Planck length from the singularity, most physicists would agree that quantum gravity will become important, so the predictions of GR are no longer a good guide. However, there are also some speculative ideas that even when $r$ is of order the horizon of the black hole, general relativity breaks down. So, a final word of warning may be that since we cannot do any experiments to really see what happens to grains of sand in a black hole, we should be humble about saying exactly what will or won't happen.

Andrew
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    Sure, a sand grain won't get spaghettified by a stellar mass (or larger) BH outside the EH. But for a Jupiter mass BH, with $r_s$ of 3 m, a 1 mm grain of silica fractures at a (Schwarzschild) distance of ~83 m. – PM 2Ring Feb 02 '23 at 03:53
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    Spagghetification occurs when the gravitational potential energy on the side of the grain of sand closer to the center of the black hole is much larger than the potential energy on the other side.” - There is no potential energy in General Relativity. For example, if two stars of the rest mass $m$ each collide under gravity, the resulting Newtonian mass (total energy) is $M=2m+P$ where $P$ is the potential energy. In General Relativity, the result is $M=2m$ with no potential energy contribution, because potential energy does not exist in reality. It is but a Newtonian artifact. – safesphere Feb 02 '23 at 08:49
  • @PM2Ring the grain could pretty much spagettify inside the event horizon (in its own time, the event won't be visible outside) – fraxinus Feb 02 '23 at 12:10
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    @fraxinus Of course. And you don't actually need to go as small as a Jupiter mass BH to get sand grains fracturing outside the event horizon. Jupiter's just a nice example object for gravity stuff because its mass is very close to 1/1000 solar mass. :) – PM 2Ring Feb 02 '23 at 12:33
  • @safesphere It seems to me that, whatever you want to call it, the kinetic energy that the stars have as a result of accelerating toward each other will add to the rest energies to become part of the total final system energy. In either theory. – RC_23 Feb 03 '23 at 01:59
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    How do you spaghettify a neutron?! Does it rip the quarks out?? – Loren Pechtel Feb 03 '23 at 05:26
  • @RC_23 In GR, the kinetic energy comes from the mass defect. The total energy of two stars with the rest mass of $m_o$ each far from each other is $E_o=2m_o$. When they collide under gravity, the mass of each (as observed by us from the same frame as before) is $m=m_o-K$ where $K$ is the kinetic energy of the star. So the total energy at the collision is $E=2(m+K)=2m_o=E_o$ per energy conservation. – safesphere Feb 03 '23 at 06:29
  • @LorenPechtel Interesting question... maybe gluon tubes form between quarks in the direction of the tidal force which may break into quark/anti-quark pairs and then the process repeats? – Michael Feb 03 '23 at 15:48
  • @RC_23 There's no disagreement about the total system energy as long as you actually look at the total system energy, not two out of three components at a time. The equation you're reacting to is wrong. In such coordinate systems as measure well-defined local kinetic energy $T$ and potential energy $U$, conserve energy: $\Sigma \Delta T + \Delta U = 0 \to E = mc^2 + T + U = mc^2+T_0+U_0$. – g s Feb 08 '23 at 23:31
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The difference in gravity between the "front" and the "rear" of a grain of sand sent towards a black hole should/must be negligible, considering how weak gravity as a force is, compared to e.g., magnetism. So how can this weak force and miniscule gravity gradient cause spaghettification on something as small as a grain of sand?

That depends entirely on the size of the black hole. Supermassive black holes are indeed so large (Schwarzschild radius grows linearly with mass!), and by consequence the gravity differential so small that you would indeed be able to fall into it crossing the event horizon fully conscious. Sadly, you won't be able to telephone home to tell us about your experience.

The Schwarzschild radius of stellar objects are in the range of tens to thousands of kilometers. The corresponding gravity differential might tear an astronaut to pieces, but would certainly leave a grain of sand intact until it crosses the horizon.

However, if you had a black hole with the weight of the moon, the Schwarzschild radius is on the order of millimeters, and your grain of sand won't stand a chance to reach it intact.

Now, "spaghettification" is a bit of a misnomer. It only means that the forces that hold your grain of sand together are not sufficient anymore for the grain to retain its shape. It will disintegrate in the same way as if you tear it apart by other means. For that grain of sand, that means it will break into smaller particles like stones do. A steel bearing ball would indeed be turned into a small wire because steel does deform before it breaks apart. But brittle materials won't get turned into spaghetti at all, they will be ripped into a stream of dust particles instead.

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To answer the "elementary particle" tag of the question which is:

Second question, if grain of sand would spaghettify in these circumstances, would something much smaller like a hydrogen atom also spaghettify? A water molecule? A free electron? Neutrinos?

A water molecule will break apart into electrons protons and neutrons, the hydrogen atom into an electron and a proton.

As Andrew says," we cannot do any experiments" near a black hole, so we have to rely on observations and mathematical models.

Our current model for hadrons , neutrons and protons, is that they are held together with the strong force which becomes infinite if the quarks are distanced, so in the current models, which do not have quantized gravity, I presume that protons remain whole while neutrons decay on the way to the singularity, into a proton a neutrino and an electron. Electrons and neutrinos are non decaying elementary particles, , and as such, retain their identity in the present models.

anna v
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    I suppose that nucleons can spahhettify into a quark-gluon plasma jet in the final nanoseconds (or less) before they hit the singularity, but that's pretty speculative without a proper theory of quantum gravity. ;) And as you say, electrons & neutrinos are point particles (in the current model), so they can't feel tidal forces. – PM 2Ring Feb 02 '23 at 08:13
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    @PM2Ring Electrons are point particles only upon detection, but in flight they are distributed waves. – safesphere Feb 02 '23 at 09:11
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    @safesphere in the mainstream theory of particle physics standard model they are always point particles. it is the probability of measurement/interactions that depends on the wavefunction solutions for the particular case – anna v Feb 02 '23 at 11:11
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    @annav It is exactly the same as what I said. Here is an example of the wave behavior using photons for simplicity. For you to see your reflection in a mirror, each photon must reflect off the entire surface of the mirror. If you detect photons at the mirror, they would appear as point particles each hitting the mirror in one point, but in this case you would no longer see your reflection. So does a particle hit a singularity at a coordinate point or over the entire light cone of the emission event? Who knows. (As a clarification, a Schwarzschild singularity is not a point, but infinite line.) – safesphere Feb 02 '23 at 15:34
  • I don't think any neutron caused by the break up of an atom would have time in its rest frame to decay; the accelerations at the point where spaghettification occurs means the particle is already traveling a significant fraction of the speed of light. – Michael Feb 03 '23 at 15:52
  • @Michael one needs detailed mathematical models, and if the neutron does not decay, then it would follow the proposed spagetification of the proton in the comment of PM2ring – anna v Feb 03 '23 at 18:45
  • "I presume that protons remain whole" not a physicist - why it couldn't cause one of the quarks to potentially separate due to gravitational force and create a quark-antiquark pair creating thus baryon and meson? – Maja Piechotka Feb 05 '23 at 03:32
  • @MaciejPiechotka To answer this one would have to apply a mathematical modeland see if it is possible. In general they would end into a quark-gluon plasma – anna v Feb 05 '23 at 04:30
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Spaghettification does not always happen outside the event horizon. You could fall through the event horizon of a supermassive black hole, like the one at the center of our galaxy, without suffering any immediate harm. But soon enough (quite soon, in fact) you would get close enough to the singularity to be spaghettified. And so would a grain of sand or a molecule.

Mark Foskey
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The difference in gravity between the "front" and the "rear"

This is exactly the point. In addition, the scribble - like everything else, has some initial speed and direction - and therefore inertia; and, having fallen under the influence of the gravity of a black hole, this, albeit a meager inertia, is the reason that multidirectional forces act on different parts of this body, which destroy it.

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A (pseudo) singularity could well be dimensionless and spaghetti could not be described, as per canonical theory mass would be indistinguishable within it, no mass no dimension, other way to see it. Only the energy transitional state of a grain of sand attracted by the "singularity" or the sand grain's wakefield could be likely described as spaghetti similar.

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