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Considering gravity acts by stretching spacetime and all forces (EM, nuclear etc...) are relative to spacetime, can we then consider that regardless of how stretched you are to an observer outside the black hole as you are being spaghettified (even though I know nothing can be practically observed from outside the black hole), you are not being spaghettified to yourself (the elevator/free fall Einstein thought experiment), and your body keeps functioning normally (making abstraction of the other falling objects that may hit you)?

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  • $\begingroup$ @DKNguyen that's not the same as free fall $\endgroup$ Apr 23, 2021 at 6:57
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    $\begingroup$ I think the most dubious part of your statements is "Considering gravity acts by stretching [...] all forces (EM, nuclear etc...)" I've not seen this anywhere. What does stretching a force even mean? You keep repeating this as if it were fact but I've not seen this anywhere and your entire argument hinges on it. $\endgroup$
    – DKNguyen
    Apr 23, 2021 at 7:20
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    $\begingroup$ I think this is a great question. Why is it downvoted? $\endgroup$
    – lalala
    Apr 23, 2021 at 7:33
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    $\begingroup$ @ohneVal the question was very open about its assumption being discussed (with arguments). it's pure silliness to close it considering the great quality of answers I got. And yes my assumption was wrong, but it took this comment to show me specifically how : physics.stackexchange.com/questions/631414/… $\endgroup$ Apr 26, 2021 at 14:31
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    $\begingroup$ I vote to keep the question open: I think it is not such an uncommon misconception that spaghettification is some kind of length dilation, only observable to a distant observer. $\endgroup$
    – doetoe
    Apr 26, 2021 at 16:47

5 Answers 5

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When you are being stretched by the tidal effect of gravity, you really are being stretched. You can estimate this already using Newtonian gravity, and then G.R. simply adjusts the details of the calculation. For example, consider a rod of length $L$ oriented vertically and falling towards a neutron star. The bottom of the rod is being told by gravity to accelerate towards the star at a rate $$ \frac{G M}{(r - L/2)^2} $$ where $M$ is the mass of the star, $r$ is the distance between the centres. Meanwhile the top of the rod is being told to accelerate towards the star at a rate $$ \frac{G M}{(r + L/2)^2} $$ So what does the rod do? Answer: it stretches. Its internal forces (from the bonding between the atoms) oppose this stretch and consequently a tension builds up in the rod. To calculate this tension, you can apply Newtonian mechanics to each section of the rod, and you find the tension increases towards the centre of the rod, where it reaches the value $$ T = \frac{1}{8} \rho A L^2 \left|\frac{dg}{dr}\right| = \frac{1}{4} \rho A L^2 \frac{GM}{r^3} $$ where $\rho$ is the density and $A$ the cross sectional area of the rod. When this $T$ exceeds the tensile strength of the rod, the rod will break.

For example, a steel cable of length 100 metres falling towards a neutron star will be snapped by this gravitational effect when it is about two thousand kilometres from the star.

General relativity alters the details but not the main qualitative result here. So an astronaut falling towards a singularity will indeed die as their body is subjected to the extreme tidal forces that arise. As another answer already mentioned, the equivalence principle is a statement that only applies on distance and time scales small enough that the tidal effects can be ignored.

An added note on reference objects

In general relativity we are at pains to note that many statements with regard to distance and time can only be made by reference to physical objects. Can spacetime itself be regarded as such a reference object? The answer is: not in any simple sense. The basic geometric idea here is that you can think of spacetime as a 4-dimensional space (the technical term is manifold) and points in the space are called events, and there is a measure of 'spacetime distance' (called interval) between neighbouring events, which can be written: $$ ds^2 = g_{\mu\nu} dx^{\mu} dx^{\nu} \tag{1} $$ This expression gives a sum over 16 components $g_{ab}$ multiplied by coordinate separations between the events. I won't spell out all the mathematical background. But for an understanding of spaghettification, one can note the following. What happens is that the worldlines (the trajectories in spacetime) of neighbouring parts of a physical object such as a falling rod draw apart. To be precise, let's consider an atom A and a neighbouring atom B in our rod. To begin with, the distance between these atoms might be $10^{-10}$ m or 1 angstrom, for example. That means that if you adopt the instantaneous rest frame of atom $A$, and then enquire about the spacetime location of atom $B$ at a time you would consider to be 'now', then you will find the interval $ds$ given by the above expression will have a magnitude equal to 1 angstrom. You could confirm this by comparing it with a handy ruler, made of diamond say, that you had with you. Now let some time go by as the rod falls. Maybe the rod is made of rubber, for example. Then at some later moment the expression above will give for the interval between atom $A$ and $B$ some larger number such as 2 angstroms. Any diamond rod you had with you will stretch by much less, so it could be used to measure this stretching of the rubber rod. If you wanted to be really accurate then you could also calculate the expected stretching of the diamond rod and allow for it. Or you could use a laser beam to set up a standard of distance and use that.

The point of this added note is to explain how the formula (1) applies to such experiments. In the case of a black hole the tidal stretching near the singularity will be strong enough to pull apart not just a rubber rod but also a diamond rod, or anything else.

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  • $\begingroup$ I did not deny the stretch, but just said it was relative to the referential. If the referential is the remote observer, you can sure see the stretch, but what if the referential is spacetime itself ? $\endgroup$ Apr 23, 2021 at 8:39
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    $\begingroup$ @ManudeHanoi If a steel cable is pulled apart into two separate lengths then every observer in every reference frame will agree that it is pulled apart and separated. The proper distance between the two parts is now non-zero and increasing. A local ruler or other measuring device will confirm this. What more can I say? I think you have not yet got a correct insight into what G.R, asserts about spacetime; calculations like the one in my answer may help; I hope it will. $\endgroup$ Apr 23, 2021 at 8:46
  • $\begingroup$ Another somewhat related example is a falling chimney, though for a different reason. $\endgroup$
    – Alchimista
    Apr 23, 2021 at 9:17
  • $\begingroup$ please vote to reopen this question you made a great effort pushing for it's resolution $\endgroup$ Apr 26, 2021 at 14:34
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    $\begingroup$ I turned that tension formula into a Python program $\endgroup$
    – PM 2Ring
    Sep 21, 2021 at 5:56
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The equivalence principle holds in the limit of very small elevators. When far enough from a very massive object, we may have a huge but almost constant gravitational potential: strong relativistic effects, but hardly any total forces or other locally measurable effects.

Where the potential varies strongly, the equivalence principle doesn't approximately describe what happens to your body as a whole anymore. You could say that in the first case different parts of your body "want to" travel on parallel geodesics, whereas in the second case they are sufficiently divergent to overcome the forces holding the different parts together.

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  • $\begingroup$ please consider earch atom of the body as an einstein observer, the forces holding the atoms together are based upon spacetime itself. So if spacetime streches, so do these forces and therefore I still expect them to hold. In that aspect gravity is a different type of force than EM forces $\endgroup$ Apr 23, 2021 at 7:30
  • $\begingroup$ take a 20cm ruler (see my edit to my op) , throw it in a bh, it will keep (locally) measuring distance accurately no matter how spaghettified it is to the outside observer $\endgroup$ Apr 23, 2021 at 7:35
  • $\begingroup$ Could it be that you are confusing spaghettification, which is a deformation of objects (not of spacetime, though it is caused by the geometry of spacetime) with something like Lorentz contraction (Lorentz dilation?) which is an actual property of spacetime? In fact, if you have two rulers, one made of rubber and another one made of diamond, they will actually have a different length when pointing in the direction of a (mild) potential gradient. $\endgroup$
    – doetoe
    Apr 23, 2021 at 7:48
  • $\begingroup$ could be im confusing the 2, but in the case of the 2 rulers i'd object that in a gravity free fall (unlike a gradient of other forces), they'd measure the same $\endgroup$ Apr 23, 2021 at 7:51
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    $\begingroup$ The whole point here is that the forces holding the body together are NOT gravitational forces and the equivalence principle does not apply to them. So, when you reach a region where there is an object bound together by non-gravitational forces, and the gravitational field varies quickly enough that you can see variation in the force across that physical body, then there will be gravitational stresses on that body, trying to tear it apart. $\endgroup$ Apr 23, 2021 at 14:20
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"Spaghettification" is a tidal effect caused by the different paths taken through spacetime by the particles of your body. Yes, you will notice this (and it ultimately will kill you). Your argument that electromagnetic forces are bent in the same way by spacetime isn't quite relevant, because it is not the bending of spacetime that causes spaghettification but the relative bending being different at your head and your feet, which is an objective thing agreed upon by all observers. Think of it this way: tides really exist on Earth and the atoms in the ocean really do move differently because of them. The tides near a black hole are tremendously stronger than those caused by the moon and the sun, strong enough that they will pull your head and feet apart just as the tidal forces near Earth pull the ocean on opposite sides of the Earth apart.

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  • $\begingroup$ best answer so far. $\endgroup$ Apr 23, 2021 at 11:49
  • $\begingroup$ consider a series of infinitely small einsteins free falling, each one of them tied to the next. drop them in a black hole. According to the equivalence principle (or the free falling thought experiment) , none of them feels the acceleration, and yet the ones at the bottom (near the bh) are getting more stretched than the ones at the top. you'd soon say there's an issue with what ties the mini einsteins together. Replace the einsteins with atoms and links with chemical bonds. The chemical bonds are held by EM force. And the EM force is stretched itself. $\endgroup$ Apr 23, 2021 at 12:04
  • $\begingroup$ But the speed of light (ie the EM force) can't change because it's absolute so isnt it possible that the chemical bonds would still hold ? $\endgroup$ Apr 23, 2021 at 12:05
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    $\begingroup$ The equivalence principle has an important caveat, it is true only "locally" (in a small enough region of spacetime). For example, free falling observers who are far apart relative to the Earth will not consider each other at rest. How far apart you have to be before this matters depends on just how curved spacetime is (indeed it is in some sense the definition of how curved spacetime is). Near a black hole "local" may be a very small distance indeed. EM forces act normally "locally", but over longer distances they do not (e.g. light is bent by the Sun's gravity). $\endgroup$
    – Eric Smith
    Apr 23, 2021 at 13:26
  • $\begingroup$ please vote to reopen this question you correctly pointed out my mistake $\endgroup$ Apr 26, 2021 at 14:33
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can we then consider that regardless of how stretched you are to an observer outside the black hole as you are being spaghettified, you are not being spaghettified to yourself

No. The curvature is an invariant fact, it is not relative.

Phenomena like length contraction lead to a geometric distortion that is stress-free. Local strain gauges or stress measurements do not detect length contraction.

That is not the case with spaghettification. With tidal forces (curvature) local strain gauges and stress measurements will detect it. When the locally-measured strains exceed your body's elastic limit your tissues will be damaged and you will die.

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  • $\begingroup$ so you mean length contraction is different than curvature ? $\endgroup$ Apr 23, 2021 at 16:36
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    $\begingroup$ Absolutely. Length contraction is a feature of flat spacetime. It has nothing to do with curvature. $\endgroup$
    – Dale
    Apr 23, 2021 at 16:57
  • $\begingroup$ please vote to reopen this question you correctly pointed out my mistake $\endgroup$ Apr 26, 2021 at 14:32
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As technical answers appear, I'll just address the "imagine it" part. Tidal forces near black hole are similar to centrifugal forces on a spinning body.

For a stellar mass black hole at 100 km, the tidal force over 2m (a man) is around 50,000g, so that would feel like spinning (piked, no tucking) at 7,000 RPM, or 100+ Hz.

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    $\begingroup$ I dont understand what you mean by "(piked, no tucking)" and you don't seem to address my points. I understand the common vision is that the gravity force tears up people, I tried to explain why I thought otherwise $\endgroup$ Apr 23, 2021 at 16:35
  • $\begingroup$ @ManudeHanoi. correction: layout. Not pike. $\endgroup$
    – JEB
    Apr 23, 2021 at 17:16
  • $\begingroup$ please vote to reopen the question thanks $\endgroup$ Apr 27, 2021 at 0:46
  • $\begingroup$ What do these terms "pike", "layout", "tucking" mean? Isn't a pike a type of fish? $\endgroup$
    – PM 2Ring
    May 19 at 4:30

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