# Why would black hole rip me apart?

According to my understanding of General Relativity, gravity is not a force and an observer which is falling freely under the influence of gravity should be considered inertial. Now, I have come across some texts about black holes that say a body approaching black hole will be eventually ripped into pieces due to large difference in the gravitational field intensity between, say head and toe.

So my question is if the observer is an inertial one and he is not experiencing any force why would his body parts be ripped apart?

• It's more accurate to say that each particle that makes up your body is in its own inertial frame. When your body is close enough to a black hole, the particles in your body will be subject to a tidal force, i.e. particles closer to the singularity will experience a stronger force than those farther away, and this differential force will eventually pull everything apart. May 19 at 15:47
• Not directly related to your question, but in general I would ignore the typical 'gravity is not a force...' conclusion popular science education often makes. Although the mathematical description (GR) is quite involved, and describes gravity vastly differently than Newton's gravity, it is still very much a force. Us being on Earth, which is a planet, and not e.g. an electrically charged bound state, orbiting the Sun should be enough phenomena to qualify as a force. May 19 at 15:51
• May 20 at 4:06
• You could read Larry Niven's "Neutron Star" for a layman's explanation. May 20 at 13:44
• @gnasher729 There is a more concise story, Neutron Tide, by Arthur C Clarke. It ends: "...the only identifiable fragment of the pride of the United States Space Navy was . . . one star-mangled spanner." May 20 at 21:14

You don't need GR to see this effect. It's due to tidal forces.

Suppose you are 2 meters tall. Then the force of the Earth on your feet is $$GMm/r^2$$, and the force on your head is $$GMm/(r+2)^2$$. The difference between the two is the tidal force you feel. Now if you calculate these two forces, you'll find that they are almost the same. It's why you aren't ripped apart.

But say the Earth was compressed to size of about $$1$$ cm (approximately the Schwarzschild radius of an Earth-mass black hole). Then the same calculation would find two vastly different forces. It's why you are ripped apart by small black holes but not by large ones.

(Of course all this can be made more precise in GR, but Newtonian mechanics suffices to answer your question.)

Edit: to answer your comment, the observer's frame is inertial as long as tidal forces aren't large enough. Once they're large enough the frame ceases to be inertial. There are two ways to make tidal forces smaller: the first is to have weak gravity, and the other is to make the observer smaller. You can see both of these in the equations above: weak gravity corresponds to large $$r$$ or smaller $$M$$, while smaller observer corresponds to you having a smaller height. Both will reduce the tidal forces, and will lead to you not getting ripped apart.

• "Suppose you are 10 meters tall"? Not that it matters much, but maybe a more realistic "suppose you are 2 meters tall" would read better.
– hft
May 19 at 22:58
• @hft Pffft, rounding errors. I'm pretty sure humans are spheres 10m in diameter. May 20 at 0:42
• I calculated in my head that earth's gravity tries to pull me apart with a force equivalent to a few milligrams if I stand upright. I can reduce that by lying down. Going to bed now to be safe :-) May 20 at 13:41
• Big black holes will still spaghettify you, they'll just wait until after you cross the event horizon meaning the rest of the universe won't see it. May 20 at 14:47
• @DanIsFiddlingByFirelight How ... considerate. May 20 at 23:22

The problem is that when you are falling, all of you can't be in the same inertial frame. That is, whilst your centre of mass might be inertial, parts of your body will be feeling accelerating forces because they are not in the same inertial frame as the centre of mass. These non-inertial forces are known as tidal forces.

The concept of an inertial frame is local. If the frame covers a large enough volume that spacetime cannot be considered flat across it, then non-inertial forces will become apparent.

A thought experiment is to consider a body made of different pieces, each falling inertially. As they fall they will become more and more separated as each follows its own geodesic through spacetime according to its own starting position. This accelerating separation can be interpreted as due to non-inertial tidal forces from the point of view of a falling frame of reference.

As a complement to the other answers:

1. Not every black hole is capable of ripping you apart tidally.
• Too small black hole will burn and blow you away you with its Hawking radiation way before you are anywhere near to feel the tidal effect
• Too big black hole (e.g. a supermassive one in a galaxy center) will swallow you as a whole because their gravity is strong but rather homogenous and you won't feel a significant tide unil deep inside.
1. There is at least one more mechanism for ripping you apart: the frame dragging of a rotating black hole. Depending on your approach direction in regard to the hole's rotation, you may get accelerated away from the black hole. The "lower" parts, however, will feel stronger "dragging" - and I am not sure weither this mechanism can dominate (maybe it also depends on how much rigid you are).

2. The magnetic fields around a rotating (and maybe electrically charged) black hole are strong enough to be capable of disintegrating the chemical bonds way before any of the above takes place.

There are nice answers by @fraxinus and @profrob, I would like to add a little side note about the balance between the forces.

It arises because the gravitational field exerted on one body by another is not constant across its parts: the nearest side is attracted more strongly than the farthest side. It is this difference that causes a body to get stretched. Thus, the tidal force is also known as the differential force, as well as a secondary effect of the gravitational field.

https://en.wikipedia.org/wiki/Tidal_force

Tidal forces are so that they are present when you are freefalling towards a massive body, may that be the Earth, the Sun or a neutron star or a black hole.

Your body is held together by the strong and EM forces (including the covalent bonds), and these forces are dominating over the tidal effects of gravity in most cases of freefall. Do you think that there are no tidal forces in play when free falling here on Earth? Yep there are. It is just that these effects are miniscule and are overwhelmed by the forces that hold your body together.

Only in extreme cases like a black hole (and as you see from fraxinus's answer, only certain black holes) are capable of creating such strong tidal effects, where the gravitational acceleration on your head is so much different from the gravitational acceleration at your feet, that this dominates over the forces holding your body together. In such cases tables turn, the balance turns in favor of the tidal effects and first the chemical bonds are overwhelmed by the tidal effects and your body stretches. But, but in certain cases the effect is so strong, that it dominates over even the EM force holding the electrons and nuclei together, ripping the atoms apart, and eventually dominates over the strong force, thus ripping the the quarks apart.

Even before that happens the gravitational tidal force will rip off the electRons and have the nucleus break up and have most of it converted to neutrons, and then rip those off and get to the quarks, and eventually fall into the singularity. That's why we say that a BH forms when there is too much gravity, nothing can withstand the gravitational effects. Not electron pressure (which holds up white dwarf stars), not nuclear forces (neutron stars), and not strong forces (quark stars, or some parts of the cores of neutron stars). Yes, the equivalence principle says everything will be accelerated (i.e. pulled) the same way, but only until the force differential between two objects in the atom see different accelerations - that's the gravitational tidal effects, caused by very strong curvatures of the spacetime due to gravity.

Is the gravitational force of Black Hole destroying Atoms?

Now as ProfRob says, here on Earth, when you are in freefall, your whole body can be considered to be in the same inertial frame, because the forces holding your body together dominate over the tidal forces, and these tidal effects can be neglected. In extreme cases, when tables turn and the tidal effects dominate over the forces holding your body together, every single particle in your body needs to be treated as having its own inertial (local) frame that is different from all other particles' frame in your body. "Local" frame in this case will become restricted to a extremely small area (for each particle). Any frame (volume of space) that corresponds to an object that consists of multiple particles cannot be considered to be a "local" frame any more. This is because the gravitational field (and acceleration) varies considerably even on the extremely small scales. I believe this is a very important side note to the answer to your question.

• The tidal force near a white dwarf is quite strong, although of course much weaker than near a neutron star or black hole. Eg, a 650 metre long rod of A36 steel (which has an ultimate tensile strength of 550 MPa) falling vertically towards Sirius B will break at a distance of ~30 km from the star's surface (~5870 km from the centre). May 21 at 12:28

It's also worth nothing that in the case of many black holes of which we have knowledge, the radiation from the accretion disk is so intense that you would be blasted into plasma long before you reached the event horizon.

TON 618, for instance, has an accretion disk that radiates energy at a rate 140 trillion times that of our Sun. (Edit:) My layman's back-of-the-envelope calculation estimates that an object orbiting TON 618 at a radius of 187 light years receives as much radiation from it as we do in our orbit around our Sun.

The inertial frame that falls towards the mass freely is only approximately an inertial frame. Around the Earth the approximation is almost perfect agreement. In the ISS, if you place marbles throughout the station, initially at rest (obviously floating) wrt to each other and the station, the marbles will stay like that almost indefinitely (ignoring friction with the atmosphere). But if we wait a long time, assuming the station can orbit long enough, we will see the marbles move away from each other slowly. The non-local tidal effects of gravity (which applies to the electric field just the same, the difference being that the frames of, say, electrons "falling" free are not inertial).

If we let the ISS circle around a black hole (say with the mass of the Sun) and apply the same procedure we see the marbles accelerate away from each other in the direction of the hole. The ISS will stay intact if the tidal force is not too big.

Now, suppose you are a collection of free marbles, falling into the hole. Your marbles would get stretched in space, towards the hole and compressed sideways, as your volume stays the same. If EM forces hold your body of marbles together you would actually feel these EM counter forces up to the point your marbles break apart... Don't jump...

• Nice explaination. Just one doubt, why does the volume remain constant? May 20 at 15:44
• @ShivamSinghAswal Good question! I got this from the comments and answer to a question I asked. Look here: physics.stackexchange.com/questions/709099/… Seems a volume falling freely keeps it volume. May 20 at 16:16
• I see. But if there are any more reliable sources regarding this, please share. May 20 at 17:29
• @Felicia The volume is preserved because the Weyl tensor only affects the shape, not the volume. The Ricci tensor affects the volume, but in free space (eg, in the vicinity of a ball of marbles of small mass, so the ball has negligible effect on the curvature) the Ricci tensor vanishes. May 20 at 22:33
• @PM2Ring You took the words right out of my mouth! So the two answer on the question I asked about freely falling volumes are wrong? May 20 at 22:54

The tidal force between your head and feet depend on the difference in $$GMm/r^2$$ for the two different r values, one about 2 metres closer than the other.

Tidal force is essentially the gradient of the gravity field (force as a function of position). A strong enough field can have a significant gradient even when your height is a small fraction of the total distance between you and its centre of mass, but of course the effects get more intense closer to a small dense object. As the wiki article approximates, tidal force varies with $$GM/R^3$$ for the gravity of a point mass (or sphere), and linearly with the length of object along the axis towards the centre.

A very dense object like a black hole lets you get very close without bumping into the surface, experiencing much higher gradients than if you approach Jupiter's surface or the Sun's surface.

@PM 2Ring calculates a steel rod 650m long falling towards the white dwarf star Sirius B will break at about ~30 km from the star's surface (~5870 km from the centre) as the tidal force exceeds its tensile strength.

A purely classical explanation of this works just fine; general relativity is useful in terms of why a black hole is black (light can't escape), not why it's a hole (hard to climb out of with a large gravity field) or why a lot of matter will stick together gravitationally and get denser.

If a cloud of matter in space doesn't collapse all the way to a black hole, electron degeneracy pressure (white dwarf) or neutron degeneracy pressure (neutron star) are what limits the collapse. Or of course fusion producing heat and thus pressure, as in a main sequence star.