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Comment by Nevermark | Hacker News Reader

If you naively apply Newton's force of gravity, between two masses at a distance, Fg = Gm1m2/d^2, you get an infinite force when a marble (mass 1) is at the center (distance = 0) of a planet (mass 2)

Of course, this singularity is just a sign the equation doesn't capture all the details.

The detail is the masses have radiuses, r1 and r2, and when the distance falls below either radius, the force of gravity for any mass's radius beyond d cancels out. When d = 0, Fg = 0.

Similarly, the idea that general relativity's black hole singularity is real is also naive.

A situation with massive gravity over tiny distances is going to require GR and QM to be united or otherwise reconciled (at a minimum!) for any chance of an accurate model.

Lesson: Our equations break down because they are incomplete. Reality doesn't break down.

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Almost all physicists have assumed that quantum effects will get rid of the singularity, as you say. The import of this paper is that there is no singularity even without accounting for quantum effects (at least for Kerr black holes and one particular definition of “singularity”).

No, you have error in your formula: "d" is vector and "F" is also vector.

If you're inside a planet, the layer of the planet that's farther from the center than you cancels out. So you need to take only the sphere that's deeper than you. By moving closer to the center, that sphere shrinks, so m2 also diminishes to 0.

So the total force magnitude smoothly reduces to zero the closer you get to the center.

Have you noticed people misapplying that to galaxies? They'll claim there is no influence on a star due to the part of the disk at greater radius, and also that the mass inside that radius can be treated as a point mass. But disks are not spheres, and rings are not shells, so those simplification do not apply. I've seen this mistake many times.

Interesting question. Given n stars/whatever at varying positions, is there always at least one point where the gravitational forces cancel out? Or is that just a solution to the simple cases of two bodies or a single body in the form of a sphere or shell?

I can't do the math to determine this more rigorously, but my intuition tells me you can do something analogous to the Hairy Ball Theorem to prove that there exists a point within the convex hull of those stars where the gravitational forces cancel out. However, as the stars move, there's no guarantee that point will remain static and stable.

So, at the center of the earth stuff is in zero gravity or micro gravity, just like above the earth?

Yes. You can think of it as: If you’re surrounded by an equal amount of mass in every direction (because you’re at the center of a set of spherical mass shells), then the gravitational force in every direction will cancel out.

Agreed.

>If you’re surrounded by an equal amount of mass in every direction

Which is a reasonable approximation for being 'out in space', but perhaps not an entirely intuitive one.

Indeed, and as a related situation. You could be inside a black holes event horizon and experience zero gravity/acceleration. You just need two blackholes to cancel out the gravity.

I feel like being in the middle of two black holes would be more like being ripped apart than zero gravity/accel.

Depends on the size of the black holes, spaghettification because of a high gravity gradients is a problem when you are close to the center of the black hole. Larger black holes make the much less of a problem, so you can cross the event horizon without any problems.

So if a cup of water were in the centre of the earth, and every layer of the earth above the centre wasn't there to compress it, then it would explode out!

Seems perfectly reasonable, if a bit useless.

If a cup of water at room temperature were on the surface of the earth and every layer of the earth above it wasn't there to compress it (so no atmospheric pressure) the water would violently boil immediately. If the cup is very tall or you're talking about a cup of very cold ice, the water's own gravitational pressure could keep the bottom stable, but to prevent the top from boiling/sublimating off, you'll need a lid to keep it pressurized.

Moving the experiment to a zero-gravity environment doesn't change the pressure considerations much.

No, it won't explode.

If you somehow made a spherical hole in the middle of the earth and placed a cup of water inside the hollow (anywhere inside, doesn't have to be the centre) it will just sit there: Newtonian gravity is 0 inside a uniform spherical shell of mass. (Einsteinian gravity is probably mostly 0 but you'll get frame dragging effects if the shell is rotating I would imagine.)

You can do the maths to prove this yourself if you want.

Yes but I think OP was referring to the lack of pressure. Without gravity, no pressure -> water will violently boil.

Fill the hole with air at 1atm & it’ll sit there very slowly evaporating though, same as it does on the ISS or on the surface of the earth. Gravity is irrelevant.

You don't seem to understand where that 1atm air pressure on the surface of the earth is coming from. Hint: on the surface of the Moon there is none.

In the ISS the air pressure is artificially kept at a certain level. To make a proper comparison, you should check what water would do outside of the ISS.

No, I understand precisely.

The context of my original comment was someone expecting that a cup of water at the centre of the earth would explode due directly to gravitational effects. This is wrong, as I pointed out.

Even on the surface of the earth it is the air pressure that prevents the water boiling off instantly, not gravity itself. On the surface of the earth that air pressure is due to the effect of gravity on the atmosphere, but you can supply that air pressure by a number of other means in other places. The air pressure inside the ISS is not there due to gravity after all! I suggest to you that if we can posit a hole at the centre of the earth, then we can fill it with air at 1atm if we want to - we are in the realm of mathematical models here, not reality after all :)

no it would be weightless, as all the gravity would cancel each other out.

Yep. You stop accelerating as you reach the center of Earth. If you were to free fall between one end of the planet and the other somehow, going directly through the center of mass, you'd oscillate until finding a resting position at the center.

Yes, exactly. The mass around you is roughly evenly distributed and so it pulls on you equally in every direction.

And never mind the pressure from all of the mass around you ;)

Every celestial sphere is actually an electromagnetic donut

Excellent. Now do the same with the Schwarzschild solution.

> Fg = Gm1m2/d^2, you get infinite force when a marble (mass 1) is at the center of a planet (mass2)

A non-point mass only approximates to a point mass when you're not inside it. The gravitational effect of being inside a shell is nothing... so as d tends to zero the mass that's not a shell also tends to zero, so you actually get zero force.

A fun fact is that even in humble fracture mechanics one gets singularities at the crack tip using standard elasticity theory. A bunch of funny mathematics is needed to get around this. And, of course, the reality is that at a certain point you hit the atomic level so the tip is always 'blunted' at least one atoms length.

> A bunch of funny mathematics is needed to get around this.

I usually use a small drill to get around it. Works every time.

TIL that drills are excellent matematicians

They're great mathematicians, too.

I thought the general determination of the singularity came from the 1916 schwartzchild solution to einsteins field equations.

I would note that reality is vacuously what reality is, regardless of model agreement. This however doesn’t mean reality isn’t weirder than we observe. That’s usually the first lesson you learn as you move beyond Newtonian physics.

The schwartzchild solution comes with a bunch of caveats!

For example, it is eternal, it's not the result of a collapse solution. As far as I know, no one has an analytic solution yet for black hole collapse of say a star in finite time.

For an outside observer, in fact I believe an infalling object into a black hole is never observed to reach the event horizon at all. Since the infalling object could communicate (using light) with the outside observer, there's a sense in which they must agree on whether the object has or hasn't crossed the event horizon -- in this case, the non-observation of crossing the event horizon must not be just apparent, because of this communication.

Since event horizons couldn't form in this case, then it should follow that a black hole cannot gain mass at all. I've yet to see a satisfactory debunk of those claims, at least at a level I could understand.

> I thought the general determination of the singularity came from the 1916 schwartzchild solution to einsteins field equations.

Only if you have a point mass. There is no reason to believe that point masses exists, it all being fields makes more sense. Einstein himself scoffed at the idea of such singularities existing, so those singularities aren't part of the core theory, just one of the possible results. (Not saying Einstein has to be right here, just that he didn't envision such singularities to exist when he created the theory)

As far as we can tell, electrons are point objects.

Why shouldn’t there be others?

Electrons are fields, not points. You can compress it to arbitrarily small spaces, but that doesn't make it a point.

>If you naively apply Newton's force of gravity, between two masses at a distance, Fg = Gm1m2/d^2, you get an infinite force when a marble (mass 1) is at the center (distance = 0) of a planet (mass 2)

and, for the people who might not have been exposed to the idea in a physics class, that marble at a distance from the planet has a potential energy, energy it would give up were it to be released and allowed to fall toward the planet. But, if instead you did work against the gravity to pull the marble farther away from the planet, that work would be saved up as an increase in the marble's potential energy. And that would continue till you pulled the marble infinitely far away, at which point the force of attraction would be 0 and the marble's potential energy would be zero. The only way this works is if the potential energy of the marble is a negative number, in which case maybe the singularity isn't such a bad idea

>at which point the force of attraction would be 0 and the marble's potential energy would be zero

That's wrong. If you actually stick to newtonian physics (because the picture changes completely in GR), then you can define the potential arbitrarily at infinite distance. Since a body with finite density will see a finite variation in gravitational potential, you could just define the potential at infinity such that it is zero at its lowest point by adding a constant. That's because forces in physics arise from the potential field's gradient - not the field itself. So any constant will just disappear in the derivative. Now, if you go to GR, the story becomes a lot more complicated, since gravitational potential energy isn't even well defined.

Do we have an inkling were the proverbial blind spots are in our theories? Like a general direction to search?

Our equations break down because they are incomplete.

But that does not mean that reality does not also break down.

IMO, reality doesn't break down, something different just happens. The information encoded in what we call "forces", "space", etc., and the rules from which those concepts emerged, continues as before, but with a different emergent form we can't predict yet. We might learn what that is, or it might be too chaotic, complex or unobservable for us to ever know.

But reality as we normally experience it breaks down!

Advances in cosmology and physics have broken generations of our perceived realities.

> But that does not mean that reality does not also break down.

What does that mean, though? Reality just is.

Our equations are just models of reality. They aren't reality.

that’s an assumption though.

An assumption in what sense?

And even if so, then what is the alternative? As far as I know, in terms of our knowledge of physics and philosophy, there is not a current alternative. We have no choice but to build models. We have no way of knowing what reality is.

Black hole is pure math though, it's about behavior of equations, not about reality. You presume there's distribution of mass smaller than its Schwarzschild radius and study how equations behave in this mathematical situation.

But the equations are the only thing even close to evidence that reality does break down. So why would you believe it does if you think the equations are incomplete in that case? There's absolutely no reason left to believe it.

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