5 ms·
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 cen
by Nevermark 3y ago
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.
- dmarchand90 3y agoA 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.
- jacquesm 3y ago> A bunch of funny mathematics is needed to get around this. I usually use a small drill to get around it. Works every time.
- GTP 3y agoTIL that drills are excellent matematicians
- otteromkram 3y agoThey're great mathematicians, too.
- fsckboy 3y ago>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
- sigmoid10 3y ago>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.
- prox 3y agoDo we have an inkling were the proverbial blind spots are in our theories? Like a general direction to search?
- deleted 3y ago[deleted]
- davelondon 3y ago> 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.
- fnordpiglet 3y agoI 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.
- Jensson 3y ago> 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)
- gnramires 3y agoThe 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.
- superposeur 3y agoAlmost 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”).
- Cacti 3y agoOur equations break down because they are incomplete. But that does not mean that reality does not also break down.
- GoblinSlayer 3y agoBlack 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.
- bmitc 3y ago> 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.
- Cacti 3y agothat’s an assumption though.
- bmitc 3y agoAn 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.
- Nevermark 3y agoIMO, 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.
- 3y ago
- deepsun 3y agoNo, 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.
- irrational 3y agoSo, at the center of the earth stuff is in zero gravity or micro gravity, just like above the earth?
- drbaba 3y agoYes. 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.
- pbhjpbhj 3y agoAgreed. >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.
- Qwertious 3y agoSo 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.
- nimchimpsky 3y agono it would be weightless, as all the gravity would cancel each other out.
- 3y ago