# Do black holes violate the uncertainty principle?

If black holes have mass but no size, does that imply zero uncertainty in position? If so, what does that imply for uncertainty in momentum?

• What do you mean that "black holes have no size"? Do you mean the singularity of the black hole? – Joshuah Heath Jul 14 '15 at 2:53
• Yes, I mean to say that the particles which were originally separate have theoretically come to occupy the same point in space. Does the uncertainty principle apply to this phenomenon? – Alex Jul 14 '15 at 2:59

General relativity is a classical theory. The Heisenberg uncertainty principle does not apply to it.

The research frontier in physics now exists in quantizing gravity and unifying it with the other three forces (strong , weak, electromagnetic). Once that is done the solution for the black hole will become a probability distribution and the Heisenberg principle will apply. The macroscopic classical solution of a point singularity will become a quantum mechanical uncertainty locus which will not change the macroscopic description. h_bar is a very small number and is already easily satisfied by the classical mechanics solutions all our constructions and engineering depend on.

• Does that mean we can say with confidence that black holes are not singularities? In other words, does the uncertainty principle forbid singularities? – Alex Jul 14 '15 at 3:54
• The uncertainty principle will make singularities irrelevant as a mathematical point, but macroscopically it will fit the general relativity mathematics of a singularity, that is when gravity is consistently quantized within a larger framework. At the moment quantization of gravity is an effective theory for specific situations/ boundary-conditions – anna v Jul 14 '15 at 5:09

If black holes have mass but no size, does that imply zero uncertainty in position? If so, what does that imply for uncertainty in momentum?

I mean to say that the particles which were originally separate have theoretically come to occupy the same point in space. Does the uncertainty principle apply to this phenomenon?

Zero size doesn't violate the uncertainty principal. It's knowing the exact location that is impossible with the uncertainty principal because you can't measure a position without moving it. Granted, black holes have enormous mass so any change in position is much smaller, but uncertainty principal still applies.

What you're describing sounds like the Pauli Exclusion Principle and that's true to an extent but there are ways that can be explained. One is that we don't really know what happens at a singularity. The math breaks down and we've never seen one, so it's a model that we admit isn't complete.

Also, one could argue that a particle in a singularity is no longer a particle, but the singularity is the particle. This approach addresses the Pauli exclusion principal problem but it raises another question, Conservation of information.

at least, that's how I look at it.

• Wouldn't a spherically symmetric black hole mean that having zero size implies knowing location with zero uncertainty? – Omry Jul 14 '15 at 15:40
• Zero size isn't the same thing as knowing location with zero uncertainty. An Electron could have zero size, but that doesn't change the fact it's clearly affected by the uncertainty principal. What makes a black holes location much more certain is it's high mass, but (at least, it makes sense to me) that the uncertainty principal still applies because you can't measure a black hole without moving it (a tiny little bit) – userLTK Jul 15 '15 at 3:17

A black hole doesn't have zero size. There are lots of parts of a black hole. There is an event horizon and there may may not be a singularity. And for an astrophysical black hole, the event horizon originally wasn't there, it formed and expanded over time. And the singularity is entirely hypothetical and hardly ever is even predicted to be of zero size anyway.

So lets look at how it forms to see the effects uncertainty and Pauli exclusion can have. So imagine you have a large star.

Each layer of the star is supported by having more pressure from below than from above. That pressure could be from density and temperature from fusion of hydrogen. But when too much hydrogen is fused there won't be enough of that fusion to provide the temperature to provide the pressure. So the star might collapse and get denser until other elements can fuse.

But gravity isn't powered by just mass, it is also affected by energy and momentum and pressure and stress. So when you try to use pressure to support your collapse then the pressure itself is something that requires more pressure to fight.

This is the true reason for collapse, when the pressure isn't enough to keep it at its current size, it collapse. And Newtonian mechanics might make you think there is always some pressure that will work regardless of how small you get, but this isn't the case in General Relativity. Sometimes more pressure.

Degeneracy pressure might support a white dwarf or a neutron star if they aren't too large.

Trying to compress it further might require less space for matter, which might require more momentum and there might not be enough momentum, so that resistance to compression is the pressure. And the pressure makes the star more massive than it otherwise would be.

So it doesn't help when the star is already too massive.

But at this point the star has matter at all points. Some matter in the center. Some matter on the outside and some matter in the layers in between. Each layer feeling some pressure. And when the pressure from below a layer is larger than the pressure from above that layer then that layer won't free fall towards the center.

There is no part with zero size, and there is no event horizon or singularity, its just a dense star. But what happens if it collapses more? What if the compression of the layers doesn't give pressure that can stop the compression?

Well, then it will compress. But how quickly? Well, free fall minus however much the pressure is slowing down the collapse (there still is some pressure, just not enough to stop the collapse).

Super. If it has to compress then it must just get smaller and smaller, right? Well, yes. But what does someone on the outside see about how quickly it gets smaller?

When a spherical shell of matter collapses the type of curvature on the outside and the inside stay the same. For instance it could be Schwarzschild of parameter $M$ on the outside and Schwarzschild of parameter $m$ on the inside, with $M\gt m.$ But these are curved spacetimes so there is time dilation for each section. And time dilation is relative.

So the time dilation of the $m$ side matches up with the time dilation of the $M$ side. But the $M$ side has more time dilation compared to the outside the deeper you go and so when the shell collapsed downwards there was more of that spacetime deeper. So you end up with greater curvature there than used to be and more time dilation.

So you, someone outside, see the images of the collapsed layer move slower and look redder the deeper it goes. And you see that for each layer. But the images you see are always from before any event horizon forms.

So all the things that happen at each part of the star (even the parts in the center) every place and every time before the horizon forms. You can see them all. And you can see non of the events that happen after the horizon forms.

So the uncertainty or exclusion could provide more pressure and that might affect ts how quickly the lager collapses. But the pressure itself can't always stop a collapse if the pressure itself contributes too much curvature.

And you never see anything get compressed into too small a region, but you can't see things that get compressed into too small a region, you instead always see the parts from back before they got that compressed.

You just see them moving slower and looking redder.