Why is a blackhole a one way ride

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Once enough mass in the form of hydrogen has been swallowed up by a black hole why cant a new nuclear reaction occur within a black hole creating enough of an outward pressure to uncollapsed it.
 

russ_watters

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Hydrogen can't exist in a black hole. It is too dense.
 

taylaron

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so everything including light can enter a black hole except hydrogen. that doesnt make much sence to me.
enlighten me
 

Hurkyl

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why is a blackhole a one way ride
By definition of black hole. It's not a black hole if something can escape its event horizon.

creating enough of an outward pressure to uncollapsed it.
Pressure contributes to gravity. More pressure = "stronger" black hole.
 

russ_watters

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so everything including light can enter a black hole except hydrogen. that doesnt make much sence to me.
enlighten me
That's not what I said. Once it enters a black hole, the gravity of the black hole crushes it and it is no longer hydrogen.

Remember what a black hole is (and isn't): A star is mostly hydrogen but a star isn't a black hole. Why? It isn't dense enough. A neutron star is one that is so dense that electrons collapsed into their nuclei and you are left with a super-dense mass of neutrons. No atoms (no hydrogen) in a neutron star, even though it formed from a normal star. But a neutron star is not a black hole either. Black holes are still denser. Black holes are so dense that the neutrons are crushed into something denser still.
 
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Black holes are so dense that the neutrons are crushed into something denser still.
This seems like an odd statement. Sure, classical GR predicts all matter in a black hole is pulled into a singularity characterised by infinite density (personally I doubt quantum theories will retain this feature), but the whole black hole (everything encompassed by the event horizon, from an external perspective) can be said to have an arbitrarily low density (given sufficiently high total mass).
 
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No matter how big any "outward pressure" would be, it still wouldn't be able to push anything out. This is because once something is inside the event horizon of the black hole, it can never get further from the hole. Your question can be converted to this one: why can't anything inside the event horizon move away from the black hole?

The answer comes from relativity: nothing can move faster than light. So if you're close enough that a light ray from your position couldn't ever leave the area near the black hole, then that means nothing can ever leave the area near the black hole. The gravitational pull of the black hole is too large.
 

russ_watters

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...but the whole black hole (everything encompassed by the event horizon, from an external perspective) can be said to have an arbitrarily low density (given sufficiently high total mass).
How can it have enough mass to be a black hole in a small enough volume to have an event horizon and not have a high density? Is this a peculiarity of supermassive black holes? The event horizon of a typical black hole is much smaller than the radius of the star that it formed from, is it not?
 
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Is this a peculiarity of supermassive black holes?
Guess you could say that.. although I don't know how dense a typical star really is to begin with. And for another fun fact: the sun produces a mere 300W per cubic meter (hence the difficulty in producing useful fusion power at a "human" scale).
 

George Jones

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How can it have enough mass to be a black hole in a small enough volume to have an event horizon and not have a high density? Is this a peculiarity of supermassive black holes?
Yes.

The following calculation is only suggestive, and it is in no way rigorous. Because of the curvature and nature of spacetime, it probably doesn't make sense to calculate the spatial volume inside the event horizon of a black hole.

Using units for which [itex]c = G = 1[/itex], the event horizon of a mass [itex]M[/itex] spherical black hole occurs at [itex]R = 2M[/itex]. Using this in the standard flat space expression for density, and treating the black hole as a sphere of radius [itex]2M[/itex], gives

[tex]
\rho = \frac{M}{V} = \frac{M}{\frac{4}{3} \pi R^3} = \frac{3}{32 \pi M^2}.
[/tex]

Notice how the "density" of a black hole rapidly decreases as its mass increases.

Even, though this isn't really the density of a black hole, this "back of the envelope" calculation is good enough to make predictions.

Setting [itex]\rho[/itex] to the average density of the sun, about 1400 kg/m^3, gives a black hole mass of about 100 million solar masses. So, if more than 100 million solar-mass stars or so (within an order of magnitude) congregate in the centre of a galaxy, they don't have to touch (initially) to form a black hole.
 

disregardthat

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This seems like an odd statement. Sure, classical GR predicts all matter in a black hole is pulled into a singularity characterised by infinite density (personally I doubt quantum theories will retain this feature), but the whole black hole (everything encompassed by the event horizon, from an external perspective) can be said to have an arbitrarily low density (given sufficiently high total mass).
If the density is infinite, how can the black hole have volume then? There is a set amount of mass in a black hole, and if the density is infinite at any volume you use, then the black hole should only be a point!
 

russ_watters

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Thanks cesium/George....

Jarle, clearly this isn't really my area of expertise, but what cesium was saying is that there is a distinction between where the mass is clustered and what the size of the black hole is. The "size" of the black hole (and thus volume for calculating average density) is measured at the event horizon. What is going on at the center may or may not be a singluarity (a point with a finite mass and no volume).
 

DaveC426913

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If the density is infinite, how can the black hole have volume then? There is a set amount of mass in a black hole, and if the density is infinite at any volume you use, then the black hole should only be a point!
The event horizon does not define the BH itself. The event horizon simply defines where light cannot escape. It could be hundreds or thousands of miles in radius.

The BH at the centre is theoretically zero volume.
 
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Chris Hillman

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Defining property of a black hole: curvature singularity or horizon?

Oh dear. Actually, it would be more correct to say that the event horizon is the defining characteristic of the notion of a "black hole". For example, if we were someday to possess a workable quantum theory of gravitation with a "low energy limit" ("effective field theory") which recovers a relativistic classical field theory which turns out to be some excellent mimic of gtr, it is conceivable that this theory might admit "black holes" which do not possess curvature singularities in the interior, but they'd still be called "black holes" if they possessed an event horizon.
 

DaveC426913

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Oh dear. Actually, it would be more correct to say that the event horizon is the defining characteristic of the notion of a "black hole".
Granted, the event horizon is functionally the part we concern ourselves with but this thread is questioning what happens inside the horizon.

Also, the horizon is not an actual physical object/location, let alone any kind of surface. It is simply an abstract construct that we use to determine where one thing (EM energy) behaves nonintuitively.
 
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Thankyou for clearing that one up, its quite obvious now Ive been given the answer.
 
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