[Lewis Thomson (OP)]

Donald is curious to find answers to this question.

"Neutrinos are noted to rarely interact with matter and mostly pass through the Earth without interacting. But how about a neutron star? Could they pass thru a black hole star?"

What do you think? Leave your thoughts in the comments below...

[Halc]

Neutrinos are noted to rarely interact with matter and mostly pass through the Earth without interacting. But how about a neutron star?

Yes, Neutrinos, like dark matter, will probably (and definitely, respectively) pass through a neutron star, which is just dense matter. The high gravitational field will significantly bend the path of the neutrino.

Could they pass thru a black hole star?

Nothing escapes a black hole, which is not a star, nor even a location in coordinate space.

[geordief]

Nothing escapes a black hole, which is not a star, nor even a location in coordinate space.

(with usual apologies  in advance for probable misunderstanding) How then can we say that there is expected to be a Black Hole  at the centre of every Galaxy?

Is there not a frame of reference involved in such a description?

[evan_au]

RE "How then can we say that there is expected to be a Black Hole  at the centre of every Galaxy?"
What we see is the event horizon - or, more accurately, the shadow of the event horizon on the accretion disk.
- This is at a particular position in our coordinates
- We do not see the singularity, which is the core of the black hole
- My simplistic understanding is that our spacetime coordinates get somewhat twisted inside the event horizon,
- A number of coordinate systems have been developed to describe black holes
- But the singularity is not at a place in our coordinate system where you could say "go there and you will see it"
 

[evan_au]

Perhaps a related question - would hypothetical gravitons be absorbed by a black hole?
- If a black hole is (say) 10km across
- And the gravitational waves have a wavelength of (say) 30,000km?

I expect that wavefunction of the gravitons allow you to calculate the probability that a particular graviton is found at a particular point in space
- There is a finite probability that an individual graviton will impact the event horizon of the black hole
- Since there are so many gravitons in a gravitational wave, some of them will impact the event horizon, and be absorbed.

The same argument applies to photons in an electromagnetic wave with a wavelength of 30,000km.

[geordief]

Perhaps a related question - would hypothetical gravitons be absorbed by a black hole?
- If a black hole is (say) 10km across
- And the gravitational waves have a wavelength of (say) 30,000km?

I expect that wavefunction of the gravitons allow you to calculate the probability that a particular graviton is found at a particular point in space
- There is a finite probability that an individual graviton will impact the event horizon of the black hole
- Since there are so many gravitons in a gravitational wave, some of them will impact the event horizon, and be absorbed.

The same argument applies to photons in an electromagnetic wave with a wavelength of 30,000km.

How might we (in theory) detect that a graviton  had interacted with the  actual singularity** of a black hole?

Could the probability of such an occurrence  exceed by very many orders the lifetime of the universe and so be considered impossible?

** are singularities mathematical objects without a physical counterpart?

[Eternal Student]

Hi.

Perhaps a related question - would hypothetical gravitons be absorbed by a black hole?

   The easiest answer is - they are hypothetical so we don't know.

    Assuming gravitons are just the force carrying particles that mediate gravity*, then they can't escape from the interior of black hole any better than any other particle.   Thus, you would never experience gravity from anything inside the event horizon while you are outside the event horizon.  (You don't need to by the way,  from your reference frame located outside the event horizon, the matter which formed the black hole still hasn't actually crossed over the event horizon yet, it's still there outside the event horizon and you can exchange gravitons with that).

* This is not the only use of the term "graviton".

I expect that wavefunction of the gravitons allow you to calculate the probability that a particular graviton is found at a particular point in space
- There is a finite probability that an individual graviton will impact the event horizon of the black hole

      There is some attempt to apply QFT (Quantum Field Theory) on a background of curved space, this is how Hawking radiation was modelled.    This is different from a full quantum theory of gravity:  In such applications gravity just isn't one of the fundamental quantum fields in the QFT, so you don't have "gravitons" and it is still possible to talk about an event horizon and curved space.   
    However, what you're talking about would seem to include gravity as a fundamental quantum field since you are talking about "gravitons" which should just be an excitation in the graviton field.   In a full quantum theory of gravity you shouldn't need to talk about curved space.  It's also just speculation on my part what a region of space previously described as an event horizon would be represented by in a quantum field theory of gravity.

Here follows a short work of fiction that might be something like a quantum theory of gravity:
..... stuff....
LATE EDITING:    There's really no need to force anyone to read these two paragraphs, I've deleted them.  This bit might be worth keeping:

    Going back to your (Evan-au) idea of there being a finite probability that the graviton is found located right on the event horizon - well not necessarily.  The event horizon location may be a node of the wave function.  Perhaps there is always 0 probability of finding it at the region of space previously described as the event horizon, instead it will always be found somewhere outside the event horizon.   In simpler quantum mechanics you might have seen the wave function that is described as modelling a "particle in a box" or a particle inside a square potential well where the potential reaches infinity outside the box.  Those wave functions are such that the amplitude is constrained to be 0 at the edges of the box,  so the particle is never found right on the edge of the box.
 - - - - - - - - - - -

How might we (in theory) detect that a graviton  had interacted with the  actual singularity** of a black hole?

  i.d.k.    See earlier speculation, the black hole might actually swell in size if the graviton is absorbed into the event horizon.  It hardly matters what happens inside the event horizon, the black hole has grown in size already.

Could the probability of such an occurrence  exceed by very many orders the lifetime of the universe and so be considered impossible?

   See earlier speculation.  The wave function for a graviton may be such that the event horizon is a node and therefore there is 0 probability of finding the graviton there at any time.  However, it may have a non-zero momentum and look like a particle that is effectively always travelling toward the event horizon region asymptotically as time progresses.

** are singularities mathematical objects without a physical counterpart?

    Possibly worth discussing in an entire thread all of it's own.  The terminology was stolen from the mathematicians.  As far as I'm concerned a singularity is a property that a function can have and not a description of some place in space.
    To formally answer your question I need to make it clear that there isn't a single agreed upon definition of the term "singularity".  Take a look at this page in Wikipedia for an idea of how many different definitions and uses of the term "singularity" exist:   https://en.wikipedia.org/wiki/Singularity .   So not all singularities are mathematical objects.

Best Wishes.

[evan_au]

Einstein's attitude to mathematical singularities was that since no physical property can really be infinite, his theory becomes inapplicable at that point.
- On the other hand, when something approaches infinity (like the mass of an electron in a particle accelerator), the theory is still well defined, and seems to make valid predictions.

Einstein's theory has proven itself quite reliable for a century. But it breaks down in the vicinity of the event horizon (as do theories of quantum gravity), so we really don't have a good theory of quantum gravity (or a nearby black hole to test).

[Eternal Student]

Hi.

Einstein's attitude to mathematical singularities was that since no physical property can really be infinite, his theory becomes inapplicable at that point.

     Yes but human beings are biased into thinking that everything that seems like a point in space must actually be a point in space.  When Schwarzschild found his solution of the E.F.E. for the situation where there is spherical symmetry and empty space in all regions for r>0 *(see note), it was soon realised that there wasn't any manifold that could be extended to include the point r=0.  From that there are two ways to go:   

   1.   Yes it is possible that the theory breaks down at r=0 and r=0 should be a valid point in space.  This is the view a lot of physicist's take.

   2.  Alternatively you can see it this way round:   There is no manifold on which a solution to the E.F.E. is defined and covers the point r=0, so  r=0 just isn't a point in anything that behaves like our spacetime.  To be clear, it could just be that r=0 isn't a point in our spacetime.

   To paraphrase it another way:  maybe Einstein's claim "my theories are inapplicable at that point" is actually quite correct if a little boastful.     His theory of GR doesn't work in places that aren't real places but everywhere else it's fine.

Best Wishes.

*LATE EDITING:  The Schwarzschild solution applies  for  r > R   where R = the radius of some mass that is the source of gravitation.  Black holes arise where  R < Schwarzschild radius.  For Schwarzschild black holes the metric is assumed to be valid for all r >0.   There is simply no way to encapsulate all of this in a short sentence.   It's just easier to say that Schwarzschild considered r>0 even though historically it started with r>R.

[alancalverd]

But R > 0 for a black hole so r= 0 can surely be located at the center of R: that is to say we can describe an approach vector from where we are to r=0. It just happens that all the physics we have observed, and hence the laws we have deduced from our observations, doesn't occur where r<R, so we can't be sure that we have actually arrived on the runway, and even if we could, we couldn't tell anyone once we have crossed the event horizon.

[Colin2B]

  Going back to your (Evan-au) idea of there being a finite probability that the graviton is found located right on the event horizon - well not necessarily.  The event horizon location may be a node of the wave function.  Perhaps there is always 0 probability of finding it at the region of space previously described as the event horizon, instead it will always be found somewhere outside the event horizon.   In simpler quantum mechanics you might have seen the wave function that is described as modelling a "particle in a box" or a particle inside a square potential well where the potential reaches infinity outside the box.  Those wave functions are such that the amplitude is constrained to be 0 at the edges of the box,  so the particle is never found right on the edge of the box.

The particle in a box is usually used to illustrate the quantisation of a particle whose wave function is described by a simple plane wave.  In order to get the standing waves, and hence harmonics (quantisations), the wave has to be 0 at the barrier. As you say, that’s a simple model. Elsewhere, you have been discussing a travelling particle modelled as a wave packet and in this situation the behaviour of the packet (and hence probability of finding the particle) is not zero at the barrier.
If the graviton is similar to the photon we might expect it to be both the force carrier and the quantisation of the gravitational wave and see some modelling similar to the particle wave packet.
I think you have discussed elsewhere that the event horizon is only seen as a barrier as measured from our perspective, but the experience of a particle in its local coordinates would be very different, not seeing this as a barrier and passing through with ease.

The terminology was stolen from the mathematicians. 

B.B. King once said, "I don't think anybody steals anything; all of us borrow."
So, we’ll let you have it back when we’ve finished with it 

     Yes but human beings are biased into thinking that everything that seems like a point in space must actually be a point in space.  When Schwarzschild found his solution of the E.F.E. for the situation where there is spherical symmetry and empty space in all regions for r>0 *(see note), it was soon realised that there wasn't any manifold that could be extended to include the point r=0.

Overlap with Alan who beat me to it. Perhaps for benefit of @geordief  it is worth saying that in our local coordinates we can plot a point which is the centre of our galaxy and about which nearby stars orbit, indicating the presence of a black hole. Like all sources of gravitational attraction the gravitational force from a black hole appears (at a suitably large distance) to ‘emanate’ from a point. Even the use of the term r implies a point from which r can be measured.
That said, the points ES makes are valid and it’s difficult to even guess whether a theory of quantum gravity would resolve the problem, when even quantum behaviour is likely to change under such extreme circumstances.

From a physics point of view I'm with Evan and Newton that examining the conditions as we approach an extreme value can be very useful.

[Eternal Student]

Hi all.

But R > 0 for a black hole

   Why and with what guarantee can you say that?   It also seems that @Colin2B had similar views.

   Suppose you had  a blob of Mass, M, before the black hole formed.  It only forms a black hole when  R becomes less than 2GM   .... but then the Schwarzschild metric should hold in at least a small region with r < 2GM  and  r > R.   Using Eddington-Finklestein co-ordinates it is seen that the light cones have tilted over for everything that might exist in that region.  If you have continuity at r=R then the boundary of the blob is also forced to progress in such a light cone.   To paraphrase this, once a blob of mass has a radius that falls under the Schwarzschild radius then further collapse to R=0 should be inevitable.
      Asserting that R>0 for a black hole is just asserting the belief that the GR and Schwarzschild metric fail at some small but definitely positive value of r.   It's a reasonable belief but still just a belief that you have.

Elsewhere, you have been discussing a travelling particle modelled as a wave packet and in this situation the behaviour of the packet (and hence probability of finding the particle) is not zero at the barrier.

      Well, let's say I was vaguely hinting at something like a travelling packet.  However, I was careful just to say that the particle the wave function represents, the graviton, could have non-zero momentum and not to directly state that the wave function would actually look like a travelling wave packet.  When you say "elsewhere" you might have been talking about another thread entirely (a recent thread about Quantum tunnelling perhaps).
    The simple QM example of the standing wave solutions for a particle in a box aren't travelling wave packets, that's certainly true.   The  wave function Ψ(x,t) doesn't appear to move from left to right.  I'll also admit that the momentum spectrum for a particle in a box is extremely complicated (See https://en.wikipedia.org/wiki/Particle_in_a_box#Momentum_wave_function, for a description) and in this case the probability density is high around 0 momentum but the probability density isn't just 0 everywhere else.   So it is possible to measure the momentum of the particle and find that it is not 0.   In this respect a wave function can have the form of a standing wave (and not a travelling wave packet) but still represent a particle that can have non-zero momentum when it is measured.  I've read this paragraph a couple of time and I know it might be as clear as mud, so I'll try and phrase it another way:    A travelling wave packet is the form of a wave function that usually shows a particle with a definite momentum,  while a standing wave solution doesn't show a particle with a definite momentum but it can still have a range of momentums and a probability of having those momentums when it is measured.

    Also, I don't think that a travelling wave packet would really produce a non-zero amplitude wave function at the edge of the box.  I was careful to mention that the box had an infinite potential outside the box.  If I've understood the situation correctly, then the wave function is still constrained to be 0 outside the box and then continuity conditions still force the requirement Ψ = 0 at the edge of the box at all times.  However, the assumption of a genuine infinite potential is arguably nonphysical and certainly controversial.  For a large but finite potential, then yes a travelling wave packet does not fall to 0 amplitude at the barrier but would tend to just fall off exponentially once you are outside the box.

    I'm not sure any of this matters too much for the original questions from @geordief or the comments raised by @evan_au .   What is important is that I'm not an expert and there wasn't any attempt to seriously present a quantum theory of gravity, just some informal discussion about what could happen.   I especially like the last comment from Colin2B:

it’s difficult to even guess whether a theory of quantum gravity would resolve the problem

    I'm not sure if r=0 is a point in space, it might be and maybe a quantum theory of gravity will tell you what happens there.  It's just that I'm prepared to consider the possibility that  (r=0 , t=anything)  isn't a point in spacetime at all.
    What would be really interesting, at least philosophically, is if both were true:   Suppose r=0 isn't a point in our spacetime but a Quantum Theory of gravity still tells you what happens there, then we have expanded our understanding to things that are genuinely oustside of our universe.

Best Wishes.

[alancalverd]

Quote from: alancalverd on Yesterday at 10:45:27
But R > 0 for a black hole
   Why and with what guarantee can you say that? 

Wikipedia lists the categories of black holes with their mass and  Schwarzchild radii. It is reasonable to assume spherical or at least axial symmetry from there inwards. If you want  to allow infinite density, then I'd modify my opening statement to R ≥ 0 but the symmetry assumption still applies until we observe a BH with an asymmetric EH.

[Halc]

Yes but human beings are biased into thinking that everything that seems like a point in space must actually be a point in space.

Indeed. See below.

*LATE EDITING:  The Schwarzschild solution applies  for  r > R   where R = the radius of some mass that is the source of gravitation.  Black holes arise where  R < Schwarzschild radius.  For Schwarzschild black holes the metric is assumed to be valid for all r >0.   There is simply no way to encapsulate all of this in a short sentence.   It's just easier to say that Schwarzschild considered r>0 even though historically it started with r>R.

This makes it sound like there's a spatial ball in there. R is a timelike worldline inside the event horizon.  Positing that there's a mass inside with nonzero radius R is to suggest that something cannot move to the future because the future is full.
The Schwarzschild solution isn't a dynamic one. It is a simple solution for a non-rotating mass with no charge and nothing falling in. Thus all the spacetime within is empty.

Wikipedia lists the categories of black holes with their mass and  Schwarzchild radii. It is reasonable to assume spherical or at least axial symmetry from there inwards.

A Schwarzschild  BH has no axis. It is thus spherically symmetric to any distance. A Kerr BH would have an axis and would not be spherically symmetric. It's physical singularity is a ring instead of a point at the center. The event horizon of any BH is just a coordinate singularity, which goes away with choice of different coordinate system.

[Eternal Student]

Hi.

This makes it sound like there's a spatial ball in there. R is a timelike worldline inside the event horizon.  Positing that there's a mass inside with nonzero radius R is to suggest that something cannot move to the future because the future is full.

    I like that quite a lot.   Absolute agreement there, the r co-ordinate is not space-like once r < Schwarzschild radius.

    However,  I don't think it would be easy to integrate that with Pop Sci ideas that r just is always just some spatial radial co-ordinate.  I reckon you'd need a lot more than just those three sentences.  I think it's another good example of human bias,   just because something seems like it should be a spatial radial co-ordinate does not mean that is actually a spatial co-ordinate.  It seems quite likely that Schwarzschild was motivated to find a solution to the EFE with co-ordiantes  (r, θ, φ, t)   where   the triple (r, θ, φ) were just ordinary spherical co-ordinates for space but that's all it was, just a motivation.  The actual proper meaning of those co-ordinates wasn't known until after the solution had been obtained.
     Unlike other theories of physics, in general relativity we simultaneously define co-ordinates and the metric as a function of those co-ordinates.   In other words, we don't know ahead of time what, for example, the radial co-ordinate r really is; we can only interpret it once the solution is in our hands.

(Page 194,  section 5.1  where the Schwarzschild metric is derived,   Spacetime and Geometry, Sean carroll)

Best Wishes.

[evan_au]

until we observe a BH with an asymmetric EH

Some computer models I have seen of black hole collisions suggest that in the last revolution before the merge, the individual black holes become rather asymmetric. (But I can't tell if that is real, or just due to the coordinates or photons they are using in the simulation).