# Event Horizon Issue

Discussion in 'Astronomy, Exobiology, & Cosmology' started by RajeshTrivedi, Feb 5, 2017.

1. ### RajeshTrivediValued Senior Member

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While a work in progress on non feasibility of formation of singularity, an interesting issue emerged, I am trying to explore the relevance of the same.

The issue: if a spherical object of mass M and radius r is just inside its schwarzschild radius (r = Rs), does it mean that all inner parts (say a concentric inner sphere of r1 where r1 < r) will also be inside their (parts') schwarzschild radius?

IMO this issue is relevant, because if the r1 sphere is not beneath its schwarzschild radius, then the escape velocity at r1 is not c.

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I've see this before asked by someone....:

The problem as I see it, anything within its Schwarzchild radius will undergo compulsory collapse as per GR: In the process of collapse, again as per GR, the sphere would be broken down into its basic constituent parts, all forces being overcome, including the strong nuclear with collapse up to or beyond the quantum/Planck level where GR effects are then not applicable..
That as I see it, is the accepted mainstream picture as to what should happen.

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Obviously I misunderstood the question.
I also E-Mailed a Professional on the question.......
Hi –

The answer is clearly no – if I have virtually all the mass in a shell between r1 and r2 and just a tiny amount between 0 and r1 then the mass between 0 and r1 does not have to be inside “its” event horizon.

General relativity is non-linear, and the horizon is not simply the sum of the event horizons of shells –

Cheers - G

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Geraint F Lewis,

Professor of Astrophysics

Sydney Institute for Astronomy

School of Physics A28

The University of Sydney

NSW 2006 Australia

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7. ### RajeshTrivediValued Senior Member

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Even Prof Lewis also seems to have either misunderstood or answered in hurry.

1. No doubt GR is non linear wrt to its solution. But schwarzschild radius can be mathematically seen as linear with mass. His reference to shells as such is irrelevant.
2. We do not need GR for Schwarzschild radius (Rs), the principle is escape velocity being c at Rs.

Now the problem is...you can refer back to Prof Lewis....that for a uniform density sphere of mass M, when it is just of its schwarzschild radius size, then the entire mass only is beneath its Event Horizon, none of its parts, not even 99.9999% part is beneath that part's Event Horizon. This is simple algebra, so I am leaving out the calculations. So Prof Lewis claim about small tiny mass being outside event horizon is loose. If we have to keep every part inside, then a designer density profile is required,which would be unrealistic. The point is that a photon emitted inside the full sphere (when it is at Rs) could travel radially away at least till Rs. Yes, as the object keeps compacting,more and more part percentage gets inside the part's event horizon, but still some fracion of innermost core will still be outside its EH.

8. ### RajeshTrivediValued Senior Member

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Another issue with Event Horizon
No need to get into the Hawking Radiation slugfest, but if we consider a very large mass Black hole, then there is no great curvature of spacetime present at Event Horizon, tidal forces also near EH are not so great. Higher the BH mass, more gravitationally comfortable would be the EH location, infact a simple algebra would show that for a test particle of mass 1 kg, the gravitational force at Event Horizon of a BH of mass M (very High), can be made as low as felt on the Earth. So what kind of QM effect we are talking about near the EH for HR to come alive? It appears dicey.

9. ### SchmelzerValued Senior Member

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Rajesh Trivedi, the event horizon is a global notion.

There is the inner part. You can live nicely in this inner part, and not be aware that the outer part even exist, you simply do not see it yet. A light ray you send out easily leaves the small inner radius $r_1$. But after this, this light ray is in the outer region of that collapsing star. And, if the global event horizon already has formed, it will never be able to leave this outer region.

What is, in GR, named "inside the event horizon" is everything which is unable to send signals to infinity. So, even if you, deep inside, observe (yet) nothing at all, you may appear to be part of something inside a (global) event horizon.

10. ### SchmelzerValued Senior Member

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The problem is that we don't know how to make relativity and QM compatible with each other. Part of this problem is the so-called "problem of time". QM and relativity use two conceptually completely different notions of time. QM time is absolute time, and unobservable. What Newton has named true time. Time of relativity is time measured with clocks. What Newton has named apparent time.

These two notions are completely different, already philosophically (already Newton found it necessary to distinguish them philosophically, without any need to distinguish them in the math of his theory). In QM, there is even a theorem that there is no clock which is able to measure time, every clock has a non-zero probability to go even backward in true time. (And, similarly, GR clock time has completely different properties than QM time.) So, the difference between true time and clock time is a QM problem.

What measures the difference? One can say that it is measured by time dilation. But time dilation becomes infinite near the horizon. So, there is something which is quite plausible a QM problem which becomes very large near the horizon.

11. ### originTrump is the best argument against a democracy.Valued Senior Member

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I think that it is true that the tidal forces can be relatively small for a large mass BH, however I think the space time curvature just outside the event horizon is so huge that a photon can only just escape, regardless of the mass of the black hole

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No, he understood perfectly...that's his job: He isn't some ego inflated CEO of some family company.

No, I don't need to refer back to him or anyone else: Your alternative approach has been invalidated.
And perhaps now thoughts should be that this is simply another ruse by yourself to legitimise your well known anti mainstream nonsense and needs to be in the appropriate section.
Let's see how we progress anyway.

13. ### RajeshTrivediValued Senior Member

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So you agree that for a photon emitted inside the global event horizon, it is not necessary that movement of this photon is towards r = 0. This creates a contradiction with GR which states that inside event horizon the time coordinate becomes spatial and movement is towards r = 0 only. Anyways I leave that.

The point which I am raising is..

1. I make an assumption that when the star compacts (after getting beneath its global EH), then beyond certain point the inner most mass gets converted into radiation energy. The validity of this assumption shall be taken up later. Just assume that it is so for time being.

2. Now assume that matter surrounding this solid surface is decreasingly super conducting (inner most being very high superconducting). We will check later if this also holds or not.

3. Now since central emitted photon (radiation) has no restriction on its trvale away, so this radiation starts travelling away from r = 0. This is the crux. But the compaction does not stop yet.

4. So now the global event horizon contains the inward travelling matter and radially away facing photons. If assumption in Sr#1 is ok, then this is what will happen.

5. Now the point is how e make it explode.

14. ### RajeshTrivediValued Senior Member

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So can I understand from your this post that you agree that a photon inside global EH can travel in non r = 0 direction also. Rather it can move away (but still staying inside the Global EH) as well.

Now I have a question, which I am attempting to resolve, suppose a huge radiation energy somehow gets released at the centre of the object (at r = 0) and the entire strcuture is highly superconducting, where will this radiation go? Just stay at the centre and form a cavity full of radiation? Travel outward but remains trapped below global horizon?

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While it is true that tidal forces are less severe the larger a BH is, nothing outside the EH will escape once it ventures inside the Photon sphere [1.5 Schwarzchild radius] or where photons will orbit.
A photon emitted just this side of the EH, will be forced to arc back secumbing to the BH's EH, unless it is emitted directly radially away, and in such cases, that photon will appear to hover forever just above the EH in that same local FoR.

16. ### RajeshTrivediValued Senior Member

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Are they not interlinked? means higher curvature implying higher tidal forces, so if the tidal force around EH for a large mass BH is relatively small, then the curvature also around EH will be relatively small?

The dependence of Schwarzschild radius on mass (Rs~M) makes the density, tidal force, g value and curvature at EH different for different masses. (when the object is just beneath its EH)

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https://en.wikipedia.org/wiki/Spaghettification

Inside or outside the event horizon

The point at which tidal forces destroy an object or kill a person will depend on the black hole's size. For a supermassive black hole, such as those found at a galaxy's center, this point lies within the event horizon, so an astronaut may cross the event horizon without noticing any squashing and pulling, although it remains only a matter of time, as once inside an event horizon, falling towards the center is inevitable. For small black holes whose Schwarzschild radius is much closer to the singularity, the tidal forces would kill even before the astronaut reaches the event horizon.[7][8] For example, for a black hole of 10 Sun masses[note 2] the above-mentioned rod breaks at a distance of 320 km, well outside the Schwarzschild radius of 30 km. For a supermassive black hole of 10,000 Sun masses it will break at a distance of 3200 km, well inside the Schwarzschild radius of 30,000 km.
much more at a reasonable WIKI link.................
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In essence, while falling into any BH will spaghettify you and shred you to your most basic components,[overcoming all forces including the strong nuclear] on your way to the Singularity region, that point for a small BH could well be outside the EH, while you may pass unhindered inside the SMBH [ignoring any radiation] and not undergo any spaghetification until well inside the EH.

Last edited: Feb 8, 2017
18. ### originTrump is the best argument against a democracy.Valued Senior Member

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The curvature of space time at the EH is the same for a large mass BH as a is for a small mass BH. Another way of saying that is the force of gravity is the same for both BHs at the EH. What is different is the gravitational gradient. For a small mass BH the gradient is large near the EH and for a large mass BH the grandient is lower at the EH.
Gravity is the curvature in space time. A gravitational gradient (tidal forces) means that there is a varying level of curvature

19. ### Q-reeusValued Senior Member

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Not correct. 'Strength of gravity' defined as the proper acceleration needed to hover in one place, is owing to a first derivative of the metric. If uniform, there is zero curvature present. I recently linked to an example of how that can be achieved. Curvature owes to 2nd derivatives and is synonymous with tidal accelerations. And, for the same level of 'g-force' experienced while hovering, is inversely proportional to BH size. Hence larger for the smaller one.
The last para here will do: http://www.einstein-online.info/spotlights/geometry_force

20. ### originTrump is the best argument against a democracy.Valued Senior Member

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OK, I am sure my terminology is poor. Lets try these 2 questions:
Is the curvature of space time due to the gravity from a black hole the same at the EH regardless of the mass of the black hole?
Do small mass black holes have larger tidal forces than large mass black holes at the EH?

21. ### Q-reeusValued Senior Member

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No to the first question. Yes to the second one. One could try plowing through say the relevant chapters here: http://www.pmaweb.caltech.edu/Courses/ph136/yr2012/1225.1.K.pdf
But a breezier read that gets the essential ideas across is this favourite go-to: https://arxiv.org/abs/gr-qc/0103044
Hope that helps.

22. ### RajeshTrivediValued Senior Member

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No.
Q_reeus has given you a metric specific counter, but since you are referring to Newtonian Force, it is very simple to deduce that force is inversely proprtional to mass of BH at EH. That is smaller force value on a test particle by a larger mass BH.

When the object is just at EH, the force on a test particle of unit mass shall be GM/R^2......but at EH R = 2 GM/C^2, so the force will be = c^4/4GM.

23. ### originTrump is the best argument against a democracy.Valued Senior Member

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First thanks for the links and the information. I do not have time right now (I will later) to go through the links but I think this will be very instructive for me. I am not a physicist and this is way outside of my expertise.
My understanding is that the EH is where the 'force' of gravity is greater than the escape velocity of a photon. I also think that gravity is the curvature of space time.
So it seemed to me that the 'force' (or acceleration) of gravity that would be greater than the escape velocity of a photon is a set number and independent of the source of the gravity, if that is true then the curvature of space to prevent the escape of a photon would also be a set number and independent of the source of the curvature.

I am very curious (no pun intended) to see where I went wrong.