Black holes may never actually form..!

Discussion in 'Physics & Math' started by RJBeery, Jul 24, 2014.

  1. Russ_Watters Not a Trump supporter... Valued Senior Member

    Busy week...
    That really doesn't make sense, but I think you're just throwing the word "temporal" in there as if it means something. Either way though, "mass never stops falling is a different and even more wrong claim than your usual one of a solid object with an infinite number of hard shells. Your jumping around doesn't make the last wrong go away, it just adds a new one -- and it's even worse:
    The only one who's ever described it as a physical surface is you. But getting away from that, you had just claimed, correctly (in post #306) that the event horizon of a collapsing star has to start at the center as almost a point, encompassing near zero volume. But you're now suggesting that you can pour an infinite amount of mass toward that near zero volume sphere and not grow the black hole. But the rest of the mass of the star piling-up just outside that point would put it far inside the Schwarzschild radius.

    You also didn't answer my question, but repeated a similar implication about what GR says...which I suppose answers my question: it is dishonest to imply that what you are saying is a correct description of the theory when you know it isn't. Or another way: It is dishonest to intentionally try to obfuscate the line between what you know the accepted theory is and what you think it should be instead.
    That's dishonest too. You can't be unaware that the mathpages article describes those models for the purpose of explaining why they don't work. Everything I described above is clearly stated in the link:
    There it is in plain English: your idea is not viable. It doesn't work.
    That's, of course, a lie as well.
    Last edited: Jan 23, 2016
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  3. Waiter_2001 Registered Senior Member

    Black holes do not exist. Just as in mathematics an unkown answer is zero the origins of black holes must also be considered so.
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  5. paddoboy Valued Senior Member

    If BH's do not exist than something even more counter intuitive and bizarre must account for the effects we observe on spacetime and matter/energy.
    Have you any idea what that might be?
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  7. paddoboy Valued Senior Member

    I also stated the reasons why that is so:[1]You were wrong in your facetious assessment of me, and I was not inferring RJ but rather Farsight and the correct inference made by PhysBang. So please cease with the false indignations.
    You are permitted to assume what you like. This thread is it seems about the existence or otherwise of BH's, and according to GR, they do exist.
    Your spacetime Singularities question is beyond GR. GR is a classical theory, and anyway, most cosmologists do not believe Singularities exist physically, rather simply a mathematical emergence where GR does not apply.
  8. QuarkHead Remedial Math Student Valued Senior Member

    So I repeat my request. Please explain how I can look at the field equations and say "Aha, so black holes must exist"
  9. paddoboy Valued Senior Member

    That was not your question: Your question was........
    My answer was
    That's the best answer I can give.
    Although I will add that GR does predict compulsory collapse once any mass reaches its Schwarzchild radius.
  10. paddoboy Valued Senior Member

    This may even explain it further for you......

    Schwarzschild realized the escape velocity from the surface of an object depends on both its mass and radius. For example, the escape velocity of the Earth is about 11.2 kilometers per second- this is the speed a rocket must attain before it can depart the Earth on a journey to the Moon or more distant planets. The Moon's escape velocity, however, is only 2.4 kilometers per second because the Moon is one fourth the size of our planet and possesses only slightly more than 1% of its mass. But, if nature can make the radius of a given mass small enough, the escape velocity will increase until it reaches the speed of light, or 300,000 kilometres (186,000 miles) per second. At that point, neither matter nor radiation can escape from the object's surface. Additionally, atomic or subatomic forces become incapable of holding the object up against its own weight. Therefore, the object collapses into an infinitesimal point- the original object disappears from view and only its gravity remains to mark its presence. As a result, it creates a bottomless pit in the fabric of space-time.

    Scientists now refer to an object with zero-volume but all of its mass as a singularity. Schwarzschild also explained that a singularity was surrounded by a spherical gravitational boundary that forever trapped anything that ventured within. This boundary was called the event horizon. He presented a formula that enabled the size of an event horizon to be calculated. This is now known as the Schwarzschild radius and it marks the edge of a bottomless pit in space-time. Venture beyond the brink and you will never return.

    The formula for the Schwarzschild radius is very straightforward: 3 times M (where "M" is the mass of the sun and the result is expressed in kilometers). For example, if the Sun were shrunk to a singularity, it's event horizon would occur at three kilometers above its surface. Interestingly, it would not disturb the orbit of our planet and we would not suddenly be sucked into oblivion! Similarly, the Schwarzschild radius of the Earth is a third of an inch: if the Earth were to be similarly compressed, the sphere of its event horizon would be about the size of a marble!

    However, scientists did not grasp its significance in the role of stellar evolution for about fifty years and have only recently realized its dramatic impact on the development of the Universe itself!

    Despite the radical predictions contained in Schwarzschild's papers, the scientific community regarded it as a curiosity rather than outrageous. The idea of a singularity troubled many scientists, including Einstein, because it flew in the face of their experience- after all, the world is finite and everything can be weighed and measured.

    Leading thinkers of that period could not imagine conditions that would create a singularity but now we know they are common throughout the Universe. Where? In the fates of massive stars and at the center of most, if not all, galaxies!
  11. Bruinthor Registered Member

    In response to the original topic of this tread:

    The paper IS interesting but relies on several highly speculative theories. Beyond that it models matter in what I believe is a non-physical manner.
    The interior of a collapsing star is not a perfect fluid nor can the equation of state it be modeled accurately as dust.

    Neutron Stars are sometimes modeled as perfect fluids but the results are less than compelling. Changes in pressure/density lead to changes in composition, ignoring this over many orders of
    magnitude is questionable (NSs are a mixture of neutrons, protons, electron and a zoo of bosons). Even if the perfect fluid model is accepted for a nearly static NS its applicability to the turbulent collapsing of a NS or Iron stellar core is even more questionable.
    Using an EoS of dust is also questionable. The central presumption of a dust model is that constituent matter is not interacting.
    Lastly the density of stellar cores and NS is extreme but nowhere near the Planck scale. Even a stellar mass black hole has bulk density of 2x10^19 kg/m^3 whereas the Planck density is 5x10^96 kg/m^3. Taking ten times the cube root of the ratio, quantum effects should begin to dominate when the majority of the mass is within the Schwartzchild Radius/10^24. The discrepancy in scale gets worse with larger masses.
    Last edited: Jan 24, 2016
  12. Farsight Valued Senior Member

    This is "cargo-cult" relativity. It contradicts Einstein, who said a gravitational field is a place where the speed of light varies:

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    And as we all know, at the event horizon, the "coordinate" speed of light is zero. This is the speed of light at some location according to observers at some different location, only Einstein didn't use the word "coordinate". He just called it the speed of light.

    As you doubtless know, you can plot the speed of light at various locations using optical clocks. What you're actually plotting is gravitational potential. The speed of light varies with gravitational potential. When clocks go slower, it's because the speed of light is lower. And the important point is this: at the event horizon, the clock stops. Clocks can't go slower than stopped, there is no more gradient in gravitational potential. So there's no further gravitational collapse.
  13. Schmelzer Valued Senior Member

    No, this is standard GR, and there is no contradiction, because what Einstein talks about is coordinate speed of light, while what those who speak about its constancy talk about is the speed of light as measured in m/s.
    No. First of all, it depends on the coordinates what is the coordinate speed, and there are very different coordinates in use in this region. Then, if we use some non-degenerate system of coordinates, there will be usually different coordinate speeds of light in different directions. If you use the radius as one coordinate, then it is only one direction - outside - which gives coordinate speed of zero for light, all others will give nonzero values.
    Nonsense, the coordinate speed of light is the speed according to the coordinates used, and not to some observers.
    Nonsense. There is no gravitational potential in GR. One can use the Newtonian one for some approximation, but this would be not about GR, but about some Newtonian approximation of it.
    krash661, paddoboy and QuarkHead like this.
  14. Farsight Valued Senior Member

    You should check what Einstein said :

    "According to this theory the metrical qualities of the continuum of space-time differ in the environment of different points of space-time, and are partly conditioned by the matter existing outside of the territory under consideration. This space-time variability of the reciprocal relations of the standards of space and time, or, perhaps, the recognition of the fact that 'empty space' in its physical relation is neither homogeneous nor isotropic, compelling us to describe its state by ten functions (the gravitation potentials gμν has, I think, finally disposed of the view that space is physically empty."

    Note the phrase gravitational potentials. And do note that at the event horizon, gravitational time dilation is infinite, and the optical clock rate is zero.
  15. Schmelzer Valued Senior Member

    Yes, ten potentials instead of a single one, which was my point.
    And you should note that there is no such animal as "the gravitational time dilation". If we have two observers, and fixed a synchronization procedure between them, then we can talk about a gravitational time dilation between them. And it depends not only on the positions of the two observers, but also on their velocities.
    krash661 likes this.
  16. PhysBang Valued Senior Member

    If one looks at how Einstein actually designed GR to work, however, we see that Farsight is merely a liar relying on those times that Einstein's language is vague.
  17. QuarkHead Remedial Math Student Valued Senior Member

    Schmelzer, I gave you a "like" for your last post, but....
    I don't think this is quite right. The problem is notation. Here's the low-down....

    Without worrying for now what it "does", define a tensor at a point \(p \in M\) (\(M\) being a 4-manifold) by

    \(G \in T^*_pM \otimes T^*_P M: T_pM \times T_pM \to \mathbb{R}\) such that \(G=\sum_{\mu\nu} g_{\mu \nu}dx^{\mu} \otimes dx^{\nu}\) where the \(g_{\mu \nu}\) are scalars and the \(dx^k\) are basis vectors.

    The \(g_{\mu \nu}\) are called the components of this tensor, and in GR there are 10 of them for each tensor at each point.

    But GR is a field theory, so in the field equations 1) the so-called "metric tensor" is a field and 2) the so-called metric tensor is a single entity \(G\) at each point each of which has the 10 scalar components \(g_{\mu \nu}\)

    Now Imagine a region of spacetime devoid of a gravitational source - \(T_{\mu \nu}=0\) Looking at the field equations, we see that the components \(R_{\mu \nu}\) must likewise be everywhere zero, and therefore of necessity, its trace. This does NOT mean that the metric field need be everywhere zero, but merely constant. This is because the curvature tensor is a second order derivative of the metric (albeit indirectly via the connection).

    By applying the Laplace equation \(\nabla^2 \phi=0\) to Newton's scalar gravitational field \(\phi\) in the absence of a source, we can convince ourselves that \(g_{\mu \nu}\) (written in the usual component form) is analogous to Newton's gravitational potential (because the Laplace equation likewise does not require that \(\phi =0\)) .

    Phew - sorry to be so long-winded!
    Last edited: Jan 24, 2016
  18. paddoboy Valued Senior Member

    I'm waiting for more revelations on his TOE he claimed to have a while back! That would be interesting.
  19. sweetpea Registered Senior Member

    I prefer the shortened version...
  20. Schmelzer Valued Senior Member

    First, even the curvature tensor does not have to be zero - this tensor has much more components, it is \(R_{ijkl}\), and only the sum \(R_{ij}= g^{kl}R_{ijkl}\) or so (too lazy to check over which indices one has to sum) has to be zero. So, even outside a star, where we have a vacuum, the spacetime remains curved.

    And, even if the second derivative is zero, the function itself does not have to be constant. With the Schwarzschild solution as an example.

    Then, of course, by naming them "gravitational potentials" I acknowledge that this means that each of them, taken separately, is in some broad sense analogical to the Newtonian potential. But these ten different functions can be different. Thus, potentials instead of a single one.

    Note also that the ten different potentials are a simplification - they would be really ten different potentials only if there are preferred coordinates. Without them, solution which look very different in terms of the functions \(g_{mn}(x)\) could define a solution which would be undistinguishable by any observation, because one would be able to transform them into each other by a coordinate transformation. According to the positivist spacetime interpretation, they would define the same solution. Once there are four coordinates to choose, which would not make a difference, there would be in fact only six independent degrees of freedom, even if all this is described by 10 different potentials.
    krash661 likes this.
  21. Farsight Valued Senior Member

    I'm not a liar. Einstein said what he said, and I referred to it. There's nothing vague about it at all. So you're the liar. And how on Earth you're given free rein to play the abusive troll here absolutely beats me.

    What's more interesting is what Einstein said. Which is not in accord with your popscience misunderstanding of GR.
  22. PhysBang Valued Senior Member

    You are a liar through omission: you pick only the vague things Einstein said and ignore the specific science that Einstein wrote.
    krash661 likes this.
  23. QuarkHead Remedial Math Student Valued Senior Member


    You know, there are few things more annoying on a forum like this than being corrected for something one did not actually say.

    I said the in the absence of a gravitational source, the Ricci tensor of necessity zero. I did NOT claim that the Ricci tensor completely determines spacetime curvature - it is well known that, whereas if spacetime is flat then the Ricci tensor is zero, it does not follow that if the Ricci tensor is zero, then spacetime must be flat. As you say, this statement is only true for the for the full Riemann tensor.

    Regarding the metric, I have no idea where you are coming from. Is it possible you don't appreciate fully the difference between a tensor space and a tensor field? I should be very surprised if this were the case.

    I repeat myself from each point in spacetime there exists a single metric tensor which has 10 scalar components. For any choice of component, the "results" must be compatible with those of any other choice. And since all other elements in the field equations depend (more or less) on the metric, this means that these equations are a set of 10 simultaneous equations which happen to be be non-linear. Which of course you already knew!

    PS PhyBang - "Liar" is an ugly epithet, even as applied to someone as irritating as Farsight. He is merely ignorant, usually mistaken and misinterprets (and overestimates his comprehension of) the relatively few sources he habitually refers to. I don't call this lying

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