Gravitational collapse

Discussion in 'Physics & Math' started by arfa brane, Jun 18, 2013.

  1. arfa brane call me arf Valued Senior Member

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    --http://principles.ou.edu/earth_figure_gravity/geoid/

    Tach's interpretation is just wrong, IOW. And still no attempt to answer this question: Where is the INFORMATION in Einstein's equation that describes the shape of a body with mass?
     
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  3. Tach Banned Banned

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    From your own quote:

    "an equipotential surface"

    Give it a rest, you do not need to demonstrate your ignorance repeatedly, with every post.

    One thing at a time, learn the basics first, the EFEs are way too advanced of a subject for you. Besides, you have confused shape and density, so you need to learn the difference first.
     
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  5. arfa brane call me arf Valued Senior Member

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    Yeah. Notice how it also says "a representation of the surface"? Notice how it implies strongly that the geoid is NOT the actual surface?
    Keep trying. You still haven't shown why the surface of the earth is an equipotential surface as you have claimed several times. Have you noticed that the quote in my post 221 says it isn't? Or are you ok with disagreeing with something a tenured professor at a recognised university posted on the 'net? Still nothing to refute my claim that Einstein's field equations don't say anything about the shape of a mass density?

    How stupid are you prepared to look?
     
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  7. arfa brane call me arf Valued Senior Member

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    No I haven't although you seem to WANT to believe I have.
    Please point to the post where I managed to confuse density and shape. Isn't mass density kilograms per cubic metre? Isn't shape something else altogether (it's related to geometry, I'm pretty sure).

    Please, please point out where I got these confused. I'd like you to look like an even bigger idiot (if you're up to it).
     
  8. Tach Banned Banned

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    Desperation time sets in : splitting hairs.

    Anybody capable of following the relatively simple paper I linked in can figure that out. Why can't you?



    Did you make up this term all by yourself?

    Nowhere as ignorant and arrogant as you.
     
  9. Tach Banned Banned

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    False: the stress-energy tensor density component (much like the pressure components) is , in the most general case, variable. So, density variations are dealt with in GR, contrary to your crank claims. A good example are Bianchi cosmologies. I can link a few papers but I know by now that they would be wasted on you.

    There is no "Ricci curvature" in Einstein equations, there is the Einstein tensor (which depends on the Ricci tensor). The EFEs , being tensorial equations, have tensors on both sides, the stress-energy tensor on one side, the Einstein tensor on the other side. You are just demonstrating your crass ignorance on the subject.
     
  10. Markus Hanke Registered Senior Member

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    I think you need to consider carefully what the EFEs do and what they don't do. A body with mass is a source of gravity, and as such it is described by the stress-energy-momentum tensor, being the source term in the field equations. This tensor is what you input into the EFEs, and what comes out of them are the 10 independent components of the metric tensor. That is all the EFEs do - they provide a system of equations which allows us to find the components of the metric tensor. Once we have that, we can then go on and examine the behaviour of our mass given that metric, e.g. using the tools of relativistic fluid dynamics. The EFEs as such do not directly contain any information about shapes of bodies, but they do give us relations between the various quantities involved; for example, in the case of the interior Schwarzschild solution, the EFEs together with the geodesics equations directly yield the Oppenheimer-Volkoff equations.

    A quick word on the curvature objects - firstly, there is the Riemann tensor. This tensor is the only object which contains all information with regards to space-time geometry at any given point. Geometrically, it can be understood as a measure of geodesic deviation, i.e. as a measure of how the separation vector between geodesics changes from point to point. More accurately, it is a direct measure of the failure of the covariant derivative to commute.

    Secondly, we have the Ricci tensor; this is a contraction of Riemann across two indices. Geometrically this object measures how the volume of a small "ball" deviates from that of a reference "ball" in flat Euclidean space. It is important to remember that Ricci curvature is a "reduced" form of curvature in that it does not contain the same information as the Riemann curvature tensor. Practically this means that a vanishing Ricci tensor does not necessarily imply a flat space-time.

    Thirdly, there is the Ricci scalar, which is a contraction of the Ricci tensor across both indices. This object measures how much the sum of angles of a small triangle in space-time deviates from 180 degrees. Once again, a vanishing Ricci scalar does not imply a flat space-time.

    The last important object ( which does not directly appear in the EFEs ) is the Weyl tensor; this tensor measures tidal accelerations, i.e. once again geodesic deviations, but, unlike the Riemann tensor, it measures only those. It therefore contains less information than the full Riemann tensor. The Weyl tensor can be understood as the information which is "missing" from the Ricci tensor to make up the full Riemann curvature. For example, in Schwarzschild vacuum around a static black hole the Ricci tensor vanishes, but the Weyl tensor does not; on the other hand, near the Big Bang singularity the Weyl tensor would have been very small, but the Ricci tensor was very large. Thus, the Weyl tensor can be used to physically distinguish between gravitational and cosmological singularities - this is known as the Weyl curvature hypothesis.

    Hopefully this makes things a little clearer.
     
  11. arfa brane call me arf Valued Senior Member

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    I wonder what kind of contradiction this is? My guess is Tach doesn't know what Ricci curvature actually is.
     
  12. Tach Banned Banned

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    You mean Ricci (curvature) tensor? Because , as already explained to you, "Ricci curvature" is not part of the EFEs.
    Considering the fact that you have already demonstrated your ignorance on the subject(s) multiple times.....Besides, I have already explained your error, so did Markus. I suggest that you take a class on the subject, "learning by posting" has already turned into an epic failure.
     
  13. arfa brane call me arf Valued Senior Member

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    That's what "you" explained to me (and anyone else who can read).

    But the Einstein tensor 'contains' R[sub]ab[/sub], the Ricci curvature tensor.
    (Look, there it is: \( R_{\mu \nu} -\frac{1}{2}R\,g_{\mu \nu} + \Lambda\,g_{\mu \nu} = {8 \pi G \over c^4} T_{\mu \nu} \) )

    You insist that this means Ricci curvature isn't part of Einstein's equation. But it is part of the Einstein tensor. Are you defining "part of" in some other way, if so can you clarify (I doubt it)?
    Correction: you think you explained my error and you think Markus did the same thing. But Markus hasn't actually stated explicitly that I'm wrong about anything. This is some of what he posted:
    Where does he say Ricci curvature is not part of the EFEs? I mean, there it is, right there in the equation the Ricci tensor (which measures this 'reduced' form of--wait for it--curvature).
    I prefer to believe that you are misinformed and that your intent is to mislead everyone. On the other hand, Markus and others with a better grasp of the subject are able to explain (their understanding) in clear, simple terms--unlike you, so what up G?


    At best your "explanations" are unclear. I would rather you didn't bother explaining, thanks. You're just wrong about too many things, so I don't trust you, and I don't see why anyone else should.

    I know how tough that must seem seeing how much you care . . .
     
  14. Tach Banned Banned

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    5,265

    Once again, there is no such thing as "Ricci curvature" in the EFEs. There is Ricci (curvature) tensor. More correctly, there is Einstein tensor. "Curvature" cannot be part of an equation. On the other hand, as explained to you several times already, "tensor" is. Clearly, you do not understand the difference between "curvature" and "tensor" . Don't worry, eventually, you'll learn.




    But you keep posting BS trying to pass it as science, just like Farsight.
     
  15. arfa brane call me arf Valued Senior Member

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    Let's have another look at Wikipedia's definition:
    "In relativity theory, the Ricci tensor is the part of the curvature of space-time that determines the degree to which matter will tend to converge or diverge in time . . . "

    Ok it says the tensor is part of spacetime curvature. I wonder what the big problem is with calling this curvature "Ricci curvature"?

    Tach is convinced (but can't of course explain why) that there is no "curvature" in the equations. Well, sure, curvature is a property of spacetime (that should be obvious). The Ricci tensor "describes" this partly; that can't be a misguided notion, surely? So Ricci curvature being "included" in the field equations isn't that bad an abuse of terminology either.

    If google really is God, then some of the roughly .5 million hits I got from "ricci curvature" would resolve this somewhat pedantic point. I guess you could give it a try . . .
     
  16. arfa brane call me arf Valued Senior Member

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    Baez's exposition at http://math.ucr.edu/home/baez/einstein/ uses this "ball of test particles" a lot. So what is a test particle? I understand this is a particle that doesn't interact with other particles, and a ball of test particles is at rest when the volume isn't changing.

    Baez uses particles of instant coffee, which are all initially 'comoving', so \( \dot V\; =\; 0\).
    Now these test particles are supposed to follow geodesics, and if the spacetime is curved, these will all converge, so the geodesic deviation is a measure of 'local' curvature. The particles aren't 'attracting' each other, or then they would be interacting (a contradiction).
     
  17. Markus Hanke Registered Senior Member

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    Actually, this isn't entirely correct; curvature is an invariant of the connection used on the manifold in question, not the manifold itself. The fact that it doesn't vanish is then a result of that manifold's geometry. One must be clear though that there are other ways to characterise the geometry of such manifolds; for example, through torsion, or a "mix" of torsion and curvature. These would correspond to a different choice of connection. In classic GR it is always understood that the Levi-Civita connection is used; the principle invariant of this connection is curvature, whereas torsion is always taken to vanish. The opposite of this would be a connection where torsion is present, but curvature vanishes ( i.e. a space-time that is curvature-flat everywhere ), a so-called Weizenboeck connection. It turns out that the physics of such a model are largely equivalent to normal GR, in terms of what they predict - this model of gravity is called teleparallelism :

    http://en.wikipedia.org/wiki/Teleparallelism

    Einstein himself has studied this in his later years. It is very interesting to realize that one can formulate a model which makes predictions similar to GR, in a space-time which is flat everywhere but possesses intrinsic torsion.
     
  18. arfa brane call me arf Valued Senior Member

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    Ok thanks for that Markus. I think though, that any distinction might come down to how we understand intrinsic curvature; I've seen the phrase "spacetime is curved by matter" a lot, so isn't it just some kind of distinction between measurement and intrinsics?

    I want to understand this better too, where Baez starts with the usual ball of test particles: "the relative velocity of the particles starts out being zero, so to first order in time the ball does not change shape at all: the change is a second-order effect."

    That looks a bit tricky, he's saying the first derivative of volume stays constant?

    Also he says it's important to understand the flow of momentum is really pressure, so how do you work that out? Baez mentions that textbooks use an ideal gas.
     
  19. brucep Valued Senior Member

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    4,098
    It's equivalent to GR as Professor Thorne points out in his great book about the history of gravitational physics. Black Holes & Time Warps. He also said that he prefers to use the Teleparallellism model when working with gravitational radiation. If it's equivalent to GR it makes the same predictions.
     
  20. arfa brane call me arf Valued Senior Member

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  21. brucep Valued Senior Member

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    4,098
    For all objects natural motion is freefall. General relativity is the theory that describes the natural path of matter through the universe [gravitational field]. That's the natural phenomena GR makes predictions about.
     
  22. arfa brane call me arf Valued Senior Member

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    I think this flies:

    Momentum has units of joule-seconds per metre, it follows that a joule-second (J.s) has the same units as momentum x distance (since J.s.m[sup]-1[/sup] x m -> J.s)
    So this idea of flow of momentum corresponding to pressure means that we need to go from J.s.m[sup]-1[/sup] (momentum units) to J.m[sup]-3[/sup] (pressure units), which means "something" with units of m[sup]-2[/sup].s[sup]-1[/sup] or "per unit area seconds".

    Well, if the time interval for momentum is the same as for pressure (in the system of interest, say an ideal gas in a container with fixed volume), intuitively these will "cancel", since they are the same physical parameter. So, in some sense, momentum becomes joules per metre (same units as Newtons of force), and we get force per unit area, as required (??)

    "Some sense" is then the integration over units of time of particle momentum.
     
  23. arfa brane call me arf Valued Senior Member

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    Just a bit more precise: if the flow of momentum is momentum per second through a unit area, then this unit area can be anywhere in an ideal gas (and oriented in any direction).
    Because the pressure is equal in every direction in a constant volume (for an ideal gas at equilibrium), the momentum flowing per second through any area must be equal in both directions.

    This must also be true for any area of the container wall, so the momentum flowing 'into' any area of the wall must be equal to the momentum flowing 'out of' the same area.

    So, we have momentum per second per ("through a") unit area with units (J.s.m[sup]-1[/sup])(s[sup]-1[/sup])(m[sup]-2[/sup]) -> J.m.[sup]-3[/sup], which is units of energy density (pressure). Hence pressure = momentum flow q.e.d..

    So, yeah, physical units are devices that make 'heuristics', or useful rules of thumb--generally applicable ones at least.
     

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