The Laws Of Cosmology May Need A Re-Write?

Discussion in 'Astronomy, Exobiology, & Cosmology' started by Ultron, Apr 18, 2016.

  1. paddoboy Valued Senior Member

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    As a newbie Schneibs, and in case you are unaware. Schmelzer could be classed as somewhat of a "maverick" who has a paper on an ether proposition that in his opinion, predicts more then the current overwhelmingly supported GR.
    It remains uncited though like thousands of other hypothetical papers.
    Queerus also has had a past thread attempting to claim that GR is less than reliable and validated moved to the fringes.
    This is the pair that you are up against.

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  3. Schneibster Registered Member

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    By claiming that energy is not conserved under GRT, that's exactly what you're saying.

    OK. So now you're claiming energy is conserved under GRT, and energy is not conserved under GRT.

    Pick a frame. Stick with it. You're logic chopping again.

    Unfortunately either you've taken this quote out of context or Wald is wrong, and the proof is that we can detect mass concentrations on and within the Earth, and verify the calculations we made with GRT, and when we look we actually see the mass concentrations where we expected to. In fact, our ability to do this is so precise and accurate that geological and archeological discoveries have been made using satellites and this technique.

    My expectation is that you've taken the quote out of context, or that Wald has left out that we can define a stress-energy tensor to the limit of our ability to calculate it, but we can't use GRT to calculate a result of infinite precision. And that makes GRT better than our best instruments, and in fact better than any possible instruments, which is good enough, I'd say.
     
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  5. Schneibster Registered Member

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    It's all good; thanks for the heads-up but I kinda already had that one figured.

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  7. Schneibster Registered Member

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    I really don't know what to do with someone who claims to know GRT but doesn't agree that you can calculate the gravity at any point to any desired precision on an irregular body using it.

    Energy is conserved under GRT to the limit of our ability to calculate it, and energy is observed to be conserved in the real world to the limit of our ability to measure it. Whether it's "exactly conserved" in some imaginary mathematical analysis or not is footless, and there is disagreement among relativists whether this is even so in the first place. GRT is good enough for the real world, and there aren't any gaps in it to insert some "god of the gaps" ether theory into. And that's that.

    Nobody's going to solve an infinite number of PDEs, and nobody's going to make any measurements with zero error bars, so this isn't science, it's philosophy.
     
    Last edited: Apr 25, 2016
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  8. paddoboy Valued Senior Member

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    Well put.
    That won't stop any of them from trying though.

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  9. Schmelzer Valued Senior Member

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    Of course not. The solution is a metric field $g_{\mu\nu}(x)$. That means, a field which defines, in every point x of the whole spacetime, a possibly different metric. But the solution of the Einstein equation is the whole set of the ten functions $g_{\mu\nu}(x)$.
    And not only on the surface, the solution tells us also what is the gravity at some distance from the asteroid, as well as inside, thus, for the whole space, and how it changes in time.
    No. Take an elementary course in GR. The Einstein equations are equations for the whole set of functions $g_{\mu\nu}(x)$ on the whole space.
    Indeed. So what? I have never made such a suggestion.
    Which is complete nonsense and has nothing to do with GR. Because in GR, you have one and only one PDE: the Einstein equations.

    You need, of course, a lot of measurements to obtain an accuracte information about the matter. But whatever you do with GR has only one PDE, the Einstein equations.

    So you've got one set of PDEs for one location on the asteroid, and a different set for another location, and so forth. If you only use one set of PDEs for the whole asteroid your results will not be of sufficient precision to check them at the resolution of our instruments, and therefore they cannot be confirmed.
    Sorry, but if you have some particular measurement devices which measure something depending on some choices of patches, fine, do it, but this has nothing to do with the use of different systems of coordinates (charts) in differential geometry and with GR. It would be better if what you do with your instruments is not in contradiction with GR. But if your measurement instruments measure something in some patches, this has nothing to do with the problems of defining local energy and momentum of the gravitational field.
    Complete nonsense. I'm not at all biased against any experiments. I take them as facts.
    Everything fine with me, except for the repetition of the nonsense that you need many PDEs. There is only one PDE in GR, the Einstein equations. (There are other for the matter, but their number is small enough too.)
    It looks like we talk in different languages. For a point particle, with a classical trajectory, one would have to solve an ordinary differential equation, ODE, not a partial differential equation, PDE, which one uses for densities, average velocities, temperatures and so on.
    Nonsense. Learn a GR course.
    And I don't know what to do with defamators, who, despite my insistence to support their inventions with quotes, invent nonsense and claim that I have claimed such nonsense.
    Learn GR. To quote a well-known established mainstream textbook about GR again: "there is no meaningful notion of the local stress-energy of the gravitational field in general relativity."
     
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  10. Schmelzer Valued Senior Member

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    5,003
    No. Learn GR. With any standard textbook.
    No, this is not a proof of your claim. Take Wald (one can find it for free in the net) and learn GR. Your claim about what we can do may be fine (even if I have not heard about archeological discoveries made in this way, but who cares), but nothing in the quote from Wald contradicts this.

    The gravitational field is defined by the metric, which is a metric field $g_{\mu\nu}(x)$, which is a collection of 10 scalar fields, which you can measure with every precision you are able to reach, and then you can compare these measurement with what the solution of the Einstein equations gives. There is no consideration of local energy of the gravitational field involved.
    The pseudo-tensor is simply a formula. If you have fixed some system of coordinates, you can compute it. With every accuracy you like. In any other system of coordinates, you can compute it too, with any accuracy you like. But the two results will not be related with each other by a usual transformation rule like for tensor fields. And it gives you nothing you can locally measure. nothing physical. All what you can locally measure is something different, and defined by some tensor.

    Learn GR, before starting argumentations about GR.
     
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  11. Schneibster Registered Member

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    390
    Convince me this isn't philosophy before we talk about it any further, Schmelzer.

    Fact is, you gotta retrieve that set of ten PDFs from GRT in order to calculate the gravity at any point, and in fact how we really do it is to assume that we only need one solution for an entire patch.

    When you tried to tell me that calculating gravity fields isn't what GRT is for, this stopped being about the real world. Sorry, man, not down with that.
     
  12. Schneibster Registered Member

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    390
    And since we're quoting textbooks, let's try Gravitation, Misner/Thorne/Wheeler, pp 386, in which the authors
    They not only show that energy-momentum is conserved under GRT, they prove it, and then they go on to show that every other law of physics applies as well, concluding:
    Underline is my emphasis; italics in the original.

    Now, I'll take Gravitation for GRT anytime you like.

    Stop telling me to "learn GRT." You are the one who needs to learn GRT.

    I'll take my 400 dollar textbook over your 39.95 paperback anytime, sport. Not to mention Kip Thorne and John Wheeler over some guy I've never heard of.

    In fact, I'd take two of the most famous relativists who ever lived writing a 400 dollar textbook as pretty much definitive of "the mainstream." So it looks like your attempt to marginalize my views has not only failed, but backfired on you.

    At this point I don't expect you'll reply any more.
     
    Last edited: Apr 26, 2016
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  13. The God Valued Senior Member

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    The issue involves interpretation of mainstream. The question is, is energy conserved in GRT ? Is use of pseudo tensor for local energy density is right approach to establish conservation of energy in GRT ? I feel technical Mod should intervene and clarify what is the mainstream interpretation on this point. As far as interpretation of mainstream is concerned, the issue should not be left open.

    IMO this is a problem_area/incompleteness of GRT as the energy conservation is a non violable law....
     
  14. paddoboy Valued Senior Member

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    27,543

    Not necessarily. Sometimes if there is more than one interpretation, both can be correct.
    I see light/photon from a distant object being lensed: I may say light is being bent.
    Or I may think slightly harder and realise that actually spacetime is warped/curved and that light is just following geodesics in curved spacetime.
    Both though are correct.
    So, no, I don't believe any problem exists with GR within its applicable parameters.
     
  15. The God Valued Senior Member

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    3,546
    Example 1
    Energy is conserved in X process.
    energy is not conserved in X process.

    Example 2
    Inflation came after BB.
    Inflation came before BB.

    These two examples contain contradictory statements, they cannot be termed as some kind of alternative interpretations, as you are suggesting. Can you say that both the statements are correct ?
     
  16. paddoboy Valued Senior Member

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    27,543
    http://math.ucr.edu/home/baez/physics/Relativity/GR/energy_gr.html
    s Energy Conserved in General Relativity?
    In special cases, yes. In general — it depends on what you mean by "energy", and what you mean by "conserved".
    Inflation is the first aspect of the evolution of spacetime we can point to with any real difinity.
    We have many aspects of Inflation, which may or may not imply: Eternal Inflation is one.
    In any circumstance though the premise that the universe evolved from a hotter, denser state and is still expanding cannot be argued with.

    http://scienceblogs.com/startswithabang/2010/01/12/q-a-did-inflation-happen-befor/
    https://www.quora.com/Did-inflation-happen-after-or-before-the-Big-Bang
     
  17. Q-reeus Banned Valued Senior Member

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    This has become an amusing saga - a poser has been continually found out but a massive ego prevents the ability to concede. It didn't end well for the hired gun Jack Wilson in Shane.
     
  18. Schmelzer Valued Senior Member

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    5,003
    Fine. Except that this "law of local energy-momentum conservation" is not a conservation law of the classical form $\partial_{\mu} T^{\mu\nu}=0$, so that you cannot integrate it to obtain something conserved.

    If you want some conservation law, one which you can integrate, that means, of the form $\partial_{\mu} T^{\mu\nu}=0$, you have to look for the pseudotensor, see MTW p. 465.
    I have no problem at all if you follow MTW. So, read and understand what is written on p.465:

    Emphasis mine. Because these have been the points I have tried to explain.

    And then learn §20.4. WHY THE ENERGY OF THE GRAVITATIONAL FIELD CANNOT BE LOCALIZED
     
    Last edited: Apr 26, 2016
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  19. paddoboy Valued Senior Member

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    Perhaps you should elaborate more on what you are so insidiously trying to say..... hired gun, oh for fuck'sake!

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  20. Schmelzer Valued Senior Member

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    Sorry, I'm unable to convince anybody who has fantasies like that I try to tell that "calculating gravity fields isn't what GRT is for" of anything.

    GR is the theory which gives us the Einstein equations, a set of PDEs which allows to define the metric $ds^2 = g_{\mu\nu}(x) dx^\mu dx^\nu$, which is the gravitational field.
     
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  21. Q-reeus Banned Valued Senior Member

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    Watch the foul language there paddoboy - it's formally against forum rules, and betrays your level of angst. Elaborate? OK: http://www.sciforums.com/posts/3373025/
     
  22. paddoboy Valued Senior Member

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    27,543
    Is it? I've seen mods also use the magic word on occasions. Obviously you need to apply logic and understand its the frequency and application of the word that may get you into trouble. My use of the word is quite within reason, and no, no angst at all....not even a teensy weensy bit.
    In fact I often get a good laugh out of the pretentious posts of some that have some inbuilt desire or need to try and change/invalidate GR from the realms of a science forum...Indeed!

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    It ain't gonna happen bud!

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    Oh, and you still havn't elaborated....names my friend, names!

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  23. Schneibster Registered Member

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    Actually, that expression is part of the proof. According to a GRT textbook. Not a 39.95 paperback.

    Let's look at this equation:

    \(\triangledown \cdot T = 0\)

    What that means is that the total energy-momentum flow into or out of any surface, under GRT, is zero.

    You can't create energy or momentum and you can't destroy it. <- energy-momentum conservation law

    We done here?
     
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