The Laws Of Cosmology May Need A Re-Write?

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

  1. Q-reeus Valued Senior Member

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    Many folks think the sun is radiating plenty of net mass-energy out from it's surface all the time. I'm such a believer. Gee, did something basic get left out of a continuity equation interpretation - like maybe the possibility of a balancing loss of stress-energy *within* any appropriate *bounding* surface (NOT 'any surface')? I think so.
     
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  3. Dr_Toad It's green! Valued Senior Member

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    But the sun isn't a closed system classically or gravitationally. The definition of bounding surface seems moot given the previous post's brevity and clarity.

    What were you objecting to?
     
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  5. Q-reeus Valued Senior Member

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    Net bounding surface flux is just *one* component in a continuity equation, even a non-covariant one. I suggest studying the basic idea behind it at e.g.
    https://en.wikipedia.org/wiki/Continuity_equation
    Section 7.2 there is relevant to GR case. Further response to that post will probably soon enough come from another respondent who will doubtless have fun with it.
     
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  7. Schmelzer Valued Senior Member

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    Once you seem to refuse to learn §20.4. WHY THE ENERGY OF THE GRAVITATIONAL FIELD CANNOT BE LOCALIZED of the book Misner, Thorne, Wheeler, Gravitation, which you seem to have accepted as describing the mainstream, we are indeed close to be done here and to give up to educate an uneducable person who refuses to read the textbooks.

    And obviously you have not even an idea about the meaning of that equation in GR, which is

    $\nabla_a T^{ab} = \partial_a T^{ab} + \Gamma^a_{ca}T^{cb} + \Gamma^b_{ca}T^{ac} = 0$

    where the $\Gamma^b_{ca}$ are the Christoffel symbols, which are quite nontrivial functions of the metric. One can easily make them zero in a single point, by introducing coordinates which define an inertial frame in this point, but having them zero only in a point will not help you to compute something conserved from the equation.

    So, this equation generalizes the conservation law for matter fields $\partial_a T^{ab} = 0$ of special relativity, and, therefore, is sloppily named "conservation law" by some physicists, but this does not make it a conservation law at all. Point. (And, note, MTW do not name (16.1e) "conservation law" but, instead, an "abstract geometric law (16.1e) in curved spacetime". Because they know it is not a conservation law.)

    And you have obviously also not understood the point of § 16. It is a quite informal prescription as how to obtain an equation for GR given an equation of SR.
     
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  8. Q-reeus Valued Senior Member

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    Strictly, in #121 I should have been referring to a divergence 'law', but the associated wording in #120 there implied a continuity relation - close enough for the situation considered.
     
  9. Dr_Toad It's green! Valued Senior Member

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    I see. Thanks.

    I look forward to more, and even hope to understand more.

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  10. Schneibster Registered Member

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    I have no time for someone who thinks a textbook contradicts itself.

    It's obvious that Misner, Thorne, and Wheeler said explicitly that energy-momentum is conserved under GRT.

    It's equally obvious that you're trying to quote mine them in order to "prove" they didn't really mean it/didn't understand it/said two different things in two places in their textbook/whatever your next excuse is.

    This is over.
     
  11. Schmelzer Valued Senior Member

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    Of course, there is no contradiction in MTW. We have a simple case of somebody who is uneducable because he refuses to learn the relevant math.

    The situation with energy and momentum conservation is, of course, subtle, in particular because we have, with the various pseudo-tensors, some conservation laws, and the problem is only that they cannot be interpreted in a physically meaningful way in the spacetime interpretation because they are not covariant.

    And, no, I do not need any excuses. I simply know that $T^{\mu\nu}_{,\mu}=\partial_\mu T^{\mu\nu}=0$ is a conservation law which can be integrated to obtain energy and momentum, but $T^{\mu\nu}_{;\mu}=\partial_\mu T^{\mu\nu} + \Gamma^\nu_{\mu\lambda}T^{\mu\lambda} +\Gamma^\mu_{\mu\lambda}T^{\lambda\nu} = 0$ is nothing you can integrate, because I have done the math. No need for me to data-mine in MTW, it will be somewhere in MTW too, no doubt.

    But against a refusal to learn one cannot do anything.
     
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  12. Schneibster Registered Member

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    It has nothing to do with math. It says in straightforward simple sentences that all the laws of physics- including energy conservation, which it proves it for- are upheld in GRT. I quoted it above.

    I have no interest in weird math education from someone who can't even read plain English.
     
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  13. paddoboy Valued Senior Member

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    Of course not! [tic mode on]

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  14. Schmelzer Valued Senior Member

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    Here is what you have quoted:
    Let's see. Is there a difference between the GR equation
    $T^{\mu\nu}_{;\mu}=\partial_\mu T^{\mu\nu} + \Gamma^\nu_{\mu\lambda}T^{\mu\lambda} +\Gamma^\mu_{\mu\lambda}T^{\lambda\nu} = 0$ and the SR equation $T^{\mu\nu}_{,\mu}=\partial_\mu T^{\mu\nu} = 0$? If one looks at the formulas, it seems. But, wait. In the SR case, where $g_{\mu\nu}(x) = \eta_{\mu\nu}$, the Christoffel symbols $\Gamma^\mu_{\nu\lambda} = \frac{g^{\mu\kappa}}{2}(\partial_\nu g_{\kappa\lambda} + \partial_\lambda g_{\kappa\nu} - \partial_\kappa g_{\lambda\nu})$ will be zero. So, in this case there is no difference between the two formulas at all. The quite different looking formula appears to be the same one in the particular situation described by the Minkowski metric of SR. But it is a covariant formula, which can be applied to all metrics.

    Does it follow that we can use the formula $T^{\mu\nu}_{;\mu}=\partial_\mu T^{\mu\nu} + \Gamma^\nu_{\mu\lambda}T^{\mu\lambda} +\Gamma^\mu_{\mu\lambda}T^{\lambda\nu} = 0$ to compute conserved integrals in the general situation? As we can in the special SR situation, where we have $T^{\mu\nu}_{,\mu}=\partial_\mu T^{\mu\nu} = 0$? Of course, not. And MTW does not make such a claim.

    Paddoboy, you want to suggest that my quote
    qualifies as "excuse"? It is not. Because these "some physicists" are not MTW. As I have mentioned:
    In fact, you could have created a more serious problem for me if you had asked me to support my accusation against "some physicists" with a quote from a serious GR textbook. I remember to have seen such things, but to find this again would be, indeed, a hard job of data mining. And I would prefer to concede that I'm actually unable to support this accusation with a quote.
     
    Last edited: Apr 27, 2016
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