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Is this the correct Lagrangian of Quantum gravity

  1. Jun 25, 2008 #1
    Tipler had proposed that the correct QM Lagrangian leading to gravity should be

    [tex] S = \int d^4x\, \sqrt{-g}(\Lambda + \frac{1}{8\pi G}R + c^2_1R^2 +c^3_1R^3 \ldots + c^2_2R_{\mu\nu}R^{\mu\nu} + \ldots + c^3_1R_{\mu\nu;\alpha}R^{\mu\nu;\alpha} + \ldots) [/tex]

    however is he right he proposed that according to a theorem of logic there would not be a great distinction between a theory with a countable (but infinite) set of axioms and a theory with a finite set of axioms

    the webpage is http://en.wikipedia.org/wiki/Omega_Point_(Tipler)

    Tipler's article can be found at : http://math.tulane.edu/~tipler/theoryofeverything.pdf and http://arxiv.org/abs/0704.3276
    Last edited by a moderator: Apr 23, 2017
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  3. Jun 25, 2008 #2


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    I have no idea what Tiplers personal theory is, but yes, thats (up to normalization and units) the lagrangian of (pure) gravity that everyone uses.

    He supresses some indices as well, but whatever.
  4. Jun 25, 2008 #3
    I thought is was the Hilbert-Einstein action that leads to Einstein's equations of general relativity by use of the Euler-Largrange equation?
  5. Jun 26, 2008 #4


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    This is the Lagrangian you'd expect with Einstein Hilbert + Renormalization, which creates all terms allowed by symmetry.
  6. Jun 26, 2008 #5
    Obviously, I'd like to see that worked out. Can you recomment a reference? Thanks.

    I skimmed the reference given below: http://arxiv.org/abs/0704.3276

    On page 26, he justifies his version of the Action integral shown in post #1 by a one sentence referring to symmetry? I don't find this very compelling. But I've not read the whole paper. Does he give more compelling reasons for it somewhere else? Thanks.
    Last edited: Jun 26, 2008
  7. Jun 26, 2008 #6


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  8. Jun 26, 2008 #7
    I'm not seeing in this link any terms higher than the Ricci scalar that are, nevertheless, shown in post #1 of this thread. I'm wondering how Tipler justifies these additional terms beyond the passing reference to symmetry which I don't find very compelling.
  9. Jun 26, 2008 #8


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    The additional terms are quantum corrections they will lead to corrections to the Einstein Equations.

    You were asking how the E-H action produces the Einstein equations, that is explained at wikipedia. The action from the original post is a universal action for the field content and symmetries of GR, all possible terms appear. From Renormalization group arguments you expect this to be the effective action you get in Quantum Gravity, including corrections.

    This is a somewhat naive argument though. Still Reuter et.al start from the action above (truncated) to quantize GR by finding a renormalization fixed point.
  10. Jun 26, 2008 #9
    Isn't it more accepted to use the E-H action in the path integral to find QG?

    I'm not sure what you end up with if you use the Euler-Lagrange equation on the Lagrangian of post #1, with all its extra terms. I'm sure it would be some modification of Einstein's equation. I wonder what that would look like, and where you'd expect to see a different with Einstein's equations.
  11. Jun 27, 2008 #10


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    Caveat: I am not a field theorist so this is to my limited understanding:

    Whether you start with the EH action or the lagrangian above is somewhat irrelevant for the case of QFT. Remember that in the path integral all paths, not just those solving the equations of motion, contribute. If you then want to write the lagrangian at a specific point of the renormalization flow you have to take all these variations into account effectively. This happens through summing them up and packing them into the infinitely many constants in front of the infinitely many terms. When you have a renormalizable field theory, the infinitely many constants will only depend on finitely many constants so you can start out with a "pure" lagrangian that contains only those pure, unrenormalized couplings the infinitely many constants will end up depending upon. But at the end of the day you could have started at any other point in the renormalization flow process, with all the terms in, and the theory would have automatically told you that it only depends on these few constants. So in a way it's more honest to start with the full lagrangian instead of the finite one. The latter is preempting results from the theory. Gravity, however, is not renormalizable, so it doesn't make sense to talk of the EH action alone quantum mechanically anyways.
  12. Jun 27, 2008 #11


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    The point is, it doesn't matter if you just write the EH by itself, or with extra terms attached, b/c even if you forget about one or n terms, the prescription will generate them at some other scale upon renormalization group flow.
  13. Jun 27, 2008 #12
    I can't say I know anything about renormalization group flows. But are you suggesting that we can come up with the same classical equations of GR by finding the minimum of the action that has all the other terms in it? I think I'd need to see that. But if this action with all the extra terms can not reproduce the same Einstein equation of GR, then I would think that this action of extra terms has beed proven wrong, since we have tested classical GR, and it agrees with experiment.
  14. Jun 28, 2008 #13


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    Einstein's field equations are reproduced only at tree level (the EH term), beyond that you have small quantum corrections that are only important when you are close to some large energy scale. (eg somewhere around Mpl). You can safely neglect or truncate all those terms when you are far away from that scale (thus satisfying experiment).

    However if you want to study a blackhole or the early quantum universe in full generality, the above action is what you want to use.
  15. Jun 28, 2008 #14
    It sounds like you are postulating that the renormalization group flows are a fundamental principle on which to build all of physics when it cannot yet be tested. This seems to me more like speculation based on the fact that it makes the mathematics of the path integral more pretty when we don't even know yet what principles gives rise to the path integral or Schrodinger's equation.
  16. Jun 28, 2008 #15


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    The renormalization group flow has been extensively tested, and is a mathematical fact about quantum field theories. Likewise, the path integral and Schroedingers equation are pretty much axioms of quantum mechanics and not necessarily derivable. But their consequences have been verified experimentally. None of particle physics would make any sense without them.

    You can derive them starting from different assumptions, but they're probably best seen as the starting points.

    Quantum gravity is defined as the process of quantizing a classical theory (General relativity), so it makes sense to apply the path integral to it and see what it tells you. You are free of course to not believe this, that somehow gravity is different than the other three forces of nature (which were successfully quantized into the standard model), but it would be a little bit weird if that was true. You are also free to hypothesize different quantization procedures --canonical quantization for instance, which is equivalent to the Path integral for the 3 aforementioned forces, but maybe isn't for gravity. But physicists usually like to take what we know works first and see it through, before we start postulating more complicated pictures.

    With the path integral, you hit a snag rather rapidly though. Namely that the theory (defined with the lagrangian above) is nonrenormalizable. It becomes unpredictive and ambigous as you reach the most extreme scales. And thats sort of where everyone was at the beginning of the 1980s, and where all the modern efforts to quantize gravity really takes off (eg strings/lqg/twistor stuff etc).

    The above is the prerequisite to knowing whats really going on though, and how its modified is at the heart of every single modern picture out there.
  17. Jun 28, 2008 #16
    Can I please ask you the reference of some books or articles giving the derivation of the Schroedingers equation from these assumptions ? Thanks in advance for the help. The question is motivated by a personal investigation going from the Lorentz Einstein Law to an equation that could be analog to the Schroedinger's one.
  18. Jun 28, 2008 #17


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    You can 'derive' the schroedinger equation for instance from the path integral. I think Wiki does it:


    Theres probably many roundabout ways to do it, maybe from the Dirac equation as well. Its a little bit ackward logically, but these things are always self referential.
  19. Jun 28, 2008 #18
    I wish I knew a bit more about renormalization group theory. I have to wonder what test was done to specifically prove the renomalization group flow theory can ALWAYS be applied.

    It seems a bit of a contradiction to say that renormalization group theory can be used to solve a non-renormalizable theory.
  20. Jun 29, 2008 #19


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    I share your basic reservation here.

    IMHO it can not be left unquestioned that one part of the difficulties to make progress in unification is that fact that there is a certain set of constructs that are assumes to be valid starting points although they lack solid understanding.

    This IMO includs the origin of the quantum logic and path integral, but also Einsteins equations. IE. before trying to compute things, and solve problems that appear when the calculations makes no sense, I would like to understand the logic of the construct we start with.

    It could be that once we understand this better, the divergences will simply not appear. I don't share the view that the postulates of QM as mentioned above are best left as starting points not to be questioned. It could be that the quantum formalism itself is just an effective formalism, and thus emergent as per some yet unkonwn logic.

    Edit: I think the need for renormalization could be related to that that QM and QFT does not respect information constrains of the observer and the possible relational nature of the entire construct. Ie it considers information, without worrying about the limit of information. If the measures are intrinsically constructed relative to the observer to start with, it seems to me the renormalization is already built in, and only observable things at the relevant observational scale is there. It should not be posible to by intrinsic constructions arrive at divergence. That fact that it happens is to me a sign that the constructions are made extrisincally to the observer view.

    Sure one can argue wether it's "speculative" to ask such questions, but one can also raise the point that it's risky behaviour to not ask them.

    Last edited: Jun 29, 2008
  21. Jun 29, 2008 #20


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    The renormalization group is part and parcel of quantum field theory and was pioneered by a bunch of people including Gellman and Lo, and people like Ken Wilson (who won the Nobel prize in related works). You really can't have one without the other, as they're intimitly linked.

    Nor does it 'solve' a nonrenormalizable theory. Instead it tells you where an effective theory is still valid and how it evolves/runs or diverges at different scales.

  22. Jun 29, 2008 #21

    According to the Bekenstein bound formula (ref. 1, ref. 2 pg. 8 eq. 1), the amount of quantum information contained in the quantum states in a Planck sphere is:
    [tex]I_p \leq \frac{2 \pi E_p r_p}{\hbar c \ln 2} = \frac{2 \pi}{\hbar c \ln 2}\sqrt{\frac{\hbar c^5}{G}} \sqrt{\frac{\hbar G}{c^3}} = \frac{2 \pi}{\ln 2} = 9.064 \; \text{Planck bits}[/tex]

    [tex]\boxed{I_p \leq \frac{2 \pi}{\ln 2}}[/tex]
    [tex]\boxed{I_p \leq 9.064 \; \text{Planck bits}}[/tex]

    General Relativity constant:
    [tex]k = \frac{c^4}{16 \pi G}[/tex]

    Spin-2 field Einstein–Hilbert action: (ref. 4)
    [tex]S = \int \mathrm{d}^4 x \sqrt{-g} \left[ k \left( R - 2 \Lambda \right) + \mathcal{L}_\mathrm{M} \right][/tex]

    The Lagrangian density for a massless spin-two field:
    [tex]\mathcal{L}_M = - \frac{1}{2} \partial_{\lambda} h_{\mu \nu} \partial^{\lambda} h^{\mu \nu} + \partial_{\lambda} h_{\mu}^{\lambda} \partial_{\nu} h^{\mu \nu} - \partial_{\mu} h^{\mu \nu} \partial_{\nu} h + \frac{1}{2} \partial_{\lambda} h \partial^{\lambda} h[/tex]

    According to Wikipedia, the spin-2 field Einstein–Hilbert action yields Einstein's field equations:
    [tex]R_{mn} - \frac{1}{2} g_{mn} R + \Lambda g_{mn}= \frac{8 \pi G}{c^4} T_{mn}[/tex]

    According to Tipler, the Einstein–Hilbert action is inconsistent with quantum field theory, because it is non-renormalizable. (ref. 2 pg. 25-26)

    Cosmological constant Lagrangian Hilbert action: (ref. 2 pg. 25 eq. 2)
    [tex]S = \int d^4 x \sqrt{-g} \left[ \Lambda + \frac{1}{8 \pi G} R \right][/tex]

    Gravitational quantum field theory general invariant action: (ref. 2, pg. 26, eq. 3)
    [tex]S = \int d^4x\, \sqrt{-g} \left[ \Lambda + \frac{1}{8\pi G}R + c^2_1R^2 +c^3_1R^3 \ldots + c^2_2R_{\mu\nu}R^{\mu\nu} + \ldots + c^3_1R_{\mu\nu;\alpha}R^{\mu\nu;\alpha} + \ldots \right][/tex]

    However, Tipler's paper (ref. 2) and Wikipedia do not state the proof of how this gravitational quantum field theory general invariant action yields Einstein's field equations and does not contain the equation term that generated the Ricci scalar expansion series and Lagrangian density for any matter fields appearing in the theory.

    Cosmological constant Lagrangian Hilbert action with Lagrangian density:
    [tex]\boxed{S = \int d^4 x \sqrt{-g} \left[ \Lambda + \frac{1}{8 \pi G} R + \mathcal{L}_M \right]}[/tex]

    http://en.wikipedia.org/wiki/Bekenstein_bound" [Broken]
    http://arxiv.org/PS_cache/arxiv/pdf/0704/0704.3276v1.pdf" [Broken]
    http://en.wikipedia.org/wiki/Omega_Point_%28Tipler%29" [Broken]
    http://en.wikipedia.org/wiki/Einstein-Hilbert_action" [Broken]
    Last edited by a moderator: May 3, 2017
  23. Jun 29, 2008 #22

    I am uncertain what symbolic term generated the Ricci scalar expansion series, this is my first attempt, I am certain someone here could state the correct symbolic term.

    Ricci scalar expansion series: ???
    [tex]\boxed{\sum_{i = 2 \; j = 1}^{\infty \; i} (c_{j}^{i} R^{i} + c_{j}^{i} R_{\mu\nu}R^{\mu\nu} ) = c^2_1R^2 +c^3_1R^3 \ldots + c^2_2R_{\mu\nu}R^{\mu\nu} + \ldots + c^3_1R_{\mu\nu;\alpha}R^{\mu\nu;\alpha} + \ldots}[/tex]
    Last edited: Jun 29, 2008
  24. Jul 22, 2008 #23

    I wonder, is there any evidence that this sequence consists of smaller and smaller corrections - like a Taylor series, perhaps? I've seen other converging power series whose first term may lead to the Hilbert-Einstein action. I wonder if it matches up to this one.
    Last edited by a moderator: Apr 23, 2017
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