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General Relativity vs. quantum gravity

  1. Jul 15, 2012 #1
    Hey all!

    So, in my study of general relativity, I've come to understand that gravity is actually what physicists would have classically called a "fictitious force", in that it is a force derived from the fact that the observer is not in an inertial reference frame, like in the case of the centrifugal force. This makes a lot of sense to me due to the geometric nature of space-time.

    So when I hear particle physicists discuss Quantum Gravity theories, I get confused. Most propose some sort of unification of general relativity and quantum field theory, i.e. there is a force carrier for gravity called the "Graviton". This makes very little sense to me in the previous context of general relativity; how can a geometric, "fictitious" force be described by a quantum field of particles? Furthermore, how could the interaction of massive objects with this graviton field cause phenomena like time dilation and space-time, which follow naturally from the underlying premise of General Relativity that seems to be in direct conflict with modern attempts to unify gravity with the other forces?

    Thanks for reading!
     
  2. jcsd
  3. Jul 15, 2012 #2

    atyy

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    There is discussion about this issue in http://arxiv.org/abs/1105.3735

    p6: "The real underlying principle of GR has nothing to do with coordinate invariance or equivalence principles or geometry, rather it is the statement: general relativity is the theory of a non-trivially interacting massless helicity 2 particle. The other properties are consequences of this statement, and the implication cannot be reversed."

    p53: "An important fact about GR is that there exists this parametrically large middle regime in which the theory becomes non-linear and yet quantum effects are still small. ... In this region, we can re-sum the linear expansion by solving the full classical Einstein equations, ignoring the higher derivative quantum corrections, and trust the results."
     
    Last edited: Jul 15, 2012
  4. Jul 15, 2012 #3

    jcsd

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    I think the basic idea behind quantum gravity is that gravity in GR is still described by a field, albeit a geometric field, and that field can be quantitized. The fact that it is a geometric field I believe is the main problem in creating a useful quantum gravity theory.

    One way of viewing it is to break the metric in to two parts: a (presumably) flat background metric and a peturbation, just like in the linearized theory. You then quantitize the field representing the peturbation of the background metric and a graviton, I believe, would be seen as a quanta of the peturbation. Some of the problems with what is called "the covariant petrubation method" is that you end up with a non-renormalizable theory and the choice of how you break your metric up is arbitary.
     
  5. Jul 15, 2012 #4

    tom.stoer

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    I understand you confusion.

    There are basically two perspective, one I would call the "gravitational" or "geometrical" one, the other is coming from elementary particle physics. You seem to take the first perspective which is somehow in-line with the perspective of e.g. loop quantum gravity. Others coming from elementary particle physics may prefer approaches to (perturbative) quantum gravity like supergravity or (some formulations of) string theory. The latter perspective seems strange, but at least these theories have GR in a certain limit, they may even be perturbatively renormalizable and they have a spin-2 graviton in its spectrum. So the "high energy physics" perspecive is not totally unreasonable.

    Personally I share your opinion, but I am not the expert here to make a judgement.
     
  6. Jul 15, 2012 #5
    As usual Tom said it best. GR can be considered the classical limit of a quantum field theory just as Maxwell's equations are the classical limit of Quantum Electrodynamics.
     
  7. Jul 15, 2012 #6
    In QFT, a difference is made between "internal" symmetries, and "external" symmetries of a field.

    Internal symmetries relate to the set of matrices acting on a fictitious space spanned by the components of the field that carry arbitrary indices. These are the gauge symmetries of the Standard Model.

    External symmetry would be the Lorentz covariance of the field. Of all the other symmetries of the field, it is assumed to hold unconditionally, and irrespective of the internal symmetries of the field. As such, it is treated on a different footing than the internal symmetries.

    Nevertheless, the mathematics of the two kinds of symmetries is the same - that of Lie algebra representations. This might suggest us that we ought to treat these symmetries on an equal footing, perhaps as manifestations of a larger symmetry.

    There are essentially 2 approaches, that I would call:
    1. "Internalizing" the external Lorentz symmetry. I think this is the approach taken in String Theories.
    2. "Externalizing" the internal symmetries. I don't know if any respectable theory had taken this approach, but I myself tried to find if [itex]SU(3) \times SU(2) \times U(1)[/itex] could be a subgroup of some [itex]SO(m, n)[/itex] Lorentz symmetry with m temporal, and n spatial dimensions. Then, the extended Lorentz symmetry gets broken down to the existing (1+3)-dimensional space-time, plus (or times) the gauge group of the Standard Model.

    There is an extra problem that I am not able to answer. Namely, can coordinate covariance be reinterpreted as local Lorentz covariance, i.e. the differentials of the coordinate vector transform by a Lorentz transformation that is different at different space-time points:
    [tex]
    d\bar{x}^{\mu} \stackrel{?}{=} \Lambda^{\mu}{}_{\nu}(x) \, dx^{\nu}
    [/tex]

    EDIT:
    I guess the necessary and sufficient condition for this would be:
    [tex]
    \Lambda^{\mu}{}_{\nu}(x) = \frac{\partial \bar{x}^{\mu}(x)}{\partial x^{\nu}} \Leftrightarrow \partial_{\rho} \, \Lambda^{\mu}{}_{\nu} = \partial_{\nu} \, \Lambda^{\mu}{}_{\rho}
    [/tex]
     
  8. Jul 15, 2012 #7

    atyy

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    I guess the relevant theorem is the Coleman-Mandula theorem, which has a loophole for supersymmetry. Apparently when supersymmetry is gauged, one gets supergravity.
     
  9. Jul 16, 2012 #8

    tom.stoer

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    I don't think that this will work.

    In the tetrad or first-order Palatini formalism one finds two different symmetries:
    1) local Lorentz symmetry which rotates the tetrads and
    2) diffeomorphism invariance.

    1) Going from one coordinate patch to another, the tetrads are rotated by an arbitrary local Lorentz transformation which results in a corresponding connection, i.e. the structure of a gauge theory on tangent space.
    2) And going from one coordinate patch to another means that coordinates are subject to diffeomorphisms ensuring consistent gluing of coordinate patches. But these diffeomorphisms do not result in any gauge-like structure.

    One can compare the SO(n-1,1) and Diff(M). The Lie algebra of Diff(M) are the n-vector fields on a manifold with a Lie bracket structure: L[f] = fa(x) ∂a; the Lie algebra of SO(n-1,1) is just so(n-1,1). Therefore Diff(M) depends on the global structure of the manifold M whereas so(n-1,1) does not; it depends on the local structure i.e. on tangent space only! In essence the two groups have different structure, different Lie algebra and different dimensions (as Lie groups).
     
  10. Jul 16, 2012 #9

    haushofer

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    No. LLT's are performed in the tangent space, while a coordinate transformation brings you from one tangent space to another.

    In gauging the Poincaré algebra one gets LLT's and local translations. Using the conventional constraint (the field strength of the translations is zero) such a local translation can be rewritten as a gct and a LLT with field dependent parameters (this constraint also allows you to solve for the spin connection). As such one obtains a soft algebra.
     
  11. Jul 16, 2012 #10
    Gravity is not a fictitious force. Gravity is equivalent to the presence of tidal forces, i.e., to a nonzero Riemannian curvature tensor, [itex]{R^{\rho}}_{\sigma\mu\nu} \neq 0[/itex]. The classical fictitious forces correspond to the special case where [itex]{R^{\rho}}_{\sigma\mu\nu} \equiv 0[/itex] (no tidal forces), but [itex]{\Gamma^{\rho}}_{\mu\nu} \neq 0 [/itex].

    Fictitious forces may be transformed away; this follows from a theorem in Riemannian geometry saying that [itex]{\Gamma^{\rho}}_{\mu\nu} \equiv 0 [/itex] (globally) may be obtained by coordinate transformation, if and only if [itex]{R^{\rho}}_{\sigma\mu\nu} \equiv 0[/itex], see for instance §11.5 in Gravitation, by Misner, Thorne, and Wheeler.

    The force of gravity cannot be transformed away, though, because [itex]{R^{\rho}}_{\sigma\mu\nu}[/itex] (unlike [itex]{\Gamma^{\rho}}_{\mu\nu}[/itex]) is a tensor; analogously, the electromagnetic field, as encoded in the Faraday tensor, [itex]F_{\mu\nu}[/itex], cannot be transformed away (by any Lorentz transformation).
     
    Last edited: Jul 16, 2012
  12. Jul 16, 2012 #11
    Ah, ok! Thanks for that bit of perspective, JustMeDK.
     
  13. Jul 16, 2012 #12
    And on the surface, that makes a lot of sense, but it just seems to me that the classical E/M field and the GR description of the gravitational field are very different. Maybe I am thinking about the quantum field for gravity a little differently than I should be (I've not yet taken a formal class in QFT) but it seems to me that it is natural to express E/M in field theory, whereas before, there was no concrete mechanism for the electromagnetic force; we only knew how to calculate the fields, not where the fields came from. This is much the same as Newton's theory of gravity, but when Einstein developed GR, he finally added a mechanism for gravity--the curvature of space-time. This, therefore, seems to be a very different force from E/M, or the nuclear forces. QFT was very successful in describing the mechanism behind forces for which there was no previous mechanism; it was previously just understood that these forces worked on certain particles. Whereas any new mechanism for quantum gravity would have to overthrow the old and very successful one. How can a field of particles create space-time?

    atyy's comment might get at it, when he quotes page 6 of the article he posted. I just don't understand how Einstein's equations can be derived from a field of massless spin-2 particles. I guess I'll have to read through the article.

    I also read something about the difference between the graviton in string theory and LQG: that in string theory, the graviton is a true particle, like the photon, W and Z bosons and the gluon, where in LQG, it's often referred to as a "pseudo-graviton"-- is this like an instanton or a quasiparticle, maybe? Or did I read it wrong? Could somebody comment on this difference?

    Sorry if I'm sorta playing devil's advocate here. I don't mean to question those who know more than I do, I'm just trying to clear up a few concepts I don't really understand.
    :smile:
     
  14. Jul 16, 2012 #13

    tom.stoer

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    I did not want to say that GR must or does necessarily follow from a QFT in the classical limit. What I wanted to sy is that the "high energy perspective" assumes that this may be the case, and that there are indications in supergravity and string theory that this may be true. I agree that the fundamental field theories are very different and that - when constructing a quantum theory - the "high energy perspective" may miss some important features; as often stressed, personally I am convinced that does miss the background independence - but I can be wrong.

    I think that some of the 'theorems' regarding (quantum) gravity rely on background-dependent methods well-known from quantum field theory, which may fail for the gravity case. The Coleman-Mandula theorem and the Weinberg–Witten theorem are examples where this may be the case.

    I agree that the graviton in LQG is not a fundamental entity but that it may be a kind of collective excitation of the underlying spin network, something like a pseudo-particle. But the semi-classical limit where something like a graviton should emerge from LQG/SF is still not fully understood.
     
  15. Jul 16, 2012 #14
    Thanks, Tom. I also found some older threads on similar QG topics that cleared a lot of things up for me. (I think you were in a few, if I'm not mistaken!) I've always felt, much like the LQG folk, that any quantum theory of gravity should share the same basic structure as GR: background independent, diffeo. invariant, curved spacetime, non-perturbative, etc. I prefer the idea of the graviton as a collective excitation as well. I do understand why particle/string physicists might disagree with me, though; force unification is rather tempting!
     
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