6
$\begingroup$

My question involves an analogy I have to point out. Consider the Lagrangian density for the a complex scalar field: \begin{equation} \mathcal{L}=\frac{1}{2}\partial_{\mu}\phi^{\dagger}\partial^{\mu}\phi-\frac{1}{2}m^{2}\phi^{\dagger}\phi \end{equation} which is invariant under the global gauge transformation $$\phi \rightarrow e^{i\alpha}\phi.$$ The divergence-less 4-current is $j=(\rho,\underline{j})$ and the associated conserved quantity is the electric charge. This illustrates a global $U(1)$ symmetry. To see the analogy we now look at spacetime translations $x \rightarrow x + a$ which give $$\phi(x) \rightarrow \phi(x+a) = e^{a^{\mu}\partial_{\mu}}\phi.$$ The Lagrangian is no longer invariant but the action: \begin{equation} S = \int \mathcal{L} d^{4}x \end{equation} is invariant due to the translation invariance of the Lebesgue measure. The Noether current is the stress-energy tensor $T^{\mu\nu}$ and the conserved quantity is the 4-momentum. Due to the role played by $T^{\mu\nu}$ in the Einstein field equations we can say that the 4-momentum plays the role of gravitational charge.

Promoting the global U(1) symmetry to a local one couples $\phi$ to the electromagnetic field. This involves modifying the Lagrangian by introducing a covariant derivative and adding the Lagrangian for the EM field. My question is: does promoting the global translation invariance of S to a local one give the correct coupling of $\phi$ to the graviton field? An action invariant under general coordinate transformations is: \begin{equation} S = \int \mathcal{L} \sqrt{- g}d^{4}x \end{equation} where $\mathcal{L}$ should also be modified appropriately and $g_{\mu\nu}$ is the metric. I rephrase and extent my question:

(1) Do general coordinate transformations include local translations $x \rightarrow x + a(x)$ or are they only $SO(1,3)$ gauge-like transformations?

(2) In view of the above analogy the gauge group for gravity is the translation group in Minkowski spacetime. The group is not compact. Shouldn't the group be compact to guarantee a positive-definite kinetic term for gravitons?

(3) The translation group is also abelian. Isn't this inconsistent with the fact that gravitons self-interact?

(4) In essence what I am saying is that the classical theory of gravity is likely incomplete. Are there any approaches to gravity that modify the underlying spacetime such that the isometry group of the new space admits a compact version of the translation group of Minkowski spacetime? Maybe the universe after formation was compact but due to expansion the compact nature is now hidden.

(5) Related to (4). Shouldn't elementary particles correspond to irreducible representations of the isometry group of the universe? If the universe is Minkowski then the isometry group is the Poincare group. Have other possibilities in this regard been explored?

(6) In c = h = 1 units Newton's constant G has mass dimension -2. The fine structure constant is given by: $$\alpha = \frac{e^{2}}{4 \pi \epsilon_{0}}$$ In gravitoelectromagnetism we have the correspondence $G \leftrightarrow \cfrac{1}{4\pi \epsilon_{0}}$. As the momentum $p$ plays the role of gravitational charge, shouldn't the coupling constant to gravity be given by $$g \sim p^{2} G = m^{2} G$$ where $m$ is mass of the particle in question? $g$ is then dimensionless. Does this imply that the covariant derivative to couple matter to gravity is given by something like $D_{\mu} = \partial_{\mu} - p^{\nu} A_{\mu\nu}$? where $A_{\mu\nu}$ is some tensor potential related to $g_{\mu\nu}$.

(7) The above ideas might imply something about spin 0 bosons and their mass renormalization. Any thoughts? Could the $\Lambda^{2}$ dependence of $m^{2}$ be just an indication that gravity was ignored?

$\endgroup$
8
$\begingroup$

Yes, the promotion of the spacetime translations to a local group – and for group-theoretical reasons, it must be the group of all coordinate transformations i.e. diffeomorphisms – produces a consistent theory of gravity with the correctly coupled metric tensor, including the nonlinear (self-interaction) terms that appear because the diffeomorphisms form a non-Abelian group.

  1. It's only the diffeomorphisms $x\to x+a(x)$ that are linked to the existence of the actual gravitational "gauge field", the metric. A description of gravity may also have the local Lorentz symmetry $SO(3,1)$ acting on "vielbeins" etc. but this extra gauge symmetry isn't what produces the gravitational force. It's just a convenience that is helpful and/or needed to couple the gravitational theory to spinors and similar types of fields.

  2. No, the translation group doesn't have to be compact and in the Minkowski space, it is not compact. In fact, the natural bilinear form on the Lie algebra is indefinite. This doesn't produce any negative-norm states in the graviton multiplet because all the components $g_{0i}$ are made unphysical by the gauge symmetry – by the diffeomorphisms. Also, there are no problems with the noncompactness effectively because the action for the gauge field starts with the second derivatives in the Ricci scalar $R$. Gravity is not special case of Yang-Mills theory in the bulk because the translations are not a Yang-Mills group. Yang-Mills groups act in individual points but diffeomorphisms mix the points. Both of them are local groups but they are not the same, so the rules are different.

  3. The self-interactions occur because the diffeomorphism group is non-Abelian. In the Yang-Mills case, the points are isolated from other points so the whole gauge group is non-Abelian if the group at one point is. But this relationship isn't true in the case of diffeomorphisms. The per-point group of translations is Abelian but the whole gauge group is not.

  4. No, the classical theory of gravity isn't incomplete and there's no reason to "demand" a compact per-point group of symmetries.

  5. Yes, but the claim is vacuous. The Universe doesn't have to have any isometries but elementary particles may still exist. Yes, they would generally come in "one-dimensional multiplets" but that doesn't mean that there could be no other way to organize them.

  6. Yes, the dimensionless gravitational coupling constant is $O(GMm)$ in the same sense as we may say it is $O(Qq/4\pi\epsilon_0)$ in electromagnetism. The dimensionless charges may be assumed to be of order one but that's not true for the masses. Your formula for the covariant derivative is wrong. The covariant derivative in the "gauge theory" called gravity is the usual covariant derivative in general relativity. It's not of Yang-Mills type because gravity isn't a Yang-Mills theory (a simple subclass of theories with local symmetries).

  7. Gravity only "intrinsically" says something about the spin-2 particles. Everything else is just "matter" and its spin is irrelevant. The corrections to the Higgs mass etc. are divergent due to non-gravitational contributions. Attempts have been made to modify gravity so much that the divergence goes away but all these attempts are either inconsistent or contradict the equivalence principle.

$\endgroup$
  • $\begingroup$ Thank you for the answer. Things seem pretty clear but I do have a few more questions: Isn't it possible to embed general relativity into a Yang-Mills type of theory? Wouldn't this help with quantisation? In this line of thought the non-abelian nature of the diffeomorphism group would follow from some generalised type of non-abelian gauge group. Or is Yang-Mills theory strictly limited to apply to spin 1 particles and no meaningful generalisations likely? $\endgroup$ – Sebby Sep 22 '13 at 9:45
  • $\begingroup$ @Sebby You're aware of the Coleman Mandula theorem and its supersymmetric loopholes? $\endgroup$ – twistor59 Sep 22 '13 at 10:46
  • $\begingroup$ Yes but not too much, from my understanding it's changing the underlying spacetime by adding the theta dimensions. If the space is changed in a less radical way wouldn't the isometry group change and the Coleman-Mandula theorem not necessarily apply? Actually the type of generalisation I have in mind should abstract away from the underlying spacetime somehow. $\endgroup$ – Sebby Sep 22 '13 at 11:30
  • $\begingroup$ Dear @Sebby, I wrote it about 3 times in the answer already but general relativity (in the bulk) is simply not equivalent to any theory of the Yang-Mills type in the same bulk. You may have AdS/CFT correspondence etc. in which gravity in the bulk is equivalent - via a very nontrivial equivalence- to a gauge theory in a lower-dimensional space (boundary). But the Yang-Mills symmetry acts on points individually while the diffeomorphisms mix points into each other. They're just qualitatively different symmetries. This is also reflected by the fact that the graviton has spin 2 and gauge bosons 1. $\endgroup$ – Luboš Motl Sep 22 '13 at 18:13
  • $\begingroup$ Yang-Mills theory means that the gauge field is a spin-one field. If it is a spin-two field, like the metric tensor, it's just not called Yang-Mills theory. Gravity is a "generalization" of a gauge theory but the word "generalization" is really needed and it means that gravity does not obey the original rules and conditions of the ordinary, non-generalized Yang-Mills theory. Frankly, I sincerely hope that I won't have to write this simple fact for the 5th time again - you don't seem to be listening at all. $\endgroup$ – Luboš Motl Sep 22 '13 at 18:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.