3
$\begingroup$

In Zee's quantum theory in a nutshell, at the end of chapter I.10, he states that

the graviton is of course the particle associated with the field $g_{\mu\nu}$.

My understanding of quantum field theory is fuzzy, but I think I understood that each field assumes its values in an irreducible respresentation of the Lorentz group, the internal state space (or a representation of its universal covering, which is equivalent to a projective representation). Such representations are classified by their restriction to the $\text{SO}(3)$ or $\text{Spin}(3) = \text{SU}(2)$ subgroup, which is a spin representation and is classified by a single number called the spin. The graviton field has spin 2.

The metric tensor $g_{\mu\nu}$ is a section of the symmetric square of the (real) tangent bundle, which can be seen as a representation of the Lorentz group by the action $g\mapsto \Lambda g\Lambda^{T}$.

  1. My first question would be: is this essentially the graviton field (as Zee seems to say, just to double check)?

The metric tensor is 6-dimensional, whereas a spin 2 field is 5-dimensional. Since for the Minkowski metric $\eta$ we have $\Lambda\eta\Lambda^T = \eta$, multiples of $\eta$ form a 1-dimensional subrepresentation, whose complement/quotient will be a 5-dimensional representation.

My other questions would be:

  1. Is this complement indeed the actual graviton field?

  2. What is its meaning or interpretation? Is it something like the non-flat part of the metric?

  3. If the preceding two were correct, does the fact that it is factored out of the graviton field mean that the graviton doesn't see unform dilations of spacetime? If so (or if not), what does that mean?

If my question is misguided, please point out why!

$\endgroup$
  • $\begingroup$ You are asking for a very long essay; Something like this math.uchicago.edu/~may/VIGRE/VIGRE2010/REUPapers/Tolish.pdf , which shows how a classical Newtonian field exists in the limit of GR. Then , once a field can be defined, the idea to quantize it and make a field theory out of it leads to the graviton and effective quantizations of gravity. See the link in this answer of mine physics.stackexchange.com/q/254352 on how string theories solve the quantization of gravity $\endgroup$ – anna v Jun 20 '16 at 3:05
  • $\begingroup$ @annav Thank you for the links. A long essay would be nice, but I would already be very happy to know if my assumptions are in the right direction and some short remarks. $\endgroup$ – doetoe Jun 20 '16 at 6:56
  • $\begingroup$ As an experimentalist, I cannot tell about your assumptions. Certainly gmunu by itself is not the graviton $\endgroup$ – anna v Jun 20 '16 at 7:21
  • $\begingroup$ @annav You say "Certainly gmunu by itself is not the graviton". Do you know how Zee's quotation at the beginning of my post should be interpreted then? $\endgroup$ – doetoe Jun 20 '16 at 11:49
  • $\begingroup$ This link might help arxiv.org/pdf/gr-qc/0607045.pdf . I think gmunu is associated with the graviton the way that Amu is assiated with the photon $\endgroup$ – anna v Jun 20 '16 at 12:57
2
$\begingroup$

A spin-2 field in 4D is not five-dimensional. The standard spin-2 object transforms in the $(1,1)$-representations of the Lorentz group where the $(m,n)$-labels are half-integers labelling an equivalent representation of $\mathfrak{su}(2)\oplus\mathfrak{su}(2)$, see the Wikipedia article on representations of the Lorentz group for more information. The spin of the $(m,n)$-representation is $m+n$ because the actual rotation algebra embeds diagonally into the two copies of $\mathfrak{su}(2)$ we are representing. A pure spin two field $h_{\mu\nu}$ is given by traceless and symmetric 2-tensor, which has 9 d.o.f. in 4D.

If we want to conceive of GR as a QFT (on ordinary flat space) in which we can speak of gravitons (an approach usually called Pauli-Fierz theory, we are essentially quantizing linearized gravity. I will not discuss the various inconsistencies and their fixes this approach needs to actually yield a somewhat consistent QFT), then you just write $g_{\mu\nu} = \eta_{\mu\nu} + h_{\mu\nu}$ and treat $h$ as the dynamical field whose associated particles are gravitons. However, not all components of $h$ are physical - it enjoys a gauge symmetry $h_{\mu\nu}\mapsto h_{\mu\nu} + \partial_\mu X_\nu + \partial_\nu X_\mu$ for an arbitrary vector field $X$ - this is the infinitesimal version of the diffeomorphism invariance of GR.

At the end of the day, the gauge symmetry means that two d.o.f. of $h$ remain as physical - corresponding to the actual graviton and its possible polarizations.


JamalS's answer to a similar question explicitly counts the d.o.f. of $h$ and describes an alternative way to deduce the spin of the particle associated to $h$ in the quantum theory by looking at the Noether currents for the Lorentz symmetry applied to the part of the Lagrangian containing $h$.

$\endgroup$
  • $\begingroup$ Shouldn't it be two degrees of freedom, because the graviton is massless? It would be five if it had mass. $\endgroup$ – Javier Jun 20 '16 at 16:21
  • $\begingroup$ @Javier: Oops, you're right. $\endgroup$ – ACuriousMind Jun 20 '16 at 16:27
  • $\begingroup$ Thanks a lot for your reply, this is very useful. I also wrote the rank of the metric tensor wrong, I meant 10 = 1 + 2 + 3 + 4 rather than 6 = 1 + 2 + 3. In that sense the 9 looks good again. Thanks for the Wikipedia link. Just to make sure: when you say "a pure spin two field $h_{\mu\nu}$ is given by traceless and symmetric 2-tensor, which has 9 d.o.f. in 4D", you are talking about a real representation and the dimension over the reals, right? $\endgroup$ – doetoe Jun 20 '16 at 22:27
  • $\begingroup$ When you write $g_{\mu\nu} = \eta_{\mu\nu} + h_{\mu\nu}$, do you first scale $g$ in such a way that $h$ is traceless? If so, does this mean that this scaling of the metric is an unphysical transformation as well in some sense? $\endgroup$ – doetoe Jun 20 '16 at 22:28
  • $\begingroup$ @doetoe: Yes, I think such a scaling is implied - it can be done using diffeomorphism invariance. When I say 9 d.o.f. I mean that the space of traceless and symmetric 4x4 matrices has real dimension 9. $\endgroup$ – ACuriousMind Jun 21 '16 at 12:26

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.