# Why aren't gravitons spin 1?

Expressing the metric as $$g_{\mu \nu} = \eta_{\mu \nu} + h_{\mu \nu}$$, assuming $$h_{\mu \nu} \ll 1$$ we can write the Einstein Hilbert action to leading order in $$h_{\mu \nu}$$ and quantize the linearized Einstein Hilbert action to construct the graviton field. Gravitons are spin 2 particles, which is easiest to see by noting that $$h_{\mu \nu}$$ has two indices. These enjoy a "gauge symmetry" corresponding to diffeomorphisms.

However, classically, gravity can be understood to be largely analogous to a gauge theory. The Christoffel symbol $$\Gamma^{\alpha}_{\beta \mu}$$ takes the place of $$A^a_\mu T^a$$ as the gauge field.

Note that $$\Gamma^{\alpha}_{\beta \mu}$$ has three indices, however the $$\alpha,\beta$$ indices can be understood as a matrix, much like the Lie algebra elements $$T^a$$ in Yang Mills theory.

If we quantize this field instead shouldn't we not be able to realize gravity as a theory mediated by spin 1 gauge particles?

(This should be especially true if we take the action to be the Kretschmann scalar, which seems to be equivalent to the Yang Mills Lagrangian.

$$\mathrm{Tr}(F_{\mu \nu} F^{\mu \nu}) \leftrightarrow R^a_{b \mu \nu} {R^b_{a}}^{\mu \nu}$$

However, this would obviously give a different theory than Einstein gravity.)

• May 12, 2019 at 18:24
• Related: physics.stackexchange.com/q/108230/2451 , physics.stackexchange.com/q/263572/2451 and links therein. May 12, 2019 at 18:31
• see my answer here physics.stackexchange.com/q/11542 .charges play a role in attraction and repulsion together with the spin. May 12, 2019 at 18:38
• It sure feels like 3 questions: a) Physically, coupling to the energy-momentum tensor dictates spin 2. b) Formally, the weak field expansion of Einstein's equations produce a metric perturbation field which is spin 2. c) Indeed, the intuitive gauge-theory simulacrum for gravity is the gauged tangent space Lorentz group, effected by the spin connection ω, related to Γ, which is not a tensor, and which is utilized in supergravity; however, possibly counterintuitively, it demonstrably leads to spin 2. You apparently want to focus on 3)? May 12, 2019 at 19:32
• ...perhaps (3.3) here might help--unless it doesn't. May 12, 2019 at 20:27

Imagine a gravitational wave traveling along the $$z$$ axis. We can choose coordinates (aka fix a gauge if you prefer that language) where the components of the metric perturbation are $$\begin{equation} h_{\mu\nu} = \left( \begin{matrix} 1 & 0 & 0 & 0 \\ 0 & h_+ & h_\times & 0 \\ 0 & - h_\times & h_+ & 0 \\ 0 & 0 & 0 & 1 \end{matrix} \right) \end{equation}$$ where $$h_+$$ and $$h_\times$$ are the amplitudes of the $$+$$ and $$\times$$ polarizations.
Under a rotation about the $$z$$ axis by an angle $$\psi$$, the time parts of the above metric perturbation $$h_{00}$$ and $$h_{0i}$$ are invariant, while the spatial part $$h_{ij}$$ transforms as $$\begin{equation} h_{ij} \rightarrow R(\psi)_{ia} h_{ab} R_{jb}(\psi) \end{equation}$$ where $$R(\psi)_{ij}$$ is a rotation matrix $$\begin{equation} R(\psi)_{ij} = \left( \begin{matrix} \cos\psi & \sin \psi & 0 \\ -\sin\psi & \cos\psi & 0 \\ 0 & 0 & 1 \end{matrix} \right) \end{equation}$$ Working through the algebra and using various trig identities, you can show this implies that $$h_{ij}$$ is invariant under a rotation by $$\pi$$, which is characteristic of a spin-2 particle. In general after a rotation you will find that the components of $$h_{ij}$$ are multiplied by functions of $$2\psi$$, another characteristic sign of a spin-2 particle.
The precise definition is in terms of helicity. An eigenstate of the helicity operator corresponds to a circularly polarized gravitational wave. A circularly polarized wave with $$h = h_+=e^{\pm i\pi/2}h_\times$$ will transform as $$h \rightarrow e^{\pm i 2\psi} h$$. The $$\pm 2$$ here are the eigenvalues of the helicity operator; a massless particle with a $$\pm 2$$ helicity eigenstates is a massless spin-2 particle.