4
$\begingroup$

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.)

$\endgroup$
8
  • 1
    $\begingroup$ Related. $\endgroup$ Commented May 12, 2019 at 18:24
  • 1
    $\begingroup$ Related: physics.stackexchange.com/q/108230/2451 , physics.stackexchange.com/q/263572/2451 and links therein. $\endgroup$
    – Qmechanic
    Commented May 12, 2019 at 18:31
  • $\begingroup$ see my answer here physics.stackexchange.com/q/11542 .charges play a role in attraction and repulsion together with the spin. $\endgroup$
    – anna v
    Commented May 12, 2019 at 18:38
  • 3
    $\begingroup$ 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)? $\endgroup$ Commented May 12, 2019 at 19:32
  • 1
    $\begingroup$ ...perhaps (3.3) here might help--unless it doesn't. $\endgroup$ Commented May 12, 2019 at 20:27

1 Answer 1

2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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