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Is it true that in canonically quantised gravity "non-traceless" and "non-transverse" gravitons are allowed to exist as virtual particles, whereas any graviton that corresponds to a "time-warping" plane wave is not allowed to exist, even as virtual particle? And if so, could this be why gravity is non-renormalisable for high energies?

Rephrasing my 2nd question:

Is perturbative quantum-gravity non-renormalisable because time-warping doesn't occur at all in the low energy limit and so cannot be accounted for through the interaction-picture?

I will explain why I would think to ask this:

I have tried to canonically quantise linearised gravity in exact parallel with the canonical quantisation of spin-1 vector fields by looking at a gauge in which the EOM (in this case the linearised EFE) become a simple wave-equation, and then using the Gupta-Bleuler technique to enforce the gauge conditions.

Using the Gupta-Bleuler approach on the gauge conditions that turn the linearised EFE into a simple-wave equation we get the requirements:

$$h^+_{0\mu} |\psi\rangle = 0 \tag{1}$$ $$h_{i}^{+i} |\psi\rangle = 0 \tag{2}$$ $$\partial^{\mu}h^{+}_{\mu\nu} |\psi\rangle = 0 \tag{3}$$

Where $$h^+_{\mu\nu} = \int\frac{d^{3}k}{(2\pi)^{3}2\omega_{\vec{k}}} \epsilon_{\mu}^r\epsilon_{\nu}^s a_{rs}(\vec{k})e^{-ikx} \tag{4}$$

Just as the gauge condition $\partial_{\mu}A^{\mu}=0$ ensures that no scalar- or longitudinally-polarised photons appear in any observables in QED, although these polarisation CAN contribute to scattering as virtual particles,

Equations (2) & (3) result in the disappearing of the earlier mentioned "non-traceless" and "non-transverse" gravitons from the observables, although it doesn't rule out their contributions to scattering amplitudes.

However equations (1) seem not to be so forgiving towards any "time warping gravitons". The conditions simply leads to the result that the creation operators associated with these "time-warping" gravitons destroy any state, preventing them from contributing to any process, even as virtual particles.

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Is it true that in canonically quantised gravity "non-traceless" and "non-transverse" gravitons are allowed to exist as virtual particles,

Yes, but this is just a way of describing a way of doing perturbation theory in some gauges, and does not carry any physical meaning.

whereas any graviton that corresponds to a "time-warping" plane wave is not allowed to exist, even as virtual particle?

I don't know what "time-warping" means exactly, but you are free to pick any gauge you want, and in some gauges $h_{0\mu}$ is not identically zero.

And if so, could this be why gravity is non-renormalisable for high energies?

No. General relativity is non-renormalizable because it has has irrelevant interactions. For example, writing $g_{\mu\nu}=\eta_{\mu\nu}+h_{\mu\nu}/M_{\rm Pl}$, the Einstein-Hilbert term has an infinite tower of interactions schematically of the form $\frac{1}{M_{\rm Pl}^n} h^n (\partial h)^2$; the fact that these interactions are multiplied by a coupling constant that has dimensions of mass to a negative power means the theory is not power-counting renormalizable.

You certainly cannot make statements about renormalizability based purely on gauge conditions in linearized gravity. Renormalization needs to be done because interactions change the physics relative to what you would expect on the basis of the free theory. If you turn off the interactions, you lose the physics that makes renormalization necessary in the first place.

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  • $\begingroup$ By time-warping gravitons I mean the ones corresponding to GW's that make the $h_{0\mu}$ (in Minkowski coordinates) oscillate. My statement about non-renormalizability does not depend on the gauge condition, it is based on the fact that no time-warping occurs in linearized gravity (an observable effect that is therefore, by definition, independent of the gauge). This fact is merely made apparent by the gauge conditions. How can we hope to recreate full gravity (which includes time-warping) without any time-warping gravitons. Is this not a problem in addition to the power counting problem? $\endgroup$
    – Quanta
    Commented Mar 5, 2022 at 16:07
  • $\begingroup$ @Quanta It is not true that $h_{0\mu}=0$ in all gauges in linearized gravity. If $h_{0x}=0$ in some gauge, then do a gauge transformation with parameter $\xi_\mu = \{0, f(t), 0, 0\}$ for some arbitrary function $f(t)$, then in the new gauge $h_{0x} = \partial_t \xi_x + \partial_x \xi_0 = \dot{f}(t)$. $\endgroup$
    – Andrew
    Commented Mar 5, 2022 at 16:18
  • $\begingroup$ The choice of gauge and renormalizabilty are two unrelated concepts. $\endgroup$
    – Andrew
    Commented Mar 5, 2022 at 16:21
  • $\begingroup$ @Quanta I also don't think time-warping is a gauge-independent observable. You've defined it in terms of the values of $h_{0\mu}$, which are gauge dependent. Can you give a gauge-invariant definition? $\endgroup$
    – Andrew
    Commented Mar 5, 2022 at 16:44

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