If we consider the weak-gravity approximation of the vacuum Einstein field equations with a positive cosmological constant then the field equations for the graviton acquire an imaginary mass term: could this be taken as an argument that gravitons may not exist?
Interesting ... No, Lambda is not a graviton mass. I have an ancient paper on this issue, see
http://iopscience.iop.org/article/10.1088/0143-0807/19/5/002
If you write down the wave equation for the Maxwell field in de Sitter, you could think similarly that Lambda is the mass of the photon. It's true that the spacetime curvature causes backscattering of low-frequency waves, as does a mass, but this creates an effective mass. But, like for photons propagating in a material, an effective mass is not a true particle mass.
I am not sure... don't we require some proper boundary conditions when we define the graviton as a field i.e. asymptotic flatness? These are violated in any spacetime with cosmological constant.
... true, with the CC the vacuum is no longer Minkowski but some de Sitter... however, this should mean that the fluctuations are no longer taken around Minkowski but around a de Sitter, but still those fluctuations should represent the graviton, right?
I seem to remember the concept of particle dos not work in this case. Have a look at the book by Birrel and Davies. I am sure the whole problem is discussed there in detail.
Excellent observation. The existence of gravitons has been the notion for many decades that each of the 4 known forces of nature should be described by exchange particles. In fact, there is nothing in QFT that demands that to be the case, as was specifically noted by Feynman. By investigating the coupling of a massless spin 2 particle with the energy momentum tensor, one finds that there is actually no coupling with the metric (with no torsion). Massless spin 2 particles describing gravity with a symmetric metric do not exist; massive gravitons on the other hand exist in the high energy short range regime. The curvature of spacetime nicely describes the effective force of gravity without the need of any exchange particles. Further investigation of the symmetric part of the relativistic quantum equation in curved spacetime reveals that GR is actually hidden in QFT down to the particle regime. Attached is an article discussing these findings. It is in the process of being re-written and extended to a new model of cosmology based on the natural dark energy tensor that appears in the modified Einstein equation. The observation that lambda must be zero to obtain the exact Einstein Hilbert action from the direct coupling of the metric with the energy momentum tensor is interesting and is directly related to your observation.
... nice, thanks for the extensive reply: I'll have a look at the attached paper and be back if I have comments/questions. Thanks again
If gravitons posses imaginary mass in frame of considered cosmological model, it means only that just this model leads to these results. If you assume that gravitons: 1) possess positive mass; 2) gravitons do not exist; etc., you can choise another model.
Gravitons don't exist. Explanation of gravity is simple and possible when we reintroduce so-called ether (in other words pre-matter) to the zoo of physical entities. Ether that could be named pre-matter as well =/= than matter due to the difference in their properties.
Interesting ... No, Lambda is not a graviton mass. I have an ancient paper on this issue, see
http://iopscience.iop.org/article/10.1088/0143-0807/19/5/002
If you write down the wave equation for the Maxwell field in de Sitter, you could think similarly that Lambda is the mass of the photon. It's true that the spacetime curvature causes backscattering of low-frequency waves, as does a mass, but this creates an effective mass. But, like for photons propagating in a material, an effective mass is not a true particle mass.
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