2019
DOI: 10.1016/j.physletb.2019.134907
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Testing the Swampland: H0 tension

Abstract: The de Sitter Swampland conjecture compels us to consider dark energy models where λ(φ) ≡ |∇ φ V |/V is bounded below by a positive constant. Moreover, it has been argued for Quintessence models that the constant λ scenario is the least constrained. Here we demonstrate that increasing λ only exacerbates existing tension in the Hubble constant H0. The identification of dark energy models that both evade observational bounds and alleviate H0 tension constitutes a robust test for the Swampland program.

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Cited by 79 publications
(43 citation statements)
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“…Aside from the lack of a compelling physical motivation, probably the biggest issue facing the RdSC ( * ) is the observed dark energy in our universe, coupled with the fact that realistic quintessence models seem to be at least as difficult to construct in string theory as metastable de Sitter vacua [64][65][66]. Furthermore, even hypothetical quintessence models that satisfy the RdSC bounds are on the verge of being ruled out experimentally [67][68][69]54].…”
Section: Quintessence Vs Cosmological Constantmentioning
confidence: 99%
“…Aside from the lack of a compelling physical motivation, probably the biggest issue facing the RdSC ( * ) is the observed dark energy in our universe, coupled with the fact that realistic quintessence models seem to be at least as difficult to construct in string theory as metastable de Sitter vacua [64][65][66]. Furthermore, even hypothetical quintessence models that satisfy the RdSC bounds are on the verge of being ruled out experimentally [67][68][69]54].…”
Section: Quintessence Vs Cosmological Constantmentioning
confidence: 99%
“…Theoretically, one possibility is modifying postrecombination physics, such as the dark energy or modified gravity models, e.g. [9][10][11][12][13][14][15][16][17][18][19][20]. Such solutions are constrained tightly by late-time observations [1].…”
mentioning
confidence: 99%
“…We end up with truerightds102=leftr2L2ds2( Mink 4)+L2dr2r2left+0.16emL2true(dα2+cos2αdψ2+sin2α(dβ2+prefixsin2βds2false(S2false))true),rightF5=left(1+10) Vol 4dr4L4,1emF1=F3=B=0,1emeϕ=c2.So the only AdS5×S2 solutions in IIB are locally italicAdS5×S5 where here S 5 is parametrised as a foliation of S1×S3. That this is all we find is consistent with the result of []. We can also restrict the period of ψ to a fraction of 2π, [], so that we have an orbifold of S 5 , and the orbifold singularity is at α=π/2.…”
Section: Examplesmentioning
confidence: 99%