# How much uncertainty has the relic graviton background?

In the paper [1], it is mentioned that inflation predicts that a relic graviton background is about 0.9 K (cf. cosmic neutrino background, 1.945 K, and cosmic microwave background, 2.73 K). How much uncertainty has that value(0.9K)? After all, in other papers I have seen a relic temperature about $$10^{-29}K=10 \mu y K$$, where $$\mu y K$$ is microyocto kelvin degrees. But I am yet confused, and I believe the latter is just the fluctuation temperature in CMB due to primordial GW, aka, relic gravitons, if they are the same (as I believe but I am not yet 100% sure of terminology here).

[1] Turner, M.S., and Wilczek, F. Relic gravitational waves and extended inflation. United States: N. p., 1990. Web. doi:10.1103/PhysRevLett.65.3080.

• Please give a reference to the paper rather than just the link. The link will eventually disappear. – Ben Crowell Jun 30 at 2:14

I think I got it. According to the paper 1968ApJ.The Astrophysical Journal,Vol. 154,December 1968. NOTES ON THE PRESENT TEMPERATURE OF PRIMORDIAL BLACK-BODY GRAVITATIONAL RADIATION, the temperature of the graviton black-body background should be $$T_g(BB, now)=T_\gamma (CMB,now) \left(\dfrac{2}{N}\right)^{1/3}$$ and where $$N$$ is the number of modes available at the time of decoupling from gravity. Weinberg, in his book on Cosmology and Gravitation, also writes this formula and gives a differente estimate of N. In the above paper, $$N\sim 13$$, while Weinberg gives $$N\sim 26$$. Accordingly to these references, the uncertainty comes from:

1. The absence of any equation of state for those temperatures and densities.
2. The uncertainty between the number of particle species (fermions counted as 7/8 of a particle). The known SM gives (counting particles and antiparticles): photon, W⁺, W⁻, Z, gluon (5 d.o.f.); (7/8)(24), since we have 24 fermions, counted with different antiparticles. Here I would highlight that if neutrino are identical to antineutrinos (if Majorana!), instead it would be (7/8)(21). Thus, assuming only SM, we would obtain: i) $$T_g\sim 1.16K$$, if neutrinos are Dirac (different to antineutrinos), and Weinberg value. ii) $$T_g\sim 1.43K$$, if neutrinos are Dirac and Alpher value. iii) $$T_g\sim 1.2K$$ IF neutrinos are Majorana, using Weinberg values.
3. Inflation. I presume that the presence of inflation (scalar fields or any other field, e.g., a 3-form, 2-form, 1-form fields or general...) alters the number $$N$$ AFTER inflation, so it could dilute completely the black-body graviton spectrum.

I think that is what I wanted to understand...Indeed, after reading all these papers...Despite this answer, am I right if I say that the non-observation of a graviton black-body spectrum is a good thing since it hints at hidden particle states at "high enough" energy?

Extra comment: using 0.8K gives about 79 (80) for N, and using 0.9K you get N=56. I wonder the meaning of this for particle physics. But, understanding the previous readings, it points out to more particle species if a low T graviton background is detected! Is this right? Is this the reason because this background is important (similarly to the neutrino cosmic background that can smell new -sterile- particle species)?

Addendum: for 10 microyoctokelvin, we get $$N\sim 4\times 10^{88}$$. what kind of theory predicts such a big number of degrees of freedom?

• I realized, I have no yet clarified myself about the 0.8 Kor 0.9 K. How does inflation guess $N$? – riemannium Mar 31 '18 at 21:59