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.

  • $\begingroup$ Please give a reference to the paper rather than just the link. The link will eventually disappear. $\endgroup$
    – user4552
    Jun 30, 2019 at 2:14

1 Answer 1


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?

  • $\begingroup$ I realized, I have no yet clarified myself about the 0.8 Kor 0.9 K. How does inflation guess $N$? $\endgroup$
    – riemannium
    Mar 31, 2018 at 21:59
  • 1
    $\begingroup$ I think the logic is right, but just a comment on this: "the non-observation of a graviton black-body spectrum is a good thing since it hints at hidden particle states at "high enough" energy?" You can't read too much into the non-observation of a graviton black-body spectrum; our current gravitational-wave detectors are not remotely sensitive enough to place interesting upper limits on it, unfortunately. $\endgroup$
    – Andrew
    Aug 8, 2021 at 3:32

Your Answer

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

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