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During inflation, the de Sitter space has a cosmological event horizon. This horizon exists because farther than the horizon distance the expansion carries light away from the observer faster than the light can travel. This cosmological horizon emits Hawking radiation.

Can we estimate the temperature of this radiation during inflation? $~$Just want to know whether this temperature negligibly small or significantly large.

Note that inflationary expansion is thought to be $10^{53}$ times faster than current expansion.

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It is called the de Sitter temperature and is given by $T = H/2\pi$, where $H$ is the Hubble parameter. It arises from the autocorrelation of superhorizon quantum fluctuations generated during inflation.

EDIT: Below, I elaborate on the above showing how the autocorrelation of field fluctuations during inflation leads to a de Sitter temperature $T=H/2\pi$.

Begin with the inflaton, $\phi$, a minimally-coupled scalar field evolving in a FRW universe: $$\ddot{\phi} + 3H \dot{\phi} – \frac{\nabla^2}{a^2}\phi + \frac{{\rm d}V(\phi)}{{\rm d}\phi} = 0$$ and consider small perturbations about its homogeneous background value, $\phi({\bf x},t) = \phi_0(t) + \delta \phi({\bf x},t)$. To first order in $\delta \phi({\bf x},t)$, the above expression becomes $$\ddot{\delta \phi} + 3H\dot{\delta \phi} -\left(\frac{\nabla^2}{a^2} – \left.\frac{{\rm d}^2V(\phi)}{{\rm d}\phi^2}\right|_{\phi = \phi_0}\right)\delta \phi = 0.$$ Next, take the Fourier transform of the fluctuation in comoving wavenumber, $k$, $$ \delta \phi({\bf x},t) = \int\frac{{\rm d}^3k}{(2\pi)^{3/2}}\delta \phi_k(t)e^{i{\bf k}\cdot{\bf r}}, $$ and use it to obtain an expression for the Fourier modes $\delta \phi_k$, $$\ddot{\delta \phi_k} + 3H\dot{\delta \phi_k} + \left(\frac{k}{a}\right)^2\delta \phi_k = 0.$$ Lastly, we can rescale the field $u_k = a\delta \phi_k$ to arrive at the more compact mode equation, $$ u_k'' + \left(k^2 – \frac{a''}{a}\right)u_k = 0, $$ where primes denote derivative wrt conformal time, ${\rm d} \tau = {\rm d}t/a$. Specializing to the case of de Sitter space for which $a(t) \propto e^{Ht}$ and $\tau = -1/aH$ we can write the mode equation as $$(k\tau)^2\frac{{\rm d}^2u_k}{{\rm d}(k\tau)^2} + \left[(k\tau)^2 – 2\right]u_k = 0.$$ This can be solved exactly in terms of Hankel functions, $$u_k(-k\tau) = \frac{1}{2}\sqrt{-k \tau}\left[c_1 H^{(1)}_{3/2}(-k\tau) + c_2H^{(2)}_{3/2}(-k\tau)\right],$$ where $H^{(1)}_{3/2} = J_{3/2} +iY_{3/2} = H^{(2)*}_{3/2}$ and $J$ and $Y$ are Bessel functions of the first and second kind.

Now, let's figure out the constants. We can get one of them by appealing to a generic feature of quantum fields in expanding spacetimes: when the frequency is high relative to the expansion rate, the field doesn't "feel" the expansion and it oscillates as a plane wave. In the short wavelength limit $k/aH = -k\tau \rightarrow \infty$, Hankel functions indeed reduce to sinusoids, $$u_k(-k\tau) = \frac{1}{\sqrt{2k}}\left(c_1 e^{-ik\tau} + c_2 e^{ik\tau}\right).$$ To recover positive frequency plane waves, we choose $c_2 = 0$.

To figure out $c_1$, we make use of the fact that $\delta \phi({\bf x},t)$ is a quantum mode, and so it must satisfy the so-called canonical commutation relation with its conjugate momentum, $\pi({\bf x},t) = a^2\delta \phi({\bf x},t)’$. This is a straight-forward but tedious calculation involving a small collection of Bessel function identities: you should find at the end that $c_1 = \sqrt{\pi/k}$.

The final expression governing the time-dependence of a quantum mode in de Sitter space can now be written down, $$ u_k(-k\tau) =-\frac{\sqrt{\pi \tau}}{2}H^{(1)}_{3/2}(-k\tau)=-\frac{1}{\sqrt{2k}}\left(1 – \frac{i}{k\tau}\right)e^{-ik\tau}.$$ This is a wonderful result, particularly the second equality which results because Bessel functions of half-integer order are just combinations of trig functions. It's incredibly insightful: the fluctuation starts as a plane wave in the distant past when its wavelength is tiny ($-k\tau \rightarrow \infty$), but then as the mode is stretched by the expansion ($-k\tau \rightarrow 0$), it evolves out of the vacuum ultimately obtaining a constant amplitude on large scales (see figure). This is called mode freezing, and is due to the quantum decoherence of fluctuations on super-horizon scales. Illustration of mode evolution in de Sitter space

Specifically, in the large wavelength limit we can use small-argument approximations of the Bessel functions to see that $H^{(1)}_{3/2}(-k\tau)$ becomes the constant $\sqrt{2/\pi}(k\tau)^{-3/2}$ and $$ |\delta \phi_k| = \frac{|u_k|}{a} = \frac{H}{\sqrt{2k^3}}.$$ This is the quantity of interest, since it gives the amplitude of the fluctuation on large scales where it is a real, classical perturbation.

The correlation function is then obtained by ensemble averaging, $$ \langle | \delta \phi_k|^2\rangle = \frac{H^4}{2k^3}. $$ To get a quantity independent of scale, it is common practice to multiply by $k^3/2\pi^2$ to get the power spectrum, $$ P(k) = \frac{H^2}{4\pi^2}. $$ This has dimensions of energy-squared, and so its square root corresponds to a temperature, the Gibbons-Hawking temperature, of $T = H/2\pi$.

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  • $\begingroup$ Is the Hubble radius during inflationary epoch smaller or larger than the current epoch? $\endgroup$ – Forge Nov 3 '18 at 21:25
  • $\begingroup$ Smaller. As long as $\dot{\rho}\geq 0$, where $\rho$ is the density, $H$ is a monotonically decreasing function of time. $\endgroup$ – bapowell Nov 4 '18 at 1:02
  • $\begingroup$ The Hubble radius was smaller during inflationary epoch than the current epoch. So that means the de Sitter temperature during inflation is smaller than the de Sitter temperature of the current epoch? $\endgroup$ – Forge Nov 4 '18 at 11:45
  • $\begingroup$ No. The temperature is proportional to $H$, not $1/H$. $\endgroup$ – bapowell Nov 4 '18 at 12:39
  • $\begingroup$ OK. So that means the de Sitter temperature during inflation is larger than the de Sitter temperature of the current epoch, right? $~$If so, can we roughly estimate this temperature during inflation? $\endgroup$ – Forge Nov 4 '18 at 15:08

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