1

I was having a conversation in another physics forum about horizons (like the event horizon of a black hole, or a cosmological horizon) emitting Hawking radiation and I mentioned that if the universe keeps an accelerating expansion there will be a radiating cosmological hoizon (as the universe approaches a de Sitter space) therefore culminating in an asymptotic non-zero temperature.

However, he wasn't sure about that, as Hawking radiation could not be measured, therefore we would have in practice a 0K temperature for the universe.

Specifically he said:

I don’t see how. The cosmological horizon is not a frame that you can boost to so I don’t see how you’d ever see particles from that perspective.

Then I found this stack exchange question (Radiation from cosmological horizons) about a similar topic, and one of the answers says that Hawking radiation from the horizon will probably be undetected (but for practical reasons, in principle you could measure it). I linked him to this and he replied:

You’re not going to get any sort of Hawking radiation because there’s no frame we could occupy to observe it. The Stack Exchange commentator is pointing out that what I’m saying is true is practice but not necessarily in principle.

However, I always thought (and I have the impression that the stack exchange question is implying) that we could never detect Hawking radiation as it would have a very low "temperature", therefore it would be difficult to detect these very low energy photons. But I never heard that we could never detect it not by a practical technological limitation, but because an observer-frame problem

So, is this guy right? Am I missing something?

Níck Aguiar Alves
  • 24,192
vengaq
  • 3,282

2 Answers2

2

I'll consider only the cosmological scenario, and this seems to be your focus and Ak2797 has already addressed the black hole case. For simplicity, I'll consider what happens in de Sitter universe, which is what our universe asymptotically tends to anyway.

In de Sitter universe, the prediction is that any inertial observer will measure a Hawking-like temperature given (in units with $c = G = \hbar = k_B = 1$) by $$T = \frac{1}{2\pi}\sqrt{\frac{\Lambda}{3}},$$ where $\Lambda$ is the cosmological constant. This expression can be found, for example, in Wald's Quantum Field Theory in Curved Spacetime and Black Hole Thermodynamics, Eq. (5.3.5), and I believe it is originally due to Gibbons and Hawking. Notice that there is no obstruction due to "observer-frame" problems. I believe your friend is trying to apply the general ideas one gets from black holes and the Unruh effect to de Sitter spacetime, when the adaptation is not exactly that straightforward.

To compute how easy it is to detect this temperature, let us crunch some numbers. First, we need to restore the units in our expression. Using that $\Lambda \approx 2.85 \times 10^{-122} M_{P}^2$ ($M_P$ is the Planck mass and the value is taken from the Planck 2018 results) $$T = \frac{1}{2\pi k_B}\sqrt{\frac{2.85 \times 10^{-122} M_{P}^2}{3 M_P^2}} E_P,$$ where $E_P$ is the Planck energy. This leads to $$T = \frac{1}{2\pi k_B}\sqrt{\frac{2.85 \times 10^{-122}}{3}} \sqrt{\frac{\hbar c^5}{G}}.$$

If we use the CODATA values for $G, \hbar, c$, and $k_B$, we get to $$T \approx 2.198 \cdot 10^{-30} \,\mathrm{K}.$$ Not only is this ridiculously small by itself, but the cosmic microwave background already has a temperature of about $2.7 \,\mathrm{K}$, which is $30$ orders of magnitude larger. Hence, in practice, I guess it is safe to say it is not experimentally feasible to measure the temperature induced by the cosmological horizons.

Níck Aguiar Alves
  • 24,192
1

You are absolutely correct. Here is a simple example. Hawking evaporation from a black hole produces all sorts of Standard Model particles with a near-blackbody spectra. The peak of this spectra is inversely proportional to the mass of the black hole. For a black hole of mass 3 Solar Mass, the temperature will be around 10$^{-12}$ eV. We obviously have not detected such a low energy photon ever! But there's more to the story. You can calculate the rate of hawking evaporation for your 3 Solar Mass black hole and what you will get is an extremely small number - effectively these black holes are not evaporating. That is the reason why Hawking evaporation is not possible to detect.

Now there is a twist in the story. As you can imagine from the earlier discussion, if we make the black hole mass extremely small then both the temperature and the evaporation rates will drastically increase. For a black hole of 10$^{-17}$ Solar mass, the temperature becomes around 1 MeV. A photon of this much energy we can obviously detect! But how can an astrophysical black hole have that low mass? Actually these are not astrophysical black holes, but rather Primordial Black Holes (PBHs). They can be formed at a very early stage of the Universe through various different mechanisms. Currently they are one of the widely discussed Dark Matter candidates. And the funny thing is that, the only way to detect these low-mass PBHs, is to detect their Hawking emission. You can read more about evaporating PBHs in this nice review: Primordial black hole constraints with Hawking radiation -- a review.

So there is a chance that we may detect Hawking evaporation! Hope this helps.

Ak2797
  • 51