3
$\begingroup$

In (Krauss & Starkman 1999) the authors critique (Dyson 1979), and in one section argues that in the very far future cooling will become impossible because thermal contact with the cosmic background radiation is lost:

As we have shown, the integrated number of CMB photons received over all time is finite. Therefore, after a certain time the probability of detecting another CBR photon, integrated over all of future history approaches zero. Thus, thermal contact with this background (and all other backgrounds) is lost.

Note also that in any case, the Dyson expression for dipole radiation, assumed above, clearly breaks down at some level, notably when the wavelength of thermal radiation becomes very large compared to the characteristic size of the radiating system. Put another way, the thermal energies will eventually become small compared to the characteristic quantized energy levels of the system, at which point radiation will by suppressed by a factor $\approx e^{-E_{char}/kT}$ compared to the estimate of Dyson. Once this occurs, further cooling will be difficult.

As far I can tell, the first paragraph is wrong. An object surrounded by 0K space will still radiate blackbody radiation as usual. That it does not get any photons back does not preclude emission. I think there is a slip here assuming there has to be reciprocal emission and absorption, but the body is not in thermal equilibrium with the space outside.

(The calculation that a surface has a finite number of received CMB photons over the future of an expanding universe model is fun to do; I get $\approx \frac{4\pi\zeta(3) }{3c^2H_0} \left(\frac{k_B T_0}{h}\right)^3 $ per square meter. However, horizon radiation in accelerated expansion makes the number diverge.)

The second paragraph on the other hand seems more reasonable. At $T_{dS}=10^{-30}$K $\lambda\approx 93$ Gpc. If one modifies the Stefan-Boltzman law to $P(T)=\sigma A T^4 e^{-E_{char}/k_BT}$ or Dyson's $T^3$ formula for a thin body similarly, there is a pretty firm cut-off below $T=E_{char}/k_B$. But as far as I can see it does not prevent radiating away energy, it is just slow. The equation $T'=-P(T)$ does have a solution that goes $T\rightarrow 0$ when $t\rightarrow \infty$. However, eventually we get to low enough temperatures that presumably one cannot integrate over a continuum of frequencies for the radiation law but instead have to do a sum over discrete energy levels. Is this the nub of the argument?

So my question is whether the two paragraphs are actually correct. Is thermal coupling to a very cold thermal background impossible, or just very weak?

$\endgroup$
0
1
$\begingroup$

The first quoted paragraph is a critique of a specific passage of Dyson's paper at the end of p. 454, beginning of p. 455 (with eq. 60), namely that CMB temperature would constrain a temperature scale of “life”. The passage does not imply that “life” could no longer dissipate heat, just that its temperature scale from some point in time would no longer be dependent on $T_\text{CMB}$ (even without cosmological constant).

Likewise, for the second quoted passage the phrase “further cooling will be difficult” does not mean that cooling would be impossible, just that Dyson's $T^3$ or even Stefan-Boltzmann $T^4$ laws would be replaced with a function behaving like $e^{-E_\text{char}/kT}$. Longer hibernation periods is a possible solution to this roadblock, but this would also make the fallibility of alarm clocks or life itself even sharper.

However, eventually we get to low enough temperatures that presumably one cannot integrate over a continuum of frequencies for the radiation law but instead have to do a sum over discrete energy levels. Is this the nub of the argument?

Yes.

In summary, both passages are addressing flaws in original Dyson's scenario without cosmological constant, making “eternal life” there even harder, presumably impossible.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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