So the inflation is evoked to solve the horizon problem so that every point in the CMB was in causal contact. Does this then contract to the calculation without inflation of the angular size of the horizon at the last scattering which is about 1 deg? That 1 deg was used to say, on scales larger than that its fluctuation is due to initial condition because nothing was in causal contact. Is this no longer valid with inflation then? Does it mean on scales larger than 1 deg there should be "acoustic peaks"?

  • $\begingroup$ Inflation doesn't solve the horizon problem, it's a possible reason for what appears to be red shift. $\endgroup$ Dec 11, 2022 at 4:53

1 Answer 1


To study the causal structure it is useful to use the conformal time $\eta$, \begin{equation} ds^2=dt^2-a^2(t)dl^2=a^2(\eta) (d\eta^2-dl^2) \end{equation} that may be found as, \begin{equation} \eta=\int^t \frac{d\tilde{t}}{a(\tilde{t})} \end{equation} The reason is that in this coordinates the lightlike trajectories $ds^2=0$ corresponds to the diagonal lines $\eta=\pm x+\mathrm{const}$.

In the hot Big Bang scenario the early universe is filled with radiation that corresponds to the equation of state $w=1/3$ and the power law for the cosmological expansion $a~t^{1/2}$. The conformal time in this case has the finite span since the Big Bang $\eta>\eta_0$ where $\eta_0$ corresponds to $t\rightarrow 0$. Thus the lightlike trajectory moving into the past ends at some finite distance $\delta x=\eta-\eta_0$. This means that the causally connected region is finite and grows with time.

However for the classical solution at the inflationary stage $w<-\frac{1}{3}$ and $\eta$ is not bounded from below even if the cosmic time $t$ sonce the Big Bang $a=0$ were finite. Therefore, the casually connected region may be enormous (formally infinite for such solution) Instead you will have a finite range for $\eta$ in the future! This reflects the accelerated expansion of the universe that drives the distant points from each other faster then they may be connected by the light signal.

You may match these two solutions at some $t_{reh}$ - the stage when inflation ends, inflation decays and reheating occurs. Because this happened when $a$ was very small the notion "causally connected since the beginning of time within the hot Big Bang scenario" and "causally connected since the end of inflation" are very close to each other. The actual horizon as we see it in CMB actually corresponds to the latter.

So what happens in the inflationary scenario? The observable universe originated from the extremely tiny region of space that was more or less a vacuum. This erased any primal inhomogeneities such as acoustic peaks you want. It was expanded extremely fast, in fact so fast that different parts stopped to influence each other and it allowed to quantum fluctuations to grow and seed future inhomogeneities. The inhomogeneities produced this way are extremely simple: almost gaussian and almost scale-invariant (though not exactly) . Then the inflation stopped. Those different parts now could communicate with each other but this process takes time and this produces the horizon evident in CMB. Not true horizon from the beginning of time, but since the end of inflation.

  • $\begingroup$ Thank you OON. "Then the inflation stopped. Those different parts now could communicate with each other but this process takes time and this produces the horizon evident in CMB. Not true horizon from the beginning of time, but since the end of inflation." makes a lot of sense. Just to make sure I absolute get this right, if I want to calculate the "horizon since end of inflation at time of last scattering", I'd still use $d_{hor}(t_{ls}) = a(t_{ls}) c \int^{t_{ls}}_{t_f}\frac{dt}{a(t)} $, but now the lower time limit should be $t_f$ at the end of inflation , rather than t=0, is this right? $\endgroup$
    – ABC
    Dec 21, 2022 at 4:05
  • $\begingroup$ And because $t_f$ is so close to 0, the result from the calculation in the comment above will still be ~0.25Mpc -- the same as horizon $\bf /since\ the\ big\ bang/$ without inflation. And this is why evoking inflation does not mess up that 1 deg peak that separates initial condition from acoustic peaks. Is this correct? $\endgroup$
    – ABC
    Dec 21, 2022 at 4:10
  • $\begingroup$ Yep. everything is correct. Just to give you the sense of the order of things. The last scattering happen when temperature was ~0.5 eV. We know from the nuclei synthesis theory successes that at $T\sim 1 MeV$ the universe was in the radiation phase with the expected number of ultrarelativistic particle species. So the inflation ended much earlier. As in the radiation phase $a\sim 1/T$ it should only affect the flictuations at $l\gg 10^8$. There is no chance to extract anything meaningful not only because of noise but also because of the processes like the Silk effect washing everything out. $\endgroup$
    – OON
    Dec 21, 2022 at 11:13
  • $\begingroup$ Great thank you! Typical textbook just calculates the horizon $\bf {with}$ inflation $\bf {since\ the\ big\ bang\ t=0}$ and says at last scattering the horizon is about 200 Mpc enough to enclose every point in the sky, without explicitly discussing the "resetting" at the end of inflation, and that's why it was confusing to me. So were photons, protons etc. created /during/ the inflation? Or were they there before inflation, diluted during inflation and "re-generated" during reheating? $\endgroup$
    – ABC
    Dec 21, 2022 at 22:30

Your Answer

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

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