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Let's find the temperature of universe for times between $$10^{-30}s-1000\;years$$ These times are bigger than inflation era but smaller than nonrelativistic matter dominance (50000 years) and smaller than light transparency era (400000 years). So, I can just use ultra relativistic assumptions: $$p=\dfrac{1}{3}\rho$$ $$\dot{\rho}+3\dfrac{\dot{a}}{a}(\rho+p)=0\Rightarrow \dfrac{d\rho}{p+\rho}=-3dln(a)\Rightarrow\rho=\dfrac{c}{a^4}$$ where a is scale factor $$\left(\dfrac{\dot{a}}{a}\right)^2=\dfrac{8\pi}{3}G\rho \Rightarrow a(t)=const\cdot t^{1/2}\Rightarrow \rho\sim 1/t^2$$ From the other side, for ultra relativistic matter I can write: $$ \rho_i=\frac{g_i}{2 \pi^2} \int \frac{E^3}{\mathrm{e}^{E / T} \mp 1} d E= \begin{cases}g_i \frac{\pi^2}{30} T^4 & - \text { bosons; } \\ \frac{7}{8} g_i \frac{\pi^2}{30} T^4 & -\text { fermions }\end{cases} $$ so $$\rho\sim 1/t^2\sim T^4\Rightarrow t\cdot T^2=known\;constant$$ is that true?

Aslan Monahov
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That works as long as the relativistic content of the universe remains constant.

More generally we can say that $$\rho(T) = \frac{\pi^2}{30}g_*(T)\, T^4,$$ where $$g_*(T) = \sum_{\text{bosons } i} \left(\frac{T_i}{T}\right)^4 g_i + \sum_{\text{fermions } i} \frac{7}{8}\left(\frac{T_i}{T}\right)^4 g_i.$$

See figure 1 of these notes for how $g_*$ evolves as a function of temperature, assuming only Standard Model relativistic content.

Moreover, if the relativistic content changes, then $a\propto T^{-1}$ (and hence $\rho\propto a^{-4}$) does not exactly hold. Instead we can say that entropy is conserved, so the entropy density, $$s=\frac{2\pi^2}{45}g_{*s}(T)\, T^3$$ with $$g_{*s}(T) = \sum_{\text{bosons } i} \left(\frac{T_i}{T}\right)^3 g_i + \sum_{\text{fermions } i} \frac{7}{8}\left(\frac{T_i}{T}\right)^3 g_i,$$ scales as $s\propto a^{-3}$. So for example, $$a(T) = \left(\frac{g_{*s}(T_0)}{g_{*s}(T)}\right)^{1/3}\frac{T_0}{T}a(T_0),$$ where $T_0$ is the temperature today (or at any other reference time). Figure 1 of the same notes also shows how $g_{*s}$ evolves as a function of temperature. You can also see this earlier question for further discussion of these points.

Now we can say that the time is $$t=\int_0^a \frac{da^\prime}{a^\prime H(a^\prime)}$$ (which follows from the definition of $H$) with $$H^2 = \frac{8\pi G}{3}\rho$$ and the relationships between $\rho$, $T$, and $a$ given above.

Sten
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    I am very happy that you answered me, you probably saved me from expulsion. So I just have to break my integral into several parts with different g in history? – Aslan Monahov Oct 07 '22 at 12:49
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    That should be a sufficient approximation. There is a tricky point around $g_{s}(T)$ being discontinuous when you want to change variables to $a$ (e.g. a jump in $g_{s}$ as a function of $T$ doesn't imply a jump in $g_{s}$ as a function of $a$, or alternatively the Jacobian $da/dT$ involves a delta function), but that probably won't matter much unless you're evaluating the time right after a phase transition (when $g_{s}$ jumps). – Sten Oct 07 '22 at 13:43
  • can I just write without integrating $$\rho=g\dfrac{\pi^2}{30}T^4;\quad \rho=\dfrac{3}{32Gt^2}$$ and so $$T^2t=\sqrt{\dfrac{3\cdot 30}{\pi^2 g\cdot 32G}}=f(g)$$ where g is constant for different time intervals? For ex: $$1GeV>T>100MeV: g(T)=61,75$$ $$100MeV>T>1MeV: g(T)=10,75$$ – Aslan Monahov Oct 07 '22 at 13:56
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    I suggest rethinking where $\rho=\frac{3}{32Gt^2}$ comes from. – Sten Oct 07 '22 at 15:21
  • Ok I probably understood it came from assumption that g=const that is not true so I have to integrate anyway – Aslan Monahov Oct 07 '22 at 16:09
  • but after integrating I am getting too big values to be real: $$\rho=g\dfrac{\pi^2}{30}T^4;\quad H^2=\dfrac{8\pi\rho G}{3}=\dfrac{8\pi^3}{90M_{pl}^2}gT^4\Rightarrow \dfrac{1}{H}=\dfrac{0,6 M_{pl}}{\sqrt{g}T^2}$$ $$t(T)=0,6M_{pl}\int_T^{\infty}\dfrac{dT}{T^3\sqrt{g}}$$ and for $M_{pl}=1,2\cdot 10^{28}eV$ and $T=10^{20}eV$ I am getting $t=10^{-13}s$ that's is a lot bigger than real $\sim10^{-30}s$ – Aslan Monahov Oct 07 '22 at 16:51
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    Neglecting $g_$, $g_{s}$, and the prefactors:

    https://www.wolframalpha.com/input?i=planck+massc%5E2+%2F+%2810%5E20+eV%29%5E2++%28planck+constant%2F%282pi%29%29

     – Sten Oct 07 '22 at 17:11
  • but in plank system $\hbar$ and $c$ equals to one anyway – Aslan Monahov Oct 07 '22 at 17:12
  • Sure but then you get the time as measured in inverse eV – Sten Oct 07 '22 at 17:14
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    amazing thanks! – Aslan Monahov Oct 07 '22 at 17:15
  • Let us continue this discussion in chat. – Aslan Monahov Oct 08 '22 at 02:11