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If I integrate all the power (radiation, matter particles, neutrinos) radiated from the cosmological horizon into the universe, what number do I get?

Is it true that the integral power/luminosity is of the order of $c^5/G = 3.6 \cdot 10^{52}$ Watt?

Has the integral horizon power been constant over the past evolution of the universe?

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First I will answer your first question, calculating a value for the radiated power.

I was unable to find any discussion of "power" in

https://en.wikipedia.org/wiki/Cosmological_horizon .

I think your concept has a flaw. When you calculate black body radiation power ($P$) in watts from an object, the object will have a surface area ($A$) and a temperature ($T$). The watts per unit area depends on temperature.

http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/stefan.html

$$P = \sigma A T^4$$

$\sigma$ is the Stephan-Boltzmann constant

https://en.wikipedia.org/wiki/Stefan%E2%80%93Boltzmann_constant

$$σ = 5.670374419 \times 10^8 \, \mathrm{ W/(m^2 K^4)}$$

I suggest you choose a value for the radius ($R_{\mathrm{ou}}$) of the observable universe and a value for the time of radiation, $t>0$. Then calculate a value for the scale factor value $a(t)$.

https://en.wikipedia.org/wiki/Observable_universe

$R_{\mathrm{ou}}\mathrm{(now)} = 46.5\, \mathrm{ Gly}$

$R_{\mathrm{ou}}\mathrm{(now)}$ is the current radius of the spherical surface that did the radiation at time $t$. You then need to calculate the radius $R_{\mathrm{ou}}(t)$.

$$R_{\mathrm{ou}}(t) = R_{\mathrm{ou}}\mathrm{(now)} / a(t).$$

The time t chosen should be well into the era when radiation mass equivalent density dominates matter (and also of course dark energy). Therefore,

$$T(t) = T(\mathrm{now}) / a(t)$$

Now calculate the power radiated.

$$R = \sigma 4 \pi R^2_{\mathrm{ou}}(t) T^4_{\mathrm{ou}}(t)$$

Now for the second question. I think this is just wrong. The radiation is watts per area, and there is no area in your calculation, presumably because $t=0$.

Now for the third question. I am not sure what you are intending to ask. There are two possible interpretations.

  1. Is the radiation from time $t$ still radiating? The answer is no. It stopped radiating immediately after time $t$. However, a little later, say at $t+dt$, there will be radiation observed at the same observation place (at the center of the observable universe sphere) at little later than "now".

  2. Is the radiation from a sphere at radius $R_{\mathrm{ou}}(t)$ detectable at the center of the sphere at all times since t, and with the same intensity? The answer is yes, it continues to be observable at all times since $t$, but the observed intensity changes with the time "now". For different times "now" the observed intensity depends upon how far the radiation travels to the observer, $R_{\mathrm{ou}}\mathrm{(now)}$. The observed intensity of radiation will be inversely proportion to the square of the distance: $1/R^2_{\mathrm{ou}}\mathrm{(now)}$.

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