# Can one quantify the total luminosity of the cosmological horizon?

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?

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

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,

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

$$R = \sigma 4 \pi R^2_{\mathrm{ou}}(t) T^4_{\mathrm{ou}}(t)$$
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 $$t=0$$.
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: $$1/R^2_{\mathrm{ou}}\mathrm{(now)}$$.