You ignored it when I did that before...
Seriously. This is not a classroom and I am not a teacher. That being said your posts are fuzzy. I am sure it may be clear to you in your head but it does not come through to me . I have not seen you yet make a coherent persuasive argument.\\The math you posted here was problematic. You posted a mix of top level variables and the base dimensions, in capitals for some reason. Threw me for a bit. The use of Pascals to measure EM energy in space claiming it applied to a volume..
What I see is someone with a science background without a lot of experience problem solving in unfamiliar ground. We all do the same thing, a little research, try a solution. A little more research and a better solution. The 'scientific method' Even seemingly simple problems can take a lot of work to go all the way to runn8ing calculations and validating the model.
Nobody usually jumps to a solution without prior experience. Most EEs use incremental model building at some level of complexity. Start off small and validate as you go.
Take it for what you will. We all go through a practical learning curve in new situations. It is not a shortcoming, it is us human beings.
Indeed it's not a classroom, and I'm not your teacher either.
OK, let's try it stepwise. If the universe is a) infinite in space, b) does not have a (recent) beginning in time, c) does not expand, and d) the visible part of it is not an extreme outlier, then we can derive the following:
The 'local', as in averaged over galaxy clusters and void over several hundred megaparsec from home, luminosity density is about 190 Million solar luminosities per megaparsec cubed (
here or
here). For some reason, astronomers are behind the rest of the world when it comes to using SI units, and while this can easily be converted to Watts per cubic meter, I'll stick with their units for the time being.
Let's for the moment ignore our own galaxy and the Local Group and pretend we're isolated by a void of 100 Mpc from the nearest galaxy, and from their assume that average density quoted above, and calculate the power output and irradiance at earth per shell, starting at 100 Mpc from home and with a thickness of 1 Mpc per shell. In effect, we're thus pretending that our closest neighbors are at 100 Mpc from home; this can of course only lead to an underestimation of the irradiance.
The total volume of the first such shell is 4/3 * pi * (101Mpc)³ - 4/3 * pi * (100Mpc)³, i. e. the volume of the sphere circumscribed by its outer limit less the volume of the sphere circumscribed by its inner limit, as given by the volume of a sphere as 4/3 * pi * r³. This comes out as
126925 Mpc^3. This volume can equivalently be derived by using the formula for the surface of a sphere with r=the (logarithmic) average of its distance, times its thickness: The volume increases in proportion to the
square of its (average)
distance.
The total power output of that volume of space is given by multiplying it with the luminosity power density of 190 million suns/cubic Megaparsec: we get 2.4 * 10^13 solar luminosities or 9.28×10^39 W total output from that one shell. With a uniform distribution of galaxies, the power output increases with volume, and since volume increases with the square of average distance for shells of equal thickness, (absolute) power output per shell increases with the square of the distance.
We can derive a lower bound by the irradiance of the stars and galaxy within that shell at Earth's surface by pretending that they're all concentrated along its outer limit, i.e. 101 Mpc away, with your 1/(r²). The result is
just under 1nW per square meter. Since the (absolute) luminosity per shell increases with r² and the relative luminosity as a function of absolute luminosity decreases with r², we expect each shell to contribute a roughly equal amount of irradiation. But let's do an explicit check for this (next paragraph).
Time for a quick sanity check: let's replace all mentions of 100Mpc or 101Mpc
with 1000/1001Mpc to verify that the sum of the apparent luminosity / energy flux at earth's surface per shell indeed is constant: The result is actually slightly higher than before. That's expected because we're calculating a lower bound by pretending that the galaxies are concentrated along the outer rim of the shell, instead of being uniformly distributed. The error this introduces grows smaller with distance.
So, how much is one nanowatt per square meter? Not a whole lot.
Sunlight is up to 1050 W/m² on a cloudless noon in the tropics when the sun's at zenith. So we'd need over a trillion such shells to equal the irradiance of a tropical noon, and a couple orders of magnitude more to get a sky almost every corner of which that's glaring as bright as the average star. At 1Mpc per shell, that's just over a trillion megaparsecs or 3.553×10^18 ly - or around 38 million times the diameter of the visible universe. But you know what's the cool thing about an infinite universe? It barely starts at 3.553×10^18 ly radius.
Please tell me where you disagree, if you do!