Does the cosmic microwave background (CMB) put a upper bound on Universe radius and size?
We know the universe is filled with a near uniform radiation at a equilibrium temperature(T) of approximately 2.725K. It's a remnant of the Big Bang. My reasoning to propose this question is, we can try to associate a density to this cosmic microwave background, as $E = mc^2$ (Einstein's equation).
The tricky part is to estimate the amount of energy in a given volume of empty space. 764
I tried to get it as follows: Imagine a spherical blackbody. Try to picture a very small section of its surface, so its size is negligible when compared to its radius. at t = 0 we see its surface emitting radiation at the same temperature of the CMB. After a very small time interval, we have the following picture.
The energy flux from the surface of a ideal blackbody (emissivity = 1) is given by <PERSON>–Boltzmann law, $Flux = \sigma T^4$, so we can write for the energy prism in the sketch:
$$ \frac{E}{V} = \frac{\sigma T^4 A \Delta t}{A c \Delta t}$$ $$ \frac{E}{V} = \frac{\sigma T^4}{c}$$ $$ \frac{m c^2}{V} = \frac{\sigma T^4}{c}$$ $$ d = \frac{\sigma T^4}{c^3}$$
I think the last expression gives us the density, in terms of matter equivalence, of this prism filled with radiation at the same temperature of the CMB.
We can try to calculate the Schwarzschild radius of a sphere of this density and volume given by $\frac{4 \pi r^3}{3}$:
$$r = \frac{2 G M}{c^2}$$ $$r = \frac{2 G d V}{c^2}$$ $$r = \frac{2 G d}{c^2}\frac{4 \pi r^3}{3}$$ $$r = \frac{2 G}{c^2}\frac{4 \pi r^3}{3}*\frac{\sigma T^4}{c^3}$$ $$r = \sqrt \frac {3c^5}{8 \pi G \sigma T^4}$$
I substituted the constants, and the result is around $3.72\times10^{28}$ meters, or $3.93\times10^{12}$ light-years, nearly four trillion light-years, if I didn't get anthing wrong.
Would even a matterless universe with this size really collapse, by sheer weight of the microwave background, forming a blackhole made of light? 649