The previous weight of light blog piece was inspired by thinking about how big the mass of the cosmic background radiation is. As per my previous post an individual photon does not have mass but a collection of photons generally will. At any rate since the cmb is isotropically distributed that means it has a net zero linear momentum and so we can find its rest mass by the simple application of the E = mc^2 law. The cmb is basically a perfect blackbody distribution so we can use the relation u = 8pi^5t^4/(15h^3c^3) Where t is the temperature in joules. Since the cmb has a temperature of about 2.7 kelvin that translates to 3.7 x 10^-23 joules. That yields an energy density of about 4 x 10^-14 J/m^3 a quick google search provides a radius for the universe at roughly 78 billion light years. A light year = 9.4605284 × 10^15 meters so we get a volume of 4/3*Pi*r^3 = 4/3*Pi*(78000000000*9.4605 x 10^15)^3 = 1.6 x 10^81 m^3 So multiplying the two gives a value for the total energy content of the cmb which is 6.4 X 10^67 J. Translating this into Kg we get m = 7.1 x 10^50 Kg.

The mass density of visible matter in the universe is estimated at 3 x 10^-28 kg/m^3 so the total mass of the visible matter using the above given value for the volume of the visible universe gives a mass of 4.8 x 10^53 Kg. So the cosmic background radiation weighs in at about a thousandth of the mass of visible matter in the universe but still a not insignificant contribution to the overall mass. Anyway I thought it would be fun to post t3h calculation.

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