CMB Flux Density: Deriving For Present Cosmology

Nabeshin
Science Advisor
Messages
2,207
Reaction score
16
Does anyone know where I can find numbers (or how to derive) the CMB flux density (W/m^2)? I'm only really interested in our present cosmological time, so a solution may assume the CMB to be at a constant temperature.
 
Last edited:
Space news on Phys.org
Nabeshin said:
Does anyone know where I can find numbers (or how to derive) the CMB flux density (W/m^2)? I'm only really interested in our present cosmological time, so a solution may assume the CMB to be at a constant temperature.
Well, the CMB is almost a perfect black body as T=2.725K. So you can compute it directly from the black body spectrum (Planck's Law):
http://en.wikipedia.org/wiki/Black_body#Planck.27s_law_of_black-body_radiation
 
First, using Planck's law would give the energy radiated per unit surface area at that temperature, but that's per unit surface area of the radiating body, isn't it? In which case the surface area is... the entire universe?

What I'm interested in is if you have, say, a 1 m^2 surface, how much energy does it absorb from the cmb?

It's relatively easy to do this for a star or a single radiating black body, like a star, by computing total energy radiated and then spreading it evenly over a sphere of radius r. I don't really know how to extend this to the cmb though...
 
Nabeshin said:
First, using Planck's law would give the energy radiated per unit surface area at that temperature,
Also the energy absorbed per unit surface area. This works for the CMB because it's isotropic (as opposed to the light from a star which only comes from a small area of the sky). So you don't multiply that value by any area to get the flux density of the CMB.
 
So just A \sigma T^4 should work for total power absorbed, then? Interesting that it should turn out to be so simple!
 
Nabeshin said:
So just A \sigma T^4 should work for total power absorbed, then? Interesting that it should turn out to be so simple!
Yup. Just bear in mind that "A" there would be dependent upon how you are doing the measurement, and would typically be the area of the beam of the detector on a telescope.
 

Similar threads

Back
Top