# [extropy-chat] Question for the Astronomy/Physics Ubergeeks

Hal Finney hal at finney.org
Sat Apr 2 19:34:48 UTC 2005

```The Avantguardian asks:
> Directional anisotropies aside, what is the average
> radiant flux density (i.e. watts/meter^2) of the CMB
> (cosmic microwave background)?

It's a little more complicated than it sounds.  Because the CMB radiation
is coming from all directions, it's as though we are immersed in a gas
of photons.  You can't derive power from this source like you could
with sunlight, because in the case of sunlight you are essentially
making use of the temperature difference between the sun and the CMB.
CMB temperature is like the effective zero temperature of the universe.
So although we can give an answer in Watts, this is not extractable power.

In practical terms, this impacts the answer by making you focus on whether
you mean the CMB to be coming from all directions; or from a hemisphere;
or from a patch of the sky.

CMB is black body radiation at 2.73 degrees Kelvin.  Black body radiation
has the nature that it does not matter how far away the emitter is.
The power from the CMB is identical to what we would get if we were
inside a sphere of any size which was held at that temperature.  There
is no "brighter" or "darker" when dealing with black body radiation.
All that matters is the temperature and the angular size of the emitter.

Black body power goes as the fourth power of the temperature.  Received
power will also be proportional to the angular size of the emitter.
This means that we can use the sun as a starting point and correct it to
earth from the sun, which is 1353 W/m^2.  We will reduce it by the fourth
power of the temperature ratio between the Sun and the CMB.  The Sun's
surface temperature is 5800 K and the CMB is 2.73 K.  The ratio is 2125
and the 4th power of that is 2.04E13.  Dividing the solar constant by
that gives 6.64E-11 W/m^2 which is the power we would receive from a
patch of CMB the same angular size as the sun.

I found a reference saying that the sun's angular size is about .001% of
the celestial hemisphere.  If you want to know the power form the CMB of a
hemisphere, like the CMB from the sky as seen on earth, we would therefore
increase the previous result by a factor of 10^4 and get 6.64E-7 W/m^2.
(Of course the radiation from earth would be far greater than that from
the CMB because the earth is much warmer.)  If you wanted the CMB power
from the entire celestial sphere, what you would see from space, you would
double this and get about 1.3E-6 W/m^2.

Hal

```