Integration of Planck's law to find photon density in space

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SUMMARY

The discussion focuses on calculating photon density in space using Planck's law and Wien's displacement law, specifically at a temperature of 3K. The integration of Planck's law yields a total energy density of approximately 9.76e-25 J/m³. By applying the formula λ_max*T = (hc)/4.9651*K, the peak wavelength is determined to be 9.68e-4 m, leading to a maximum frequency of 3.1e11 Hz. The resulting photon density is calculated to be 0.0047 protons/m³, which is comparable to the average density of hydrogen atoms in the universe.

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  • Understanding of Planck's law and its application in thermodynamics
  • Familiarity with Wien's displacement law for peak wavelength calculations
  • Basic knowledge of integration techniques in calculus
  • Concepts of photon density and energy density in physics
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Homework Statement


The universe is filled with EM radiation emanating from the Big Bang. This radiation was initially unimaginably hot but, as the universe has expanded, it has cooled to 3K. The distribution of the energy density of these photons in frequency (or wavelength) is given by the Planck formula. At this temperature, what is the wavelength of the photons at the peak of the Planck distribution? You can use the Planck formula to figure out how much total energy there is per unit volume by integrating over frequency. By dividing by hv and integrating over frequency, you can figure out how many photons there are per unit volume. Use your result to estimate how many Big Bang relic protons there are per cubic meter of intergalactic space. For comparison, the average density of hydrogen atoms in the universe is just one per cubic meter, you will find that there are a lot more protons.


Homework Equations


from Wien displacement law:
λ_max*T = (hc)/4.9651*K
Planck's law:
du = (8*pi/c^3)[hv/(exp(hv/kT)-1)]v^2*dv
where h = Planck's constant, k = Boltzmann's constant; c = speed of light,
v = frequency

The Attempt at a Solution


integrate Planck's law from 0 to infinity
replace hv/kT with x
and integrate from 0 to infinity
[(k^3*T^3)/(h^2)]*[8*pi/c^3][x^3/(exp(x)-1)]dx]
U = [(k^3*T^3)/(h^2)]*[8*pi/c^3](gamma(4)zeta(4))
= 9.76e-25 J/m^3
Use λ_max*T = (hc)/4.9651*K at T = 3K to find
λ_max = 9.68e-4m
v_max = c/λ_max = 3.1e11 Hz
photon density = U/hv = U/h*v_max (which is the highest distribution frequency for v)
= 0.0047 protons/m^3
But this isn't much larger than the number of hydrogen atoms per cubic meter, as the question says.
 
Physics news on Phys.org
Hi!

How did you get:

du = (8*pi/c^3)[hv/(exp(hv/kT)-1)]v^2*dv

?

This is very urgent please :)
 

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