Problem Set 10 - Problem 3

Energy density of blackbody spectrum:

energydensity  = 
      8 Pi h c/lambda^5 1/(Exp[h c/(lambda k T)]  - 1)

Energy of the individual photons is h c/lambda.  Therefore the number density is:

f  = energydensity/(h c/lambda)

The total number of photons is:

nphotons = Integrate[f, {lambda, 0, Infinity}][[2]]

Aside--The Zeta function is defined as Zeta[x]=Sum[n^x,{n=1,Infinity}]and looks like:

Plot[Zeta[x], {x, 0, 5}, PlotRange -> {-10, 10}]

In our case:

N[Zeta[3]]

The other constants in mks units are:

constants = {k -> 1.381*10^(-23), c -> 2.99*10^8, T -> 2.7, 
        h -> 6.626*10^(-34)}

Therefore, the total number of photons/m^3 is:

N[nphotons /. constants]

A huge number when compared to 1 proton/m^3.  The total energy density is:

energy  =  
      Integrate[energydensity, {lambda, 0, Infinity}][[2]]

And numerically (in units of joules/m^3):

N[energy /. constants]

The energy density due to the protons is m c^2:

protonenergydensity  =  
      1 *1.007825* 1.66*10^(-27) (3 10^8)^2

And the ratio of proton energy density to photon energy density is:

protondensity/N[energy /. constants]

Problem Set 10 Problem 4

Given the proton density from above, the mass needed to close the universe is 11 time greater:

needtoclose = 11 * protonenergydensity

The mass of an individual neutrino would be in kg/c^2:

massnu  =  needtoclose/N[nphotons /. constants]

Converting to eV/c^2:

massnu/(1.6 10^(-19))

Present laboratory limits on neutrino masses are:
	m nu_e < 15 eV/c^2
	m nu_mu < 170 keV/c^2
	m nu_tau < 24 MeV/c^2

Thus both the nu_mu and the nu_tau are candidates for objects to close the universe.