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Homework Help: Energy density of photons and matter

  1. Apr 27, 2010 #1
    hi

    im doing a question at the moment and am having issues understanding the question, and i cant ask my lecturer as he is stuck abroad with no internet.

    the question asks to calculate the ratio of current energy density of CMBR photons to that of baryonic matter.
    the present density of baryonic matter in the universe is pm,0=2.56x10-27kg/m3
    CMBR Temp = 2.725K

    ok so i calculated energy density of the CMBR = 4/c * [sigma]T4 = 0.417x10-13J/m3
    sigma=stefans constant
    i'm not quite sure how to turn the pm,0 of baryonic matter into an energy density. but i can see it needs to be multiplied by dimensions (m/s)^2

    thanks for any hints

    also what does pm,0 mean?
     
  2. jcsd
  3. Apr 27, 2010 #2
    The problem states what [tex] \rho_{m,0} [/tex] means:
    The present density of baryonic matter in the universe. This is the measured mass of baryons in our universe per cubic meter.
     
  4. Apr 27, 2010 #3
    Hint: Where does the matter get its energy from??
     
  5. Apr 27, 2010 #4
    i am thinking that i times the baryonic mass density by c^2
    this would give 1.44Gev/m^3
    so the ratio would be 0.261Mev/m3 / 1.44Gev/m^3 = 1.8125x10-4
     
  6. Apr 27, 2010 #5
    But why? I know the units work and that is what I got as well. Where does energy from mass come form?
     
  7. Apr 27, 2010 #6
    oh from the E=mc^2 relation of mass to energy.
     
  8. Apr 27, 2010 #7
    Yep and since energy density [tex] \epsilon = \frac{E}{V} [/tex]


    [tex] \epsilon = \frac{mc^2}{\frac{m}{\rho_0}} = \rho_0c^2 [/tex]
     
  9. Apr 27, 2010 #8
    ok thanks for that the reasoning behind that makes more sense now

    btw the next part says to calculate th redshift at which the energy density of matter = that of radiation (i.e. the CMBR photons)

    i tried:

    e = energy density

    e[rad] / e[matter] = 1 when equal and this is also proportional to a-4/a-3 where a is cosmological scale factor. and i am told a ~ (z+1)

    this would mean however that 1= 1/a = 1/(z+1) and so z=0
    i know this is wrong

    i think it should be ~3600 this value i found many times in my reading but didnt understand how it was found
     
  10. Apr 27, 2010 #9
    ok i just tried

    z+1 ~ (1.8x10-4)-1
    so z is ~5526

    i know this doesnt take into account 3 types of neutrinos
    do i times 1.8x10-4 by 1.68 to take them into account?
     
  11. Apr 27, 2010 #10
    [tex] \epsilon_m = \epsilon_{m,0}(1+z)^3 [/tex] and

    [tex] \epsilon_{rad} = \epsilon_{rad, 0}(1+z)^4 [/tex]
     
  12. Apr 27, 2010 #11
    ok so em0/erad0 = 1+z

    z= (1.8x10-4)^-1 - 1 = 5526.0066027

    this gives a temp of the CMBR => T=1/(z+1) = 1.809x10-4 K
    this doesnt seem right. surely temp should be larger than now?
     
  13. Apr 27, 2010 #12
    No, [tex]

    \frac{\epsilon_{rad}}{\epsilon_{m}} = \frac{\epsilon_{rad,0}}{\epsilon_{m,0}}(1+z) [/tex]
     
  14. Apr 27, 2010 #13
    but surely
    erad/em = 1 [cause im looking for equality]
    and so
    erad/em = 1 = erad0(z+1)/em0
    so
    erad0/em0 = 1/(z+1)
    therefore
    z+1 = em0/erad0
    which is what i put
     
  15. Apr 27, 2010 #14
    O right, my bad.
     
  16. Apr 27, 2010 #15
    thats ok, its just that it doesnt seem right.
    is it correct to get the temp by saying T~1/(z+1)?
    cause this just gives values i would have thought as being too small. or is this the overall temp of radiation and matter? and i only want the CMBR temp
     
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