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Astrophysics - How much CMB-photons do you absorb per second

  1. Apr 24, 2008 #1
    1. The problem statement, all variables and given/known data
    1) Calculate approx. how many Cosmic Microwave Background-photons that you would absorb per second if you were in space.

    3. The attempt at a solution
    Ok, I know that the CMB-photons fit with a blackbody spectrum with T = 2.725 K. So the total amount of photons I find by using

    n = beta * T^3, where beta = 2.03*10^7 m^{-3} * K^{-3}.

    This is the amount of photons per cubicmeter, which is also the amount that would pass through my body (assumed spherical) per second.

    Am I correct?
     
    Last edited: Apr 24, 2008
  2. jcsd
  3. Apr 24, 2008 #2
    No suggestions?
     
  4. Apr 24, 2008 #3
    Are you not given any more details?

    Edit:

    I misread,

    If you use the Stefan-Boltzmann law to find the corresponding energy flux density, you can then find the energy/s incident on the surface. It's then simple to convert this energy to photons/s.
     
    Last edited: Apr 24, 2008
  5. Apr 24, 2008 #4
    The Stefan-Boltzmann law

    \epsilon_{gamma} = alfa * T^4.

    The temperature I use is T = 2.750 K. Then I get a number with units J/m^3. How do I get from this to a time-unit involving s?
     
  6. Apr 24, 2008 #5
    It'll give you a number with units: [tex]{Js^{-1}m^{-2}}[/tex]

    So you'll need to multiply by the surface area of body absorbing the energy to give you the total amount of energy absorbed per second. Dividing by the energy of the photon will then give you the number of photons/s absorbed.
     
  7. Apr 24, 2008 #6
    Hmm, in my book the constant alfa = 7.56*10^-16 J * m^-3 * K^-4?
     
  8. Apr 24, 2008 #7
  9. Apr 24, 2008 #8

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    Where did you get this formula? Show it with h, c etc. so that we can understand. (At least, I don't have this ready made value.)

    Why do you have to be spherical? Are you very fat?

    EDIT: After I submitted, I see that astrorob has given a link. But you need the number density, not the energy density. You have to divide E(ν)dν by hν before integrating.
     
    Last edited: Apr 24, 2008
  10. Apr 25, 2008 #9
    Ehh, now I'm confused.

    In my book I have the energy density of photons, which when integrated gives E = alfa*T^4. But this is not correct, since we want the number density of the photons. So I divide with the energy of one photon, and then I integrate?

    EDIT: The equation in my book is the one on the bottom of this page: http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/raddens.html#c1

    And the thing about me possibly being fat (which I'm actually not.. I'm quite skinny) - I guess that doesn't matter. All I need is my area, right?
     
    Last edited: Apr 25, 2008
  11. Apr 25, 2008 #10

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    That sounds reasonable, doesn't it? In fact, the energy density per unit frequency width was obtained by multiplying the number density per unit frequency width by [itex]h\nu[/itex].

    Yes, then you'll have to find the number of photons incident on your body per m² per sec.
     
  12. Apr 26, 2008 #11
    Ok, I found an expression for the number density of photons in blackbody radiation:

    n = beta * T^3, where

    beta = 2.03 *10^7 m^-3 * K^-3.

    But again, I don't get seconds from anywhere?
     
  13. Apr 27, 2008 #12

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    Sorry for the delay in replying. Since you have solved it in the mean time, I hope you got the value as 3*10^16/m²/s.
     
  14. Apr 30, 2008 #13
    I did.

    Thanks for helping! :-)
     
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