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Temperature of Universe

  1. Jun 25, 2016 #1
    1. The problem statement, all variables and given/known data
    Two scientists detected the cosmic microwave background radiation at a frequency of 160 GHz. What is the temperature of the universe?

    2. Relevant equations
    peak wavelength x temperature = 2.898 x 10^-3
    c = f x wavelength

    3. The attempt at a solution
    I calculated the wavelength of the microwave radiation to be 1.875x10^-3m. So the temperature is 1.5456K.
    But the answer is 3K... How come??
     
  2. jcsd
  3. Jun 25, 2016 #2
    The cosmic microwave background radiation has a thermal black body spectrum at a temperature of 2.72548±0.00057 K.
    The spectral radiance dEν/dν peaks at 160.23 GHz.

    see detail at
    https://en.wikipedia.org/wiki/Cosmic_microwave_background
     
  4. Jun 25, 2016 #3

    phyzguy

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    Science Advisor

    The peak of the wavelength distribution is not at the same place as the peak of the frequency distribution, because the blackbody spectrum is not linear. So you cannot say: λpeak = c / νpeak. Try reading this link. Or try taking the blackbody distribution I(λ) and differentiate to find the peak, then compare it to the peak of I(ν). You will then see why they are different.
     
  5. Jun 25, 2016 #4
    Thanks for replying!
    From the link you gave me, "However the form of the law remains the same: the peak wavelength is inversely proportional to temperature (or the peak frequency is directly proportional to temperature)." Why couldn't I use the v=fλ to find peak frequency?
     
  6. Jun 25, 2016 #5

    phyzguy

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    Science Advisor

    It's true that, "the peak wavelength is inversely proportional to temperature (or the peak frequency is directly proportional to temperature)". But the proportionality constants are different and the peaks occur in different places, so the peak frequency is not related to the peak wavelength by λpeak = c / νpeak. The reason is that the differential dν is not linearly related to the differential dλ. I'm going to repeat what I said before. Try finding the peaks of the two distributions and you will see why λpeak = c / νpeak does not work.
     
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