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Homework Help: Doppler shift of object in expanding universe

  1. Sep 16, 2013 #1
    1. The problem statement, all variables and given/known data
    What's the doppler shift (Δf) of the object moving away from us if we measure a wavelength of light λ=1.55μm emitted from it and it's at a distance of 10 megaparsecs? And is the object red or blue shifted?
    Added information to the problem is for each megaparsec the object is away from us, it's moving 74 km/s faster.

    2. Relevant equations

    For source and receiver receding from each other.

    For source and receiver approaching each other. (shouldn't need this one)

    β = [itex]\frac{v}{c}[/itex]

    3. The attempt at a solution

    I find frequency first, by using equation #1, and converting the micrometers of the wavelength to meters. I find it to be 5.1666667x10-15.
    Then I plug that into the second equation to find f0, which is 5.1794269x10-15

    f0 is larger than f. Does f0 stand for initial frequency? Well, the difference in frequencies is f0-f, which is 1.276x10-17

    So I assume that means the frequency was stretched? Which means it was redshifted?
    I guess if it was compressed, I would have received a negative number, and that would have meant it was blueshifted?

  2. jcsd
  3. Sep 17, 2013 #2


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    Staff: Mentor

    That number has missing units, but as a frequency it is certainly wrong. Is it the inverse value, if you add units?



    By the way: 10 MPc is too close to ignore the local motion of objects. But that is an issue of the problem statement, you can ignore it here.
  4. Sep 17, 2013 #3
    From the first equation, f ends up with 1/s. Hertz is the unit of frequency, and hertz is 1/s, right? And I just figured out one of the things I did wrong. How the heck I ended up with such a small number from equation 1, I have no idea.

    So f is actually 1.935x10^14 and f0 = 1.94x10^14

    Which means delta f = 4.78x10^11 s^-1

    Since since it's a positive number, that means it's redshifted? Did I fix my mistakes?

  5. Sep 17, 2013 #4


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    Staff: Mentor

    That looks more realistic (and those are the inverse values of the numbers in post 1)

    Right. You don't need the sign to know it is redshifted - it is moving away, so it is redshifted, done.
  6. Sep 17, 2013 #5
    Thanks. Yeah, I knew it was redshifted by the fact that it was moving away, but the way the question is worded, it makes it sound like I need to use the value I obtained to determine whether it was redshifted or blueshifted. Thanks again.
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