Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Photon wavelength in expanding universe

  1. Apr 22, 2007 #1
    Let us say we have a universe permeated with a monochromatic photon gas of wavelength L0 at time t0 and the universe is expanding (say for simplicity with a constant rate H, Hubble's Constant). If I sit there and measure the wavelength of the photons as a function of time, what does it look like?

    From a dimensional analysis arguement I'd expect something like
    L = L0 (1 + H(t-t0))

    Is this correct? And how would I calculate this directly?

    Also, this suggests all frequencies are adjusted by just a multiplicative constant. If so, this seems to suggest a photon gas initially in a blackbody thermal distribution would not stay so (it's distribution would have a different form at a later time). But that can't be right, as (for instance) the CMB is a very nice blackbody distribution. So what am I doing wrong here?
    Last edited: Apr 22, 2007
  2. jcsd
  3. Apr 22, 2007 #2


    User Avatar
    Science Advisor

    Let a be the scale factor. In an expanding universe, a is a function of time, and the wavelength of photons is proportional to a.

    Also, the distance between galaxies "at rest" in the cosmos (co-moving) remains proportional to a. The value H is equal to (da/dt)/a.

    Here are several simple models for expansion.
    1. a = H0t (empty universe model)
    2. a = (3H0t/2)2/3 (flat matter filled universe)
    3. a = eH0t (inflationary universe)

    The real universe is a bit more complicated, and you need to solve differential equations for a. The parameters for the current simplest consensus model are the amount of matter, and dark energy, and curvature; but in all cases the wavelength scales by a factor, and so a blackbody spectrum remains a blackbody.

    Note that H varies with time. H0 is a fixed values, being the value of H in the present epoch, when a = 1. The time for the current epoch in these simple models is
    1. t0 = 1/H0
    2. t0 = 2/3H0
    3. t0 = 1/H0

    A photon "now" with wavelength L0 will in general have wavelength at time t of a*L0

    Solving for the three simple models
    1. L = H0L0t
    2. L = (3H0t/2)2/3L0
    3. L = L0eH0t

    Cheers -- Sylas
  4. Apr 22, 2007 #3
    Thank you so much for your response.

    I am still having trouble with this though:
    [tex]I(\lambda,T)=\frac{2hc^2}{\lambda^5}\frac{1}{\exp(\frac{hc}{\lambda k T}) -1}[/tex]

    Now if at a later time all wavelength's are scaled by the same factor, it doesn't seem to be the same form anymore.

    Let's say [tex]\lambda \rightarrow \alpha \lambda[/tex]

    [tex]I(\lambda,T)=\frac{2hc^2}{\lambda^5}\frac{1}{\alpha^5(\exp(\frac{hc}{\alpha \lambda k T}) -1)}[/tex]

    Now for it to have the same form as before, we should be able to write:

    [tex]\frac{1}{\alpha^5(\exp(\frac{hc}{\alpha \lambda k T}) -1)}=\frac{1}{\exp(\frac{hc}{\lambda k T'}) -1}[/tex]

    and we should be able to solve for T' as a constant (independent of [tex]\lambda[/tex]) otherwise it will no longer look like a blackbody distribution.

    Trying to solve for T' I get:
    [tex]T' = \frac{hc}{\lambda k \ln [1+\alpha^5(\exp(\frac{hc}{\alpha \lambda k T}) -1)]}[/tex]

    Which does not give a T' independent of [tex]\lambda[/tex]. So it appears to not stay a blackbody distribution.

    I'm obviously still doing something wrong, so if you could point out what I'm missing it would be appreciated.
    Last edited: Apr 22, 2007
  5. Apr 22, 2007 #4


    User Avatar
    Science Advisor

    I is the spectral radiance; energy per unit time per unit area per unit sold angle per unit wavelength. Your transformation is a calculation for energy of a particular group of shifted photons; not the new radiance function. To get the new radiance function you need to think in terms of a particular range of photons and track their new contribution to the radiance.

    A more useful term is the spectral energy density. The energy is coming to you from every corner of the sky ([itex]4\pi[/itex] steradians) and passing through a the speed of light. This means the spectral energy density expected if every part of the sky is a black body at temperature T is given by
    [tex]\mu(\lambda,T)=\frac{8{\pi}hc}{\lambda^5}\frac{1}{\exp(\frac{hc}{\lambda k T}) -1}[/tex]

    The units here are energy per unit volume per unit wavelength.

    Now... consider a small spectral range, wavelength range n to m. The energy of all the photons per unit volume in that range of wavelength is [itex]\int_n^m I(\lambda, T) d\lambda[/itex]

    What will their energy be after the scale factor changes by [itex]\alpha[/itex]?

    The wavelength is shifted by [itex]\alpha[/itex], so the photons are now in the range [itex]n\alpha[/itex] to [itex]m\alpha[/itex]. The density of those photons drops by [itex]\alpha^3[/itex] since they are within a larger volume. Each photon is less energetic by a factor [itex]\alpha[/itex]. Hence those photons now have an energy density of [itex]\alpha^{-4}\int_n^m I(\lambda, T) d\lambda[/itex].

    Substitute [itex]x = \lambda\alpha[/itex] as the integration variable

    The integral becomes [itex]\int_{n\alpha}^{m\alpha} I(x/\alpha, T) \alpha^{-1} dx[/itex], which simplifies to [itex]\alpha^4\int_{n\alpha}^{m\alpha} I(x, T/\alpha) dx[/itex]

    But recall that we were calculating this energy originally over the range n to m; and now we integrate those same photons over a shifted range, and so we do indeed have a perfect blackbody with temperature scaled down by [itex]\alpha[/itex].

    Cheers -- Sylas
    Last edited: Apr 22, 2007
  6. Apr 23, 2007 #5
    Thank you for your response. Yes, I can see now that I wasn't "deriving" the new spectral radiance correctly.

    While I can see your arguements and they make sense, something still feels missing to me. I'm having trouble expressing it concretely, but I've finally figured out a way to word part of what seems weird to me in a mathematical way. Maybe helping me see this will allow me clear my understanding better.

    Let us consider an observer moving at a constant speed with respect to the center of momentum frame of the photon gas. To this observer the photons reaching him from behind are redshifted (wavelength changed by a multiplicative factor as before). This should look like nothing more than a blackbody distribution from an object stationary in his frame (because that is what we see with the CMB). The doppler shift on the distribution should just make it look like the temperature of radiation coming from that direction dropped slightly. We can apply your arguements as before ...

    Except now we shouldn't be able to use the arguement "The density of those photons drops by [itex]\alpha^3[/itex] since they are within a larger volume."

    So it looks like a doppler shifted spectrum will no longer look like a blackbody spectrum ... but we know it must. So what is missing here?
    Last edited: Apr 23, 2007
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook