The flux F of an event at redshift $z$ is related to its luminosity L as $$F=\dfrac{L}{4\pi d_L^2}\,,$$ where $d_L\equiv d_L(z)$ is the luminosity distance between us (the observer) and the event (the emitter).

We also know that a photon emitted at a wavelength $\lambda_{em}$ is observed at $$\lambda_{obs}=(1+z)\lambda_{em}\,.$$

We might want to know how does the observed spectra of the event $f_{\lambda}$ is related to the spectral energy distribution (SED) of the source $l_{\lambda}$. We can then deduce that $$\dfrac{dF}{d\lambda_{obs}}=\dfrac{1}{4\pi d_L^2}\dfrac{dL}{d\lambda_{obs}}=\dfrac{1}{4\pi d_L^2}\dfrac{dL}{d\lambda_{em}}\dfrac{d\lambda_{em}}{d\lambda_{obs}}$$ $$f_{\lambda}(\lambda)=\dfrac{1}{4\pi d_L^2}l_{\lambda}\left(\dfrac{\lambda}{1+z}\right)\dfrac{1}{1+z}$$ $$f_{\lambda}(\lambda)=\dfrac{1}{4\pi d_L^2 (1+z)}l_{\lambda}\left(\dfrac{\lambda}{1+z}\right)$$

Although I understand that the factor $(1+z)^{-1}$ comes from the change of variable between emitted and observed wavelength, what is the physical meaning of it? As an example, the $d_L^{-2}\propto (1+z)^{-2}$ and those two factors arises from the photons losing energy because of its wavelength being stretched and arriving further apart because the space between them grew.

  • 3
    $\begingroup$ +1 just for typing all of that MathJax, wish I could +1 separately for the well-written question :-) $\endgroup$
    – uhoh
    Mar 5, 2020 at 0:19

1 Answer 1


I think you can understand that factor as follows:

When photons lose energy, they're spread out over a larger wavelength range. Since there is a fixed number of photons, that means that the number of photons per observed wavelength bin decreases by a factor of (1+z), and hence the flux density, which is really what $f_\lambda$ is, decreases by this factor.


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