<p>I think you can understand that factor as follows:</p>

<p>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 <em>density</em>, which is really what <span class="math-container">$f_\lambda$</span> is, decreases by this factor.</p>

CC BY-SA 4.0
2020-03-04T18:56:26.300Z
2020-03-04T18:56:26.300Z
pela
<p>The flux F of an event at redshift <span class="math-container">$z$</span> is related to its luminosity L as
<span class="math-container">$$F=\dfrac{L}{4\pi d_L^2}\,,$$</span> 
where <span class="math-container">$d_L\equiv d_L(z)$</span> is the luminosity distance between us (the observer) and the event (the emitter).</p>

<p>We also know that a photon emitted at a wavelength <span class="math-container">$\lambda_{em}$</span> is observed at
<span class="math-container">$$\lambda_{obs}=(1+z)\lambda_{em}\,.$$</span></p>

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

<p>Although I understand that the factor <span class="math-container">$(1+z)^{-1}$</span> comes from the change of variable between emitted and observed wavelength, what is the <strong>physical</strong> meaning of it?
As an example, the <span class="math-container">$d_L^{-2}\propto (1+z)^{-2}$</span> 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.</p>

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2020-03-05T00:19:45.653Z
`+1` just for typing all of that MathJax, wish I could `+1` separately for the well-written question :-)
uhoh
<spectroscopy><spectra><mathematics>
What is the physical interpretation of the derivative of the emitted and observed wavelength?
Catarina Alves
# What is the physical interpretation of the derivative of the emitted and observed wavelength?

(Asked by: username_0 on 2020-03-04)

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.

**Tags:** spectroscopy, spectra, mathematics
> **Comments:**
> **username_1**: `+1` just for typing all of that MathJax, wish I could `+1` separately for the well-written question :-)  

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## 1 answers

### Answer by username_2 on 2020-03-04. Score: 5 | Accepted answer

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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