<p>For the sake of clarity, I want to note that flux density and radiance are two different parameters. Radiance is the power per (projected) unit area per unit solid angle, and spectral radiance is the radiance per unit frequency; these are also known as the intensity and specific intensity, respectively. Flux density is the power per (projected) unit area per unit frequency. Flux density depends on the distance to the source, while radiance and spectral radiance are conserved along any particular path.</p>
<p>One way to see this is to consider that <a href="https://astronomy.stackexchange.com/q/48382/2153">specific intensity is conserved</a>. We then look at the following relation for flux density:
<span class="math-container">$$S_{\nu}=\int_{\mathrm{source}} I_{\nu}(\theta,\phi)\cos\theta\;\mathrm{d}\Omega$$</span>
where <span class="math-container">$I_{\nu}(\theta,\phi)$</span> is the specific intensity. Now, if a source is farther away, the solid angle it covers is smaller; therefore, we integrate over a smaller region of the sky. The flux density is then correspondingly smaller.</p>
<p>As an example, say we have a small source far away with constant specific intensity across a circle in the sky. In this case, we can write the flux density as
<span class="math-container">$$S_{\nu}\simeq I_{\nu}\Omega$$</span>
Using the small-angle approximation, the angular radius should scale approximately as <span class="math-container">$\theta\approx R/d$</span>, with <span class="math-container">$R$</span> the physical radius and <span class="math-container">$d$</span> the distance to the source. The solid angle scales as <span class="math-container">$\Omega\sim\theta^2\propto d^{-2}$</span>, so the flux density scales according to the inverse-square law, as you might expect.</p>
<p>As pela notes, this doesn't hold on cosmological scales, thanks to cosmological redshift decreasing photon energy and photon number per bin of frequency/wavelength.</p>

CC BY-SA 4.0
2022-04-04T12:48:15.933Z
Perhaps you should add that this is only true for "local" sources, but not for cosmological distances. Not only because redshift decreases the energy of a photon, but also because the number of photons per wavelength bin — i.e. the flux density — decreases as the spectrum is shifted toward longer wavelengths.
pela
CC BY-SA 4.0
2022-04-05T16:29:06.793Z
@pela Great point, thanks!
HDE 226868
CC BY-SA 4.0
2022-04-03T14:31:29.380Z
2022-04-05T16:28:57.950Z
2022-04-05T16:28:57.950Z
HDE 226868
<p>A particular observatory is providing the flux density of an astronomical source and only from this information, is it implicated that this particular parameter is unaffected by the distance between Earth and the source?</p>
<p>Edit: Removed a wikipedia link</p>

<observational-astronomy><radio-astronomy><flux>
Is the flux density independent of source-observer distance?
darthV
# Is the flux density independent of source-observer distance?

(Asked by: username_0 on 2022-04-03)

A particular observatory is providing the flux density of an astronomical source and only from this information, is it implicated that this particular parameter is unaffected by the distance between Earth and the source?

Edit: Removed a wikipedia link

**Tags:** observational-astronomy, radio-astronomy, flux

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## 1 answers

### Answer by username_1 on 2022-04-03. Score: 3

For the sake of clarity, I want to note that flux density and radiance are two different parameters. Radiance is the power per (projected) unit area per unit solid angle, and spectral radiance is the radiance per unit frequency; these are also known as the intensity and specific intensity, respectively. Flux density is the power per (projected) unit area per unit frequency. Flux density depends on the distance to the source, while radiance and spectral radiance are conserved along any particular path.

One way to see this is to consider that [specific intensity is conserved](https://astronomy.stackexchange.com/q/48382/2153). We then look at the following relation for flux density:
$$S\_{\nu}=\int\_{\mathrm{source}} I\_{\nu}(\theta,\phi)\cos\theta\;\mathrm{d}\Omega$$
where $I\_{\nu}(\theta,\phi)$ is the specific intensity. Now, if a source is farther away, the solid angle it covers is smaller; therefore, we integrate over a smaller region of the sky. The flux density is then correspondingly smaller.

As an example, say we have a small source far away with constant specific intensity across a circle in the sky. In this case, we can write the flux density as
$$S\_{\nu}\simeq I\_{\nu}\Omega$$
Using the small-angle approximation, the angular radius should scale approximately as $\theta\approx R/d$, with $R$ the physical radius and $d$ the distance to the source. The solid angle scales as $\Omega\sim\theta^2\propto d^{-2}$, so the flux density scales according to the inverse-square law, as you might expect.

As pela notes, this doesn't hold on cosmological scales, thanks to cosmological redshift decreasing photon energy and photon number per bin of frequency/wavelength.

> **Comments:**
> **username_2**: Perhaps you should add that this is only true for "local" sources, but not for cosmological distances. Not only because redshift decreases the energy of a photon, but also because the number of photons per wavelength bin — i.e. the flux density — decreases as the spectrum is shifted toward longer wavelengths.  
> **username_1**: @pela Great point, thanks!  

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