<p>Short answer.</p>

<p>Great question! Ask yourself what absorbs better, something that is white, or black.Then ask yourself what radiates better. The answer is the same thing. It turns out that absorptivity and emissivity are linked pretty closely.</p>

<p><a href="https://en.wikipedia.org/wiki/Kirchhoff%27s_law_of_thermal_radiation" rel="nofollow noreferrer">Kirchov's Law of Thermal Radiation</a> can be stated as:</p>

<blockquote>
  <p>For an arbitrary body emitting and absorbing thermal radiation in thermodynamic equilibrium, the emissivity is equal to the absorptivity.</p>
</blockquote>

<p>If you know that the atmospheric absorption at a given wavelength is 0.01, then you <em>could estimate</em> the emissivity to be 0.01 as well. It doesn't mean it's exactly right, but it's a great start. </p>

<p>If you then say the average temperature of the atmosphere in the region where absorption (and thermal radiation) at 4080 MHz is 230K, then the <a href="https://en.wikipedia.org/wiki/Noise-equivalent_power" rel="nofollow noreferrer">Noise Equivalent Power</a> (NEP) will be 0.01 time 230 K or 2.3 K.</p>

<h2>How can we express power in terms of a temperature?</h2>

<p>Wait a minute, thermal radiated power varies as something like temperature to the <em>fourth power!</em> How can we add two temperatures together, or multiply them by 0.01?</p>

<p>It's because the energy of the photons we're talking about is way down on the left edge of Planck's blackbody distribution. <strong>See for example <a href="https://space.stackexchange.com/a/25744/12102">this answer</a></strong> to the question <em>Why doesn't thermal radio emission from a DSN “hot dish” completely swamp the benefits of a cold LNA?</em> Down at photon energies well below <span class="math-container">$k_B T$</span> we can use the <a href="https://en.wikipedia.org/wiki/Rayleigh%E2%80%93Jeans_law" rel="nofollow noreferrer">Rayleigh-Jeans law</a> where the power at a given frequency depends linearly on temperature.</p>

<p>Plotted on a log-log graph, you can see that the peak heights increase three times faster than the straight-line low-frequency segments do, showing the <span class="math-container">$T^3$</span> dependence of the spectral density peak height versus the linear behavior in the Rayleigh-Jeans regime. If you integrated the total area under the curve, you'd get the <span class="math-container">$T^4$</span> dependence of the total power radiated by a blackbody.</p>

<p><strong>below:</strong> From New Jersey Institute of Technology's Dr. Dale Gary's nicely-written Physics 728 Radio Astronomy; <a href="https://web.njit.edu/~gary/728/Lecture1.html" rel="nofollow noreferrer">Lecture #1</a> notes:</p>

<p><a href="https://i.sstatic.net/aZ7MW.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/aZ7MW.jpg" alt="enter image description here"></a></p>

CC BY-SA 4.0
2019-04-08T15:35:13.210Z
2019-04-09T06:16:18.810Z
2019-04-09T06:16:18.810Z
uhoh
<p>The classic paper <a href="http://adsabs.harvard.edu/full/1965ApJ...142..419P" rel="nofollow noreferrer">A MEASUREMENT OF EXCESS ANTENNA TEMPERATURE AT 4080 Mc/s</a> begins:</p>
<blockquote>
<p>Measurements of the effective zenith noise temperature of the 20-foot horn-reflector antenna (Crawford, Hogg, and Hunt 1961) at the Crawford Hill Laboratory, Holmdel, New Jersey, at 4080 Mc/s have yielded a value about 3.5° K higher than expected. This excess temperature is, within the limits of our observations, isotropic, unpolarized, and free from seasonal variations (July, 1964-April, 1965). A possible explanation for the observed excess noise temperature is the one given by Dicke, Peebles, Roll, and Wilkinson (1965) in a companion letter in this issue.</p>
<p><strong>The total antenna temperature measured at the zenith is 6.7° K of which 2.3° K is due to atmospheric absorption.</strong> The calculated contribution due to ohmic losses in the antenna and back-lobe response is 0.9° K.</p>
</blockquote>
<p>So, why would it increase? And, what is the significance of zenith in this sentence?</p>

<radio-astronomy><cosmic-microwave-background><radio-telescope><atmospheric-effects>
Why would atmospheric absorption increase rather than decrease the antenna temperature in Penzias and Wilson's famous CMB observation?
Youngsub Yoon
# Why would atmospheric absorption increase rather than decrease the antenna temperature in Penzias and Wilson's famous CMB observation?

(Asked by: username_0 on 2019-04-08)

The classic paper [A MEASUREMENT OF EXCESS ANTENNA TEMPERATURE AT 4080 Mc/s](http://adsabs.harvard.edu/full/1965ApJ...142..419P) begins:

> Measurements of the effective zenith noise temperature of the 20-foot horn-reflector antenna (Crawford, Hogg, and Hunt 1961) at the Crawford Hill Laboratory, Holmdel, New Jersey, at 4080 Mc/s have yielded a value about 3.5° K higher than expected. This excess temperature is, within the limits of our observations, isotropic, unpolarized, and free from seasonal variations (July, 1964-April, 1965). A possible explanation for the observed excess noise temperature is the one given by Dicke, Peebles, Roll, and Wilkinson (1965) in a companion letter in this issue.
>
> **The total antenna temperature measured at the zenith is 6.7° K of which 2.3° K is due to atmospheric absorption.** The calculated contribution due to ohmic losses in the antenna and back-lobe response is 0.9° K.

So, why would it increase? And, what is the significance of zenith in this sentence?

**Tags:** radio-astronomy, cosmic-microwave-background, radio-telescope, atmospheric-effects


---

## 1 answers

### Answer by username_1 on 2019-04-08. Score: 2 | Accepted answer

Short answer.

Great question! Ask yourself what absorbs better, something that is white, or black.Then ask yourself what radiates better. The answer is the same thing. It turns out that absorptivity and emissivity are linked pretty closely.

[Kirchov's Law of Thermal Radiation](https://en.wikipedia.org/wiki/Kirchhoff%27s_law_of_thermal_radiation) can be stated as:

> For an arbitrary body emitting and absorbing thermal radiation in thermodynamic equilibrium, the emissivity is equal to the absorptivity.

If you know that the atmospheric absorption at a given wavelength is 0.01, then you *could estimate* the emissivity to be 0.01 as well. It doesn't mean it's exactly right, but it's a great start.

If you then say the average temperature of the atmosphere in the region where absorption (and thermal radiation) at 4080 MHz is 230K, then the [Noise Equivalent Power](https://en.wikipedia.org/wiki/Noise-equivalent_power) (NEP) will be 0.01 time 230 K or 2.3 K.

## How can we express power in terms of a temperature?

Wait a minute, thermal radiated power varies as something like temperature to the *fourth power!* How can we add two temperatures together, or multiply them by 0.01?

It's because the energy of the photons we're talking about is way down on the left edge of Planck's blackbody distribution. **See for example [this answer](https://space.stackexchange.com/a/25744/12102)** to the question *Why doesn't thermal radio emission from a DSN “hot dish” completely swamp the benefits of a cold LNA?* Down at photon energies well below $k\_B T$ we can use the [Rayleigh-Jeans law](https://en.wikipedia.org/wiki/Rayleigh%E2%80%93Jeans_law) where the power at a given frequency depends linearly on temperature.

Plotted on a log-log graph, you can see that the peak heights increase three times faster than the straight-line low-frequency segments do, showing the $T^3$ dependence of the spectral density peak height versus the linear behavior in the Rayleigh-Jeans regime. If you integrated the total area under the curve, you'd get the $T^4$ dependence of the total power radiated by a blackbody.

**below:** From New Jersey Institute of Technology's Dr. Dale Gary's nicely-written Physics 728 Radio Astronomy; [Lecture #1](https://web.njit.edu/~gary/728/Lecture1.html) notes:

[![enter image description here](https://i.sstatic.net/aZ7MW.jpg)](https://i.sstatic.net/aZ7MW.jpg)


---

