<p>I have been reading the paper by <a href="https://www.nature.com/articles/184844a0" rel="nofollow noreferrer">Cocconi and Morrison</a> on interstellar communication and I have trouble understanding their formulas.
The paper begins by stating, that a star similar to the quiet sun will emit power, which produces at a distance <span class="math-container">$R$</span> a flux of</p>
<p><span class="math-container">$$
10^{-15} \frac{f^2}{R^2}.
$$</span>
I think that the <span class="math-container">$f$</span> should be in the denominator, since the unit is given to be <span class="math-container">$\frac{W}{m^2 Hz}$</span>. Actually they use <span class="math-container">$\frac{c.}{s.}$</span> instead of <span class="math-container">$Hz$</span>.</p>
<p>They continue with the statement, that the galactic background yields received power:</p>
<p><span class="math-container">$$
\frac{10^{-12.5} }{f} \left(\frac{\lambda}{l_d}\right)^2 (l_d)^2
$$</span></p>
<p>where <span class="math-container">$l_d$</span> is the diameter of the radio telescope. The above formula yields a value, having the units <span class="math-container">$\frac{W}{Hz}$</span> They explain, that the <span class="math-container">$\left(\frac{\lambda}{l_d}\right)^2 (l_d)^2$</span> part, as being resolution times the area of the mirror (actually, shouldn't it be the radius, since the area is given by <span class="math-container">$\pi \left(\frac{l_d}{2}\right)^2$</span>). However why the resolution is squared?
They continue to say, that if at the source a mirror with diameter <span class="math-container">$l_s$</span> is used (I guess that says that the source's radiation is collimated), then the source power required for to generate in the detector a signal, as strong as the galactic background, at distance <span class="math-container">$R$</span> is:</p>
<p><span class="math-container">$$
\frac{d W_b}{d Sd\Omega d f} \left(\frac{\lambda}{l_s}\right)^2 \left(\frac{\lambda}{l_d}\right) R^2 =10^{-24.2} \frac{R^2}{l_s^2} l_d^2
$$</span></p>
<p>Above again the same factors representing the angular resolution (the second is linear) and a magic constant. I guess that the numeric constant is just some measurement value.</p>
<p>I shall really appreciate if someone would explain the above formulas.</p>

CC BY-SA 4.0
2023-05-06T19:10:00.687Z
A rather low quality scan of the paper http://www.astrosurf.com/luxorion/Documents/seti-cocconi-morrison-nature-1959.pdf
James K
CC BY-SA 4.0
2023-05-06T19:26:07.737Z
Thank you, for the link. This will help the readers of my question.
Alexander Cska
CC BY-SA 4.0
2023-05-06T19:57:25.280Z
By the way my reading of the first equation is $10^{-15} f^2/R^2\quad  \mathrm{W.m.^{-2}(c{.}/s{.})^{-2}}$
James K
CC BY-SA 4.0
2023-05-06T20:45:44.157Z
You are correct, i just checked the paper. Actually the unit is $Wm^{-2}(c./s.)^{-1}$. the last power is to the -1.
Alexander Cska
<radio-astronomy><radio-telescope>
Questions about the Cocconi Morrison paper on interstellar communication
Alexander Cska
# Questions about the Cocconi Morrison paper on interstellar communication

(Asked by: username_0 on 2023-05-06)

I have been reading the paper by [Cocconi and Morrison](https://www.nature.com/articles/184844a0) on interstellar communication and I have trouble understanding their formulas.
The paper begins by stating, that a star similar to the quiet sun will emit power, which produces at a distance $R$ a flux of

$$
10^{-15} \frac{f^2}{R^2}.
$$
I think that the $f$ should be in the denominator, since the unit is given to be $\frac{W}{m^2 Hz}$. Actually they use $\frac{c.}{s.}$ instead of $Hz$.

They continue with the statement, that the galactic background yields received power:

$$
\frac{10^{-12.5} }{f} \left(\frac{\lambda}{l\_d}\right)^2 (l\_d)^2
$$

where $l\_d$ is the diameter of the radio telescope. The above formula yields a value, having the units $\frac{W}{Hz}$ They explain, that the $\left(\frac{\lambda}{l\_d}\right)^2 (l\_d)^2$ part, as being resolution times the area of the mirror (actually, shouldn't it be the radius, since the area is given by $\pi \left(\frac{l\_d}{2}\right)^2$). However why the resolution is squared?
They continue to say, that if at the source a mirror with diameter $l\_s$ is used (I guess that says that the source's radiation is collimated), then the source power required for to generate in the detector a signal, as strong as the galactic background, at distance $R$ is:

$$
\frac{d W\_b}{d Sd\Omega d f} \left(\frac{\lambda}{l\_s}\right)^2 \left(\frac{\lambda}{l\_d}\right) R^2 =10^{-24.2} \frac{R^2}{l\_s^2} l\_d^2
$$

Above again the same factors representing the angular resolution (the second is linear) and a magic constant. I guess that the numeric constant is just some measurement value.

I shall really appreciate if someone would explain the above formulas.

**Tags:** radio-astronomy, radio-telescope

> **Comments:**
> **username_1**: A rather low quality scan of the paper http://www.astrosurf.com/luxorion/Documents/seti-cocconi-morrison-nature-1959.pdf  
> **username_0**: Thank you, for the link. This will help the readers of my question.  
> **username_1**: By the way my reading of the first equation is $10^{-15} f^2/R^2\quad  \mathrm{W.m.^{-2}(c{.}/s{.})^{-2}}$  
> **username_0**: You are correct, i just checked the paper. Actually the unit is $Wm^{-2}(c./s.)^{-1}$. the last power is to the -1.  

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## 0 answers

