# Can a telescope detect a blackbody?

I'd like to determine the distance at which a telescope with a certain set of parameters (sensitivity, collector area) can detect a blackbody with certain other parameters (temperature, emitting area).

For example, let us suppose we have an infrared telescope with a sensitivity of '1Jy/sr' and a collector area of 10m^2 at a distance D from a 300m^2 perfectly black body at a temperature of 30 Kelvin.

How do I determine D? What additional information do I need to perform the calculation?

• You need to specify the geometry of the emitter and it's geometric relationship with the receiver. – Rob Jeffries Apr 3 '18 at 6:22
• I was going to suggest calculating the telescope parameters to observe gravitational lensing, which indirectly might identify a black hole. So much for reading comprehension. – Carl Witthoft Apr 3 '18 at 15:46

Let's assume a uniform laminar emitter, oriented so that it is at right angles to the line to the detector.

The specific intensity emitted by a blackbody, at the surface of the blackbody is $$B_{\nu} = \frac{2h\nu^3}{c^2} \frac{1}{\exp[h\nu/kT] -1}\ {\rm Wm}^{-2}{\rm Hz}^{-1},$$ where $B_{\nu}$ is the "Planck function".

Using the definition of specific intensity we can calculate the flux (energy per unit time, per unit frequency) received at the detector as $$f_{\nu} \simeq B_{\nu}A \Delta \Omega \cos \theta\,$$ where $\Delta \Omega$ is the solid angle of the detector subtended at the laminar surface, $A$ is the area of the laminar surface, and $\theta$ is the angle with respect to a normal from the surface and here, $\cos \theta \simeq 1$.

The solid angle subtended by the detector at the surface is $\Delta \Omega = a/D^2$, where $a$ is the area of the detector, but we can also define a solid angle $\Delta \omega = A/D^2$, which is the solid angle of the surface at the detector. Thus $$\frac{f_{\nu}}{\Delta \omega} = B_{\nu}a$$

If the detector has a fixed sensitivity per unit solid angle then unless the object is unresolved, then it does not matter how far away it is; the value of $D$ does not come into it. It is a well-known fact in astronomy that resolved, extended objects have an observed surface brightness (flux density per unit solid angle) that is independent of distance.

• I find this rather hard to understand. According to that relationship, I read that a 30K surface of any size at any distance can be detected by telescope with a very low sensitivity of 3.3 million MJy/sr...?! – Matterbeam Apr 4 '18 at 22:07
• Read carefully @user140629 - it says that "resolved, extended objects have an observed surface brightness that is independent of distance". If you want the answer for an unresolved "point" source then more information is required, or the sensitivity of your telescope should be given in Jy. – Rob Jeffries Apr 4 '18 at 23:07
• OK, what if we used the data from a document such as this one (arxiv.org/ftp/arxiv/papers/0707/0707.0883.pdf) detailing the predicted performance of the 10m SAFIR telescope? How could we use that data? – Matterbeam Apr 5 '18 at 0:34

So the peak spectral radiance (This from Wikipedia) is given by $$\displaystyle \nu _{\max }=T\times 1.04\times 10^{11}\ \mathrm {Hz} /\mathrm {K}$$ This works out to about $3\times 10^{12} Hz$. The and the spectral radiance will be given by Planck's Law:

$${\displaystyle B_{\nu }(T)={\frac {2h\nu ^{3}}{c^{2}}}{\frac {1}{e^{\frac {h\nu }{kT}}-1}},}$$

giving about $3\times 10^{-15} W/m^2/Hz$ so total power emitted by the $300 m^2$ black body is about $10^{-12} W/Hz$. That is spread over a spherical surface of radius $D$ for an intensity of $10^{-13}/D^2$ of which you are capturing $10^{-12}/D^2 W/Hz$ or about $10^{14}/D^2 Jy$. So you are fine if $D< 10^7 m$ more or less.

• Any chance you could include the formulas that you used? – Donald.McLean Apr 3 '18 at 14:32
• The power emitted per unit area from a black body surface is $\pi B_{\nu}$, not $B_{\nu}$. What is "a cubic surface of radius $D$"? – Rob Jeffries Apr 4 '18 at 12:37
• The flaw in your answer is not only my previous comment, but that the sensitivity is defined in Jy/sr. i.e. The received $10^{14}/D^2$ Jy is received over a solid angle of $300/D^2$. Thus the received flux density/sr is $10^{14}/300$ Jy/sr and is independent of distance. – Rob Jeffries Apr 4 '18 at 13:57
• Working through those numbers, I get 9.918e-13W/Hz, which is very close. Why would you assume a cubic surface though? Does heat spread in all directions, or in this case, into a hemisphere or surface area 2*3.14*R^2? Also, doesn't the telescope's collector area increase this distance? – Matterbeam Apr 4 '18 at 22:12