Could anyone direct me to papers with info on minimal flux that can be registered with radiotelescopes?

  • $\begingroup$ Have a look at the answer I've posted. If it is not what you are expecting or you have further questions, please feel free to add a comment. I'm not an expert, but I'll do my best. $\endgroup$
    – uhoh
    May 1 '19 at 9:06
  • $\begingroup$ The links I've provided may not direct you to the best papers on the topic, I think probably others more familliar with the literature will add answers as well. $\endgroup$
    – uhoh
    May 1 '19 at 9:26

I'll try a quick estimate.

The strength of a radio source $S$ is often measured in Jansky's and 1 Jy is $1 \times 10^{-26}$ W/m^2/Hz. That means every square meter of collection area is exposed to $1 \times 10^{-26}$ per Hz of bandwidth.

Total power received would then be

$$P_S = S A \Delta \nu$$

where $A$ is the receiving area of the dish and $\Delta \nu$ is the bandwidth in Hz.

The noise limit of an RF receiving system is often expressed as an equivalent temperature. A typical number might be 50 Kelvin, which includes a receiver temperature, sky temperature, spillover (thermal radio emission radiated from the ground) etc. Though it seems counterintuitive, we can add temperatures as a proxy for adding powers because we are in the Rayleigh-Jeans regime where $\nu << k_B T/h$ as explained in this answer

You convert the noise temperature to a power in Watts:

$$P_N = k_B T \Delta \nu$$

where $k_B$ is the Boltzmann constant which is about $1.381 \times 10^{-23}$ Joules/K, and $\Delta \nu$ is again bandwidth in Hz.

Let's set the minimum strength easily detected to be equal to the noise, and equate the two powers:

$$P_S = P_T $$

$$S A \Delta \nu = k_B T \Delta \nu$$

$$S A = k_B T$$

$$S = k_B T / A$$

Note that the bandwidth drops out since both the thermal noise and most radio sources are fairly broadband. Even if the original line is fairly narrow, you'll tend set your minimum bandwidth of a specific measurement to cover the spectral region of interest.

Ballpark Estimate

The Arecibo dish has a radius of about 150 meters, so the area $A$ is about 70,700 square meters. A typical radio astronomy system temperature is 50 Kelvin (it can be significantly higher or lower depending on many factors, NASA deep space 70 meter dish liquid helium cooled front-end is 20K (see here and here), so plugging that into the equation above, we get $ = 9.8 \times 10^{-27}$ watts, which is (probably not coincidentally) almost exactly 1 Jansky!

Confirm that

According to Joeri van Leeuwen's Radio Astronomy lecture 5, The Radio Telescope at the bottom of page 79, the System equivalent Flux Density or SEFD of Arecibo is 3 Jansky, so a basic back of the envelope calculation gets us quite close!

Other radio telescopes will very roughly scale based on area, a smaller area dish will have a higher SEFD or signal level equivalent to the system noise.

Another example, VLA

A VLA dish is 25 meters in diameter, which is about 150 times less area. My back-of-the-envelope calculation would then be 150 Jansky.

According to Table 3.2.1 on the NRAO page Performance of the VLA during the Next Semester; Sensitivity at its lowest, a VLA dish's best SEFD is 250 Jansky. I've plotted the table below. Still checks out.

enter image description here


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