Expanding a bit on this question: Is it possible to communicate in space while the sun is between parties?

Is there any wavelength usable by our presently available technologies at which the Sun is both quiet and transparent? If so, could a signal on such a frequency propagate through without distortion?

It is known that the Sun produces electromagnetic energy of various wavelengths, but is it actually quiet in some bands?

If it is quiet in a given band, is it transparent? Or is it opaque to any EM energy because it is a giant ball of plasma?

If it does happen to be transparent in some wavelength, does it refract, scatter, or otherwise disorganize EM energy to the point that no external signal could propagate through it intact?


2 Answers 2


It would only be transparent to radiation of wavelength longer than the characteristic scale of the sun. That would be in the 0.1 Hz range, even lower frequency than ELF. It's not practical to transmit or receive on such frequencies (you would need an antenna larger than the Earth), and even if you could, there's so little bandwidth that the data rates would be incredibly slow (perhaps one byte per minute). A relay station is a better option.

  • 1
    $\begingroup$ I guess a fair number of neutrinos also go straight through. $\endgroup$
    – gerrit
    Jun 29, 2017 at 11:06
  • 2
    $\begingroup$ But the sun is not quiet in neutrinos. $\endgroup$
    – CCTO
    Sep 9, 2019 at 17:12
  • 1
    $\begingroup$ @CamilleGoudeseune Morse code is digital. $\endgroup$
    – RonJohn
    Sep 9, 2019 at 20:13
  • $\begingroup$ @CamilleGoudeseune digital data modulation is not restricted to datarates larger than1 bit per second, it is possible too with bits per minute or even bits per hour. You just have to wait a very long time for a few bits to be transmitted. Slow transmission is not restricted to Morse code only. $\endgroup$
    – Uwe
    Sep 9, 2019 at 20:20
  • 1
    $\begingroup$ Neutrinos are also not electromagnetic radiation, which was specified in the question title. $\endgroup$ Sep 9, 2019 at 20:35

I somewhat disagree with this answer and here's why:

For the purposes of the question the Sun can be thought of as a ball of plasma. A huge amount of charges. The Earth's ionosphere is a plasma, and the carriers in a ball of metal can be thought of as a plasma as well.

For the Earth's ionosphere and a ball of metal, there's no frequency low enough that it becomes transparent to electromagnetic radiation. They are opaque to low frequencies.

So you should model the Sun as an opaque obstruction, and instead consider diffraction around it.

You can get a point-to-point concentration using the Arago spot. It's not a tight, proper focus, but a locus of concentration. But you can estimate a characteristic focal length $L$ (from Wikipedia) to be about $d^2/ \lambda$ where $d$ would be the diameter of the Sun.

Using a simple relation

$$\frac{1}{F} = \frac{1}{1 \text{AU}} + \frac{1}{1.5 \text{AU}}$$

to calculate a "focal length" for the Sun's Arago spot, we get 0.6 AU. That corresponds to a wavelength of 5300 km or a frequency of 57 Hz!!

Perhaps the 52-hertz whale was trying to tell us something, or have I been watching too much Star Trek?

This means that if Earth happened to be "leaky" at a line frequency of 50 or 60 Hz, if you were deliriously low on oxygen on Mars, you could imagine listening for it by building a giant radio telescope and waiting for the Earth to go behind the Sun and pass through the Sun's conjugate Arago spot.

Good luck!

Arago spot, from A photograph of the Arago spot

enter image description here

  • $\begingroup$ It is not true that the metal is always opaque to all low frequencies. C.f. the skin depth formula for a good conductor en.wikipedia.org/wiki/Skin_effect#Formula. $\endgroup$
    – Hans
    Sep 12, 2019 at 17:57
  • $\begingroup$ @Hans it's correct that any material that is opaque can be made so thin that something leaks through, and for copper or aluminum at 57 Hz it's going to be surprisingly thick; a large fraction of a centimeter, but it scales as $t \sim \sqrt{1/f}$, $f \sim 1/t^2$ so for the skin depth to reach 1 meter it's going to be 0.0057 Hz and to reach 1 km the frequency would have to be 0.0000000057 Hz. For the skin depth of a star to reach the size of the star itself the frequency would have to be quite uselessly low. $\endgroup$
    – uhoh
    Sep 12, 2019 at 22:25
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    $\begingroup$ To make your statement "frequency would have to be quite uselessly low" more specific, I translate that to the following one: the penetrating thickness $t\sim\sqrt\lambda$ grows much slower than the antenna length $\sim\lambda$ needed to receive the signal. Nice answer. +1 $\endgroup$
    – Hans
    Sep 12, 2019 at 23:10

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