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As of 2024, according to https://voyager.jpl.nasa.gov/ , Voyager 1 is around one light·day away from Earth and still in radio contact. When Voyager 1 sends messages to Earth, roughly how many photons are (1) transmitted and (2) received per bit?

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    $\begingroup$ There is now a long thread on HackerNews about this post: news.ycombinator.com/item?id=40561872 $\endgroup$
    – Punnerud
    Commented Jun 3 at 18:21
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    $\begingroup$ Does it make sense to speak of photons when it comes to radio waves? $\endgroup$ Commented Jun 4 at 23:14
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    $\begingroup$ @PeterMortensen as much as it does at other wavelengths. And judging by Jos's answer we're getting into the realm where shot noise (the "lumpiness" in the arrival rate of signal photons, considered as a Poisson process) starts to become a factor, so we probably should speak about them. $\endgroup$
    – hobbs
    Commented Jun 5 at 0:11

2 Answers 2

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For an exact calculation we need to address a few choices: (you can change them, the answer will not be tremendously affected)

  1. What is the receiver? Let's assume a 70 m dish, like this one [CDSCC] in the Deep Space Network.
  2. [Voyager 1] can transmit at $2.3 {\rm GHz}$ or $8.4 {\rm GHz}$. Let's assume $8.4 {\rm GHz}$, for better beam forming (but probably it can only use the lowest frequency at the highest power, so this could be too optimistic).
  3. Does "received" mean all photons hitting the antenna dish, or only those entering the electronic circuit of the first LNA? A similar question can be asked for the transmitter in the space craft. We'll ignore this here since losses related to illuminators or Cassegrain construction will not even be one order of magnitude, insignificant compared with the rest.

Answers:
A) Voyager sends $160$ bits/second with $23{\rm W}$. Using $8.3 {\rm GHz}$ this is $4 \cdot 10^{24}$ photons per second, or $2.6 \cdot 10^{22}$ per bit, because for frequency $f$ the energy per photon is only $$E_\phi=\hbar\, \omega=2\pi\hbar f=5.5\cdot10^{-24}{\rm J} \ \ \text{or} \ \ 5.5 \ \text{yJ (yoctojoule)}.$$

B) The beam forming by Voyager's $d=3.7{\rm m}$ dish will direct them predominantly to Earth, with $(\pi d/\lambda)^2$ antenna gain, but still, at the current distance of $R=23.5$ billion kilometers, this only results in $3.4\cdot10^{-22}$ Watt per square meter reaching Earth, so a receiver with a $D=70{\rm m}$ dish will collect only $1.3$ attowatt ($1.3\cdot 10^{-18}{\rm W}$), summarized by: $$ P_{\rm received} = P_{\rm transmit}\ \Big(\frac{\pi d}{\lambda}\Big)^2 \ \frac1{4\pi R^2}\ \frac{\pi D^2}4 $$ Dividing by $E_\phi$ we see that this power then still corresponds to c. $240000$ photons per second, or $1500$ photons per bit. If we assume $f=2.3{\rm GHz}$ this becomes $415$ photons per bit. And if we introduce some realistic losses here and there perhaps only half of that.

C) (Although not asked in the question) how many photons per bit are needed? The [Shannon limit] $ C=B \, \log_2(1+{\large\frac{S}{N}})$, relates bandwidth $B$, and $S/N$ ratio to maximum channel capacity. It follows that with only thermal noise $N=k\, T_{\rm noise}\,B$, the required energy per bit is: $$ E_{\rm bit} = \frac S C = k\, T_{\rm noise} \ \frac{2^{\,C/B}-1}{C/B} \ \Rightarrow \ \lim_{C\ll B}\ E_{\rm bit} = k\, T_{\rm noise} \log 2, $$ where $C\ll B$ is the so-called "ultimate" Shannon limit. With only the CMB $(T_{\rm noise}\!=\!3{\rm K})$ we would then need $41 {\rm yJ}$, or $41 \cdot 10^{-24}\,{\rm J}$, per bit. That's only $7.5$ photons at $8.3 {\rm GHz}$. But additional atmospheric noise and circuit noise, even with a good cryogenic receiver, could easily raise $T_{\rm noise}$ to about $10{\rm K}$ and then we need $25$ photons per bit at $8.3 {\rm GHz}$, and even $91$ at $2.3 {\rm GHz}$. So clearly there is not much margin.

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    $\begingroup$ Wow, beside the excellent answer, I have also learnt that the Voyager signal is received on the Earth. I thought it is received by a network of receivers in the space. $\endgroup$
    – peterh
    Commented Jun 2 at 2:24
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    $\begingroup$ @peterh: We can build better antennas on the ground than we can possibly launch due to being not size constrained. $\endgroup$
    – Joshua
    Commented Jun 2 at 14:24
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    $\begingroup$ @Joshua Yes but the atmosphere is a sea of quadrillions of photons, how do you find the 400 representing a bit? $\endgroup$
    – peterh
    Commented Jun 2 at 19:54
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    $\begingroup$ @peterh because almost all of the photons in the atmosphere are the wrong frequency or coming from the wrong direction. The thermal noise Jos discusses at the end of the answer is the 'random' photons from the environment that are absorbed by the antenna. Deep space probes operate in parts of the spectrum no one else is allowed to radiate in to avoid manmade noise. $\endgroup$ Commented Jun 3 at 3:16
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    $\begingroup$ FYI this post has been screenshotted and posted on Twitter, without attribution. x.com/Duderichy/status/1797633261625393225 $\endgroup$
    – Criggie
    Commented Jun 4 at 2:21
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If my memory serves me correctly, the Voyager dishes are made from carbon fibre reinforced plastic (CFRP), and in an attempt to save weight, they were not metallised. That reduces the efficiency of the dish surfaces to around 25%, depending on the relative permittivity of the CFRP, so the margin is perhaps 3-5 dB less than the calculations suggest.

Engineering traffic is sent at 40 bits/sec, so it has a higher margin and could outlast the 160 bits/sec science stream if the RTG power doesn't give out first.

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  • $\begingroup$ Also the "aperture efficiency" gives probably at least 3dB extra loss. (For both Tx and Rx dish, en.wikipedia.org/wiki/Parabolic_antenna#Gain ). And other things could be worse than assumed. This makes sense, because clearly we would not spend time on our biggest radio telescopes if there was enough margin to do it with smaller ones! $\endgroup$ Commented Jun 4 at 6:16

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