https://physics.stackexchange.com/questions/816698/how-many-photons-are-received-per-bit-transmitted-from-voyager-1 Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange [ ] Loading... 1. + Tour Start here for a quick overview of the site + Help Center Detailed answers to any questions you might have + Meta Discuss the workings and policies of this site + About Us Learn more about Stack Overflow the company, and our products 2. 3. current community + Physics help chat + Physics Meta your communities Sign up or log in to customize your list. more stack exchange communities company blog 4. 5. Log in 6. Sign up Physics 1. 1. Home 2. Questions 3. Tags 4. 5. Users 6. Unanswered 2. Teams [teams-promo] Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Explore Teams Create a free Team 3. Teams 4. Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Explore Teams Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams How many photons are received per bit transmitted from Voyager 1? Ask Question Asked 2 days ago Modified today Viewed 42k times 77 $\begingroup$ 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? * electromagnetic-radiation * estimation * antennas * radio * space-travel Share Cite Improve this question Follow edited 2 days ago Qmechanic's user avatar Qmechanic 204k4747 gold badges559559 silver badges2.3k2.3k bronze badges asked 2 days ago Craig Gidney's user avatar Craig GidneyCraig Gidney 6,54222 gold badges3030 silver badges3939 bronze badges $\endgroup$ 1 * 1 $\begingroup$ There is now a long thread on HackerNews about this post: news.ycombinator.com/item?id=40561872 $\endgroup$ - Punnerud 4 hours ago Add a comment | 1 Answer 1 Sorted by: Reset to default [Highest score (default) ] 118 $\begingroup$ 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. Share Cite Improve this answer Follow edited yesterday answered 2 days ago Jos Bergervoet's user avatar Jos BergervoetJos Bergervoet 3,75011 gold badge1010 silver badges2424 bronze badges $\endgroup$ 12 * 4 $\begingroup$ 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 yesterday * 8 $\begingroup$ @peterh: We can build better antennas on the ground than we can possibly launch due to being not size constrained. $\ endgroup$ - Joshua yesterday * 3 $\begingroup$ @Joshua Yes but the atmosphere is a sea of quadrillions of photons, how do you find the 400 representing a bit? $\endgroup$ - peterh yesterday * 15 $\begingroup$ @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. $\endgroup$ - Dan Is Fiddling By Firelight 19 hours ago * 1 $\begingroup$ On "bad weather days" the atmosphere can of course add more thermal noise than the value I used in the example. Also the cryogenic LNA's can be somewhat worse. The total result can then easily be a factor 2 or 3 more pessimistic, see the complicated bookkeeping discussion around page 55 in : descanso.jpl.nasa.gov/monograph/series10/02_Reid_chapt2.pdf $\ endgroup$ - Jos Bergervoet 16 hours ago | Show 7 more comments Your Answer [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] Thanks for contributing an answer to Physics Stack Exchange! * Please be sure to answer the question. Provide details and share your research! But avoid ... * Asking for help, clarification, or responding to other answers. * Making statements based on opinion; back them up with references or personal experience. Use MathJax to format equations. MathJax reference. To learn more, see our tips on writing great answers. 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