return to table of content

How many photons are received per bit transmitted from Voyager 1?

Strilanc
21 replies
4h18m

Wasn't expecting my question to hit top of HN. I guess I'll give some context for why I asked it.

I work in quantum error correction, and was trying to collect interesting and quantitative examples of repetition codes being used implicitly in classical systems. Stuff like DRAM storing a 0 or 1 via the presence or absence of 40K electrons [1], undersea cables sending X photons per bit (don't know that one yet), some kind of number for a transistor switching (haven't even decided on the number for that one yet), etc.

A key reason quantum computing is so hard is that by default repetition makes things worse instead of better, because every repetition is another chance for an unintended measurement. So protecting a qubit tends to require special physical properties, like the energy gap of a superconductor, or complex error correction strategies like surface codes. A surface code can easily use 1000 physical qubits to store 1 logical qubit [2], and I wanted to contrast that with the sizes of implicit repetition codes used in classical computing.

1: https://web.mit.edu/rec/www/dramfaq/DRAMFAQ.html

2: https://arxiv.org/abs/1208.0928

grog454
5 replies
4h4m

by default repetition makes things worse instead of better

Can you elaborate on this a bit? My intuition is that, by default, statistical models benefit from larger N. But I have no experience in quantum physics.

Strilanc
2 replies
3h30m

It's because unintended measurement is a type of error in a quantum computer. Like, if an electron passing near your qubit would get pushed left if your qubit was 0 and right if was 1, then you will see errors when electrons pass by. Repeating the 0 or 1 a thousand times just means there's 1000x more places that electrons passing by would cause a problem. That kind of redundancy makes that kind of error mechanism worse instead of better.

There are ways of repeating quantum information that protect against accidental measurement errors. For example, if your logical 0 is |000> + |110> + |011> + |101> and your logical 1 is |111> + |001> + |100> + |010> then can recover from one accidental measurement. And there are more complex states that protect against both bitflip errors and accidental measurements simultaneously. They're just more complicated to describe (and implement!) than "use 0000000 instead of 0 and 1111111 instead of 1".

nomel
0 replies
24m

Is this the correct interpretation?

Classical systems: You measure some state, with the measurement containing some error. Averaging the measurement error usually gets closer to the actual value.

Quantum systems: Your measurement influences/can influence the state, which can cause an error in the state itself. Multiple measurements means more possible influence.

Kerbonut
0 replies
2h9m

If there's interference, could you do something like when using 7 repetition for each bit, take whatever 5 of 7 is, e.g. 1111100 is 1 and 1100000 is 0.

ziofill
0 replies
3h35m

It actually depends how this sentence is intended. There exist quantum repetition codes: the Shor code is the simplest example that uses 9 physical qubits per logical qubit. Since the information is quantum it needs majority voting over two independent bases (hence 3x3=9 qubits to encode a logical one).

Sniffnoy
0 replies
34m

You might be making the mistake of thinking that quantum mechanics runs on probabilities, which work in the way you are used to, when in fact it runs on amplitudes, which work quite differently.

nico
4 replies
4h12m

Very cool. It’s interesting to realize that at some level, every system is a quantum system if you “zoom in” enough

empyrrhicist
1 replies
3h48m

I think the point is the model though - if a system's behavior can be modeled/described classically, it's a bit silly to to call it a "quantum" system in the same way that it's reductive to say Biology is just applied particle physics. Sure, but that's not a very useful level of abstraction.

jessriedel
0 replies
59m

If you want to understand the transition between a fundamental theory and its effective description in some limiting regime, you need to be able to describe a system in the limiting regime using the fundamental theory. It's not "silly" to talk about an atom having a gravitational field even if its currently unmeasurably small.

fsckboy
0 replies
2h55m

at some level, every system is a quantum system

if we consider "quantum" to mean our quantum theory, at the level of general relativity, gravity is not a quantum system. and the qualifier "yet" is also not known.

Ringz
0 replies
3h59m

I would spontaneously respond that you are right and at the same time have no problem if someone explains to me that it is not so.

cycomanic
4 replies
4h0m

Subsea cables don't use repetition codes (they are very much suboptimal), but typically use large overhead (20%) LDPC codes (as do satellite comms systems for that matter (the dvb-s2 standard is a good example). Generally to get anywhere close to Shannon we always need sophisticated coding.

Regarding the sensitivity of Subsea systems they are still significantly above 1 photon/bit, the highest sensitivity experiments have been done for optical space comms (look e.g. for the work from Mit Lincoln Labs, David Geisler, David Kaplan and Bryan Robinson are some of the people to look for.

Strilanc
3 replies
3h39m

I think you're picturing a different level of the network stack than I had in mind. Yes, above the physical level they will be explicitly using very sophisticated codes. But I think physically it is the case that messages are transmitted using pulses of photons, where a pulse will contain many photons and will lose ~5% of its photons per kilometer when travelling through fiber (which is why amplifiers are needed along the way). In this case the "repetition code" is the number of photons in a pulse.

cycomanic
2 replies
1h46m

But we are classical, so I think it's wrong (or at least confusing) to talk about the many photons as repetition codes. Then we might as well start to call all classical phenomena repetition codes. Also how would you define SNR when doing this?

Repetition codes have a very clearly defined meaning in communication theory, using them to mean something else is very confusing.

jessriedel
0 replies
1h9m

Then we might as well start to call all classical phenomena repetition codes

All classical phenomena are repetition codes (e.g., https://arxiv.org/abs/0903.5082 ). And this is perfectly compatible with the meaning in communication theory, except that the symbols we're talking about are the states of the fundamental physical degrees of freedom.

In the exact same sense, the von Neumann entropy of a density matrix is the Shannon entropy of its spectrum, and no one says "we shouldn't call that the Shannon entropy because Shannon originally intended to apply it to macroscopic signals on a communication line".

Strilanc
0 replies
50m

Yeah, I agree it's unusual to describe "increased brightness" as "bigger distance repetition code". But I think it'll be a useful analogy in context, and I'd of course explain that.

resters
1 replies
1h38m

Isn't sending more than one photon always "repetition" in that sense? Classical systems probably don't do that because of the engineering complexity of sending a single photon at a time -- we had oscillators and switches, not single photon emitters.

jessriedel
0 replies
1h4m

Isn't sending more than one photon always "repetition" in that sense?

Yes. But regardless of whether its feasible to send single quanta in any given circumstance, the redundant nature of the signals is key to understanding its much higher degree of robustness relative to quantum signals.

And to be clear, you can absolutely send a classical signal with individual quanta.

fsckboy
1 replies
2h53m

How many photons are received per bit transmitted from Voyager 1?

wouldn't you also want to know how many photons are transmitted and how many bits transmitted are received?

stracer
0 replies
2h50m

All transmitted bits are also received, at least when everything works as intended.

s1dev
0 replies
3m

I believe that a classical radio receiver is measuring a coherent state. This is a much lower level notion than people normally think about in QEC since the physical DoF are usually already fixed (and assumed to be a qubit!) in QEC. The closest analogue might be different choices of qubit encodings in a bosonic code.

In general, I'm not sure that the classical information theory toolkit allows us to compare a coherent state with some average occupation number N to say, M (not necessarily coherent) states with average occupation number N' such that N' * M = N. For example, you could use a state that is definitely not "classical" / a coherent state or you could use photon number resolving measurements.

A tangential remark: The classical information theory field uses this notion of "energy per bit" to be able to compare more universally between information transmission schemes. So they would ask something like "How many bits can I transmit with X bandwidth and Y transmission power?"

cycomanic
18 replies
4h14m

Actually the limit predicted by Shannon can be significantly beaten, because Shannon assumes gaussian noise, but if we use photon counting receivers we need to use a poisson distribution. This is the Gordon-Holevo limit.

To beat Shannon you need PPM formats and photon counters (single photon detectors).

One can do significantly better than the numbers from voyager in the article using optics even without photon cpunting. Our group has shown 1 photon/bit at 10 Gbit/s [1] but others have shown even higher sensitivity (albeit at much lower data rates).

[1] https://www.nature.com/articles/s41377-020-00389-2

stracer
5 replies
1h58m

Is there some fundamental limit to the number of bits per photon that can be communicated via EM radiation? I think it does not exist, because photons aren't all equal, we can use very high frequency and X-ray quantum can probably carry much more information than RF quantum.

resters
2 replies
1h32m

this is called the Shannon limit. To discern signal from noise, a minimum sample rate of 2x the frequency of the signal is required. A signal is something that can be turned on or off to send a bit.

Higher frequencies can carry more data as you infer but the engineering challenges of designing transmitters and receivers create tradeoffs in practical systems.

ganzuul
1 replies
1h9m

In addition to wavelength EM also has several polarization modes and near/far field characteristics that can carry information.

resters
0 replies
1h0m

Can individual photons be measured for polarization and phase or is there a similar limit that requires more than one photon to do so? I suppose both are relative to some previous polarization or phase?

im3w1l
0 replies
5m

Send three photons A B C. They arrive at times ta, tb, tc. Compute fraction (tc - tb) / (tb - ta). This can encode any positive real number with arbitrary precision. But clearly you need either very precise measurements or send the photons at a very slow rate.

ganzuul
0 replies
1h13m

Good induction to thinking about quantum gravity.

aptitude_moo
4 replies
3h35m

Very interesting, I studied telecommunications and I thought the Shannon limit was the absolute limit. I wonder now if this Gordon Holevo limit is applicable for "traditional" telecommunications (like 5G) as opposed to photon counting a deep space probe

EDIT: This paper seems to answer my question [1]

[1] https://opg.optica.org/directpdfaccess/8711ab35-bbc2-4d51-8e...

stracer
2 replies
1h53m

Can you please post another link? This one does not work.

adrian_b
0 replies
1h42m

As also explained in the conclusion of the paper linked by you, "photon-counting" detectors are possible only when the energy of one photon is high enough, which happens only for infrared light or for higher frequencies.

"Photon-counting" methods cannot be implemented at frequencies so low as used in 5G networks or in any other traditional radio communications.

ramraj07
2 replies
4h1m

Can’t you calculate the CRLB for any given distribution if you wanted? That’s what my lab did for microscopy anyway. Saying you’re beating the Shannon limit is like saying you’re beating the second law of thermodynamics to me.. but I could be wrong.

cycomanic
0 replies
1h53m

Shannon theory assumes Gaussian noise, however in the very low power regime that's just not true. I agree it's unintuitive. Have a look at the Gordon paper I posted earlier.

Sanzig
0 replies
1h55m

You are correct. People often say "Shannon limit" (the general case) when they are really referring to the "Shannon-Hartley Limit" (the simplified case of an additive white Gaussian noise channel).

For example, MIMO appears to "break" the Shannon-Hartley limit because it does exceed the theoretical AWGN capacity for a simple channel. However, when you apply Shannon's theory to reformulate the problem for the case of a multipath channel with defined mutual coupling, you find that there is a higher limit you are still bounded by.

nico
2 replies
4h8m

Interesting. Is that related to compressed sensing? I wonder if compress sensing could be used for something like the Voyager signals

It seems there might be multiple ways to go beyond Shannon’s limit, depending on what you are trying to do

cycomanic
1 replies
3h57m

I don't think compressed sensing is really extracting more information than Shannon, it simply exploits the fact that the signal we are interested in is sparse so we don't need to sample "everything". But this is somewhat outside my area of expertise so my understanding could be wrong.

nico
0 replies
2h56m

Maybe I’m mixing Shannon’s limit with the sampling rate imposed by the Nyquist-Shannon Sampling theorem

Around 2004, Emmanuel Candès, Justin Romberg, Terence Tao, and David Donoho proved that given knowledge about a signal's sparsity, the signal may be reconstructed with even fewer samples than the sampling theorem requires.[4][5] This idea is the basis of compressed sensing

However, if further restrictions are imposed on the signal, then the Nyquist criterion may no longer be a necessary condition. A non-trivial example of exploiting extra assumptions about the signal is given by the recent field of compressed sensing, which allows for full reconstruction with a sub-Nyquist sampling rate. Specifically, this applies to signals that are sparse (or compressible) in some domain

From: https://en.m.wikipedia.org/wiki/Nyquist%E2%80%93Shannon_samp...

sansseriff
0 replies
1h24m

The Deep Space Optical Communication (Dsoc) between earth and the psyche spacecraft uses large-M PPM for this reason! This mission is currently ongoing.

They send optical pulses in one of up to 128 possible time slots, thereby carrying 7 bits each. And each optical pulse on earth may only be received by 5-10 photons.

superposeur
14 replies
3h32m

The overwhelming loss in this calculation is from the antenna’s radiated energy spreading out over a larger and larger area (despite the directional “gain” factor).

I’m wondering: would a probe launched today instead employ a laser to communicate? This would seem to offer many orders of magnitude improvement in the directionality of the signal.

londons_explore
5 replies
3h14m

The main challenge is the earth to probe comms for distant probes, since the earth is often very close (in an angular sense) to the sun from the probes perspective, and the sun gives out a lot of black body radiation.

However, due to the shape of the black body radiation curve, the sun gives out relatively less microwave radiation than it does visible light, which might outweigh the advantages of more directionality given by using a laser.

superposeur
2 replies
2h49m

Ok what about using a maser instead of a laser?

superposeur
0 replies
15m

Cool use of maser for receive.

Not having thought this through before, I see now that while a transmit maser may have efficiency advantages, it may not improve directionality relative to a standard parabolic radio transmitter. All methods of producing microwaves will have basically the same diffraction-limited gain for a given “aperture” (dish) size. That darn uncertainty principle! (However, an optical laser would still give way better directionality.)

sebzim4500
0 replies
2h43m

Presumably though it would be useful to have a high bandwidth link back to earth even if we had to use conventional microwave transmitters to send data back.

We want to download high resolution images/spectrographs whereas we only want to upload code/instructions.

jjk166
0 replies
3h0m

Further, we're good at building really big radio transceivers here on Earth, we don't have nearly the same technical experience with lasers of that scale.

deelowe
3 replies
3h23m

I imagine it'd certainly employ some type of beamforming at the least.

wongarsu
2 replies
3h2m

Assuming you don't need fast steering, is a 3.7m transmitter array doing beamforming really better than a 3.7m dish transmitting at the same power?

My intuition would have been that you are better off using a fairly standard transceiver and spending your engineering budget either increasing power or getting a bigger dish (either by launching on a wider rocket or with a folding design).

Lasers might interesting for the downlink, but receiving a laser signal on the probe sounds difficult (earth is pretty bright).

outworlder
0 replies
32m

There's some value in getting rid of mechanical devices (or reducing the need to rotate the entire spacecraft).

cycomanic
0 replies
1h27m

Diffraction scales inversely proportional to wavelength so you gain significantly by going to optics, i.e. you can use a much smaller aperature in optics.

superposeur
0 replies
2h50m

Interesting about the JPL program and I’m amazed this prototype was only launched last year! Apparently the answer to comms laser use is “not yet but soon”.

mordae
0 replies
1h28m

Improving directionality also makes aiming much harder.

cycomanic
0 replies
1h38m

All space agencies have optical comms in their road maps. Largely they are thinking about inter satellite communications (the atmosphere causes significant issues when going back to earth). So the main application is to have some relay satellite that can then transmit to earth via RF. The application is not mainly deep space ropes but Leo or meo satellites, the typically only have very short transit times over the ground stations, so can't get all their measurement data down. By using e.g. a geo relay they can transmit lots of data optically and the geo relay can more slowly transmit the data to earth until the leo satellite comes back in view.

lordnacho
14 replies
5h29m

I love these kinds of questions. So what does that conclusion mean about when the probe will be so far away that we are below the Shannon limit?

And can we beat the Shannon limit somehow, eg collect for longer, put the dish outside the atmosphere, and so on?

ethbr1
5 replies
5h14m

Not my field, but assuming transmitting hardware (including beam forming) is constant and that atmosphere can mostly be ignored (see comments about it usually being a non-impact in the transmission frequencies), two approaches would suggest:

1. Increase the effective receiving dish size, to capture more of the signal. Essentially, this would be effective in direct proportion to beam spread (the more beam spread, the bigger dish you can use to capture signal).

In practice, this would use multiple geographically-displaced dishes to construct a virtually-larger dish, to allow for better noise-cancellation magic (and at lower cost than one huge dish). I believe the deep space network (DSN) already does this? Edit: It certainly has arrayed antennae [0], though not sure how many are Voyager-tasked.

2. Increase the resilience of the signal, via encoding. The math is talking about bits and photons, but not encoded information. By trading lower bit-efficiency for increased error tolerance (i.e. including redundant information) we can extract a coherent signal even accounting for losses.

Someone please point out if I'm wrong, but afaik the Shannon–Hartley limit speaks to "lower" in the physical stack than error coding. I.e. one can layer arbitrary error coding on top of it to push limits (at the expense of rate)?

If the above understanding is correct, is there a way to calculate maximum signal distance assuming a theoretically maximally efficient error coding (is that a thing?) ? Or is that distance effectively infinite, assuming you're willing to accept an increasingly slow bit receiving rate?

[0] https://en.m.wikipedia.org/wiki/NASA_Deep_Space_Network#Ante...

retrac
1 replies
4h50m

There's another major factor. I suppose it falls under the encoding. Frequency stability. In a certain sense, having an extremely precise oscillator at both the receiver and the transmitter, is the same thing as just having a better more frequency-stable antenna, or less noise in the channel (because you know what the signal you're listening for should look like).

I'm no physicist here so take this with a major grain of salt. I think the limit might ultimately arise from the uncertainty principle? Eventually the signal becomes so weak that measuring it, overwhelms the signal. This is why the receiver of space telescopes is cooled down with liquid helium. The thermally-generated background RF noise (black bodies radiate right down into the radio spectrum) would drown everything else out otherwise.

Along those lines, while I'm still not quite sure where the limit is, things become discrete at the micro level, and the smallest possible physical state change appears to be discrete in nature: https://en.wikipedia.org/wiki/Landauer%27s_principle Enough work physically must occur to induce a state change of some kind at the receiver, or no communication can occur. (But this interpretation is disputed!)

lxgr
0 replies
4h43m

Coding counters the uncertainty principle by allowing multiple measurements, which can then be averaged. Thant counters the contribution of random noise.

There are practical signals we use every day that are "below the noise floor" before we decode them.

So while there is an ultimate limit of the maximum coding rate for a given signal-to-noise ratio, this is expressed in terms of a data rate (i.e. bits per second). If you're fine with lowering your data rate, there is no fundamental theoretical limit, as far as I understand.

shagie
0 replies
4h47m

The resilience of the signal part...

https://www.allaboutcircuits.com/news/voyager-mission-annive...

The uplink carrier frequency of Voyager 1 is 2114.676697 MHz and 2113.312500 MHz for Voyager 2. The uplink carrier can be modulated with command and/or ranging data. Commands are 16-bps, Manchester-encoded, biphase-modulated onto a 512 HZ square wave subcarrier.

The "Manchester encoding" brings us to https://www.allaboutcircuits.com/technical-articles/manchest...

https://en.wikipedia.org/wiki/Manchester_code

Note that "16 bps" while the system runs at 160 bps. This suggests that the data is repeated ten times and xor'ed with a clock running at 10 HZ.

While there's no VOY set up now, https://eyes.nasa.gov/dsn/dsn.html will occasionally show it. When that happens, you will likely see two set up for it. I've not seen them set up across multiple facilities - the facilities are 120° apart and only one has a spacecraft above the horizon for any given length of time.

---

In the "sensitivity to photons" category, I'll also mention https://en.wikipedia.org/wiki/Lunar_Laser_Ranging_experiment...

At the Moon's surface, the beam is about 6.5 kilometers (4.0 mi) wide[24][i] and scientists liken the task of aiming the beam to using a rifle to hit a moving dime 3 kilometers (1.9 mi) away. The reflected light is too weak to see with the human eye. Out of a pulse of 3×10^17 photons aimed at the reflector, only about 1–5 are received back on Earth, even under good conditions. They can be identified as originating from the laser because the laser is highly monochromatic.

While there's no signal there, we're still looking at very sensitive equipment.

nsguy
0 replies
3h20m

TIL from another comment that the Shannon limit assumes Gaussian noise so it's not actually always the theoretical limit.

You can't work around the Shannon limit by using encoding. It's the theoretical information content limit. But you can keep reducing the bandwidth and one way of doing that is adding error correction. So intuitively I'd say yes to your question, the distance can go to infinity as long as you're willing to accept an increasingly low receive bit rate. What's less clear to me is whether error correction on its own can be used to approach the Shannon limit for a given S/N ratio - I think the answer is no because you're not able to use the entire underlying bandwidth. But you can still extract a digital signal from noise given enough of a signal...

EDIT: There is a generalization of the Shannon limit to non-white Gaussian noise here: https://dsp.stackexchange.com/a/82840

RetroTechie
0 replies
4h22m

3. Use relays.

Of course that would mean sending (a) giant receiver dish(es) in the general direction a probe is sent. On the flip side, if using a single relay it could travel at roughly 1/2 the speed of the probe.

Note that signal strength weakens with distance^2. So if eg. you'd have 2 relays (1/3 and 2/3 between Earth & the probe), each relay would receive 9x stronger signal.

No doubt the 'logistics' (trajectory, gravity assist options, mission cost etc) make this impractical. But it is an option.

fsmv
1 replies
5h25m

Seems like we can just build a bigger receiving dish

rcxdude
0 replies
4h53m

There's not a lot of appetite for that, though. the 70m receiver is already one of the largest ever built, and for most use cases it's looking better to use an array of smaller ones. Which works fine for receiving but not so much for transmitting (while there are multiple sites in the world capable of receiving from voyager 2, there's only one dish which can actually transmit to it)

(I recall seeing a video on that dish, and the director seemed confident there was enough noise margin left that voyager's power would fail before they lost contact with it)

mpreda
0 replies
5h11m

Maybe multiple receiving antennas could be used, spaced widely apart (e.g. one on Earth, one on the moon, one in orbit somewhere); each would receive some signal affected by noise, but the noise would be different for each (such as the atmosphere affecting only the on-ground antenna). Also given that we know the precise location of the transmitter, we can work-out the exact phase difference between the antennas. Some digital post-processing would combine the raw signal+noise from the multiple receivers and extract the signal.

In fact, I assume that at this distance, even a very narrow signal would spread wide enough to illuminate more than just the Earth diameter.

mlok
0 replies
5h12m

Maybe some kind of relay in space (maybe it could follow Voyager 1, slightly faster) which amplifies the signal and retransmits to Earth ?

kibwen
0 replies
4h22m

> So what does that conclusion mean about when the probe will be so far away that we are below the Shannon limit?

I think the practical limit right now is that the Voyagers are losing power.

"The radioisotope thermoelectric generator on each spacecraft puts out 4 watts less each year. [...] The two Voyager spacecraft could remain in the range of the Deep Space Network through about 2036, depending on how much power the spacecraft still have to transmit a signal back to Earth."

https://voyager.jpl.nasa.gov/frequently-asked-questions/

Symmetry
0 replies
5h2m

The Shannon Limit is fundamental so you can't get around it directly. But yes, there are things you can do to receive information from further away.

1) A bigger dish is the most obvious one.

2) Use a lower bitrate to send more energy per bit at the same transmitter power.

3) Reduce the effective receiver temperature. This is a case where putting it outside the atmosphere might help in reducing noise.

Asraelite
0 replies
5h17m

I guess we could just program Voyager 1 to lower the data bitrate, adding more redundancy / error correction. I think there's no real limit to how far it could get if we keep doing that.

7373737373
0 replies
4h34m

If the current number of photons per bit is 1500 and the effective limit at 8.3GHz is 25, that is a factor of 60.

With each doubling of distance the number decreases by the inverse square law, so with the current setup we'd have a maximum distance of log2(60) = 5.9 times the current distance (about 163 astronomical units (AU)) which is 961 AU.

In comparison, the closest star to our sun, Proxima Centauri, is 268774 AU away!

Which means we would need something 268774/961 ~= 280 times SQUARED = 78222 times more sensitive than the current setup at the Shannon limit to communicate with it if it managed to get that far.

pcdoodle
12 replies
4h57m

I am confused. I thought photons were just visible light but I guess these little buggers are everywhere. Also very surprised voyager is using 2.3ghz, that's crazy saturated on earth due to wifi. How these engineers make this all work, is magic to me.

nilamo
3 replies
4h51m

Also very surprised voyager is using 2.3ghz, that's crazy saturated on earth due to wifi

Wifi didn't exist when Voyager was launched...

thsksbd
2 replies
4h16m

But the band was free for use, wasn't it? (Obviously not crowded)

ianburrell
0 replies
3h5m

The ISM band is 2.4-2.5 GHz. Voyager at 2.3Ghz is outside the band.

anotherhue
0 replies
4h8m

Available for microwaves since 1947, intentional emission came later in the 80s.

noneeeed
2 replies
4h40m

Nope. It's one of those things that can take a bit to get used to, but everything on the electromagnetic spectrum is just light in the general sense. The only difference between radio-waves, x-rays, infra-red and (human) visible light is the frequency/wavelength.

If the frequency is high enough then the waves of light can be detected by things as small as cells in the back of your eye, or the pixels in a camera sensor. If it is too low then you need much larger detectors.

Other animals have detectors for different frequencies/wavelengths, allowing them to see either infra-red (mosquitos) or ultraviolet (bees, butterflies etc).

What we call "visible light" is just the particular range that our eyes can detect (about 400 to 800THz). If we were the size of a planet, and our eye cells were the size of a radio-telescope dish we would be able to "see" in those wavelengths. In fact, when we see images taken by radio telescopes, those have been essentially pitch-shifted up to something we can see, like the reverse of what we do when listening for bat clicks (where the pitch is downshifted to our hearing range).

The wikipedia article has a nice little diagram putting the wavelengths into perspective. https://en.wikipedia.org/wiki/Electromagnetic_spectrum

pcdoodle
1 replies
4h26m

Thanks for the reply.

This makes me think of the dual slit experiment. Does the universe treat everything as a wave to save CPU cycles or something?

If we think of light as little balls (at our size i think that would make sense). If we were much bigger, we would think of these longer waves as balls too?

SJC_Hacker
0 replies
35m

Wave-particle duality/quantum mechanics have different interpretations, but since "the math works" and there's nothing better, thats what they go with.

One way of thinking of it - everything is a wave until you make a measurement. Then you get a "collapse" (localization) of the wavefunction. Which leads to the question of why the wave function collapses - i.e. how does nature know we want to make a measurement.

Which leads to all sorts of crazy ideas like the simulation hypothesis. Which is not a scientific theory, because you can't falsify it, but even very educated people like Neil DeGrasse Tyson have remarked on it.

pythonguython
0 replies
4h32m

Photons and waves both model electromagnetism. Photons are just the quantization of electromagnetic radiation, where E=hv. This is the whole idea of wave-particle duality. We often describe radio frequencies with waves because they act more like waves than particles (Diffraction, spherical propagation, have an easily measureable wavelength)

mpreda
0 replies
4h52m

Electromagnetic radiation includes visible light, radio spectrum, X-rays, etc. Photons.

einsteinx2
0 replies
4h20m

Photons are just the quanta or particles of electromagnetic radiation, of which visible light is a small portion of the overall spectrum. So you can have photons of microwaves as well, such as in this case. Or photons of X-rays or gamma rays or infrared light or ultraviolet light or whatever. It’s pretty wild actually just how small a section of the EM spectrum our eyes are sensitive to!

drmpeg
0 replies
4h40m

WiFi is at 2.4 GHz. LTE band 30, satellite radio (XM/Sirius) and aeronautical telemetry all exist between the deep space downlink at 2290 to 2300 MHz and WiFi at 2400 MHz.

BenjiWiebe
0 replies
3h59m

2.3 GHz is not saturated due to WiFi. Wifi is 2.4GHz and up. Even the harmonics will be above 2.4 GHz (by definition).

amirhirsch
5 replies
4h59m

TLDR; 4e22 photons per second 2.6e22 per bit.

For comparison, ~2e26 photons will be received through your iris in your life

rcxdude
1 replies
4h49m

That's how many are sent by Voyager. Only about 1500 or 400 photons per bit are actually received by the radio dish (depending on which frequency is being used).

croemer
0 replies
1h22m

Maybe OP sends photons from their eyes in addition to receiving them

amelius
1 replies
4h55m

How many of them come from Voyager 1?

amirhirsch
0 replies
4h50m

Someone’s asking the hard questions! According to the oracle, 1 in 4 people will experience one photon from voyager in their lifetime.

KeplerBoy
0 replies
4h55m

Does this number account for all photons or just those in the rather narrow optical band?

Then again RF photons just don't fit through the pupils and will get backscattered, i guess.

jsjohnst
4 replies
5h34m

Didn’t realize the math would be that straightforward. Is there something the author isn’t taking into account or is that a decent plausible range?

magicalhippo
1 replies
4h28m

One thing that seems missing to me is that while the probe might send 160 bits/sec of useful data, those bits are not sent directly as such[1]:

The TMU encodes the high rate data stream with a convolutional code having constraint length of 7 and a symbol rate equal to twice the bit rate (k=7, r=1/2).

So the effective symbol rate is 320 baud[2], and thus a factor of two should be included in the calculations from what I can gather.

Note that the error correction was changed after Jupiter to use Reed-Solomon[3] (255,223) to lower the effective bit error rate, so effectively I guess the data rate is more like 140 bps.

[1]: https://web.archive.org/web/20130215195832/http://descanso.j...

[2]: https://destevez.net/2021/09/decoding-voyager-1/

[3]: https://destevez.net/2021/12/voyager-1-and-reed-solomon/

ooterness
0 replies
2h32m

For the "photons per bit" question, the useful throughput of 140 bps is the only thing that matters.

For the "how many photons are needed" question, I agree that 320 baud (i.e., the effective analog bandwidth of 320 Hz) should have been used for the Shannon-Hartley calculations.

rcxdude
0 replies
4h56m

It looks like a pretty reasonable order of magnitude estimate to me. Energy arguments tend to be quite neat for that because if the efficiency is at all reasonable they constrain things well with fairly simple calculations. The antenna directionality is also reasonably well understood and characterised. The exact noise level discussed later on is probably where things get a bit more uncertain (but aren't directly needed to answer the question).

krylon
0 replies
5h8m

I was surprised, too. Knowing little about physics, this was a pleasant surprise, however.

bandrami
4 replies
4h40m

It's crazy to me how many theoretical limits Shannon predicted way before the hardware was there.

ziofill
3 replies
4h23m

That’s because his results are about pure information (and in the limit for infinite string lengths), so sooner or later some hardware will hit onto those limits or tend to them.

bandrami
2 replies
4h11m

Agreed, but: he's still understudied. I think in retrospect any 21st-century math course has to include Shannon, and they don't all, yet.

sebzim4500
1 replies
2h41m

On some level he's a victim of his own success. He invents information theory in the same paper that proves the most interesting results, so who else will work on it?

aidenn0
0 replies
2h22m

Even the practical work was done surprisingly early. I have a book on error correcting codes from the 1950s and it's missing very little (Most notably trellis codes and LDPC; the former being invented in the '70s and the latter in 1963).

ziofill
2 replies
4h14m

What a lovely question. The estimate is 10-100 photons/bit (minimum).

If you’re curious about how many bits a single photon can carry, in controlled settings (tabletop quantum optics) a single photon can carry log(n) bits where n is the size of the state space of the photon, which theoretically is infinite and in practice it can reach into the hundreds/thousands.

mnw21cam
0 replies
3h44m

Visible light is different, because each photon has a lot more energy than in the 2.3GHz range. Your average decent consumer-level camera has a sensor that can nominally just about detect single photons some proportion of the time (as in, some of them bounce off instead of being detected) though it can't technically count them. The graininess on digital camera images is more from the Poisson noise of the incoming photons than it is from the applied noise of the sensor itself.

layer8
0 replies
3h54m

No, the estimate is around 750 or 200 photons/bit received, depending on the transmission frequency. The answer to the question is B, not C. Your numbers are the estimated minimum needed, not the actual amount received, which is what the question was asking.

somat
1 replies
1h38m

An interesting thing about photons (which may not be true, I just enjoy this stuff amateurishly, that is, without the effort or rigor to actually understand it.) is that they might not exist. the em field is not quantized, or at least is not quantized at the level of photons. A "photon" only exists where the em field interacts with matter, where the electrons that create the disturbance can only pulse in discrete levels.

https://www.youtube.com/watch?v=ExhSqq1jysg

Not that this changes anything, we can only detect or create light with matter. but it does make me curious about single photon experiments and what they are actually measuring.

leetrout
0 replies
52m

Thanks for the link. I never conceptualized photons outside of the visual spectrum so the headline made me take a step back and get nerd sniped in the process.

I stumbled upon this before seeing your comment:

https://physics.stackexchange.com/questions/90646/what-is-th...

moffkalast
1 replies
4h43m

You know, I never really thought of lower wavelengths than light as being carried by photons, but I suppose it's all EM. Antennas are technically just really red light bulbs.

gjstein
0 replies
4h39m

This is true enough, though remember that material properties change dramatically when you start moving through wavelengths by orders of magnitude. Silicon is transparent in the mid-infrared, which is what makes silicon photonics possible [1]

[1] https://en.wikipedia.org/wiki/Silicon_photonics

hammock
1 replies
3h23m

Voyager sends 160 bits/second

This makes me wonder, are the bits = the power turned on for exactly 1/320th sec, every 1/160th sec? Or is the power on/power off ratio something different? Does it vary by protocol? What are the pros and cons?

danbruc
0 replies
2h54m

Without looking up what kind of encoding and modulation they are using, I would assume that they are sending a continuous sine wave at the carrier frequency that has the bits - probably after encoding the raw data bits with some error correction code - modulated onto it by changing frequency, amplitude, phase, or a combination of them depending on the value of each bit or group of bits.

wwarner
0 replies
1h0m

23 watts.

tb0ne
0 replies
32m

Super interesting! But I feel like there is a bit of a conclusion missing for me.

So 1500 Photons hit the receiver per bit send, but this is obviously way to few to keep processing the signal and it will just be drowned out by noise? Where do we go from here? Does voyager repeat its signal gazillions of times so we can average out the noise on our end? Where can I find more information on what is done with these few photons?

notorandit
0 replies
5h34m

It's really nice! Both the question and the answer!

mooktakim
0 replies
1h54m

Why didn't they send out new relays as Voyager travelled out.

ks2048
0 replies
2h5m

Nice question. Does anyone know what exactly data is being sent? What kind of compression it is using? etc

cycomanic
0 replies
3h52m

For anyone who is interested in the ultimate limits to communications the seminal paper by Jim Gordon is quite easy to understand even without a physics degree (unlike the Holevo paper IMO). He was incredibly good at writing in an accessible manner (apart from probably being the person who most deserved a Nobel prize but didn't get it).

https://doi.org/10.1109%2FJRPROC.1962.288169

croemer
0 replies
1h23m

That's why I love Physics and was enamoured with it in my late teens.

HarHarVeryFunny
0 replies
4h14m

The fact that we can communicate with Voyager, and in both directions, blows my mind. It's completely counter-intuitive.

At least for Voyager->earth we can use giant radio telescopes to detect the faint signal, but how do we manage to focus on those few hundreds of photons per bit coming from a pinpoint source a light day away?!

In the earth->Voyager direction it seems even less intuitive - sure we can broadcast a powerful signal, but it's being received by a 12' wide antenna 15 billion miles away. WTF?

I guess radio communications in general is magic, a bit like (in nature of counter-intuition) quantum entanglement of particles arbitrarily far apart. It seems there is something deeply wrong about our mental models of space and time.