# What would it take to get a message to another star?

Let's assume that we've somehow detected the existence of an extraterrestrial civilization through "passive" means, such as detecting a suspicious exoplanetary atmospheric signature. We're now hoping to send that civilization a "hello, neighbor!" message - more than just a flash, something that actually communicates information.

1. Given our existing communication hardware, without having to build more, what's the furthest distance we could hope to send a signal before it gets lost in background noise?
2. Assuming we could gather extraordinary resources (this could be first contact, after all...) what's the most powerful message that we could send, and how far could it reach?
• – called2voyage Aug 13 '19 at 20:43
• I fielded a similar question in an answer to How far away would an alien civilization need to be for us to not notice them? on Worldbuilding three years and change ago. (If someone spots a mistake in that answer, please do comment over there.) – user Aug 14 '19 at 8:08
• Do the aliens know what kind of message we'd be sending? There are many encoding algorithms that make a message extremely resistant to noise, but which are very difficult to detect if you aren't looking for them. Also, what bitrate is acceptable? 5 bits/hour is easier than 5 bits/second, which is easier than 5 Kb/s. – forest Aug 15 '19 at 3:03
• ‎‎A‎‎ ‎‎t‎‎o‎‎w‎‎e‎‎‎l‎‎.‎‎‏‏‏‏‏‏‏‏ – dotancohen Aug 15 '19 at 13:26
• "Without building more". Done. Voyager 1 and 2 will intercept a number of celestial objects over the next couple billion years. nationalgeographic.com/news/2017/09/… . So the answer is: nothing and time. Like everything in space travel the answer is a trade off between time and effort (delta V); on the Earth side budget matters to escape the gravity well. – David J Eddy Sep 4 '19 at 18:15

The Arecibo raqdio telescope has a $$300\ \mathrm m$$ diameter mirror. Let's consider a radio wavelength of $$3\ \mathrm{cm}$$ ($$10\ \mathrm{GHz}$$) for convenience of arithmetic. That gives a diffraction limited beam width of $$100\ \mathrm{µrad}$$, so at 100 light years, the signal would be spread over an area $$10^{14}\ \mathrm m$$ across.

The Arecibo signal was transmitted at $$450\ \mathrm{kW}$$, so supposing the data rate was $$1\ \mathrm{bit/s}$$, so that the bandwidth is just $$1\ \mathrm{Hz}$$, the signal flux is the power per square meter, per steradian (of source width) per hertz.

So that is $$450\ \mathrm{kW}$$ divided by the beam area (roughly $$10^{28}\ \mathrm{m^2}$$) divided by the receiving antenna beam solid angle ($$10^{-8}\ \mathrm{sr}$$) divided by the bandwidth ($$1\ \mathrm{Hz}$$). This comes to $$4.5\times 10^{-15}\ \mathrm{W\ m^{-2}\ sr^{-1}\ Hz^{-1}}$$ or about half a trillion janskys. A decent radio telescope can detect a flux of $$1\ \mathrm{Jy}$$ over a period of an hour or less, so this signal will stand out like a sore thumb, once the correct frequency has been detected. In fact, you could probably up the data rate to $$1\ \mathrm{kHz}$$ or more.

It's not too hard to see that the effects of data rate and distance are both quadratic, so one can generalise this argument to say that an Arecibo-like telescope could talk to a copy of itself $$d$$ light years away at a data rate of about $$\frac{10^5}{d} \mathrm{Hz}$$, provided that nothing in between was absorbing the signal and that nothing along the direction of the beam (as seen by the recipient) was contributing unusual noise.

• I just spent 45 minutes confirming that this is the correct answer! ;-) – uhoh Aug 13 '19 at 23:53
• Answers that introduce to me a new unit (Janskys in this case) are rare! For others: Jansky: Unit of spectral flux density. Equivalent to 10^−26 watts per square metre per hertz . – dotancohen Aug 14 '19 at 7:32
• This is all wonderful, but it only gets data to the star. Given that it's a civilization previously unknown to us, how would they decipher the data we sent? – Ben Aug 14 '19 at 15:24
• @Ben different question, I think. There has been lots of speculative work on this. – Steve Linton Aug 14 '19 at 15:29
• @Ben getting a message to someone, and getting a message they can read to someone are technically different. I could signify you "as the chosen one" by biffing you upside the head with a wet pool noodle... The fact that you would probably interpret it as meaning I'm a childish jerk rather than being able to understand my message is technically irrelevant in a way. – TheLuckless Aug 14 '19 at 18:02

@SteveLinton's answer is excellent and I'll just confirm below that its logic and numbers are correct. Then I'll show that you can do it optically as well, but with 10 meter telescopes instead of Arecibos you run into a challenge because each individual light photon carries most of the total received power per second.

One standard way to estimate how well signals can be sent between points is to use a link budget calculation, where things are in a standardized format so Engineers can understand each part of the link separately, and to share the information with each other.

Since the calculation is a series of multiplications and division, when you use dB, these become addition and subtraction of logarithms. I'm going to leave out the smaller corrections from the big equation shown here since this is an approximate calculation.

$$P_{RX} = P_{TX} + G_{TX} - L_{FS} + G_{RX}$$

• $$P_{RX}$$: Received Power
• $$P_{TX}$$: Transmitted Power
• $$G_{TX}$$: Gain of Transmitting antenna (compared to isotropic)
• $$L_{FS}$$: "Free space Loss", what we usually call $$1/r^2$$ (but also has $$R^2 / \lambda^2$$) because receive gain is relative to isotropic)
• $$G_{RX}$$: Gain of Earth's Receiving antenna (compared to isotropic)

$$L_{FS} = 20 \times \log_{10}\left( 4 \pi \frac{R}{\lambda} \right)$$

$$G_{Dish} \sim 20 \times \log_{10}\left( \frac{\pi d}{\lambda} \right)$$

Using 300 meters and 3 cm for an Arecibo antenna at each end as mentioned in the other answer, the gain (over an isotropic antenna) at each end is about 90 dB. The transmit power of 450 kW is 56.5 dBW. 100 light years is 9.5E+17 meters, so $$L_{FS}$$ is 412 dB.

This gives the Arecibo to Arecibo at 100 Ly, 3 cm, 450 kW received power as

$$P_{RX} = 56.5 + 90 - 412 + 90 = -175.5 \text{dBW}.$$

Assuming a bandwidth $$\Delta f$$ of 1 Hz as in the other answer, and a receiver front-end temperature of 20 Kelvin (typical for practical Deep Space Network dishes) the NEP (Noise Equivalent Power) would be $$k_B T \times \Delta f$$ (where $$k_B$$ is the Boltzmann constant or 1.381E-23 J/K) is only -215.6 dBW, and would be -185.6 dBW for roughy 1 kHz, so @SteveLinton's answer is spot-on!

## Optical Transmission

note: After writing this section I realized that the Sun is going to drown out your signal unless you can find a narrow wavelength range where the Sun's emission is extremely dark. You use a very stable laser wavelength, and you hope that the people 100 light years away use a filter that isolates your laser wavelength taking into account the Doppler shift due to all of the motion between your planet and their planet.

This is extremely unlikely to work, whereas the Sun is going to be much dimmer in a narrow radio band, giving you more room to work with. For more on that see answers to How far have individual stars been seen by radio telescopes?

You can apply the same calculation to an optical link. Using a 10 meter telescope at each end, a 10W laser and a wavelength of 500 nm, you now get gains of 156 dB, a power of 10 dBW, and a path loss of 507.6 dB. The received power is then

$$P_{RX} = 10 + 156 - 507.6 + 156 = -185.6 \text{dBW}.$$

That's surprisingly similar to the radio received power. If you used a temperature-based bolometer to measure the optical signal, you might think that you could do a similar comparison to NEP, but there's a problem because each visible photon carries so much energy.

Doing photon counting and using $$E = hc/ \lambda$$, the photon energy is about 4E-19 Joules means that -185.6 dBW (about 2.8E-19 Joules/sec) is going to be only about 1.3 photons per second.

This means that if you were simply counting photons per 1 second bin, you wouldn't be able to do 1 kHz, and even 1 Hz would require a lot of statistical analysis.