# How much power would an Alpha Centauri probe require for communication?

Project Longshot report (1988) says:

A laser with an input power of 250 kilowatts would allow for a data rate of 1000 bits per second at maximum range."

Yet Breakthrough Starshot website (2016) says:

Images of the target planet could be transmitted by a 1Watt laser onboard the nanocraft

Why do these estimates differ by 5 orders of magnitude? Is one (or two) of them incorrect, or did the technology change so much in 30 years? What are the physical principles behind this?

• Note that the 2nd does not mention a data rate.. – Andrew Thompson Aug 2 '16 at 20:07
• The lower the data rate, the less signal (to noise ratio) you need. See Shannon's theorem – Nick T Aug 2 '16 at 20:08
• Re: the Starshot page, I'm not sure if I'm calculating it right, but even with a 1 km focuser at 1.3 parsecs, the diffraction-limited spot size would be several earth diameters. – Nick T Aug 2 '16 at 20:23
• Continuous power or peak power? For lasers, that makes a huge difference. – gerrit Aug 3 '16 at 11:29
• Page 49 states they would use SIX 250 kW lasers. – Andrew W. Aug 3 '16 at 21:51

Super cool question! It combines a lot of my interests / career areas of focus: concept space mission analysis, communications theory, optics, and deep space exploration.

How much power would an Alpha Centauri probe require for communication?

To fully answer this I'll need to work through the math myself and provide some example solutions based on various assumptions. Stand by for the math... in the meantime here is a quick response.

Is one (or two) of them incorrect?

Longshot provides detailed calculations on page 48. I haven't double checked the math but the logic used makes sense. I can do the math and post it here when I get home if requested. Starshot doesn't provide any math which makes me doubt their claim.

The biggest assumption made by Starshot is that they can create a Fresnel lens with enough accuracy to improve the focus beyond the $3.25*10^{-7}rad=0.067''$ Longshot arrives at. Back of the envelope calculation: Note that they mention using 6 250 kW lasers on page 49, giving 1.5 MW of lazing power. To increase the power received by 1.5 million times the area subtended by the Earth over the total beam area would need to increase by $\sqrt{1,500,000}=1225$ (area of a circle). A focal length of infinity is theoretically possible, but would be dependent on the technology of the sail. Further, a thin lens would require a high number of grooves, reducing the efficiency of the lens. Lastly, they don't mention a data rate, so data may be able to be transmitted if sent VERY slowly.

did the technology change so much in 30 years?

Yes and no. Physics didn't change. The biggest change I can think of is the invention of Turbo Codes, which revolutionized the communications world in 1993. Low-light imaging technology has improved (not sure by how much, but a lot). The size of large ground based telescopes has increased a bit, and now use adaptive optics to counter atmospheric distortion.

• I can't make sense of this claim from Starshot: "For a 4m sail, for example, the diffraction limit spot size on Earth would be on order of 1000m." - solving for wavelength, $\lambda = angle\cdot aperture/1.22 = \frac{1000 m}{4.109\cdot 10^{16} m}\cdot 4 m / 1.22 \approx 8\cdot 10^{-14}$, which is ridiculously short (disclaimer: I don't know what I'm doing, learned this formula 10 minutes ago from wikipedia). – al13n Aug 4 '16 at 19:49
• Ok, I think I figured it out: if "10-14" in "A kilometer-scale receiving array would intercept 10-14 of the transmitted signal." is taken to mean $10^{-14}$, it implies $10^7$ meter spot radius, contrary to the previous sentence about 1000m spot size. This makes the wavelength on the order of 0.8 micrometers, close to Longshot's 0.532. So probably they assume an extremely low data rate, like tens or hundreds bits per day. It makes some sense considering they're proposing to send lots of probes at once. – al13n Aug 4 '16 at 20:03