Really Wild Stuff - Lighting a Star with a Flashlight


Nothing practical, but - if I shine a flashlight into the night sky, pointed at a large but dim star, how many photons per second reaches that star, when the light beam finally arrives?

A cheap 9 LED flashlight produces about 25 lumens of "white" light; probably the combined output of red, green, and blue LEDs. Assume the average energy per visible photon is 2.3 electron volts, about 3.7e-J/photon. A lumen is 683 watts, so the flashlight produces about 37 milliwatts of light, about 1e17 photons per second. The beam is about 50 centimeters wide at 150 centimeters distance, about 0.1 steradian solid angle, so it emits 1e18 photons per steradian per second.

Arcturus is an orange giant, not as intense as a G star like the sun but large. It is 36.7 light years away, D = 3.5e17 meters, and 25 times wider than the sun, a radius r of 1.8e7 kilometers. The star's solid angle from earth is π(r/D)2, 8e-21 steradians. 8e-3 photons from my flashlight hits the star every second, about one photon every two minutes ... 36.7 years after I shine my flashlight at it. Surprisingly, more than zero, though impossible to detect compared to the power output of Arcturus.

How about the most powerful pulsed laser, with excellent optics? The uV laser at the National Ignition Facility produces a 500 terawatt, 1.85 megajoule pulse - 37 microseconds. If we assume 10 meter output optics and 351 nm uV (and some magic way to send that up through atmosphere), The beamwidth would be about 1.22 * 351 nm / 10m radians or 4.3e-8 radians, producing a beam solid angle of of 1.5e-15 steradians. 351 nm is 3.5 eV or per photon, so 1.85 megajoules is 3.2e24 photons, or about 2e39 photons per steradian, with 3e24 of the photons reaching the star. 351 nm is 8.5e14 Hz.

Arcturus surface temperature is 4300K, and Boltzmann's constant is 8.62e-5 eV/K, so kT = 0.37 eV. The black body formula is P/hz-m2 = 2 ( h \nu) / ( \lambda^2 (e^{h \nu / kT} - 1 ) ) = 2 * 5.6e-19 J / ( (3.51e-7 m)2 * 12800 ) = 1.6 nW/m2-Hz , or 1.6e12 W / hz for the entire disk. Our pulse is 37 microseconds, so the bandwidth of a gated detector would be 27 kHz, so the noise power of Arcturus over that band is 4e16 W, but during our gate period it is 1.6e12 joules. The energy of our pulse is a mere 1.8 MJ, and the fraction of that reaching Arcturus is 8e-21/1.5e-15 or 5.5 ppm, about 10 joules. Worse, since the portion of Arcturus facing us is a half-sphere, not a disk, about 60 light seconds in radius, the reflected pulse would be stretched over 60 seconds. We could average many pulses, perhaps, but we might need trillions of years to send them all.

Ah, well, no lidar to Arcturus, even with our best "flashlight".

Real Communication to a Cold Dish is MUCH Easier

Let's try for usable communication. Presume two large, precisely figured parabolic dishes focused on each other, 36.7 light years or D=3.5e17 meters apart. Assume the dishes and detectors are shielded from sun/starlight and very cold, perhaps 20K. Assume 351 nm as before - the photons are higher energy and expensive to make, but they focus better, which matters at these huge distances. Assume the dishes are far enough from the nearby sun/star to be optically distinguishable - we'll compute that. If we are not competing with the light of the star, we can get by with much less light.

If the dish radius is r , then the beam spread is 1.12 \lambda / r radians. The solid angle is approximately ( \lambda / r )^2 steradians. The solid angle of the receiving dish is \pi ( r / D )^2 , so the fraction of the energy received is \pi r^4 / ( D \lambda )^2 . Assuming we want to receive pulses of 20 photons per bit, and send a gigabit per second, will need to receive 2e7 * 3.5 * 1.6e-19 = 11 picowatts. We must transmit 1.1e-12 ( D \lambda )2 / π r4 = 1.1e-12 ( 3.5e17*3.5e-7 )2/π * (1/r4) watts, or 1.3e10 W-m2 / r4. If r = 100 meters, power = 130 watts. Neither is very difficult in nanogravity - compared to moving a dish and transmitter 37 light years!

How far must we be from the star, to put it in the Airy null of the other transmitter? We can cheat - we can make the dishes very wide in the plane defined by the other dish and star. Presume we make the dishes 10 meters "high" and 1000 meters "wide", the first Airy null will be 1.1*3.5e-7/1000 or 4e-10 radians separation. Times 3.5E17 meters distance, that is a mere 140,000 kilometers. If we orbit the dishes 100 AU out, 1.5e13 meters, then we are 100,000 Airy fringes away from the star, very easy to separate with good optics.

Our sun is 5780K, ( kT = 0.5 eV ) a higher temperature than Arcturus, so (given the exponential in the Planck equation) it is brighter in the UV per square meter. The sun produces 2 ( h \nu) / ( \lambda^2 (e^{h \nu / kT} - 1 ) ) = 2 * 5.6e-19 J / ( (3.51e-7 m)2 * 1123 ) = 8 nW/m2-Hz, 5 times brighter per unit area though 625 times smaller area, resulting in 0.004 of the UV compared to the much larger star. The total UV emission of all of Arcturus over a 1 GHz bandwidth is 1.6e-9W/m2 * ( 1.8e10m )2 * 1000 = 5.2e14 W per steradian. The solid angle of the far receiving dish is \pi ( r / D )^2 or π ( 100/3.5e17)2 = 4e-29 steradians, for an (off axis) power level of 2e-14 W, 2e-3 of the received signal power. We do not even need good optics to discriminate that from our signal.

If we know exactly where to point, and can do so very accurately, and use multiple redundant dishes "just in case", we can do high bandwidth interstellar communication over large distances.

Round trip signal time to Arcturus will be 73 years; if the time value of money is 10% per year, then the value of data returned in response to a request will be 1100 times less valuable than a request to an in-system asset. Even if we have some means to send the far transciever at nearly the speed of light, the value of the bits returned must be enormously high to justify the cost with an adequate return on investment.

FlashlightStar (last edited 2015-01-12 04:48:39 by KeithLofstrom)