How I wonder what you are

Often when I’m gazing at the night sky I will focus on a very dim star and marvel at its ability to shine steadily. No matter how long I look, and no matter whether I move a few inches to the left or right, it keeps right on shining. (It may twinkle a bit, but as you probably know, that’s due to our fluid, shifting atmosphere causing the starlight to refract slightly differently from one moment to the next, not due to anything about the star.) Photons emitted by that star hundreds of years ago poured out in such numbers that, even as they fanned out across unimaginable distances, there are still enough of them landing continuously in the tiny target of my eyes that the star remains a constant point of light.

Knowing the distances involved, it seems implausible that enough photons would be headed in my exact direction — of all the other places in the universe they could have gone! How implausible? Let’s figure it out.

We’ll start by estimating that the area of my dilated, night-sky-gazing pupils is about one square centimeter, or (to keep all measurements using the same units) one ten-thousandth of a square meter.

Thanks to Wikipedia I know that a single photon of visible light carries about 4×10^-19 joules of energy which is just enough to excite a photoreceptor in my eye. How many must arrive each second for me to perceive a continuous image? I know that movies, projected at 24 frames per second, are good enough to trigger the persistence-of-vision effect. Since a distant star has far less detail than a frame of movie film, I’m guessing that it can appear steady at even fewer than 24 “frames” per second. Let’s guess 10 photons per second must arrive in my eyes to perceive a steadily shining star. 10 photons delivering 4×10^-19 joules every second is 4×10^-18 watts (because one watt is equal to one joule per second).

Now, how far is that star? A bit of googling reveals that a typical faintly visible star is 200 light years away (though many are much closer and many are much farther). 200 light years is 1.89×10¹⁸ meters.

From the perspective of that star, what fraction of its “sky” is taken up by my eyes? If you imagine a gigantic spherical shell centered on and surrounding that star, with a radius of 200 light years and my eyes on the inner surface, then the total area of that inner surface — the star’s “sky” — is 4×π×200light-years², which is 503,000 square light-years, which is 4.5×10³⁷ square meters. My pupils, one ten-thousandth of a square meter, comprise 1/2.22×10⁴² of that “sky.” That’s two millionths of a trillionth of a trillionth of a trillionth — not like that’s going to make my big ego feel small or anything.

So if 4×10^-18 watts of visible light is delivered to 1/2.22×10⁴² of the star’s sky, then the total power delivered to its full sky is that first number divided by that second, which is 1.8×10²⁴ watts, or 1.8 trillion terawatts. Is that a plausible number?

Yes it is. Our own sun has a visible-light output of 3.8×10²⁶ watts. Which means, incidentally, that (using my guesses from above) it would be visible as a steadily shining star to a distance of 2.7×10¹⁹ meters, or about 2,900 light-years. Future criminals exiled from our galaxy will lose sight of their home star while barely a thousandth of the way to their new home in Andromeda.