16 November 2001

Little Discoveries Using Stellar Magnitudes
Part 1

by Art Winfree

My 9 November column about curved lines in the sky, features a date, 23 November. Follow-up before that date would deprive you of the chance to explore this matter on your own. Meanwhile I provide supplemental material following up Alan MacRobert’s primer on "The Stellar Magnitude System" in the same issue of E-Bulletin.

"The Stellar Magnitude System" introduced us to the logarithmic scale of visual magnitudes used by astronomers for two millennia. Here are some exercises that should help to install the magnitude scale in our personal toolboxes. In these log units the Sun’s intensity is called "magnitude -27", roughly. Similar stars show up with more positive magnitudes (i.e., they look dimmer) in our sky because they are farther away. Twice as far away, four times dimmer. To put it in terms of the magnitude scale, 100^1/5 times farther away, 100^2/5 times dimmer: 2.51 times farther away, 2 magnitudes dimmer. Or sqrt(2.51) = 1.58 times farther away, 1 magnitude dimmer. Of course, stars also differ also by intrinsic Magnitude but not systematically with distance from here, so let's just think in terms of Sun-like stars. How many more can we see by looking deeper into space with a larger-aperture telescope that accordingly picks up fainter stars? Multiply the aperture by 1.58 so you collect 2.51 time more light and a source 2.51 times dimmer looks the same, which might be the same source 1.58 times further away. Doubling the aperture merely doubles the depth of view, but it gives us 8 times as many stars, if they are uniformly distributed in volume. We will come back to this below.

Now let's compare the magnitudes of things illumined by the Sun, such as the air itself, the planets, and the Earth's Moon and Iridium satellites.

For the air, this is hard to calculate, but a glance at a shadow reveals that some of the brightness hitting the ground comes from skylight. I tried to quantify this today with a 35 mm camera whose exposure needle was a bit jumpy. You can check me and probably do better: At noon, center the light-meter needle over a patch of sunlit surface, then move into a shadow on the same surface and re-center the needle by increasing the shutter time or opening up the aperture: you may need something like 4 clicks of the dial, i.e., 16 times longer exposure. (Shutter speeds double per click, and so does diaphragm aperture area, but the numbers written on the aperture ring are tricky:  f_stop is diaphragm aperture diameter = focal_length/f_stop, so the ratio of f_stop is the ratio of diameters, not of areas. Two clicks doubles the f_stop and so quadruples the area, equivalent to two clicks of speed.) This says at least 1/16 of the sunlight (mostly blue, and maybe more than 1/16, considering that some of it scattered back into space) has scattered out during transit of the air over our heads. 

(A more systematic effort found in Minnaert's book on "Light and Color" (1954) shows how starlight magnitude depends on angle of view and so thickness of passage through the atmosphere. It concludes that 0.2 magnitudes = 1/6 is lost at one thickness straight overhead.)

Can we turn this into an estimation of the brightness of a patch of blue sky, per (arc minute)² of area? In a hemisphere there are 2p (360/2p)² = 20,000 square degrees, each of 60² (arc min)² which is about 100^20/5. We saw that altogether they contribute about 16x less light than direct sunlight, i.e., 100^3/5 less. So each contributes 20 +3 = 22 magnitudes less than the Sun, i.e., 27 -22 = -5 magnitudes. Since with naked-eye at 20/20 we can resolve no lesser detail than 1 arc min, a point source of light would have to exceed magnitude -5 to stand out visually in the noon-day sky by doubling the intensity in that "pixel" (as blue sky plus the point source). Even Venus doesn't quite qualify at noon, though I have seen it with naked eye an hour before sunset, when my camera would have recorded less light in the shadow on the ground.

You can check the "pixel" size of your own vision by putting dots 1 mm apart on paper then walking away until you can no longer resolve them. Without glasses I had to go about 2.4 meters, so the angle was 1/2400 radian = (360/2p)/2400 degree = 3420/2400 minute = 1.5 min of arc.

What would happen if you used binoculars? Binoculars are identified by two numbers, like 20x 80 mm. The first means all angles look larger by 20x magnification. The second means the aperture is 80 mm: compared to 5 mm pupil, this is a 16-fold gain, thus P = 256-fold increase of photon capture, so intensification by 6 magnitudes. Because all intensities are upped in the same proportion you might think the contrast remains unchanged and still inadequate. But think on it more. Increased aperture may bring really faint points up to visibility by gathering enough photons to score with a retinal cell, but where contrast is the problem it doesn't help. This is where magnification helps a point source (only). There is a little Discovery to make here. If the magnified Venus remained a point source for purposes of naked-eye resolution (it would, at magnification M < about 5x, independent of aperture), then it would continue to effectively occupy only the usual 1 arc min pixel of visual angle. But looking through the binoculars that visual angle grabs only 1/M arc min of blue sky, thus M² less area, bringing PM² more photons to the eye per least resolvable pixel, in contrast to hardly more than P more from the pixel containing Venus (neglecting the added small contribution from blue air in that smaller patch of sky.) The contrast has changed M²-fold (independent of P) in favor of seeing Venus. With M=5 this like naked-eye viewing with the sky dimmed relative to Venus by 3.5 magnitudes (100^3.5 /5 = 5²). 

M is actually greater in real binoculars, e.g., 20x in the example above, but I didn't use that in the calculation because it makes Venus into a discernible disk; this is a further advantage, but without taking it, this way of estimating remains good for Sirius, too, never more than a blurred point source. Sirius is our brightest star at magnitude -1.5. For daytime visibility at magnitude -5 it needs 3.5 magnitudes of contrast enhancement, which we found that 5x magnification or better would provide. It follows (correctly or not?) that if you knew exactly where to look, or used a go-to automated small telescope, you could presumably see Sirius at lunch-time with any eyepiece providing magnification > 5. Just now Sirius sets at 9 AM so this test will have to wait several months. Arcturus is a better candidate for lunch-time in November. With magnitude 0 it should need a boost of 5 magnitudes = 100x, available with 10x or better eyepiece. I donut have a telescope capable of pointing to a chosen star without some help from a skilled celestial navigator, but you might try it and report back, likely correcting the starter notions presented here. Remember, the "philosophy" of this Column is not to Promulgate Truth, but to present engaging riddles that may lead you to experience of personal Discovery. Here is one such opportunity.

What about a more challenging project now? The next brightest object in the sky is Venus, represented in old times by the symbol of a the goddess’ hand-mirror. Let's see how well it reflects sunlight. When we see the sunlit face of Venus flat on, it looks too near the Sun and is lost in the glare. But at either elongation of orbit, as morning or evening star, we see half of the sunlit face. At all times Venus is 2/3 the Earth's distance ("Astronomical Unit", AU) from the Sun, and at that time it is sqrt( (1)² + (2/3)²) = 1.2 AU distant from us (see below).
 

Click image to enlarge

It visibly subtends about 1/3 arc min of angle in the sky, so we see in this hypothetical mirror not the full half-degree disk of the Sun but only 1/2 (1/3 / 60)² = 1/2 10^-4 of that area, and not at standard 1 AU distance but rather attenuated (2/3 + 1.2)² = 3.5 —fold by the greater distance, (2/3 + 1.2) AU, along this bent reflection path. In terms of powers of the unit of magnitude, 100^1/5, these are diminutions by 10.5 magnitude units and by 1.4 more, thus from -27 to -15. Venus looks nowhere near that bright, not even -5 in reality. So we "Discover" that, unlike its astronomical symbol, Venus is not a hand-mirror held at a great distance. Not even an old tarnished one: the ten-magnitude discrepancy is 10,000-fold. Yes, true, we already knew Venus is not a mirror, at least by hearsay. The point here is only that in assimilating the notion of stellar magnitudes we learn this for the first time by our own personal observation, from the mere fact of Venus’ brightness. The intensity of that point of light in the sky is no longer arbitrary and meaningless, and so our experience is enriched.

While looking with binoculars to find out that it subtends 1/3 arc min, you may also have seen its shininess as oblong, like a mirror tilted to reflect the Sun --- but this notion is excluded now --- or maybe like a sphere illuminated from the side and diffusely glowing in that distant sunlight. Maybe that sphere absorbs light and scatters it in all directions, unlike a mirror. Aha. Of course. Were it a flat mirror we would catch a glint of bright sunlight only occasionally, when the direction is exactly right. In point of fact, we can see Venus continuously month after month while the direction changes through tens of degrees, vastly exceeding 1/2 degree. So the two facts of its constancy and of its dimness, which we have noticed life-long but maybe without further thought, consistently lead to scrapping the mirror analogy. Corroboration is good, especially since, at least in my own experience, even the most satisfyingly plausible notions are almost always wrong.

But here's a nice thing about mistakes: they are often the right answer for some other problem. We will encounter one such when we consider Iridium satellite flashes next time.

Back to the sphere idea. Will it fare any better? Dunno until we re-calculate on this new basis. Suppose a hemisphere illuminated at some uniform intensity. Suppose it re-radiates all that power uniformly in all directions of the hemisphere, so the problem is the same as though the full spherical surface were uniformly illuminated and radiating. At our distance that power is spread over a larger sphere of radius equal to our distance, so it is attenuated by a dimensionless factor, the square of the distance ratio (Venus radius / distance radius). We donut know either of those quantities, but do we need to? Only their ratio matters. And that is just half the observed visual angle (in radians) subtended in the sky: 1/3 arc min = 1/2 10^-4 radians (at 360/2p degrees per radians). Squaring, we expect 4 10⁸ -fold attenuation, which in terms of multiples of 100^1/5 per magnitude is 21.5 magnitude units. 

And what is the uniform sunlight intensity in Venus’ clouds that will get so attenuated by dispersal over the larger sphere? First impulse is to use standard intensity at Earth / (2/3)² because Venus is that much closer to the Sun: about one magnitude brighter. What's wrong with this? Two things: 

1. By "standard" I guess we mean noon-time. But the clouds are lit that bright only in the small part of Venus directly facing the Sun. Illumination elsewhere is less in proportion to the cosine of an angle in expanding circles around that place, because there the sunlight arrives at a slant, and
2. We see Venus from the side, so we see mostly those dimmer afternoon clouds (and the darkness of night.) 

So instead of adding a magnitude brighter, let's guestimate a magnitude dimmer. Then we expect to see not the Sun's magnitude -27 at Earth but rather -27+21.5 = -5.5 plus 1: about -4.5 or even dimmer because (we can guess) not all of the sunlight is re-radiated from Venus’ clouds at visible wavelengths: some of it warms the planet as it is degraded to IR that does not register for purposes of visual magnitude. We then expect something a bit dimmer than -4.5 and that is indeed what we get: go look … well, put a sticky-note on your calendar to look next summer. Just now Venus is falling behind the Sun. The important thing is that Venus should not look brighter than the maximum possible under these hypotheses: if it does, it is not a passively illuminated planet but star radiating on its own. Or the mirror image of one, viz., the Sun, but we excluded that version on grounds of a 10-magnitude discrepancy. From the viewpoint of independent personal exploration of the world, this can be another small Discovery: Venus radiates passively.

A good puzzle: can you work through the geometry of seeing Venus at diverse phases, distances, and elongations (angle from Sun, which determines its phase), to estimate magnitude at each? How close does your result come to the observed fact that magnitude hardly varies?

And at this point we are beginning to notice the importance of Alan MacRobert’s remarks last week that a scale of relative apparent visual magnitude ratios gets into trouble when the wavelength distribution of the light changes. Magnitudes as measured by a radiometer (total energy) would not be the same as those estimated visually by humans, or by human with cataract or with some "color blindness". So none of what we are doing here can be very precise. But it can be precise enough to make some plausible inferences if we take the trouble to think on our everyday observations.

Enough for today. Next week these exercises continue with Jupiter, the stars, and Iridium satellites. Can you guess beforehand what it might be possible to infer about them from thinking in magnitudes?