Wikipedia:Reference desk/Archives/Mathematics/2009 September 27

= September 27 =

Fictional Solar Eclipse Calculation
This is an inquiry related to the calculations for solar eclipses. I am pretty good with scientific concepts but not so good with the actual math. Any help is appreciated.

I know the basics of how solar eclipses cause lunar shadows (umbra and penumbra) to fall on small portions of the Earth. I also know that the distance between the sun, moon, and earth affect these shadows. As a kid I learned that I could create an quasi-eclipse (ie: "blot out the sun") with a simple penny if I held it the right distance between my eye and the sun. This is because of something known as "apparent size" I think.

Assuming (just for the sake of simplifying the discussion) that it is possible to change the size and position of the moon without affecting gravity here is what I would like to know:

1. If I wanted to create a penumbra big enough to cover the whole Earth (well, the sun facing side of course) how big would the moon need to be? What about for an umbra that big?

2. Like the penny, would moving the moon closer or farther from the Earth help make the umbra/penumbra bigger? If so what is the optimal combination of size (smallest diameter) and distance (between the sun and earth) to put the moon so the whole planet is in shadow?

Of course I would love to see the math involved though I make no promises that I will understand it. Thanks again. 66.102.205.169 (talk) 06:04, 27 September 2009 (UTC


 * This is simple geometry. Draw a line (proportional to) the diameter of the sun. Draw a line parallel to it the diameter of the earth at the right distance. Draw 4 lines from the both sides of the sun to both sides of the earth. Choose a position for the moon and the outermost two lines determine the minimum diameter of the moon for an umbra that is visible every where on the facing earth simultaneously. (It will be more practical to draw it not to scale and to use equations...) -- SGBailey (talk) 11:09, 27 September 2009 (UTC)


 * SGBailey, I appreciate the graphic description but such a diagram is already present on Wikipedia here. As I mentioned I am good at concepts but weak in mathematics. Diagrams are a nice way to bypass the math but diagrams can't produce accurate numbers without equations. What I am looking for are the equations you mention. Conceptually I know the solar system is three dimensional with varied planes but to keep the equations simple let's just use a solar system where all bodies are centered on the same plane (no need to get into the complications of three dimensional math for this). Likewise I realize these bodies are in various orbital motions but again lets use fixed objects to simplify the math. Considering that the distances between the three bodies is not constant I figure I will probably need to run the equations several times with some different variables. Any help is appreciated. 66.102.205.169 (talk) 19:11, 28 September 2009 (UTC)
 * Ok, but that figure misses the body that you want to be covered. So it doesn't show all you are looking for.
 * It does however answer your question 2 fairly directly.
 * To get the answer to your first question imagine that the earth in the picture was actually the moon, where whould you place the earth so that it was completely covered by the penumbra? It helps to draw lines perpendicular to the diameter of the earth, since you then get a set of similar triangles.
 * For the umbra, in order for the umbra to completetly cover the earth the moon would have to be bigger than the earth. So it's perhaps better to ask when will the moon be completely in the umbra of the earth?  As before simply place the moon into the picture at those points where this would be satisfied, then work out the triangles in question.  Taemyr (talk) 19:36, 28 September 2009 (UTC)

Let's think about perfect circles on a piece of paper. Assume that the sun has radius s, the moon has radius m and that the centre of the sun and the centre of the moon are distance d apart. We can parametrise the perimeter of the sun by S(t) = (s.cos(t), s.sin(t)) and the perimeter of the moon by M(&theta;) = (d + m.cos(&theta;), m.sin(&theta;)). The tangent line to the perimeter of the sun at S(t) is given by cos(t).x + sin(t).y = s. The tangent line to the perimeter of the moon at M(&theta;) is given by cos(&theta;).x + sin(&theta;).y = m + d.cos(&theta;).

Consider the red lines in this picture. Drawing some pictures will show you that the red lines are tangent to the sun's perimeter and the moons perimeter, and that happens when t = &tau; and &theta; = &tau;, for some &tau; &isin; [0,2&pi;). Consider the blue lines in this picture . Drawing some pictures will show you that the blue lines are tangent to the sun's perimeter and the moons perimeter, and that happens when t = &tau; and &theta; = &tau; + &pi;, for some &tau; &isin; [0,2&pi;).

Equating the equations of the tangent lines with the added conditions that &theta; = t or &theta; = t + &pi; gives solutions for t. Substituting these solutions into the equations for the tangent lines to the perimeter of the sun give the following: the equations of the blue and red lines are given by
 * $$ \pm (s-m)x \pm \left(\sqrt{(m-s+d)(s-m+d)}\right)y = sd \ . $$

Next you work out the equation of the line passing through the centre of the sun and the intersection of the top blue and red lines. This equation is horrible and I won't write it here. This gives the centre of maximal circles fitting inside the upper-penumbra region; just make the circles bigger until they are tangent to the top blue and red lines. The bottom red line comes into play after a while and so the centres of circles will kink after a while.

The equations get more and more ugly. If you wanted an exact solution then I could give you one, but it would make little sense to a human being. Dr Dec ( Talk )    20:04, 28 September 2009 (UTC)


 * The distances between the objects are quite far compared to the radii so we can approximate the profiles of the bodies as lines instead of circles. Under that approximation, let RE be the radius of the Earth, RS the radius of the Sun, D the distance between the Earth and Sun.
 * For the Moon to cast a penumbra that covers the whole Earth when it's at a distance of d away from the Earth, its radius must be RM ≥ (1 - d/D)RE - (d/D)RS. Note that when d exceeds DRE/(RE + RS), the radius can be any size no matter how small.
 * For the Moon to cast an umbra that covers the whole Earth when it's at a distance of d away from the Earth, its radius must be RM ≥ (1 - d/D)RE + (d/D)RS. Note that in this case the Moon has to be larger than the Earth and the farther away, the larger it has to be.
 * Rckrone (talk) 06:07, 29 September 2009 (UTC) Edit: On thinking about it, these formulas are correct with no approximations necessary.


 * Are you sure that you've not missed something? First of all, if the centres of the sun, moon and earth are colinear then part of the earth is always going to be in either the umbra or the antumbra regions. Let RS = 4, RM = 2, D = 24, d = 12. Your first inequality becomes RE &le; 8. I've made a plot of this with RE = 8. The sun is yellow, the moon is black and the earth is green. Notice that part of the earth is in the umbra region. Also we need the centres of the earth, the moon and the sun to be colinear. If they're not then the earth won't even be contained inside the region bounded by the two blue lines (the upper limit of one penumbra region, and the lower limit of the other penumbra region}. The calculations depend upon the angle of the chord joining the centre of the earth to the centre of the sun and the chord joining the centre of the moon to the centre of the sun. Even if you get the earth inside the upper penumbra region (bounded by the upper blue and red lines) as it moves further away the antumbra region pops up and the upper penumbra region is then bounded by the upper blue and the lower red lines.  Dr Dec  ( Talk )    09:28, 29 September 2009 (UTC)
 * According to NASA's definition a body that is in the umbra is also in the penumbra. Ie. they define the penumbra as the region where some or all light is blocked. Taemyr (talk) 12:29, 29 September 2009 (UTC)
 * Are you sure about that? This picture is wrong then? And this picture ? NASA seem to distinguish between being in the umbra and the penumbra in this document . The Wikipedia article says that "The umbra, penumbra and antumbra are the names given to three distinct parts of a shadow". I think the original question related to this picture . Otherwise the solution would be simple: as Rckrone's beautiful inequalities show.  Dr Dec  ( Talk )    16:32, 29 September 2009 (UTC)
 * No not completely, I am merely parroting Penumbra, and I should be more precise since the definition I quote is specifically from the Navigation and Ancillary Information Facility. The definition I use is sourced to, see page 5.  Neither of the two pictures, nor the PDF, you provide contradicts the statement "the region covered by the umbra is a subregion of the region covered by the penumbra", although in the first picture one of the red lines is superflous.  Taemyr (talk) 18:31, 29 September 2009 (UTC)


 * Hmmm, interesting. The Wikipedia article says that NAIF's definition is an aletenative definition. The experience of an observer would be totally different when in different regions of the shadow. We even have different names for different eclipses: a total eclipse when in the umbra region, a partial eclipse when in the penumbra antumbra region and an annular eclipse when in the penumbra region. In fact they are topologically distinct: in the umbra region the visible sun would be the empty set, in the penumbra regions the visible sun would topologically be a disk, and in the penumbra region the visible sun would be an annulus. These are really quite different phenomena and, IMHO, the umbra, penumbra and antumbra regions should be mutually exclusive. Again, our own solar eclipse article seems to make a distinction. To maintain an interesting mathematical problem we should keep this distinction too. Given the different radii, the different distances and the angle between the sun-moon chord and the sun-earth chord, what are the conditions for all of the daytime part of the earth to see a partial eclipse, all of the daytime part of the earth to see a total eclipse, and all of the daytime part of the earth to see an annular eclipse?   Dr Dec  ( Talk )    18:55, 29 September 2009 (UTC)


 * In any case, if you want the Earth entirely in the penumbra in the strict sense (no part in the umbra or antumbra), then there are two cases:
 * The distance between the Earth and Moon satisfies d ≥ DRE/(RE + RS) and the radius of the moon satisfies RM ≥ (d/D)RS.
 * The distance and radius exactly satisfy the equation RM = (d/D)RS, and as before RM ≥ (1 - d/D)RE - (d/D)RS, which in this case becomes RM ≥ (1 - d/D)RE/2.
 * If the penumbra and antumbra are allowed, but the umbra is excluded, then there are also two cases:
 * The distance between the Earth and Moon satisfies d ≥ DRE/(RE + RS), and there are no restrictions on the radius.
 * The radius of the moon is in the range (1 - d/D)RE - (d/D)RS ≤ RM ≤ (d/D)RS. Note that this can only be satisfied for d ≥ DRE/(RE + 2RS).
 * These formulas do make use of the approximation of the distances being much larger than the radii. Rckrone (talk) 19:11, 29 September 2009 (UTC)


 * For the Moon to cast an antumbra that covers the whole Earth when it's at a distance of d away from the Earth, its radius must be RM ≤ (d/D)RS - (1 - d/D)RE. Note that d must be at least DRE/(RE + RS) to satisfy this. Rckrone (talk) 19:23, 29 September 2009 (UTC)


 * (ec) But you're missing the key problem again, the problem I originally raised. The angle between the sun-moon chord and the sun-earth chord matters. You need to include this in your solution. Your formulae only involve the sizes and distances of the objects. You have not considered their relative positions. You haven't imposed any upper bounds on d or D, so it is possible that the earth could be on the other side of the sun to the moon. In that case no part of the earth would experience any eclipse.  Dr Dec  ( Talk )    19:29, 29 September 2009 (UTC)
 * p.s. Don't get me wrong. I think that your inequalities are very beautiful and are a step in the right direction. You just need to iron out the creases. I have some exact solutions (taking into account the angles of relative position and the kink that we get when the lower bound of the upper penumbra region changes from being the umbra to the penumbra) on Maple, but like I said: they're of no use to a human being.  Dr Dec  ( Talk )    19:38, 29 September 2009 (UTC)


 * I understand what you're doing, but I don't think it's really necessary. Given that D/RS ~ 200, the angles are very small (less than 1°) and the error introduced into most of these distances is on the order of less than 10-4, which is far less than concerns like the irregularity of the Earth's orbit.  Obviously it would become an issue for arbitrary RS, RE and D, but that's not really what the OP asked about.  Also I think a good approximation is better than nothing, since as you say the exact version is ugly.
 * The implicit bounds on d are that it's between 0 and D except where more restrictive bounds are mentioned. I'm finding the distances and radii of the moon that allow the specified effects to happen given the right alignment.  I'm not too concerned with finding formulas for the alignments that allow these to happen.  Rckrone (talk) 19:50, 29 September 2009 (UTC)


 * Fair enought. Sorry, I didn't realise that you were finding approximations. When you said "On thinking about it, these formulas are correct with no approximations necessary." I assumed that your inequalities were exact. Thinking about all of this makes me wonder: how do the guys (and girls) at NASA come up with document like this . I guess that they must use computer approximations.  Dr Dec  ( Talk )    20:08, 29 September 2009 (UTC)
 * The inequalities in the first post do happen to be exact (somewhat luckily), but the set in the second post aren't. The antumbra inequality is though. Rckrone (talk) 22:25, 29 September 2009 (UTC)

Well folks, the amount of mathematical discussion on what I thought was a simple question has absolutely blown me away. I will probably spend the next several weeks trying to make heads or tails of all this but I **really** appreciate the feedback. 66.102.206.234 (talk) 03:13, 5 October 2009 (UTC)

Derivative of a complex function
I'm reading through a book on complex analysis, and have got stuck on one of the first problems of the text ("Introduction to Complex Analysis" by Nehari). It reads as follows:


 * "If the function F(z) is regular in a domain D, show that the function Re[F(z)] has a derivative only at those points of D at which F'(z)=0."

The term "regular" may be taken to mean "differentiable". According to my following "proof", however, one has G'(z)=Re[F'(z)], where G=Re[F]. Indeed, let F(z)=u(z)+iv(z), where u and v are real functions; as F is regular on its domain D, for any z in D it must be that u and v are differentiable at z and hence F'(z)=u'(z)+iv'(z). Then one has


 * $$\lim_{h\to0}\frac{G(z+h)-G(z)}{h}=\lim_{h\to0}\frac{u(z+h)-u(z)}{h}=u'(z);$$

the above seems to hold without any need whatsoever for F'(z)=0 (which would then imply that u'(z)=v'(z)=0). Either the book is wrong or I'm missing some fundamental piece of information here; any help on ascertaining which of the two is so would be appreciated. -- Nm420 (talk) 17:14, 27 September 2009 (UTC)


 * He (Nehari) means (...) the function Re[F(z)] has a complex derivative only at those points of D at which F'(z)=0. At any point z, say with F'(z)=a, as in your computation you have, for all h:
 * ReF(z+h) = ReF(z) + Re(ah) + o(h).
 * This only shows that the map Re F(z) is R-differentiable and the R-differential is the R-linear map taking the complex h into Re(ah): but you want it C-differentiable, that is, C-linear (and not only R-linear), which simply means Re(ah)=Re(a)h even for h=i : clearly, this happens if and only if a=0 . --pma (talk) 18:40, 27 September 2009 (UTC)
 * Alright, that makes more sense. Also, just to make sure I'm fully getting this, the Cauchy-Riemann equations will also assert this, correct? For, if z=x+iy, we then need $$u_x=v_y$$ and $$u_y=-v_x$$ (where u and v are now the real and complex parts of G=Re[F], so that v=0); as v=0, it must be that $$u_x=u_y=0$$ (and clearly $$v_x=v_y=0$$). -- Nm420 (talk) 16:18, 28 September 2009 (UTC)