1. Introduction
2. Rotations vs. Translations on the Sphere?
3. Definitions in the Three Geometries
4. Parallelograms on the Sphere?
5. Possible Solutions to the Parallelogram Problem
6. How Spherical Rotations Differ from Euclid
7. Glide Reflections on the Sphere
8. Feynman, Pages 71-160
9. Old Charter School Update
10. Conclusion
Introduction
As I wrote in my last post, I was undecided over whether to cover rotations or translations in this post, after covering reflections in my last post. It was a very tough decision.
In the end, I chose rotations. We can use the U of Chicago definition of rotation, from Lesson 6-3:
- A rotation is the composite of two reflections over intersecting lines.
This definition works in both Euclidean and spherical -- hence in natural -- geometry. The following theorem is also valid in both geometries:
Two Reflection Theorem for Rotations:
The rotation r_m o r_l, where m intersects l, "turns" figures twice the non-obtuse angle between l and m, measured from l to m, about the point of intersection of the lines.
Since pairs of lines on the sphere intersect in antipodal points, either can be considered to be the center of the rotation. The easiest rotations to envision are centered at the poles. The image of any point under such a rotation is easy to find -- it has the same latitude as the preimage, while the change in longitude is given by the magnitude of the rotation. Note that a counterclockwise rotation as viewed from the North Pole becomes clockwise as viewed from the South Pole -- both of these refer to an eastward rotation.
I could have covered translations before rotations -- this would have adhered to the U of Chicago text, where translations appear in Lesson 6-2, followed immediately by rotations. On the other hand, Berkeley mathematician teaches rotations first (and his order is closely followed by EngageNY), but we can't do that using the U of Chicago definition, where rotations are defined in terms of reflections.
Thus in a way, our order -- reflections, rotations, translations -- is a compromise between the U of Chicago and Wu-EngageNY orders. It's also the order I tried to follow the first two years of this blog, before I switched to the pure U of Chicago order.
Rotations vs. Translations on the Sphere?
And indeed, we wonder whether translations even exist on the sphere. The U of Chicago defines a translation as the composite of reflections over parallel lines. Since there are no parallel lines on the sphere, hence there are no translations either.
But some authors not only consider translations to exist on the sphere, but they consider all rotations to be translations! For example, we notice the following link:
https://mathstat.slu.edu/escher/index.php/Spherical_Geometry
Symmetries in Spherical Geometry
Rotations are Translations: Spherical rotations involve spinning the sphere around an axis line that goes through the center of the sphere. A spherical rotation has two points that don’t move, where the rotation axis hits the sphere at a pair of antipodal points. For example, the Earth (idealized a bit) rotates on its axis, and the North and South poles don’t move. Translations on the sphere are exactly the same as rotations. A translation should slide along a geodesic. The geodesics are great circles, and if you slide along a great circle the sphere rotates around an axis. Picture the Earth’s equator, and as the world turns it appears that points near the equator are being translated east.
Notice that this website is titled "Math & the Art of MC Escher." Hmm, Escher -- that name surely sounds familiar. Yes, Hofstadter strikes again! Here the artist Escher is mentioned because his abstract art reminds some people of non-Euclidean geometry.
But if no parallel lines exist on the sphere, and translations are the composite of reflections in parallel mirrors, then how can any rotations be translations, much less all of them?
Well, there is a definition of "translation" that allows all spherical rotations to be translations, without allowing any Euclidean rotations to be translations (unless you count the identity as a zero-degree rotation and a zero-vector translation). But to see how this definition works, we must look at the third type of geometry -- hyperbolic geometry.
In spherical geometry, there is only one isometry that's the composite of two reflections. In Euclidean geometry, there are two such isometries. In hyperbolic geometry, there are three such isometries, with the third isometry known as "horolation":
- A rotation is the composite of two reflections in intersecting mirrors.
- A translation is the composite of two reflections in ultraparallel mirrors.
- A horolation is the composite of two reflections in horoparallel mirrors.
It's easy to define "ultraparallel" here -- two lines are ultraparallel if there exists a line perpendicular to both of them. By this definition, all parallel lines are ultraparallel in Euclidean geometry, but there exist parallel lines that aren't ultraparallel in hyperbolic geometry -- these are horoparallel instead.
Recall that my vision of a high school course is to teach Euclidean geometry most of the year, but then introduce non-Euclidean geometry at the end of the year. Such a course could look like this:
- Point-Line-Plane Postulate
- Reflections
- Rotations
- Translations
...
- Horolations
Like most of the other units, the final unit would be named after a transformation, except that this one would be called "horolations," to indicate that this is hyperbolic geometry. The geometry taught would be neutral as long as possible, then Playfair would be introduced when needed. When the hyperbolic unit is reached, we state that everything proved before Playfair holds in both geometries.
Definitions in the Three Geometries
But the problem is that we don't want the non-Euclidean geometry introduced at the end of the year to be hyperbolic -- we want it to be spherical. This time, it isn't easy to name the last unit, as spherical geometry has one fewer, not one more, translation.
Anyway, in hyperbolic geometry, a translation is defined as the composite of two reflections in ultraparallel mirrors, and ultraparallel lines are defined as having a common perpendicular. But what if, in spherical geometry, we define "ultraparallel" as "having a common perpendicular"?
Then, it turns out, all pairs of lines are ultraparallel on the sphere, even while no lines are parallel. To find the common perpendicular of any two lines, we begin by thinking of the two lines -- which intersect at an antipodal pair of points, as the two sides of a lune. Then their common perpendicular bisects both sides of the lune.
If the points of intersection are the North Pole and the South Pole, then the common perpendicular is the Equator. This the situation described at the Escher site listed above -- nearest the Equator, the common perpendicular, is where the rotation looks the most like a translation.
Therefore, we conclude by using the hyperbolic definitions of "ultraparallel" and "translation," all pairs of lines are ultraparallel, and all rotations are translations, thus vindicating the Escher site. But here's the big question, do we really want to define "ultraparallel" on the sphere in this manner just to make all rotations into translations?
On one hand, I like following the U of Chicago text and using inclusive definitions -- for example, the inclusive definition of trapezoid which allows a parallelogram to be a trapezoid. Extending the idea of inclusiveness to non-Euclidean geometry implies that if there's a term for a Euclidean object that doesn't exist in a non-Euclidean geometry, it's better to redefine the term, rather than leave it undefined, in the new geometry. This is why, for example, some authors define "rectangle" to mean "equiangular quadrilateral" in non-Euclidean geometry, since such quads exist in all geometries.
On the other hand, redefining terms can be confusing, especially for students. After all, if we're going to define "ultraparallel" so that all lines on the sphere are ultraparallel, why don't we just skip the middle man and define "parallel" itself to mean "having a common perpendicular"? The reason, of course, is that defining all lines on the sphere to be parallel is misleading, since all lines on the sphere in fact intersect.
We must keep in mind that our students will only see spherical geometry at the end of the year -- the rest of the year is purely Euclidean. It's actually not terrible to tell students, at the end of the year, "On the sphere, no lines are parallel, but all lines are ultraparallel." We then point out that some previous (Euclidean) proofs are based on ultraparallelism rather than parallelism, and to that extent, such proofs are valid on the sphere.
One U of Chicago proof based on ultraparallelism is the Two Reflection Theorem for Translations:
Two Reflection Theorem for Translations:
If m | | l, the translation r_m o r_l slides figures two times the distance between l and m, in the direction from l to m perpendicular to the lines.
This theorem takes us from the "two mirrors" concept of translation to the "vector" concept, since it provides us with a distance and a direction. The direction is stated as being perpendicular to both lines, so the lines must have a common perpendicular (that is, be ultraparallel).
Notice that while the theorem is valid in all three geometries, only in Euclidean geometry are there many common perpendiculars -- indeed, every point in the plane lies on such a point. In the other two geometries, there's at most one common perpendicular to any pair of lines (otherwise all four lines would form the sides of a rectangle, with four 90-degree angles). Only for points along that that common perpendicular does the Two Reflection Theorem for Translations apply -- other points are slid a different distance in a different direction.
Likewise, we might choose to define "rectangle" as "equiangular quad" in the last unit of the year, especially if the previous (Euclidean) proofs involving rectangles use only the congruence of their angles and not their 90-degree measures. Then those proofs would also be valid on the sphere.
Parallelograms on the Sphere?
Even if we consider all rotations on the sphere to be translations, we will still have separate units for rotations and translations. Up to and including the rotations unit, all proofs will be natural -- that is, they hold in both Euclidean and spherical geometry. Then "translations" as a separate unit from "rotations" implies that we're now in Euclidean geometry, where translations are a separate transformation from rotations. At that point, the proofs will no longer be valid on the sphere.
But now we must ask, what sort of objects might we see in a rotations unit? Well, let's think back to the reflections unit. The simplest polygon with reflection symmetry is the isosceles triangle. Thus this unit focuses on triangles, specifically isosceles triangles. And we also see other natural theorems involving triangles in this unit, such as SSS, SAS, and ASA Congruence.
Now the simplest polygon with rotational symmetry is the equilateral triangle. But the only relevant theorems involving equilateral triangles (that is, all equilateral triangles are equiangular and vice versa) are just corollaries of the Isosceles Triangle Theorem (and its converse). Hence we don't wait until the rotations unit to teach equilateral triangles.
So instead, we move up to four-sided polygons. (We could also move down to two-sided polygons, since lunes have both reflection and rotational symmetry. But since lunes don't truly exist in Euclid, they shouldn't be discussed in natural geometry, which should hold for both Euclid and the sphere.)
Some quadrilaterals have reflection symmetry -- these are kites and isosceles trapezoids. These symmetries are discussed in Lessons 5-4 and 5-5, respectively, of the U of Chicago text (which is prior to translations and rotations in Chapter 6).
But only one type of quad has rotational symmetry in Euclidean geometry -- the parallelogram. This figure has 180-degree symmetry about the point where its diagonals intersect. It's because of this symmetry that its opposite sides and angles are congruent, and that its diagonals bisect each other.
So far, if we were to define "rectangle" as "equiangular quad," then we'd be redefining the term to refer to a quad with the same symmetries as a Euclidean rectangle. Likewise, a spherical "square" is a quadrilateral with the same symmetries as a Euclidean square -- that is, a quad that is equilateral as well as equiangular.
But it's awkward to use the term "parallelogram" to refer to a spherical quad with the same symmetries as a Euclidean pgram. This is because unlike other terms for quads such as "square," the word "parallelogram" begins with the word "parallel." Thus if no parallel lines exist on the sphere, then neither can parallelograms.
Of course, quads with the same 180-degree symmetry exist on the sphere. In fact, we can generate one simply by taking a triangle and rotating it 180 degrees about its midpoint. But its opposite sides won't be parallel, since there are no parallel lines on the sphere.
Notice that some proofs about these quadrilaterals work in both Euclidean and spherical (and hence natural) geometry. For example, we can easily prove:
On the other hand, redefining terms can be confusing, especially for students. After all, if we're going to define "ultraparallel" so that all lines on the sphere are ultraparallel, why don't we just skip the middle man and define "parallel" itself to mean "having a common perpendicular"? The reason, of course, is that defining all lines on the sphere to be parallel is misleading, since all lines on the sphere in fact intersect.
We must keep in mind that our students will only see spherical geometry at the end of the year -- the rest of the year is purely Euclidean. It's actually not terrible to tell students, at the end of the year, "On the sphere, no lines are parallel, but all lines are ultraparallel." We then point out that some previous (Euclidean) proofs are based on ultraparallelism rather than parallelism, and to that extent, such proofs are valid on the sphere.
One U of Chicago proof based on ultraparallelism is the Two Reflection Theorem for Translations:
Two Reflection Theorem for Translations:
If m | | l, the translation r_m o r_l slides figures two times the distance between l and m, in the direction from l to m perpendicular to the lines.
This theorem takes us from the "two mirrors" concept of translation to the "vector" concept, since it provides us with a distance and a direction. The direction is stated as being perpendicular to both lines, so the lines must have a common perpendicular (that is, be ultraparallel).
Notice that while the theorem is valid in all three geometries, only in Euclidean geometry are there many common perpendiculars -- indeed, every point in the plane lies on such a point. In the other two geometries, there's at most one common perpendicular to any pair of lines (otherwise all four lines would form the sides of a rectangle, with four 90-degree angles). Only for points along that that common perpendicular does the Two Reflection Theorem for Translations apply -- other points are slid a different distance in a different direction.
Likewise, we might choose to define "rectangle" as "equiangular quad" in the last unit of the year, especially if the previous (Euclidean) proofs involving rectangles use only the congruence of their angles and not their 90-degree measures. Then those proofs would also be valid on the sphere.
Parallelograms on the Sphere?
Even if we consider all rotations on the sphere to be translations, we will still have separate units for rotations and translations. Up to and including the rotations unit, all proofs will be natural -- that is, they hold in both Euclidean and spherical geometry. Then "translations" as a separate unit from "rotations" implies that we're now in Euclidean geometry, where translations are a separate transformation from rotations. At that point, the proofs will no longer be valid on the sphere.
But now we must ask, what sort of objects might we see in a rotations unit? Well, let's think back to the reflections unit. The simplest polygon with reflection symmetry is the isosceles triangle. Thus this unit focuses on triangles, specifically isosceles triangles. And we also see other natural theorems involving triangles in this unit, such as SSS, SAS, and ASA Congruence.
Now the simplest polygon with rotational symmetry is the equilateral triangle. But the only relevant theorems involving equilateral triangles (that is, all equilateral triangles are equiangular and vice versa) are just corollaries of the Isosceles Triangle Theorem (and its converse). Hence we don't wait until the rotations unit to teach equilateral triangles.
So instead, we move up to four-sided polygons. (We could also move down to two-sided polygons, since lunes have both reflection and rotational symmetry. But since lunes don't truly exist in Euclid, they shouldn't be discussed in natural geometry, which should hold for both Euclid and the sphere.)
Some quadrilaterals have reflection symmetry -- these are kites and isosceles trapezoids. These symmetries are discussed in Lessons 5-4 and 5-5, respectively, of the U of Chicago text (which is prior to translations and rotations in Chapter 6).
But only one type of quad has rotational symmetry in Euclidean geometry -- the parallelogram. This figure has 180-degree symmetry about the point where its diagonals intersect. It's because of this symmetry that its opposite sides and angles are congruent, and that its diagonals bisect each other.
So far, if we were to define "rectangle" as "equiangular quad," then we'd be redefining the term to refer to a quad with the same symmetries as a Euclidean rectangle. Likewise, a spherical "square" is a quadrilateral with the same symmetries as a Euclidean square -- that is, a quad that is equilateral as well as equiangular.
But it's awkward to use the term "parallelogram" to refer to a spherical quad with the same symmetries as a Euclidean pgram. This is because unlike other terms for quads such as "square," the word "parallelogram" begins with the word "parallel." Thus if no parallel lines exist on the sphere, then neither can parallelograms.
Of course, quads with the same 180-degree symmetry exist on the sphere. In fact, we can generate one simply by taking a triangle and rotating it 180 degrees about its midpoint. But its opposite sides won't be parallel, since there are no parallel lines on the sphere.
Notice that some proofs about these quadrilaterals work in both Euclidean and spherical (and hence natural) geometry. For example, we can easily prove:
- If the opposite sides of a quad are congruent, then so are its opposite angles.
Proof:
Draw in one of its diagonals. This divides the quad into two triangles that are then easily proved congruent by SSS. Then by CPCTC, corresponding angles of these triangles are congruent, which then forces the opposite angles of the original quad to be congruent as well. QED
A full two-column proof would entail labeling the vertices of the quad as ABCD and then proving, say, that Triangles ABC and CDA are congruent.
Even though this proof is naturally valid, we'd almost never prove it this way in Euclid. Instead, we'd simply write:
Proof:
Since its opposite sides are congruent, the quad is a pgram. And since it is a pgram, its opposite angles are congruent. QED
That is, we teach high school students to think in terms of parallelograms -- pgrams have certain properties, and any quad with these properties is a pgram.
Possible Solutions to the Parallelogram Problem
We might argue that we could just call them "parallelograms" when teaching them for the first time, then suddenly switching to a new name in the spherical unit, such as "upsidedowngrams" (since due to their 180-degree symmetry, they look the same upside-down). Then spherical "upsidedowngrams" have all the same properties as Euclidean "upsidedowngrams" (pgrams) -- except, of course, for the parallelism of their opposite sides (and those referring to angle sums).
But at some point, we'd like to give the standard Euclidean proof of the pgram properties. In order for our earlier proof to work:
Proof:
Since its opposite sides are congruent, the quad is a pgram. And since it is a pgram, its opposite angles are congruent. QED
Then we must have previously proved two pgram theorems:
- If the opposite sides of a quad are congruent, then the quad is a pgram (that is, those opposite sides are also parallel).
- If a quad is a pgram (its opposite sides are parallel), then its opposite angles are congruent.
The second statement is technically natural (holding vacuously on the sphere), but the first one definitely fails on the sphere. As I wrote above, there exist spherical quads whose opposite sides are congruent, yet those sides definitely aren't parallel.
Oh, and replacing "parallel" with "ultraparallel" doesn't help us here. Yes, the first statement then becomes valid in natural geometry (since all lines are "ultraparallel" on the sphere), but then the second statement becomes invalid (as all lines are "ultraparallel," all quads would then become pgrams, yet not all quads have congruent opposite angles).
By the way, we might notice that here I'm suddenly insisting that "parallel" means "parallel" (so that no parallel lines exist on the sphere), and that "parallelogram" means "parallelogram" (so that no pgrams exist on the sphere) since it contains the word "parallel." Yet "ultraparallel" doesn't mean "parallel" (so that all lines are ultraparallel on the sphere) despite containing the word "parallel."
Then again, we could introduce the idea of ultraparallelism without using that word. Instead, we might use something like "coperpendicular" to mean "having a common perpendicular" (in analogy with "collinear" to mean "lying on a common line"). Besides, so far the only proof so far that requires lines to be coperpendicular is the Two Reflection Theorem for Translations -- and that appears in the translations unit, by which we expect geometry to be fully Euclidean anyway. So it might not even be worth mentioning "ultraparallel" at all.
The important thing here is, how can we incorporate our traditional Euclidean pgram proofs here? We could simply wait until the translations unit and Euclidean geometry. But then this would leave our rotations unit mostly devoid of content.
We could also move parallel lines and Euclidean geometry down into the rotations unit. Recall that I'm still not quite sure how well our proofs on translations and parallel lines will work anyway. We might prefer to use the Wu/EngageNY proofs based on 180-degree rotations. Then this would fit better in the rotations unit anyway.
How Spherical Rotations Differ From Euclid
Wu uses 180-degree rotations to prove that parallel lines exist on the Euclidean plane. But 180-degree rotations also exist on the sphere. Why, then, doesn't this prove that parallel lines exist on the sphere as well? (That is, why isn't that proof natural?)
The reason is that spherical rotations lack a key property of Euclidean rotations:
Theorem (Euclidean geometry):
Let R be a rotation of magnitude theta. Then if l' is the image of line l under rotation R, and lines l and l' intersect, then the angle at that intersection must be exactly theta.
It's this property that forces 180-degree images to be parallel. For when theta = 180, the lines l and l' would have to intersect at 180 degrees -- but since a 180-degree angle is a straight line, this would force l and l' to be the same line. Thus unless l is invariant, this forces l and l' to be parallel.
But it's clear that this theorem is false on the sphere even for angles other than 180. For example, we consider the case theta = 90. We start with a line l and rotate it thrice under this rotation, to obtain the three image lines l', l", and l'". The fourth image would be l itself.
Now if the claim above held in spherical geometry, then l, l', l", and l'" would be the sides of a rectangle with four right angles. But this doesn't exist on the sphere. Hence the angle between a line and its image under a 90-degree rotation can't possibly be 90 (unless the center of rotation lies on l, in which case we obtain two lines l and l' intersecting at that center, not a rectangle). Notice that since isometries preserve angle measure, the quad formed by l, l', l", and l'" must be equiangular (indeed equilateral, since they also preserve distance).
Glide Reflections on the Sphere
On the Euclidean plane, there are four types of isometries -- reflections, rotations, translations, and glide reflections. This last isometry is the most complex, as well as the most forgotten.
Do glide reflections exist on the sphere? Well, we usually think of a glide reflection as the composite of a reflection and a translation. If we ignore the "ultraparallel" loophole, then translations don't exist on the sphere since parallel lines don't exist. Thus glide reflections wouldn't exist on the sphere either.
But there definitely exist isometries which are the composite of three reflections yet aren't equivalent to a single reflection. On the Euclidean plane, we call these isometries "glide reflections." Thus it seems logical to call these glide reflections on the sphere as well.
Once again, we turn to the third geometry -- hyperbolic geometry -- for help. In hyperbolic geometry there are three isometries that preserve orientation (rotations, translations, and horolations) yet only two that reverse orientation (reflections and glide reflections). Thus each of the three geometries has the same number of indirect isometries despite having different numbers of direct isometries.
What is the composite of a reflection and a rotation, in either Euclidean or hyperbolic geometry? The answer is, unless we're lucky, a glide reflection. And in the hyperbolic plane, the composite of a reflection and a horolation is usually a glide reflection as well. In both neutral geometries, if we take the composite of a reflection and any direct isometry, the composite is usually a glide reflection. That is, in all cases, there exists a translation (even if the original direct isometry is a rotation or horolation) and a mirror, in the same direction as the translation, such that the composite of a reflection in that new mirror and that translation is equivalent to the composite of the reflection in the original mirror and the original direct isometry. (Once in a while, the vector of the translation works out to be the zero vector, and so the composite is merely a reflection instead of a glide reflection.)
Thus returning to the sphere, there's nothing wrong with referring to the composite of a reflection and a rotation as a glide reflection, even if you don't consider rotations to be translations on the sphere. In any of the three geometries, the composite of a reflection and a rotation is a glide reflection (reducing to merely a reflection if the center of the rotation lies on the mirror of the reflection).
I once considered having a glide reflections unit in our geometry course, but this unit might be even more difficult to fill. There is only one glide reflection theorem in the entire text, in Lesson 6-6:
Glide Reflection Theorem:
In a glide reflection, the midpoint of the segment connecting a point to its image lies on the glide-reflection line (that is, the mirror of the reflection component of the glide reflection).
The proof given here is valid only in Euclidean geometry, since it uses the Midpoint Connector Theorem (which rules out both non-Euclidean geometries).
The theorem assumes that there is a unique "mirror" of the glide reflection -- that is, that there exists only one possible mirror and one possible translation in the direction of the mirror, such that the composite of that reflection and that translation is the given glide reflection. We can think of this as the "canonical representation" of the glide reflection.
On the sphere, it's not quite obvious what a translation (or rotation) in the same direction as the mirror even means. If we follow Euclidean geometry, it ultimately means that the mirror of the reflection is a common perpendicular of the two mirrors whose composite is the translation. (In other words, "ultraparallel" strikes again!)
Then on the sphere, I believe that all glide reflections have canonical representations except for one very special glide reflection -- the antipodal map (that is, the isometry that maps every point on the sphere to its antipodal point). In fact, the composite of any three reflections in mutually perpendicular mirrors is the antipodal map. And the composite of any reflection and the 180-degree rotation centered at the poles of the mirror is the antipodal map.
Feynman, Pages 71-160
It's time for us to continue our summer side-along graphic novel, Jim Ottaviani's Feynman. We left off with our hero at Los Alamos to begin work on the Manhattan Project, but somehow he's been sneaking off and back on the base. Let's find out how he does it:
Richard: Glad you noticed, but I already told you people -- there's a hole in the fence around the side here and you can walk right through. Ya know, there's a better solution here. How about instead of arresting me you fix the hole!
Military Leader: I have a better use for your talents, Feynman. Bob Christy is sick, so we need you to go to Oak Ridge, Tennessee, and fix some important holes.
Richard: And do what?
Military Leader: They need help making their plant safe.
Richard: How'm I gonna do that?
Military Leader: There's a lot of military brass there, but also some technically able people among them. Make sure they're at the meeting I've called.
Richard: And what if they're not? Or what if they don't listen?
Fortunately, these guys are all too eager to work with Feynman:
Guy #1: We'll try it!
Guy #2: Double shifts!
Guy #3: Yeah, I want me some overtime!
"I knew my guys would say that," he narrates. "I had them start right away (June 16th, 1945, 9:22 P.M.) -- all other problems were out. And that's when I got the call."
Caller: You'd better come down here right away.
"Her (my sick wife Arline's) condition had become much worse. I had arranged ahead of time with my friend Klaus to borrow his car in case of an emergency." (The author explains in a footnote that Klaus Fuchs turned out to be a Russian spy, and he's used this same car to take secrets out of Los Alamos. But nobody knew that then.) "I picked up a couple of hitchhikers. You know, to help me in case something happened. We hitchhiked most of the way."
But unfortunately, it was too late. Arline passed away of TB shortly thereafter.
I must include the following scene in this post, because it takes place on today's date -- July 16th, 1945, at exactly 05:29:45 Mountain "War Time" (that is, Year-Round Daylight Saving Time used during World War II). This is the day, 74 years ago today, that the Manhattan Project is completed:
Richard: Hey! What's going on? What's with all the cards? I told you you're not supposed to do more than one problem -- only one problem!
Guy #1: Get out!
Guy #2: Get out!
Guy #3: We'll explain later.
Guy #1: Leave us alone!
Guy #2: We'll explain later!
"You know how it is: they must've had trouble with the calculation and figured out a way to fix it," he narrates, "then the boss comes walking in. I left them alone and the calculation worked out fine."
Someone: Richy, there's a telegram.
"I was home on a small vacation, fooling around on a shortwave, when I got the message telling me that the time for calculations was over. It said, 'The baby is expected on July 16th.' So I rushed back, arriving just before the buses left for the test. We drove to the Trinity test site -- actually, they dropped us off about 20 miles away. We settled in for the wait. I checked on the other group to make sure their radios worked..."
Anyway, at 5:30 A.M., the "baby" is delivered -- the first atomic bomb. It's sad what happens to the civilians of Hiroshima and Nagasaki, but this post is all about Feynman and how his team ultimately wins the war for the Allies.
After the project, Richard moves on to Tennessee, where he learns more about safecracking:
"So I began to work in earnest, and in about two minutes...I told the colonel how I did it, and about the danger of leaving the doors open so anybody could pick off the last two numbers. His solution?"
And the colonel types:
Attn: All Oak Ridge Personnel
Re: Possible security breach
If Mr. Feynman has at any time been in your office, or near your office, or walking through your office, please change the combination of your safe.
"That was it: I was the danger. It's a pain in the neck to remember a new combination, so the next time I visited, nobody was happy to see me."
Back in New Mexico, Feynman works with the forerunners of modern computing machines in 1945:
"As head of T4, I had an army special engineering attachment under me. Clever high school boys with engineering ability who they sent to Los Alamos, put in barracks, and..."
Richard: ...they got told nothing! It's like at Oak Ridge, only worse."
After the war, Feynman is hired as a professor at Cornell, but he doesn't do real research there:
"Now that I'm burned out and won't ever accomplish anything," he narrates, "I'm going to play with physics, whenever I want to..."
Richard (spinning a plate): Hey, watch this!
"...without worrying about whether it's important."
Richard: I wonder why the medallion goes around slower than the wobble?
"Since I had nothing better to do, I started to figure out the motion of the rotating plate."
Hey, this blog post is titled "rotations," so of course I had to mention something rotating here! And of course, the physicist eventually finds the equations to describe the rotation of the plate.
The author shows us "A Little Hooey," a rivalry between Feynman and another physicist:
"So it turned out this way," he narrates. "Julian Schwinger and I climbed the same theoretical mountain, but from different sides. We were trying to correct...well...complete...work done by Dirac, who determined the strength of the magnetic moment in an electron to be 1 in certain units. But nobody could calculate it more accurately without getting screwy answers."
Dirac (flashback): It seems that some essentially new physical ideas are here needed.
"So I had some. And I started making all these pictures to visualize the stuff going on in the equations."
Freeman Dyson: Dick, it's me, Dyson.
Japanese scientists have been working on the same work during the war, but not until 1948 do they attempt to contact Feynman or any American scientists:
Scientist #1 (in Japanese): Yes, well. Now that we have saved enough money for postage, I would like to get the opinion of the Americans on the work I did during the war.
Scientist #2: Who will you send it to?
Scientist #1: I have no idea who might be working on this. Nobody, perhaps. But the famous Mr. J. Robert Oppenheimer will know.
And of course, Oppenheimer sends the material to Feynman. He and Dyson go on a road trip through Oklahoma to attend another meeting to discuss what they had learned from the Japanese.
Richard: Heya, what's goin' on?
Cop: Big floods ahead. The road isn't passable beyond town.
Freeman: This makes the hardest rains in England look like a drizzle. Perhaps we should find a room to kip in?
Richard: Aw, c'mon. Let's see if we can make it.
"We drove another 10 miles or so," he narrates, "and then reached an arroyo."
Freeman: Well?
Richard: Yeah, okay. Even for me that's too much.
"We headed back to town."
Richard: Don't worry, happens all the time. I'll find us a place.
As a Californian, I must of course show the scene where Feynman takes "the scenic route" as he moves from Cornell to Caltech:
"Cornell had some good people," he narrates, "but also some really inane departments, like 'domestic science.' The weather wasn't so hot either."
Richard: I can give you a hand with that physics homework if you like.
Girl: Hey, I've heard of you. You're not a student. You're Professor Feynman!
"And...my cover was blown. The coeds had caught on. So I figured there must be a part of the world that didn't have these problems. Professor Bacher had gone from Cornell to Caltech, and I'd visited him a couple of times."
Bacher: Here, borrow my car and have a look at Hollywood and the Sunset Strip.
"He knew me well. Caltech had good weather and a college with a 100% scientific focus. Perfect for a one-sided guy like me."
Before settling at Caltech, though, Feynman takes a side trip to Brazil:
"We (my Brazilian samba friend and I) walked at random all over Copacabana," he narrates. "It was wonderful. There was a competition just before Carnaval, and we entered it."
Friend: Professor Feynman, the samba schools of the beaches compete tonight. Copacabana, Ipanema, and Leblon beaches will all be represented.
Richard: Ah, I'm kinda busy tonight. I may not make it.
Friend: But you must see it -- it is not like...well, these who just pass through. It is tipico brasileiro.
"He said I'd love it so much, and was sorry I'd miss it. So we farcantes joined the march, which eventually passed in front of my hotel.
Friend: O professoooor!!
"We won the competition, and I was a real farcante."
Feynman surely likes to travel, doesn't he?. Still before making it to Pasadena, he goes to Japan:
Richard: And when I ask you to do it, the word "solve" is different?
Japanese scientist: Yes, that's so.
Richard: But when I solve it, and when you solve it, it's the same damn thing!
Japanese scientist: Yes, but it's more polite that way. (in Japanese) I told him, but...
"I gave up," he narrates. "This stuff about when it's me it's miserable, when it's you it's elegant? Not the language for me."
Feynmann arrives at Caltech in 1952. "As a full professor I had more sabbaticals, but Caltech was so perfect for me that sometimes I stayed on campus for them."
In 1957, Feynman meets an artist named Jirayr at a party. The pictures show him drawing some art on his body -- with toothpaste:
Jirayr: Whoo! Whoo hooo!
"I liked this guy right away," he narrates.
When Feynman meets an English au pair girl, he discusses his cooking skills -- or lack thereof:
Richard: There's only nine permutations of meals from that, and then it repeats! Boring!
Gweneth: Well, I cannot really say that the idea of America had crossed my mind. I was thinking Australia would be my next port of call, actually.
"I persuaded her," he narrates, "though it took a while and then a while longer to get her a visa. It wouldn't do to have me be the sponsor though, so my friend Matt Sands helped out. She lived in the other half of my duplex. For over a year Gweneth dated other guys, and I had a few dates myself. And then, we sort of discovered each other. We got married on September 4th, 1960."
Today's reading section concludes with Feynman discussing his famous quantum theory -- and whether it can be used to create much smaller quantum computers:
"If we wanted to make a computer having this marvelous quantitative ability," he explains, "we'd have to make it, perhaps, the size of the Pentagon. This has disadvantages. First, it requires too much stuff. There's also the problem of heat generation, and power consumption. And because information can't travel faster than the speed of light, to make our computers faster and more elaborate, we'll have to make them smaller."
Richard: But there's nothing I can see in the physical laws that says they can't be made enormously smaller than they are now. There's plenty of room. Now, you might say, "Who will do this and why should they?" Well, the reason I might do it is just for fun. Perhaps only economics will excite you, though.
"So I offered $1,000 to the first guy to take the page of a book and shrink it 25,000 times in such a manner that it can be read by an electron microscope. And I offered another $1,000 prize to the first guy who made an operating electric motor only 1/64 inch on each side."
So I guess we'll have to wait until a later post to find out whether anyone is able to claim either of the physicist's thousand-dollar prizes.
Old Charter School Update
Late last week, I visited my old charter school once again. I'm still looking for clues as to whether my old school is appealing the charter renewal denial with the county. As I wrote in my last post, by now it appears that the sister charter has school already given up and is closed completely.
Anyway, here's what I spotted that day -- the Clutter truck was no longer there, and signs advertising the charter were still there. Even though the front door was locked, there definitely appeared to be someone inside! So I immediately dialed their phone number -- but once again, all I got was a voicemail message.
Once again, the school seems to be stuck in this holding pattern. If the charter has been denied and the county isn't jumping in to save the school, then why are the signs still up, rather than taken down as quickly as those at the sister charter? If, on the other hand, the school will reopen in the fall, then why is the front door locked and no one is taking my phone calls? Considering the advertisements, it's possible that some parents might wish to enroll their kindergartners for the fall, but no one at the school is responding to visitors.
Perhaps the school has appealed to the county, but the county hasn't responded to the appeal yet. I suppose the county doesn't have to make a decision until the night before the first day of school -- but then that leaves students and parents in a tough position.
I plan on making another visit to the school in a few days -- and this time, I have another plan to extract some information about the charter situation.
I've also been revisiting my blog posts from the year that I worked at the old charter school. I've still looking for my most successful lessons from my most important class -- the seventh grade. But as I admitted in an earlier post, most of what I blogged was about the eighth grade class. And so I must read between the lines -- in my own posts -- to recall how my seventh graders did.
One particularly successful week that year was January 30th-February 3rd -- since I did write in my February 3rd post that the seventh graders were successful on the quiz that day. So we ask, what were the seventh graders studying that week, and what made the lesson so successful?
Well, the lesson that day was on the following standard:
CCSS.MATH.CONTENT.7.EE.A.1
Apply properties of operations as strategies to add, subtract, factor, and expand linear expressions with rational coefficients.
And I believe what made the students so successful on this lesson is that these linear expressions didn't contain any negative signs! In other words, the reason that so many seventh graders struggle on this standard is not because they don't understand combining like terms or the distributive property, but that they're still struggling with integer operations.
In my class, I was confused by the order in which to teach the lessons in the Illinois State text -- first going by "STEM projects" rather than the order intended by Illinois State (all the RP standards, then all the NS standards, then all the EE's, etc.). Based on the STEM projects, I ended up not teaching integers until January/February, when I was told which standards I needed to cover. Somehow, I still ended up covering EE1 before the NS standards on integers (which I finally covered a week later).
High school students still struggle with signs, of course. Some lessons in Algebra I are completely dependent on negative numbers -- consider the formulas used in that class that contain negative signs (slope, vertex of a parabola, and ultimately the Quadratic Formula).
The sixth graders that week studied division of fractions. Of course, fractions are another subject where students struggle. But we can avoid fractions in Algebra I (for example, multiplying both sides of an equation to clear the fractions) in a way that we can't avoid negative numbers.
Thus even though my lesson that week was superficially successful, in reality I needed to teach negative numbers more effectively so that those seventh graders would go on to do well in eighth grade and high school math.
Conclusion
As I wrote earlier in this post, today is the 74th anniversary of the Manhattan Project. It is also another much more publicized anniversary -- 50 years since the launch of Apollo 11. (The actual moon landing was four days later, on July 20th, 1969, but July 16th was the launch.) And there just so happens to be a full moon today. Here in Southern California, there's an Apollo 11 exhibit at a local museum -- the Columbia Memorial Space Center in Downey:
https://www.columbiaspacescience.org/apollo
(Yes, this means that on those Calendar Reform calendars based on the year of the moon landing, we're now officially in Year 50.)
It's actually on a spherical body without oceans, such as the moon, where the rules of spherical geometry become most evident. For example, some people wonder why we insist that only great circles on the sphere are considered to be lines. Thus, for example, the states of Colorado and Wyoming aren't rectangles because two of their sides aren't lines.
On the moon, if we move (in a lunar rover, say), without using a steering wheel, then we will go automatically along a great circle only. Thus no quads with four right angles can exist on the surface, since we can't move four rovers along the sides of such a quad without steering.
On the Euclidean plane, there are four types of isometries -- reflections, rotations, translations, and glide reflections. This last isometry is the most complex, as well as the most forgotten.
Do glide reflections exist on the sphere? Well, we usually think of a glide reflection as the composite of a reflection and a translation. If we ignore the "ultraparallel" loophole, then translations don't exist on the sphere since parallel lines don't exist. Thus glide reflections wouldn't exist on the sphere either.
But there definitely exist isometries which are the composite of three reflections yet aren't equivalent to a single reflection. On the Euclidean plane, we call these isometries "glide reflections." Thus it seems logical to call these glide reflections on the sphere as well.
Once again, we turn to the third geometry -- hyperbolic geometry -- for help. In hyperbolic geometry there are three isometries that preserve orientation (rotations, translations, and horolations) yet only two that reverse orientation (reflections and glide reflections). Thus each of the three geometries has the same number of indirect isometries despite having different numbers of direct isometries.
What is the composite of a reflection and a rotation, in either Euclidean or hyperbolic geometry? The answer is, unless we're lucky, a glide reflection. And in the hyperbolic plane, the composite of a reflection and a horolation is usually a glide reflection as well. In both neutral geometries, if we take the composite of a reflection and any direct isometry, the composite is usually a glide reflection. That is, in all cases, there exists a translation (even if the original direct isometry is a rotation or horolation) and a mirror, in the same direction as the translation, such that the composite of a reflection in that new mirror and that translation is equivalent to the composite of the reflection in the original mirror and the original direct isometry. (Once in a while, the vector of the translation works out to be the zero vector, and so the composite is merely a reflection instead of a glide reflection.)
Thus returning to the sphere, there's nothing wrong with referring to the composite of a reflection and a rotation as a glide reflection, even if you don't consider rotations to be translations on the sphere. In any of the three geometries, the composite of a reflection and a rotation is a glide reflection (reducing to merely a reflection if the center of the rotation lies on the mirror of the reflection).
I once considered having a glide reflections unit in our geometry course, but this unit might be even more difficult to fill. There is only one glide reflection theorem in the entire text, in Lesson 6-6:
Glide Reflection Theorem:
In a glide reflection, the midpoint of the segment connecting a point to its image lies on the glide-reflection line (that is, the mirror of the reflection component of the glide reflection).
The proof given here is valid only in Euclidean geometry, since it uses the Midpoint Connector Theorem (which rules out both non-Euclidean geometries).
The theorem assumes that there is a unique "mirror" of the glide reflection -- that is, that there exists only one possible mirror and one possible translation in the direction of the mirror, such that the composite of that reflection and that translation is the given glide reflection. We can think of this as the "canonical representation" of the glide reflection.
On the sphere, it's not quite obvious what a translation (or rotation) in the same direction as the mirror even means. If we follow Euclidean geometry, it ultimately means that the mirror of the reflection is a common perpendicular of the two mirrors whose composite is the translation. (In other words, "ultraparallel" strikes again!)
Then on the sphere, I believe that all glide reflections have canonical representations except for one very special glide reflection -- the antipodal map (that is, the isometry that maps every point on the sphere to its antipodal point). In fact, the composite of any three reflections in mutually perpendicular mirrors is the antipodal map. And the composite of any reflection and the 180-degree rotation centered at the poles of the mirror is the antipodal map.
Feynman, Pages 71-160
It's time for us to continue our summer side-along graphic novel, Jim Ottaviani's Feynman. We left off with our hero at Los Alamos to begin work on the Manhattan Project, but somehow he's been sneaking off and back on the base. Let's find out how he does it:
Richard: Glad you noticed, but I already told you people -- there's a hole in the fence around the side here and you can walk right through. Ya know, there's a better solution here. How about instead of arresting me you fix the hole!
Military Leader: I have a better use for your talents, Feynman. Bob Christy is sick, so we need you to go to Oak Ridge, Tennessee, and fix some important holes.
Richard: And do what?
Military Leader: They need help making their plant safe.
Richard: How'm I gonna do that?
Military Leader: There's a lot of military brass there, but also some technically able people among them. Make sure they're at the meeting I've called.
Richard: And what if they're not? Or what if they don't listen?
Fortunately, these guys are all too eager to work with Feynman:
Guy #1: We'll try it!
Guy #2: Double shifts!
Guy #3: Yeah, I want me some overtime!
"I knew my guys would say that," he narrates. "I had them start right away (June 16th, 1945, 9:22 P.M.) -- all other problems were out. And that's when I got the call."
Caller: You'd better come down here right away.
"Her (my sick wife Arline's) condition had become much worse. I had arranged ahead of time with my friend Klaus to borrow his car in case of an emergency." (The author explains in a footnote that Klaus Fuchs turned out to be a Russian spy, and he's used this same car to take secrets out of Los Alamos. But nobody knew that then.) "I picked up a couple of hitchhikers. You know, to help me in case something happened. We hitchhiked most of the way."
But unfortunately, it was too late. Arline passed away of TB shortly thereafter.
I must include the following scene in this post, because it takes place on today's date -- July 16th, 1945, at exactly 05:29:45 Mountain "War Time" (that is, Year-Round Daylight Saving Time used during World War II). This is the day, 74 years ago today, that the Manhattan Project is completed:
Richard: Hey! What's going on? What's with all the cards? I told you you're not supposed to do more than one problem -- only one problem!
Guy #1: Get out!
Guy #2: Get out!
Guy #3: We'll explain later.
Guy #1: Leave us alone!
Guy #2: We'll explain later!
"You know how it is: they must've had trouble with the calculation and figured out a way to fix it," he narrates, "then the boss comes walking in. I left them alone and the calculation worked out fine."
Someone: Richy, there's a telegram.
"I was home on a small vacation, fooling around on a shortwave, when I got the message telling me that the time for calculations was over. It said, 'The baby is expected on July 16th.' So I rushed back, arriving just before the buses left for the test. We drove to the Trinity test site -- actually, they dropped us off about 20 miles away. We settled in for the wait. I checked on the other group to make sure their radios worked..."
Anyway, at 5:30 A.M., the "baby" is delivered -- the first atomic bomb. It's sad what happens to the civilians of Hiroshima and Nagasaki, but this post is all about Feynman and how his team ultimately wins the war for the Allies.
After the project, Richard moves on to Tennessee, where he learns more about safecracking:
"So I began to work in earnest, and in about two minutes...I told the colonel how I did it, and about the danger of leaving the doors open so anybody could pick off the last two numbers. His solution?"
And the colonel types:
Attn: All Oak Ridge Personnel
Re: Possible security breach
If Mr. Feynman has at any time been in your office, or near your office, or walking through your office, please change the combination of your safe.
"That was it: I was the danger. It's a pain in the neck to remember a new combination, so the next time I visited, nobody was happy to see me."
Back in New Mexico, Feynman works with the forerunners of modern computing machines in 1945:
"As head of T4, I had an army special engineering attachment under me. Clever high school boys with engineering ability who they sent to Los Alamos, put in barracks, and..."
Richard: ...they got told nothing! It's like at Oak Ridge, only worse."
After the war, Feynman is hired as a professor at Cornell, but he doesn't do real research there:
"Now that I'm burned out and won't ever accomplish anything," he narrates, "I'm going to play with physics, whenever I want to..."
Richard (spinning a plate): Hey, watch this!
"...without worrying about whether it's important."
Richard: I wonder why the medallion goes around slower than the wobble?
"Since I had nothing better to do, I started to figure out the motion of the rotating plate."
Hey, this blog post is titled "rotations," so of course I had to mention something rotating here! And of course, the physicist eventually finds the equations to describe the rotation of the plate.
The author shows us "A Little Hooey," a rivalry between Feynman and another physicist:
"So it turned out this way," he narrates. "Julian Schwinger and I climbed the same theoretical mountain, but from different sides. We were trying to correct...well...complete...work done by Dirac, who determined the strength of the magnetic moment in an electron to be 1 in certain units. But nobody could calculate it more accurately without getting screwy answers."
Dirac (flashback): It seems that some essentially new physical ideas are here needed.
"So I had some. And I started making all these pictures to visualize the stuff going on in the equations."
Freeman Dyson: Dick, it's me, Dyson.
Japanese scientists have been working on the same work during the war, but not until 1948 do they attempt to contact Feynman or any American scientists:
Scientist #1 (in Japanese): Yes, well. Now that we have saved enough money for postage, I would like to get the opinion of the Americans on the work I did during the war.
Scientist #2: Who will you send it to?
Scientist #1: I have no idea who might be working on this. Nobody, perhaps. But the famous Mr. J. Robert Oppenheimer will know.
And of course, Oppenheimer sends the material to Feynman. He and Dyson go on a road trip through Oklahoma to attend another meeting to discuss what they had learned from the Japanese.
Richard: Heya, what's goin' on?
Cop: Big floods ahead. The road isn't passable beyond town.
Freeman: This makes the hardest rains in England look like a drizzle. Perhaps we should find a room to kip in?
Richard: Aw, c'mon. Let's see if we can make it.
"We drove another 10 miles or so," he narrates, "and then reached an arroyo."
Freeman: Well?
Richard: Yeah, okay. Even for me that's too much.
"We headed back to town."
Richard: Don't worry, happens all the time. I'll find us a place.
As a Californian, I must of course show the scene where Feynman takes "the scenic route" as he moves from Cornell to Caltech:
"Cornell had some good people," he narrates, "but also some really inane departments, like 'domestic science.' The weather wasn't so hot either."
Richard: I can give you a hand with that physics homework if you like.
Girl: Hey, I've heard of you. You're not a student. You're Professor Feynman!
"And...my cover was blown. The coeds had caught on. So I figured there must be a part of the world that didn't have these problems. Professor Bacher had gone from Cornell to Caltech, and I'd visited him a couple of times."
Bacher: Here, borrow my car and have a look at Hollywood and the Sunset Strip.
"He knew me well. Caltech had good weather and a college with a 100% scientific focus. Perfect for a one-sided guy like me."
Before settling at Caltech, though, Feynman takes a side trip to Brazil:
"We (my Brazilian samba friend and I) walked at random all over Copacabana," he narrates. "It was wonderful. There was a competition just before Carnaval, and we entered it."
Friend: Professor Feynman, the samba schools of the beaches compete tonight. Copacabana, Ipanema, and Leblon beaches will all be represented.
Richard: Ah, I'm kinda busy tonight. I may not make it.
Friend: But you must see it -- it is not like...well, these who just pass through. It is tipico brasileiro.
"He said I'd love it so much, and was sorry I'd miss it. So we farcantes joined the march, which eventually passed in front of my hotel.
Friend: O professoooor!!
"We won the competition, and I was a real farcante."
Feynman surely likes to travel, doesn't he?. Still before making it to Pasadena, he goes to Japan:
Richard: And when I ask you to do it, the word "solve" is different?
Japanese scientist: Yes, that's so.
Richard: But when I solve it, and when you solve it, it's the same damn thing!
Japanese scientist: Yes, but it's more polite that way. (in Japanese) I told him, but...
"I gave up," he narrates. "This stuff about when it's me it's miserable, when it's you it's elegant? Not the language for me."
Feynmann arrives at Caltech in 1952. "As a full professor I had more sabbaticals, but Caltech was so perfect for me that sometimes I stayed on campus for them."
In 1957, Feynman meets an artist named Jirayr at a party. The pictures show him drawing some art on his body -- with toothpaste:
Jirayr: Whoo! Whoo hooo!
"I liked this guy right away," he narrates.
When Feynman meets an English au pair girl, he discusses his cooking skills -- or lack thereof:
Richard: There's only nine permutations of meals from that, and then it repeats! Boring!
Gweneth: Well, I cannot really say that the idea of America had crossed my mind. I was thinking Australia would be my next port of call, actually.
"I persuaded her," he narrates, "though it took a while and then a while longer to get her a visa. It wouldn't do to have me be the sponsor though, so my friend Matt Sands helped out. She lived in the other half of my duplex. For over a year Gweneth dated other guys, and I had a few dates myself. And then, we sort of discovered each other. We got married on September 4th, 1960."
Today's reading section concludes with Feynman discussing his famous quantum theory -- and whether it can be used to create much smaller quantum computers:
"If we wanted to make a computer having this marvelous quantitative ability," he explains, "we'd have to make it, perhaps, the size of the Pentagon. This has disadvantages. First, it requires too much stuff. There's also the problem of heat generation, and power consumption. And because information can't travel faster than the speed of light, to make our computers faster and more elaborate, we'll have to make them smaller."
Richard: But there's nothing I can see in the physical laws that says they can't be made enormously smaller than they are now. There's plenty of room. Now, you might say, "Who will do this and why should they?" Well, the reason I might do it is just for fun. Perhaps only economics will excite you, though.
"So I offered $1,000 to the first guy to take the page of a book and shrink it 25,000 times in such a manner that it can be read by an electron microscope. And I offered another $1,000 prize to the first guy who made an operating electric motor only 1/64 inch on each side."
So I guess we'll have to wait until a later post to find out whether anyone is able to claim either of the physicist's thousand-dollar prizes.
Old Charter School Update
Late last week, I visited my old charter school once again. I'm still looking for clues as to whether my old school is appealing the charter renewal denial with the county. As I wrote in my last post, by now it appears that the sister charter has school already given up and is closed completely.
Anyway, here's what I spotted that day -- the Clutter truck was no longer there, and signs advertising the charter were still there. Even though the front door was locked, there definitely appeared to be someone inside! So I immediately dialed their phone number -- but once again, all I got was a voicemail message.
Once again, the school seems to be stuck in this holding pattern. If the charter has been denied and the county isn't jumping in to save the school, then why are the signs still up, rather than taken down as quickly as those at the sister charter? If, on the other hand, the school will reopen in the fall, then why is the front door locked and no one is taking my phone calls? Considering the advertisements, it's possible that some parents might wish to enroll their kindergartners for the fall, but no one at the school is responding to visitors.
Perhaps the school has appealed to the county, but the county hasn't responded to the appeal yet. I suppose the county doesn't have to make a decision until the night before the first day of school -- but then that leaves students and parents in a tough position.
I plan on making another visit to the school in a few days -- and this time, I have another plan to extract some information about the charter situation.
I've also been revisiting my blog posts from the year that I worked at the old charter school. I've still looking for my most successful lessons from my most important class -- the seventh grade. But as I admitted in an earlier post, most of what I blogged was about the eighth grade class. And so I must read between the lines -- in my own posts -- to recall how my seventh graders did.
One particularly successful week that year was January 30th-February 3rd -- since I did write in my February 3rd post that the seventh graders were successful on the quiz that day. So we ask, what were the seventh graders studying that week, and what made the lesson so successful?
Well, the lesson that day was on the following standard:
CCSS.MATH.CONTENT.7.EE.A.1
Apply properties of operations as strategies to add, subtract, factor, and expand linear expressions with rational coefficients.
And I believe what made the students so successful on this lesson is that these linear expressions didn't contain any negative signs! In other words, the reason that so many seventh graders struggle on this standard is not because they don't understand combining like terms or the distributive property, but that they're still struggling with integer operations.
In my class, I was confused by the order in which to teach the lessons in the Illinois State text -- first going by "STEM projects" rather than the order intended by Illinois State (all the RP standards, then all the NS standards, then all the EE's, etc.). Based on the STEM projects, I ended up not teaching integers until January/February, when I was told which standards I needed to cover. Somehow, I still ended up covering EE1 before the NS standards on integers (which I finally covered a week later).
High school students still struggle with signs, of course. Some lessons in Algebra I are completely dependent on negative numbers -- consider the formulas used in that class that contain negative signs (slope, vertex of a parabola, and ultimately the Quadratic Formula).
The sixth graders that week studied division of fractions. Of course, fractions are another subject where students struggle. But we can avoid fractions in Algebra I (for example, multiplying both sides of an equation to clear the fractions) in a way that we can't avoid negative numbers.
Thus even though my lesson that week was superficially successful, in reality I needed to teach negative numbers more effectively so that those seventh graders would go on to do well in eighth grade and high school math.
Conclusion
As I wrote earlier in this post, today is the 74th anniversary of the Manhattan Project. It is also another much more publicized anniversary -- 50 years since the launch of Apollo 11. (The actual moon landing was four days later, on July 20th, 1969, but July 16th was the launch.) And there just so happens to be a full moon today. Here in Southern California, there's an Apollo 11 exhibit at a local museum -- the Columbia Memorial Space Center in Downey:
https://www.columbiaspacescience.org/apollo
(Yes, this means that on those Calendar Reform calendars based on the year of the moon landing, we're now officially in Year 50.)
It's actually on a spherical body without oceans, such as the moon, where the rules of spherical geometry become most evident. For example, some people wonder why we insist that only great circles on the sphere are considered to be lines. Thus, for example, the states of Colorado and Wyoming aren't rectangles because two of their sides aren't lines.
On the moon, if we move (in a lunar rover, say), without using a steering wheel, then we will go automatically along a great circle only. Thus no quads with four right angles can exist on the surface, since we can't move four rovers along the sides of such a quad without steering.
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