Crystals, Non-Crystals and Quasicrystals

Quasicrystal1 — **Quasicrystalline ordering** of a aluminum-palladium-manganese alloy. Image by J.W.Evans via Wikipedia.

100px-Halite — Regular ordering of a halite **crystal**. The atoms that make up salt crystals are arranged in a cubic shape. The smaller grey atoms are sodium (Na), and the larger green ones are chlorine (Cl).

Daniel Shechtman was just awarded the Nobel Prize in Chemistry (2011). He discovered that matter can exist not only as crystals, which have a regular geometric arrangement of atoms, and amorphous non-crystals that do not, but also as quasicrystals which have a different type of atomic ordering.

The Guardian has an interesting article on Schechtman, whose discovery was roundly disbelieved by other scientists and he was ridiculed for years.

In an interview this year with the Israeli newspaper, Haaretz, Shechtman said: “People just laughed at me.” He recalled how Linus Pauling, a colossus of science and a double Nobel laureate, mounted a frightening “crusade” against him. After telling Shechtman to go back and read a crystallography textbook, the head of his research group asked him to leave for “bringing disgrace” on the team. “I felt rejected,” Shachtman said.

— Sample (2011): Nobel Prize in Chemistry for dogged work on ‘impossible’ quasicrystals in The Guardian

Hat tip to M. Eisenberg for this link.

93 Ways to Prove Pythagoras’ Theorem

Pythagoras-2a-300 — Geometric proof of the Pythagorean Theorem by rearrangemention from Wikimedia Commons' user Joaquim Alves Gat. Animaspar.

Elegant in its simplicity but profound in its application, the Pythagorean Theorem is one of the fundamentals of geometry. Mathematician Alexander Bogomolny has dedicated a page to cataloging 93 ways of proving the theorem (he also has, on a separate page, six wrong proofs).

Some of the proofs are simple and elegant. Others are quite elaborate, but the page is a nice place to skim through, and Bogomolny has some neat, interactive applets for demonstrations. The Wikipedia article on the theorem also has some nice animated gifs that are worth a look.

Cut the Knot is also a great website to peruse. Bogomolny is quite distraught about the state of math education, and this is his attempt to do something about it. He lays this out in his manifesto. Included in this remarkable window into the mind of a mathematician are some wonderful anecdotes about free vs. pedantic thinking and a collection of quotes that address the question, “Is math beautiful?”

Mathematics, rightly viewed, possesses not only truth, but supreme beauty — a beauty cold and austere, like that of sculpture, without appeal to any part of our weaker nature, without the gorgeous trappings of painting or music, yet sublimely pure, and capable of a stern perfection such as only the greatest art can show.

– Bertrand Russell (1872-1970), The Study of Mathematics via Cut the Knot.

Tau not Pi

This one’s for the nerds. It’s hard to argue against the elegance of using tau instead of pi. Natalie Wolchover explains, while Kevin Houston makes the argument in video:

It’s 10 PM and the Moat is Empty

My students and I had a great chance to use the our recent geometry work when we figured out how long it would take to drain the new moat in front of the school.

It’s not really a moat, it’s going to be a flower bed that will soak up some of the runoff that tries to seep into the school’s doors every time a spring or fall thunderstorm sweeps through.

The hole was dug on Thursday evening and filled with rainwater with water, half a meter deep, by Friday morning’s rain. At least we know now that the new beds are in the right place to attract runoff.

But to fill the trenches with gravel, sand and soil, we needed to drain the water. With a small electrical pump it seemed like it would take forever; except that we could do the math.

The pump emptied water through a long hose that runs around the back of the building where the topography is lower. I sent two students with a pitcher and a timer (an iPod Touch actually) to get the flow rate.

They came back with a time of 18.9 seconds to fill 4 liters. I sent them back to take another measurement, and had them average the to numbers to get the more reliable value of 18.65 seconds.

Then one of the students got out the meter-stick and measured the depth of the moat at a few locations. The measurements ranged from 46 cm to about 36 cm and we guesstimated that we could model the moat as having two parts, both sloping. After measuring the length (~6 m) and width (2 m), we went inside to do the math.

moat-volume — Rough sketch of the volume of water in the moat.

With the help of two of my students who tend to take the advanced math option every cycle, we calculated the volume of water (in cm³) and the flow rate of the pumped water (0.2145 cm³/s). Then we could work out the time it would take to drain the water, which turned out to be a pretty large number of seconds. We converted to minutes and then hours. The final result was about 7 hours, which would mean that the pump would need to run until 10 pm.

And it did.

Tsunami Geometry: Calculating the Height of a Tsunami using Basic Geometry

Since we’re working on geometry this cycle, I thought it would be an interesting exercise to think about how we could use geometry to think about how the strength of tsunamis decreases with distance from the source.

Of course, we’ll have to do this using some intense simplification so we can actually apply the tools we have available. The first is to approximate the tsunami as a circular wall of water centered on the epicenter of the earthquake.

tsunami-geometry — Simplified tsunami geometry.

This lets us figure out the volume of the wave pretty easily because we know that the volume of a cylinder is:

(1) $! V_c = \pi r^2 h$

The size of the circular water wall we approximate from the reports from Japan. The maximum height of the wave at landfall was somewhere in the range of 14 m along the northern Japanese coast, which was about 80 km from the epicenter. Just as a wild guess, I’m assuming that the “effective” width of the wave is 1 km.

Typically, in deep water, a tsunami can have a wavelength greater than 500 km (Nelson, 2010; note that our width is half the wavelength), but a wave height of only 1 m (USSRTF). When they reach the shallow water the wave height increases. The Japanese tsunami’s maximum height was reportedly about 14 m.

At any rate, we can figure out the volume of our wall of water by calculating the volume of a cylinder with the middle cut out of it. The radius of our inner cylinder is 80 km, and the radius of the outer cylinder is 80 km plus the width of the wave, which we say here is 1 km.

Calculating the volume of the wave

However, for the sake of algebra, we’ll call the radius of the inner cylinder, r_i and the width of the wave as w. Therefore the inner cylinder has a volume of:

(2) $! V_i = \pi r_i^2 h$

So the radius of the outer cylinder is the radius of the inner cylinder plus the width of the wave:

(3) $! r_o = r_i + w$

which means that the volume of the outer cylinder is:

(4) $! V_o = \pi (r_i + w)^2 h$

So now we can figure out the volume of the wave, which is the volume of the outer cylinder minus the volume of the inner cylinder:

(5) $! V_w = V_o - V_i$

(6) $! V_w = \pi (r_i + w)^2 h - \pi r_i^2 h$

Now to simplify, let’s expand the first term on the right side of the equation:

(7) $! V_w = \pi (r_i ^2 + 2 r_i w + w^2) h - \pi r_i^2 h$

Now let’s collect terms:

(8) $! V_w = \pi h \left( (r_i ^2 + 2 r_i w + w^2) - r_i^2 \right)$

Take away the inner parentheses:

(9) $! V_w = \pi h (r_i ^2 + 2 r_i w + w^2 - r_i^2)$

and subtract similar terms to get the equation:

(10) $! V_w = \pi h ( 2 r_i w + w^2 )$

Volume of the wave

Now we can just plug in our estimates of width and height to get the volume of water in the wave. We’re going to assume, later on, that the volume of water does not change as the wave propagates across the ocean.

(11) $! V_w = \pi (14) ( 2 r_i (1000) + (1000)^2 )$

rearrange so the coefficients are in front of the variables:

(12) $! V_w = 14 \pi ( 2000 r_i + 1000000 )$

So, at 80 km, the volume of water in our wave is:

(13) $! V_w = 14 \pi ( 2000 (80000) + 1000000 )$

(14) $! V_w = 7081149841 m^3$

Height of the Tsunami

Okay, now we want to know what the height of the tsunami will be at any distance from the epicenter of the earthquake. We’re assuming that the volume of water in the wave remains the same, and that the width of the wave also remains the same. The radius and circumference will certainly change, however.

We take equation (10) and rearrange it to solve for h by first dividing by rearranging all the terms on the right hand side so h is at the end of the equation (this is mostly for clarity):

(15) $! V_w = \pi ( 2 r_i w + w^2 ) h$

Now we can divide by all the other terms on the right hand side to isolate h:

(16) $! \frac{V_w}{\pi ( 2 r_i w + w^2 )} = \frac{\pi ( 2 r_i w + w^2 ) h}{\pi ( 2 r_i w + w^2 )}$

so:

(17) $! \frac{V_w}{\pi ( 2 r_i w + w^2 )} = h$

which when reversed looks like:

(18) $! h = \frac{V_w}{\pi ( 2 r_i w + w^2 )}$

This is our most general equation. We can use it for any width, or radius of wave that we want, which is great. Anyone else who wants to calculate wave heights for other situations would probably start with this equation (and equation (15)).

Double checking our algebra

So we can now figure out the height of the wave at any radius from the epicenter of the earthquake. To double check our algebra, however, let’s plug in the volume we calculated, and the numbers we started off with, and see if we get the same height (14 m).

First, we’ll use all our initial approximations so we get an equation with only two variables: height (h) and radial distance (r_i). Remember our initial conditions:

w = 1000 m
r_i = 80,000 m
h_i = 14 m

we used these numbers in equation (10) to calculate the volume of water in the wave:

V_w = 7081149841 m³

Now using these same numbers in equation (18) we get:

(19) $! h = \frac{7081149841}{\pi ( 2 (r_i) (1000) + (1000)^2 )}$

which simplifies to:

(20) $! h = \frac{7081149841}{ 2000 r_i \pi + 1000000 \pi }$

So, to double-check we try the radius of 80 km (80,000 m) and we get:

h = 14 m

Aha! it works.

Across the Pacific

Now, what about Hawaii? Well it’s about 6000 km away from the earthquake, so taking that as our radius (in meters of course), in equation (20) we get:

(21) $! h = \frac{7,081,149,841}{ 2,000 (6,000,000) \pi + 1,000,000 \pi }$

which is:

h = 0.19 m

This is just 19 cm!

All the way across the Pacific, Lima, Peru, is approximately 9,000 km away, which, using equation (20) gives:

h = 0.13 m

So now I’m curious about just how fast the 14 meters drops off to less than 20 cm. So I bring up Excel and put together a spreadsheet of tsunami height at different distances. Plot on a graph we get:

So the height of the tsunami drops off relatively fast. Within 1000 km of the earthquake the height has dropped by 90%.

How good is this model

This is all very nice, a cute little exercise in algebra, but is it useful? Does it come anywhere close to reality? We can check by comparing it to actual measurements; the same ones used by NOAA to compare to their model (see here).

tsunami-model-data-honolulu — The red line is the tsunami's water height predicted by the NOAA computer models for Honolulu, Hawaii, while the black line is the actual water height, measured at a tidal gauge. Other comparisons can be found here.

680_20110311-TsunamiWaveHeight — Tsunami wave heights in the Pacific, as modeled by NOAA. Notice how the force of the tsunami is focused across the center of the Pacific.

The graph shows a maximum height of about 60 cm, which is about three times larger than our model. NOAA’s estimate is within 20% of the actual maximum heights, but they’ve spent a bit more time on this problem, so they should be a little better than us. You can find all the gruesome details on NOAA’s Center for Tsunami Research site’s Tsunami Forecasting page.

Notes

1. The maximum height of a tsunami depends on how much up-and-down motion was caused by the earthquake. ScienceDaily reports on a 2007 article that tried to figure out if you could predict the size of a tsunami based on the type of earthquake that caused it.

2. Using buoys in the area, NOAA was able to detect and warn about the Japanese earthquake in about 9 minutes. How do they know where to place the buoys? Plate tectonics.

buoys-warning — The locations of the buoys in NOAA's tsunami warning system.

Update

The equations starting with (7) did not have the 2 on the r_iw term. That has been corrected. Note that the numerical calculations were correct so they have not changed. – Thanks to Spencer and Claude for helping me catch that.

And Pythagoras Scores!

centerpoint — Locating the center point of the soccer field.

Our school, Lamplighter, has started up a couple soccer teams to play in the local under-8 and under-6 leagues this year. I’m now one of the under-6 coaches, and the curious similarities between them and the middle schoolers is going to have to be the topic of another post; Montessori observed some interesting parallels between the first and third planes of development that are worth getting into. However, since the teams are new, we did not have lines on the practice field. And teaching throw-ins is kinda tricky with imaginary lines.

One of the parents/coaches of the under-8 team, Mr. Surbrook, offered draw out the lines. He also volunteered to give a lesson on geometry and let the middle school (and upper elementary) students help.

pythogoras — Refreshing ourselves on Pythagoras' Theorem.

To prepare the middle schoolers I did a quick review of Pythagoras’ Theorem using the 3×3, 4×4 and 5×5 squares (see above).

$! a^2 + b^2 = c^2$

The lesson was interesting because the 7th graders had had a more recent exposure to the equation but, unlike the 8th graders, have not had any algebra yet, so there were some puzzled looks when I rearranged the equation.

geom-lesson-soccer — Lesson on the geometry of rectangles.

That was in the morning. After lunch Mr. Surbrook came in and showed us how to use Pythagoras’ Theorem to make right angles and locate the center of the field. If stretch out six pieces of string, four for the sides and two for the diagonals (calculated with the equation,) at their fullest extent you have a rectangle with decent right angles.

corner-angles — Corner of the soccer field. Note the nice 90 degree angle.

After figuring out the theory inside, we went out to the field and help cut the string and lay out the lines. The kids were a bit disappointed they did not get to actually paint the lines, but we’d run out of time for the day.

Fortunately, they’ll get another chance at surveying when Dr. Houghton brings her class out to map the topography of the campus.

I very much liked how the whole procedure went, with my preparatory lesson first, then Mr. Surbrook practical lesson, and finally the actual practical application. We did something similar when we laid out the greenhouse the first time. That time we threw the kids in without a guide and without the practical lesson. It was a bit of a team-building exercise. It also took quite a bit longer.