Spaghetti Science

Ever run out of soup noodles and relied on breaking up spaghetti into small 1 to 2 cm pieces? That’s what I just did, and I never realized how many 1-2 mm fragments are generated with each and every break. In fact some bits were even smaller than a millimeter. I realized it by fluke because I was breaking the spaghetti with my hands inside a strainer, and the “dust” trickled through the sieve’s 3 mm holes, decorating the counter top. Here’s what it looked like.

It turns out that this has been common knowledge. Scientists have looked into it as well. According to a Smithsonian article,

To break it requires bending it into a bow shape. Eventually the force of the bend snaps the rod in the middle where it is most curved. But the physics doesn’t end there. That break releases energy back down the pieces of spaghetti in a “snap-back” wave or vibration. There’s enough energy in those waves to break off smaller lengths of pasta.

Jason Daley August 16, 2018

If you first twist the spaghetti, apparently, the “twist wave” travels faster than the snap, dissipating its energy. And the spaghetti breaks cleanly. But it would have taken a lot of impractical twisting in my case, given that I broke each strand several times while clumping several strands together to break them faster.

What goes on when spaghetti cooks? The chemistry part.

There are basically two parts to the process occurring between 55 and 85 oC. Water moves into the starch granules, causing them to expand. But for the pasta to fully cook, its protein has to react. If you have egg noodles, an insoluble network of egg and flour proteins form, trapping the swelling starch granules. A pH of 6, according to molecular gastronomist, Herve This, helps the proteins bind the starch even more firmly. One year I had my students test this notion by acidifying the boiling water with a tablespoon of lemon juice. Surely, enough it prevented the pasta from becoming sticky.

What if you have regular noodles? The same thing happens to the starch. But heat converts the flour’s globular proteins into relaxed chains. If overcooked the chains don’t trap the expanding starch, and its amylopectin diffuses out. As it clings to the surface of different strands, it binds them together. You end up with messy lumps of spaghetti.

Oddly, it never takes me the 10 minutes of recommended cooking time. Five seem to be sufficient to create a slightly al dente spaghetti. What’s also noteworthy is how the same recipe but different shapes of pasta creates a different taste, probably because of the role that texture plays in taste and because different shapes have different surface to volume ratios. Thus varying amounts of sauce cling to each noodle.

Recent research (2021) at the University of Parma in Italy confirmed that pasta is a medium to low-GI food, (glycemic index = GI). That’s related to the fact that the starch granules remain trapped in the network, and so they are not completely hydrolyzed in the small intestine. The GI was lowest for those pastas that were enriched with legumes or other plant based products.

Sources:

Herve This. Molecular Gastronomy. Columbia University Press. 2006

Exploratorium. Soaking Pasta. https://www.exploratorium.edu/food/soaking-pasta

Jason Daley. Physics Reveals How to Break Spaghetti Cleanly In Two. Smithsonian. 16/08/2016

Giuseppe Di Pede and al. Glycemic Index Values of Pasta Products: An Overview. Foods 202110(11), 2541; https://doi.org/10.3390/foods10112541

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Euler’s Phi Function from Wells’ Curious and Interesting Numbers

In 1986, David Wells, a mathematician specializing in number theory, wrote a delicious little book, entitled The Penguin Dictionary of Curious and Interesting Numbers. If you enjoy dabbling in math, each numerical entry, although brief, is a gateway into all sorts of realms of an incredibly diverse field. Although it goes without saying that without mathematics, physics, chemistry, astronomy, computer science, engineering, banks and insurance companies would all be crippled, math is beautiful in its own right. It is a source of infinite puzzles to distract us from pain, taxes and death.

Under his entry for the number 87, Wells first wrote  σ[ϕ(87)]= σ(87). The neat thing is that σ[ϕ(n)]= σ(n) is only true for very few numbers. But what is σ(n) ? It is simply the sum of all the factors of n. In fact, capital sigma, Σ, is routinely used to designate sums of all sorts of things, but lower-case sigma, σ, has the specialized role of adding up only the factors of a number. So σ(87)=1+3+29+87= 120.

What is ϕ(n), the so-called, Euler’s phi function ? It’s the total number of positive integer values that are less than n and which are relatively prime to n. Relatively prime in turn means that the two numbers only have a common factor of 1. To obtain ϕ(87) the long, painful way, we begin by subtracting two from 86 ( 2 because of the two factors, 3 and 29, which are not relatively prime to 87; and 86 because the definition of ϕ(n) is based on integers less than 87). But then all other multiples of 3 and 29 are also to be excluded because they will have something in common with 87. Thus subtract the 27 other multiples of three that are less than 87, given that we’ve already counted the number 3; and finally the 1 other multiple of 29 < 87, which is 58. Hence ϕ(87) = 86 – 2 – 27 -1 = 56.

There are actually a few formulas, easily derived, that make the calculation of ϕ(n)less tedious. For any prime number, p, ϕ(p)=p-1. If you raise p to an integer power of k, the number of relatively prime numbers will be pk-1 lowered by (pk-1-1). Simplified that yields ϕ(pk)=pk(p-1)/p. For a non prime number, you can use the basis of the prime number formula to come up with:

p1, p2 and pk are the prime factors of n

To quickly evaluate ϕ(87),

Now that we know that ϕ(87)=56, we can get σ[ϕ(87)] = σ[56] = 1 + 2+ 4 +7+8+14+28+56 = 120. We saw earlier that σ(87)= 120, so indeed σ[ϕ(87)]= σ(87) =120. You can write a very simple computer program(see footnote) to discover that σ[ϕ(n)]= σ(n) is also true for 362, 5002 and 9374 and for two other numbers revealed by these images. This will give me an excuse to go off on artistic and scientific trivia-tangents. 🙂

The first image is a Delacroix painting of the great pianist-composer, Chopin. The artist, Eugene Delacroix, who believed that “nothing can be compared with the emotion caused by music” was born in 1798. σ(1798) = 2880, the year when asteroid 1950AD will come close to Earth. When it first made the news, the risk of an impact was reported to be higher than it is currently believed to be. Its trajectory can be difficult to predict due to the Yarkovsky effect in which unequal heating of an asteroid’s surface cause emitting photons to slightly propel it. ϕ(1798) = 1798[(2-1)/2*(29-1)/29*(31-1)/31]=1*28*30=840.

σ(840)=2880= σ(1798) , so σ(840)=σ(ϕ(1798)) = σ(1798)

The 3rd picture is that of the Indonesian volcano Mount Samalas, which erupted in 1257 in one of the world’s most intense eruptions in the last 12 millennia. The injection of aerosols high into the atmosphere cooled the earth for a few years. Ozone was even affected, given that halogen gases accompanied the release of the sulfur compounds by a ratio of almost 3:2. If you’ve ever wondered if global warming could have fluky, positive effects in being able to offset natural cooling from large volcanic eruptions, this is a reminder that the effects would only be very temporary. If you have mass volcanism in mind, such events are even rarer, with hundreds of thousands of years elapsing without them. If one would occur, the magnitude of the devastation would be so large, that it would overshadow anything else.

σ(ϕ(1257))=σ(836)=1680 =σ(1257), the year in which the long-tailed C/1680 V1 comet was observed. It’s also known as The Great Comet or Newton’s or Kirch’s Comet. It was bright enough to be seen during the day. The image is the 1680 Dutch painting, Staartster (komeet) boven Rotterdam(Tail star (comet) above Rotterdam) by Lieve Verchuier. The artist normally painted marine scenes; ships being built and the occasional landscape. But it’s no surprise that he chose to paint the unusual sighting of such a stunning comet, and imagine if he realized that σ(ϕ(1257))=σ(1257)=1680. 🙂

Sources:

David Wells. The Penguin Dictionary of Curious and Interesting Numbers. Penguin. 1986

Vidal, C. M. et al. The 1257 Samalas eruption (Lombok, Indonesia): the single greatest stratospheric gas release of the Common Era. Sci. Rep. 6, 34868; doi: 10.1038/srep34868 (2016).

George Beekman. The Nearly Forgotten Scientist Osipovich Yarkovsky. J. Br. Astron. Assoc. 115, 4, 2005 https://articles.adsabs.harvard.edu/pdf/2005JBAA..115..207B

NASA. Sentry Earth Impact Monitoring. https://ssd.jpl.nasa.gov/tools/sbdb_lookup.html#/?sstr=2029075&view=OPC

Donald K. Yeomans. Great Comets in History. Jet Propulsion Laboratory. April, 2007

Maple Program

The phi and sigma functions are built-into Maple, which makes it easy. Also note that the syntax will be slightly different in newer versions of the software. Mine, version V, is 27 years old!

with(numtheory):

for w from 2 to 10000 by 1 do; if sigma(phi(w))= sigma(w) then print(w,sigma(w));fi;od;

And here is the output, where the first number of each pair is what we’re looking for, while the 2nd number is the sigma of that number.

                           87,120
                           362, 546
                          1257, 1680
                          1798, 2880
                          5002, 7812
                          9374, 14520

6 out of 10000, or 0.06% meet the criteria. How many solutions are there between 10 000 and 100 000?

with(numtheory):
for w from 10000 to 100000 by 1 do; if sigma(phi(w))= sigma(w) then print(w,sigma(w));fi;od;

The results:

21982, 34200
22436, 40320
25978, 40320
35306, 53760
38372, 68796
41559, 63360
50398, 76608
51706, 78624
53098, 80640
53314, 89280
56679, 86400
65307, 95040
68037, 90720
89067, 129600

There are only 14 out of 90 000 ( 0.016%). So do the solutions become consistently scarcer as the numbers grow? Between 100 000 and a million, σ[ϕ(n)]= σ(n) is true for only about 0.0093% of the numbers. So apparently yes.

How Solving Cubic Equations Led to the Discovery of Imaginary Numbers

In math, a complex number has a real and an imaginary component combined in the form of a + bi.

a and b are real numbers, which means that they can include whole numbers, repeating decimals, terminating decimals and those that go on forever without patterns– the transcendental numbers. Examples of transcendental numbers include e, π, eπ, probably Catalan’s constant ( 1 – 1/9 + 1/25 – 1/49 +1/64- …) and ii. i without its exponent is an imaginary number, the square root of -1.

Above, to the left, Gerolamo Cardano (1501 – 1576) was a Renaissance mathematician and inventor of the universal joint, the cardan. He published Tartaglia’s and Ferrari’s solution for 3rd(cubic) and 4th degree equations. The first European to use negative numbers, he also realized, while investigating the cubic, that imaginary numbers had to be considered. To the right, Raffaele Bombelli(1526 – 15) first treated the square root of -1 as a variable. The Italian mathematician also worked as a hydraulic engineer.

Many math students have their 1st encounter with i typically when solving second degree differential equations. They learn that a seemingly nonsensical entity becomes very useful in coming up with real solutions to real scientific and engineering problems. But it’s when trying to solve somewhat simpler equations that a 16th century Italian mathematicians first stumbled upon i.

Consider x3 + 3x2 – 2x – 2 = 0. This is a 3rd degree equation, dubbed cubic because its highest power is three. It can’t be solved by isolating the variable. Factoring won’t work either. Well, sort of. In this particular case, by fluke, you can use trial and error and quickly discover that x = 1 is one of the solutions. Then, as we will show later, you can subsequently use long division and find the other two solutions. But mathematicians wanted a more systematic technique because most numbers are not whole, so trial and error will, for cubic equations, fail miserably most of the time.

It turns out that the above can be reduced to something simpler if we could somehow eliminate the power of 2 term. How? Let’s say we replaced each x with a y+k term, where k is some constant that we will find.

x = y + k

Now consider a general case where we have ax3 + bx2 + …, If we replace each x with y + k then

a(y + k)3 + b(y + k)2 +….. = 0. Expanding this we will get
y3 + 3y2k + 3yk2 + k3 + by2 + 2byk + bk2 +…..

Since k will just be a number, we don’t care if k itself has a power of 2. But we do want to eliminate the square of any variable—here, terms with y2. Essentially we want 3y2k +by2 = 0, and that will happen if, in our substitution, we choose k=-b/3. Why? Well, it’s the value of k if we solve the above equation:

3y2k +by2 = 0. The y2 cancels; then 3k + b = 0, so k=-b/3.

Let’s apply it to x3 + 3x2 – 2x – 2 = 0. We let x = y + k, where k =-b/3= -3/3 =-1.

Hence, x = y -1 and we substitute it into the equation:

(y -1)3 + 3(y -1)2 – 2(y -1) – 2 = 0

Simplifying we get a 3rd degree equation free of a squared term, what’s called a depressed cubic:

y3 – 5y =- 2. But how do we solve that?

(Instead of y3+py=q, above is another way of representing a depressed cubic. )

We begin with the observation that if we expand the cube of difference of any two numbers, we obtain the following:

(s-t)3 = s3 – 3s2t +3st2 – t3. If we factor the st terms and leave the individual cubes by themselves on the right hand side of the equation, we obtain:

(s -t)3+3st(s -t) = s3 – t3. Now compare this to our equation:

y3 -5y = -2.

We notice that y = s – t; 3st=-5 and that s3 – t3 = -2.

We eliminate one variable, s. Since s =-5/(3t), we substitute it into s3 – t3 = -2 and after eliminating the variable from the denominator, we obtain

t6 – 2t3 + 125/27 = 0. If we let u = t3, we will get a simple quadratic equation!

u2– 2u + 125/27 = 0. With the quadratic formula we get u= 1± (7/9)√(-6).

Cardano knew that an expression such as u= 1± (7/9)√(-6) was meaningful and that it was somehow connected to the real solution, but he didn’t know what to do. It’s thanks to another Italian mathematician, Raffaele Bombelli who realized that we can do the following: let the square root of -6 be expressed as √(6)*√(-1), where√(-1) was defined by Bombelli to be the imaginary number i.

So u= 1± (7/9)i√(6). Now it seems that the solution to our problem does not have a real answer, but it’s merely an illusion. If we work with i, we can work our way backwards to the value of x, and we will discover that it gives us the trial-and-error real solution of x=1. How?

Recall that u = t3. Thus t = u(1/3), which we discovered is t= [ 1± (7/9)i√(6) ]1/3

That’s a real monstrosity, so there must be a simpler way to express the complex number, t.

Let t = a +bi =[ 1± (7/9)i√(6) ]1/3

That means that (a +bi)3 = 1 ± (7/9)i√(6). Now expand the left side and we get

a3 + 3a2bi -3ab2 – b2i = 1 ± (7/9)i√(6). Let’s separate the real and imaginary parts from the left hand side of the equation and factor:

a(a2-3b2) +bi (3a2-b) = 1 ± (7/9)i√(6).

If we focus on the real part of each equation we notice that a(a2-3b2)= 1. Could we be so lucky as to have a = 1 and a2-3b2= 1, given that 1(1) =1? Well, no. If we replace a with 1 and solve for b, we would get a b value of 0, which is impossible since we already know that t is a complex number. Maybe a= -1 and a2-3b2= -1, given that -1(-1) =1. If that’s the case, solving for b yields b=√(2/3) =(1/3)√(6). To check if these values make sense, see if they also satisfy the imaginary part: bi (3a2-b2) =(7/9)i√(6). Cancel the i‘s. Substitute in a = -1 and b=(1/3)√(6):

(1/3)√(6)[3(-1)2-(1/3)2√(6)2]=√(6)-(1/27)(6)√(6)=√(6)-(1/9)(2)√(6)= (7/9)√(6), which checks out as being the coefficient for the i in our expression!

Given, then, that t = a +bi = -1 +(1/3)√(6)i

Working backwards we had s =-5/(3t) =-5/(3[-1 +(1/3)√(6)i]) = 1 +(1/3)√(6)i

y = s -t = 1 +(1/3)√(6)i – [-1 +(1/3)√(6)i] = 2

The imaginary terms have cancelled out and led us to a real answer!

Finally we had x = y -1, so x = 2-1 =1.

We’re almost done.

If we consider the roots ( x intercepts) of the graph of y = x3 + 3x2 – 2x – 2, we need to set the x terms to zero and thus we’re back to our problem. y = x3 + 3x2 – 2x – 2 = 0. A third degree equation can have up to 3 x intercepts. This one does indeed have 3.

To obtain the remaining two, use the fact that if our first root is x=1, then x-1 is a factor of x3 + 3x2 – 2x – 2. By using long division, we can get that x3 + 3x2 – 2x – 2 =(x-1)(x2 + 4x + 2) = 0. Use the quadratic formula to solve x2 + 4x + 2 and we find that x = -2 +√(2)~ = -0.586 and x =-2 -√(2)~ = -3. 414, as confirmed by the graph.

Source for leading image: Matheepan Panchalingam, via Flickr. Biographical pictures are form Wikipedia.