An Eighty-Atom Fullerene Molecule

c80-fullerene

The fullerene molecule that gets the most attention is C60, so I’m giving C80 a little bit of the spotlight, for balance. I made this using polyhedral modeling software called Stella 4d; you can try it for yourself at this website.

A “Thumbs Up” for Google Classroom

This is my 22nd year of teaching, but my first year using Google Classroom. We’re finding it to be a useful tool. This, for example, is the diagram for the Atwood’s machine lab we are doing in Pre-AP Physical Science, beginning today. My students will find this waiting for them in their virtual classroom (on Chromebooks my school district provides), with discussion-prompts to get us started:

atwoods-machine-diagram

I had no idea that four years of blogging, here on WordPress, had been preparing me to use this teaching tool. However, active blogging does require one to develop some transferable skills, especially in fields (such as what I teach) which are similar to the topics of one’s blog, as is the case here.

One Possible Definition of Physics

Visual definition of physics

With my metaphorical “mathematics of sets” hat on, this is what physics looks like, to me. The further you go in the field, the more challenging the mathematics gets; also, the better (and more expensive) the toys become.

We’re Going Back to School Tomorrow, and I’m the Teacher.

Trunc Icosa

This is my 22nd year teaching. This year, I teach in only one department. This is nice; I’ve spent much of my career in multiple departments, simply because I am certified in multiple subject areas. This year, in my building, I am one of three science teachers. Our high school has become so large that the 9th grade has been “spun off” to a new freshman campus, while remaining part of the high school, and I’m one of the teachers who gets to go to the new campus. This provides my students, my colleagues (especially at the new campus), and myself the opportunity for a fresh start, to a greater degree than is usually the case when a new school year begins.

My students are in just two subjects, this year: Physical Science, and Pre-AP Physical Science. I don’t want the students in the class without the “Pre-AP” prefix to feel that they are in a “lesser” class, in any sense of that word, so I am renaming “Physical Science,” slightly: “High School Physical Science.” It is my hope that this change in wording serves to communicate high expectations, and 9th grade is the first year of high school — which, in the USA’s public school systems, means 9th grade students must pass courses to earn credits toward graduation, usually for the first time.

In the other class, Pre-AP Physical Science, I am teaching that version of the course for the first time, but I feel well-prepared by the extensive training I had this Summer, and last school year, through my university, the school where the Summer training was held, and the College Board. Both classes will challenge students, but it is also true that the two classes will be different, for Pre-AP Physical Science have to leave students prepared to function effectively, later, in other Pre-AP and/or AP science courses. 

Physical Science is an introduction to two sciences: physics, and then chemistry, at least in my school district. It helps me that I have experience teaching both subjects as higher-level, “stand-alone” classes. In this class (both versions), we also touch on some other sciences which are also physical sciences, such as geology, astronomy, and the science of climate change. However, those sciences do not dominate these courses, as physics and chemistry do. The image above is from chemistry (and was created with Stella 4d, which you can try here), and shows a model of a sixty-atom all-carbon molecule called buckminsterfullerene, one of a class of roughly-spherical carbon allotropes called fullerenes. Mathematicians call this particular fullerene’s shape a “truncated icosahedron,” and, in sports, this same shape is known as the (non-American) “football” or “soccer ball.” Physical modes of this shape may be made with molecular model sets of various kinds, Zometools, and other materials. In both versions of my science classes this year, building models of this molecule will be one of many lab activities we will do; one of my goals this year is for my students to spend a third of their time doing labs. The legal requirement for science class time spent in lab, in my state, is at least one-fifth, so more than that is fine. Science classes helped me learn both science and mathematics, but what I remember the most is the labs. I don’t think that’s just me, either; students learn more effectively, I have observed, by conducting scientific experiments themselves, than by being “lectured at” for extended periods of time.

I’m looking forward to a good year — for all of us.

For Science Teachers: A Safer Alternative to Liquid Mercury

Liquid mercury, in schools, poses three major problems:

  1. It is extremely toxic,
  2. It has a high vapor pressure, so you can be poisoned by invisible mercury vapor leaving any exposed surface of liquid mercury, and
  3. Playing with liquid mercury is a lot of fun.

These are compelling reasons to leave use of mercury to those at the college level, or beyond. In the opinion of this science teacher, use of liquid mercury in science classes, up through high school chemistry, inside or outside thermometers, is a bad idea. If the bulb at the bottom of a thermometer, as well as the colored stripe, looks silvery, as in the picture below (found on Wikipedia), then that silvery liquid is mercury, and that thermometer should not be used in labs for high school, let alone with younger children. Your local poison control center can help you find the proper thing to do with mercury in your area; it should definitely not just be thrown away, for we do not need this serious environmental toxin in landfills, where it will eventually reach, and poison, water. Red-stripe thermometers without any silvery line, on the other hand, are far safer, although broken glass can still cause injury.

Maximum_thermometer_close_up_2

I turned ten years old in 1978, and, by that time, I had already spent many hours playing (unsupervised) with liquid mercury, pouring it hand-to-hand, etc., so I know exactly how irresistible a “plaything” mercury can be, to children. Luck was on my side, and I suffered no ill effects, but I can state from experience that children should not be tempted with highly-toxic “mercury as a toy,” for it’s not a toy at all. Mercury spills require special “hazmat” training to clean up safely; anyone encountering such a spill who does not have such training should simply notify the proper authorities. In the USA, this means evacuating the area immediately, and then calling 911 — from far enough away to keep the caller from breathing invisible mercury vapor.

Fortunately, there is a safe alternative which can give students a chance to experiment with a room-temperature metal: an alloy of three parts gallium to one part indium, by mass. Gallium’s melting point is between normal human body temperature and room temperature, so it can literally melt in your hand (although a hot plate is faster). Indium, on the other hand, has a melting point of 156.6°C. For this reason, I will not buy a hot plate unless it can reach higher that that temperature. (Note: use appropriate caution and safety equipment, such as goggles and insulated gloves, with hot plates, and the things heated with them, to avoid burns.)

Once both elements are massed, in the proportions given above, they can then be melted in the same container. When they melt and mix together, they form an alloy which remains liquid at room temperature.

Some might wonder how mixing two elements can create an alloy with a melting point below the melting points of either of the two ingredients, and the key to that puzzle is related to atomic size. Solids have atoms which vibrate back and forth, but don’t move around each other. In liquids, the atoms are more disordered (and faster), and easily slip around each other. In solid, room-temperature gallium, all the atoms are of one size, helping the solid stay solid. Warm it a little, and it melts. With pure indium, this applies, also, but you have to heat it up a lot more to get it to melt. If the two metals are melted and thoroughly mixed, though, and then frozen (a normal freezer is cold enough), the fact that the atoms are of different sizes (indium atoms are larger than gallium atoms) means the atoms will be in a relatively disordered state, compared to single-element solids. In liquids, atoms are even more disordered (that is, they possess more entropy). Therefore, a frozen gallium/indium alloy, with two sizes of atoms, is already closer to a disordered, liquid state, in terms of entropy, than pure, solid gallium or indium at the same temperature. This is why the gallium-indium mixture has a melting point below either individual element — it requires a lower temperature to get the individual atoms to flow past each other, if they are already different atoms, with different sizes.

liquid metals

Those who have experience with actual liquid mercury will notice some important differences between it and this gallium-indium alloy, although both do appear to be silver-colored liquids. (This is why mercury is sometimes called “quicksilver.”) For one thing, their densities are different. A quarter, made of copper and nickel, will float on liquid mercury, for the quarter’s density is less than that of mercury. However, a quarter will sink in liquid 3:1 gallium-indium alloy. To float a metal on this alloy, one would need to use a less-dense metal, such as aluminum or magnesium, both of which sink in water, but float in liquid Ga/In alloy.

Other differences include surface tension; mercury’s is very high, causing small amounts of it on a floor to form little liquid balls which are difficult (and dangerous) to recapture. Gallium-indium alloy, by contrast, has much less surface tension. As a result, unlike mercury, this alloy does not “ball up,” and it will wet glass — and doing that turns the other side of the glass into a mirror. Actual mercury will not wet glass.

The most important differences, of course, is that indium and gallium are far less toxic than mercury, and that this alloy of those two elements has a much lower vapor pressure than that of mercury. Gallium and indium are not completely non-toxic, though. Neither indium nor gallium should be consumed, of course, and standard laboratory safety equipment, such as goggles and gloves, should be worn when doing laboratory experiments with these two elements.

George Carlin, on Change

change machine

On numerous occasions, I have repeated this experiment, in keeping with the scientific method. I have obtained the same null result as Carlin obtained, each and every time.

Is This What’s Going On? A Set of Questions of Global Concern.

Is This Whats Going On

I have a set of conjectures, and want input from my friends and blog-followers about them. How much of this has actually happened over the past months, weeks, and days?
 
1. The Chinese have been buying huge amounts of silver, thus driving up its price, because…
 
2. The political and business leaders in Greater China are, themselves, sick of living in an environmental nightmare based on decades of high consumption of oil and dirty coal, and are working on building enormous numbers of solar panels to get away from fossil fuel consumption, using lots of silver, which has the highest reflectivity of any element. China’s silver buying-spree is being misinterpreted, globally, because China is not well-understood, outside China.
 
3. These leaders of China have to breathe the same air, for one thing, as many Chinese people with much less power, and going green is the pragmatic thing to do. It is quite Chinese to be pragmatic. Living in Shanghai, Beijing, Hong Kong, Taipei, or other population centers, air quality is a major issue, as is global warming and other environmental concerns — all issues which many Americans are in the habit of ignoring.
 
4. As the Chinese phase themselves out of the human addiction to fossil fuels, total global oil consumption drops. Evidence: gasoline prices fell. I was buying for under $2 a gallon a week ago.
 
5. Falling oil prices have led to severe economic problems in the oil-producing countries of the Middle East. Higher-than-usual amounts of political stability have rippled through the Middle East through the last five years, and this has intensified further in recent months. The latest such development has been in Turkey, often seen as the most politically stable country in the Muslim world, is going through an attempted(?) coup, on the far side of the Middle East from China.
 
6. In the USA, one of the people running for president is a reactionary xenophobe, as well as a populist demagogue, and is running against an opponent with little to no ethical principles who is winning by default because she’s running against Trump. Donald Trump and his people (and he has a lot of people) have been spewing Islamophobia and Sinophobia, and they’ve been doing it loudly.
 
7. Many people all over the world are reacting to the Trump Trumpet o’ Hate, and freaking out. Various end-of-the-world scenarios are been floated publicly, especially in cyberspace. People are getting “off the grid” if they can, either because it’s a good idea, or because they’re panicked. In some places, efforts are actually being made to use the force of government to stop people from weaning themselves off the services of utility companies.
 
8. Few people realize that a lot of this is a set of unintended consequences of China (of all nations) leading the charge to do the right thing regarding oil addiction, from an environmental and ecological point of view, plus having a lunatic run for the White House.
 
9. The rising price of silver, panic-in-advance about a widely-expected coming collapse of fiat currencies, and the pronouncements and predictions of Ron Paul and his ilk, are all feeding off each other, in an accelerating spiral. In the meantime, the political instability in Turkey is capping off a slight rise in gas prices over recent lows, just in the last week.
 
10. Most Americans don’t know much about a lot of this because we’re at a point in the current, nasty election cycle that America as a people has simply forgotten (again) that the world outside the United States actually exists. Ignorance about the Middle East, economics, environmental science, and Greater China is widespread in the best of times. Thanks to (a) the “Donald and Hillary Show” playing 24/7 on cable news, (b) civil unrest at home (brutality on the part of some, but not all, police), and (c) a backlash against Black Lives Matter, with horrible behavior from some, but not all, of the protesters on all sides, and (d) an anti-or re-backlash against BLM is in “full throttle” right now, and (e) unrest abroad (Turkey, etc.), these certainly aren’t the best of times.
 
I invite anyone to weigh in on the subject of which of the above conjectures are valid, and which are invalid. I have deliberately cited no sources, yet, because I am asking for independent peer review, and so do not wish to suggest sources at this point. In addition to “Which of these statements are correct, and which are wrong?” I am also asking, “What am I missing?”

Have you noticed what silver’s been doing lately? The price of silver is literally on fire!

silver is literally on fire

Because of the price of silver being literally on fire, they will not be buying and selling troy ounces of metallic silver when the markets open in New York tomorrow morning. Instead, they will be selling “oxide ounces” of silver oxide, in sealed-plastic capsules of this black powder, with an oxide ounce of silver oxide being defined as that amount of silver oxide which contains one troy ounce of silver.

silver oxide capsule

A troy ounce of silver is 31.1 grams of that element, which has a molar mass of 107.868 g/mole. Therefore, a troy ounce of silver contains (31.1 g)(1 mol/107.868 g) = 0.288 moles of silver. An oxide ounce of silver oxide would also contain oxygen, of course, and the formula on the front side of a silver oxide capsule (shown above; information on the back of the capsule gives the number of oxide ounces, which can vary from one capsule to another) is all that is needed to know that the number of moles of oxygen atoms (not molecules) is half the number of moles of silver, or (0.288 mol)/2 = 0.144 moles of oxygen atoms. Oxygen’s non-molecular molar mass is 15.9994 g, so this is (0.144 mol)(15.9994 g/mol) = 2.30 g of oxygen. Add that to the 31.1 g of silver in an oxide ounce of silver oxide, and you have 31.1 g + 2.30 g = 33.4 grams of silver oxide in an oxide ounce of that compound.

In practice, however, silver oxide (a black powder) is much less human-friendly than metallic silver bars, coins, or rounds. As you can easily verify for yourself using Google, silver oxide powder can, and has, caused health problems in humans, especially when inhaled. This is the reason for encapsulation in plastic, and the plastic, for health reasons, must be far more substantial than a mere plastic bag. For encapsulated silver oxide, the new industry standard will be to use exactly 6.6 g of hard plastic per oxide ounce of silver oxide, and this standard will be maintained when they begin manufacturing bars, rounds, and coins of silver oxide powder enclosed in hard plastic. This has created a new unit of measure — the “encapsulated ounce” — which is the total mass of one oxide ounce of silver oxide, plus the hard plastic surrounding it on all sides, for a total of 33.4 g + 6.6 g = 40.0 grams, which will certainly be a convenient number to use, compared to its predecessor-units.  

# # #

[This is not from The Onion. We promise. It is, rather, a production of the Committee to Give Up on Getting People to Ever Understand the Meaning of the Word “Literally,” or CGUGPEUMWL, which is fun to try to pronounce.]

 

 

A Zome Torus, Before and After Adding Dodecahedra, As a Model for a Pulsar’s Accretion Disk and Radiation Jets

zome torus

I’ve been using Zometools, available at http://www.zometool.com, to build interesting geometrical shapes since long before I started this blog. I recently found this: a 2011 photograph of myself, holding a twisting Zome torus. While I don’t remember who was holding the camera, I do remember that the torus is made of adjacent parallelopipeds.

After building this torus, I imagined it as an accretion disk surrounding a neutron star — and now I am imagining it as a neutron star on the verge of gaining enough mass, from the accretion disk, to become a black hole. Such an object would emit intense jets of high-energy radiation in opposite directions, along the rotational axis of this neutron star. These jets of radiation are perpendicular to the plane in which the rotation takes place, and these two opposite directions are made visible in this manner, below, as two dodecahedra pointing out, on opposite sides of the torus — at least if my model is held at just the right angle, relative to the direction the camera is pointing, as shown below, to create an illusion of perpendicularity. The two photographs were taken on the same day. 

zome torus with dodecahedra 2011

In reality, of course, these jets of radiation would be much narrower than this photograph suggests, and the accretion disk would be flatter and wider. When one of the radiation jets from such neutron stars just happens to periodically point at us, often at thousands of times per second, we call such rapidly-rotating objects pulsars. Fortunately for us, there are no pulsars near Earth.

It would take an extremely long time for a black hole to form, from a neutron star, in this manner. This is because most of the incoming mass and energy (mostly mass, from the accretion disk) leaves this thermodynamic system as outgoing mass and energy (mostly energy, in the radiation jets), mass and energy being equivalent via the most famous formula in all of science: E = mc².

An Asymmetrical Static Equilibrium Physics Problem Involving Pulleys and Hanging Masses

An interesting phenomenon in physics, and physics education, is the simplicity of symmetric situations, compared to the complexity of similar situations which are, instead, asymmetrical. Students generally learn the symmetrical versions first, such as this static equilibrium problem, with the hanging masses on both left and right equal. 

static equilibrium pulley setup

The problem is to find the measures of the three angles shown above, with values given for all three masses. Here is the setup, using physical objects, rather than a diagram.

100_170_100

The masses on the left and right are each 100 g, or 0.100 kg, while the central masses total 170 g, or 0.170 kg. Since all hanging masses are in static equilibrium, the forces pulling at the central point (at the common vertex of angles λ, θ, and ρ) must be balanced. Specifically, downward tension in the strings must be balanced by upward tension, and the same is true of tension forces to the left and to the right. In the diagram below (deliberately asymmetrical, since that’s coming soon), these forces are shown, along with the vertical and horizontal components of the tension forces held in the diagonal strings.

static equilibrium pulley setup force diagram

Because the horizontal forces are in balance, Tlx = Trx, so Mlgcosλ = Mrgcosρ — which is not useful now, but it will become important later. In the symmetrical situation, all that is really needed to solve the problem is the fact that the vertical forces are in balance. For this reason, Tc = Tly + Try, so Mcg = Mlgsinλ + Mrgsinρ. Since, due to symmetry, Ml = Mand λ = ρ,  Mr may be substituted for Ml, and ρ may be substituted for  λ, in the previous equation Mcg = Mlgsinλ + Mrgsinρ, yielding Mcg = Mrgsinρ + Mrgsinρ, which simplifies to Mcg =2Mrgsinρ. Cancelling “g” from each side, and substituting in the actual masses used, this becomes 0.170 kg = 2(0.100 kg)sinρ, which simplifies to 0.170 kg = (0.200 kg)sinρ, then 0.170/0.200 = sinρ. Therefore, angle ρ = sin-1(0.170/0.200) = 58°, which, by symmetry, must also equal λ. Since all three angles add up to 180º, the central angle θ = 180° – 58° – 58° = 64°. These answers can then be checked against the physical apparatus.

53_theta_53

When actually checked with a protractor, the angles on left and right are each about 53° — which is off from the predicted value of 58° by about 9%. The central angle, of course, is larger, at [180 – (2)53]° = 74°, to make up the difference in the two smaller angles. The error here could be caused by several factors, such as the mass of the string itself (neglected in the calculations above), friction in the pulleys, or possibly the fact that the pulleys did not hang straight down from the hooks which held them, but hung instead at a slight diagonal, as can be seen in the second image in this post. This is testable, of course, by using thinner, less massive string, as well as rigidly-fixed, lower-friction pulleys. However, reducing the error in a lab experiment is not my objective here — it is, rather, use of a simple change to turn a relatively easy problem into one which is much more challenging to solve. 

In this case, the simple change I am choosing is to add 50 grams to the 100 g already on the right side, while leaving the central and left sides unchanged. This causes the angles where the strings meet to change, until the situation is once more in static equilibrium, with both horizontal and vertical forces balanced. With the mass on the left remaining at 0.100 kg, the central mass at 0.170 kg, and the mass on the right now 0.150 kg, what was an easy static equilibrium problem (finding the same three angles) becomes a formidable challenge. 

100_170_150

For the same reasons as before (balancing forces), it remains true that Mlgcosλ = Mrgcosρ (force left = force right), and, this time, that equation will be needed. It also remains true that Mcg = Mlgsinλ + Mrgsinρ (downward force = sum of the two upward forces). The increased difficulty is caused by the newly-introduced asymmetry, for now Ml ≠ Mr, and λ ≠ ρ as well. It remains true, of course, that  λ + θ + ρ° = 180.

In both the vertical and horizontal equations, “g,” the acceleration due to gravity, cancels, so Mlgcosλ = Mrgcosρ becomes Mlcosλ = Mrcosρ, and Mcg = Mlgsinλ + Mrgsinρ becomes Mc = Mlsinλ + Mrsinρ. The simplified horizontal equation, Mlcosλ = Mrcosρ, becomes Ml²cos²λ = Mr²cos²ρ when both sides are squared, in order to set up a substitution based on the trigonometric identity, which works for any angle φ, which states that sin²φ + cos²φ = 1. Rearranged to solve it for cos²φ, this identity states that  cos²φ = 1 – sin²φ. Using this rearranged identity to make substitutions on both sides of the previous equation Ml²cos²λ = Mr²cos²ρ yields the new equation Ml²(1 – sin²λ) = Mr²(1 – sin²ρ). Applying the distributive property yields the equation Ml² – Ml²sin²λ = Mr² – Mr²sin²ρ. By addition, this then becomes -Ml²sin²λ = Mr² – Ml² – Mr²sin²ρ. Solving this for sin²λ turns it into sin²λ = (Mr² – Ml² – Mr²sin²ρ)/(-Ml²).

Next, Mc = Mlsinλ + Mrsinρ (the simplied version of the vertical-force-balance equation, from above), when solved for sinλ, becomes  sinλ = (Mrsinρ – Mc)/(- Ml). Squaring both sides of this equation turns it into sin²λ = (Mr²sin²ρ – 2MrMcsinρ + Mc²)/(- Ml.

There are now two equations solved for sin²λ, each shown in bold at the end of one of the previous two paragraphs. Setting the two expressions shown equal to sin²λ equal to each other yields the new equation (Mr² – Ml² – Mr²sin²ρ)/(-Ml²) = (Mr²sin²ρ – 2MrMcsinρ + Mc²)/(- Ml)², which then becomes (Mr² – Ml² – Mr²sin²ρ)/(-Ml²) = (Mr²sin²ρ – 2MrMcsinρ + Mc²)/(Ml)², and then, by multiplying both sides by -Ml², this simplifies to Mr² – Ml² – Mr²sin²ρ = – (Mr²sin²ρ – 2MrMcsinρ + Mc²), and then Mr² – Ml² – Mr²sin²ρ = – Mr²sin²ρ + 2MrMcsinρ – Mc². Since this equation has the term – Mr²sin²ρ on both sides, cancelling it simplifies this to  Mr² – Ml² = 2MrMcsinρ – Mc², which then becomes Mr² – Ml² + Mc² = 2MrMcsinρ, and then sinρ = (Mr² – Ml² + Mc²)/2MrM= [(0.150 kg)² – (0.100 kg)² + (0.170 kg)²]/[2(0.150 kg)(0.170 kg)] = (0.0225 – 0.0100 + 0.0289)/0.0510 = 0.0414/0.510 = 0.812. The inverse sine of this value gives us ρ = 54°.

Having one angle’s measure, of course, makes it far easier to find the others. Two paragraphs up, an equation in italics stated that sinλ = (Mrsinρ – Mc)/(- Ml). It follows that λ = sin-1[(Mrsinρ – Mc)/(- Ml)] = sin-1[(0.150kg)sinρ – 0.170kg)/(-0.100kg)] = 29°. These two angles sum to 83°, leaving 180° – 83° = 97° as the value of θ.

31_theta_58.png

As can be seen above, these derived values are close to demonstrated experimental values. The first angle found, ρ, measures ~58°, which differs from the theoretical value of 54° by approximately 7%. The next, λ, measures ~31°, also differing from the theoretical value, 29°, by about 7%.The experimental value for θ is (180 – 58 – 31)° = 91°, which is off from the theoretical value of 97° by ~6%. All of these errors are smaller than the 9% error found for both λ and ρ in the easier, symmetrical version of this problem, and the causes of this error should be the same as before.