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.
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.
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 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.
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.
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.
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 = Mr and 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.
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.
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)².
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)/2MrMc = [(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.
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.
Because there’s nothing wrong with mixing a little Rolling Stones with your physics, that’s why.
In the Summer of 2014, with many other science teachers, I took a four-day-long A.P. Physics training session, which was definitely a valuable experience, for me, as a teacher. On the last day of this training, though, in the late afternoon, as the trainer and trainees were winding things up, some of us, including me, started getting a little silly. Physics teachers, of course, have their own version of silly behavior. Here’s what happened.
The trainer: “Let’s see how well you understand the different forces which can serve as centripetal forces, in different situations. When I twirl a ball, on a string, in a horizontal circle, what is the centripetal force?”
The class of trainees, in unison: “Tension!”
Trainer: “In the Bohr model of a hydrogen atom, the force keeping the electron traveling in a circle around the proton is the . . . ?”
Class: “Electromagnetic force!”
Trainer: “What force serves as the centripetal force keeping the Earth in orbit around the Sun?”
Me, loudly, before any of my classmates could answer: “God’s will!”
I was, remember, surrounded by physics teachers. It took the trainer several minutes to restore order, after that.