Happy Holidays

I'm not selling anything

It’s Christmas Eve. I haven’t done much in the past two weeks, since I took my final exam.

I’ve never worked harder than I did this fall. I didn’t have time to play my guitar or paint very much. I wrote a lot, though. “Measurement of the Band Gap in a Silicon Diode” was a hit, as was “Defending Functionalism: Qualia as Cognitive Shortcuts.”

I’ve been playing guitar and singing again, though, because I’m playing two shows in January.

January 4th, 4pm
Omar and Friends: Fiddles, Strings and Stories
The Arts Center
400 Roberson Street, Carrboro, NC 27510
Get tickets: https://app.arts-people.com/index.php?actions=4&p=1

January 11th, 5pm
The Velvet Feast Song Swap: Angel Snow, Jonathan Byrd, Jess Klein
Eno House
903 Eno Street, Hillsborough, NC 27278
Get tickets: https://www.eventbrite.com/e/the-velvet-feast-song-swap-angel-snow-jonathan-byrd-jess-klein-tickets-1974948840145

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The physics final had two types of answers. Some questions gave enough information that, if you could figure out how, you could arrive at an exact-ish answer. Others, you just had to do the best you could. Maybe the answer is a collection of unsolved variables, or even a guess that shows you understand what’s going on.

The first type: “What will be the distance of closest approach to a gold (Z=79) nucleus for an alpha particle of energy 5.0 MeV?”

Z=79 means a gold nucleus has seventy-nine protons. An alpha particle is two protons and two neutrons, also known as a helium nucleus. The number of protons in each mass tells us the electrostatic charge. Like repels like, and at some point, they will push each other away (there’s not nearly enough energy for fusion).

Alpha particles are emitted by the radioactive decay of elements like uranium, and that’s where the helium in your party balloon comes from. Notice this is a gold nucleus, not an atom, and a helium nucleus, not an atom. Electrons would complicate things, and the question is quietly telling us that there are none.

An “eV” is an electron volt. It’s a convenient unit of energy for tiny things, and if you know how, you can get a lot of information out of it.

The potential energy between two charges is called the Coulomb potential, and it is:

\(\Delta U=\frac{q_1 q_2}{4\pi \epsilon_0 r}\Longrightarrow\quad\frac{(2e)(79e)}{4\pi \epsilon_0 r}=\frac{158e^2}{4\pi \epsilon_0 r}\)

The q’s are the charges of each mass. The “epsilon naught” in the denominator is the vacuum permittivity, meaning how much of an insulator ‘nothing’ is, which is all that lies between our two masses as the alpha particle closes in. ‘r’ is the distance between the charges. None of this is in eV yet, so my strategy was to calculate the square of e (the positive charge of a proton, or ‘elementary charge’), multiply everything out except for r, and convert to eV.

\(e^2=(1.602\times10^{-19})^2\approx2.5664\times10^{-28} \,\,\text{Coulombs}^2\)

\(\frac{2.5664\times10^{-28}}{4\cdot \pi\cdot 8.85\times10^{-12}}=2.3077\times10^{-28}\,\,\text{Joules x meters}\)

\(\text{convert to eV}\rightarrow\frac{2.3077\times10^{-28}}{1.602\times10^{-19}}\approx 1.44\times10^{-9} eV\cdot m\)

If I convert these nano-eVs to Mega-eVs, then the meters will become femtometers:

\(10^{-9}\cdot 1=10^6\cdot 10^{-15}\)

Then r needs to be in femtometers to cancel units and give me back MeV. The particle’s energy is 5.0MeV, and now I can use that in the equation to get my answer.

\(5.0\,\,MeV=\frac{1.44(2\cdot79)}{r_{fm}}\Longrightarrow\quad r_{fm}=\frac{1.44(158)}{5.0}=\frac{227.52}{5.0}=45.504\,fm\)

We were given two significant figures, so I answered that the distance of closest approach was 46 femtometers. Amazingly, the energy also tells us how fast the alpha particle is traveling, and you’re about to learn how strong the electrostatic force actually is—

36 million mph, or 58 million kph. And the gold nucleus says, “Talk to the hand.”

The second type: This question took me the entire exam to answer. I kept coming back and thinking about it. “If a particle decays in 10⁻⁸ seconds in its own frame and travels 30m in the lab frame, what is its Lorentz factor in the lab frame?”

The Lorentz factor tells you how much the weird relativistic effects happen at high speeds. The thing about this problem is, we are given the time elapsed in the particle’s frame, but the distance traveled in the lab frame. I tried systems of equations and got answers in terms of variables, but I suspected this wasn’t that kind of problem.

Eventually, I realized that a decaying particle that lived long enough to travel 30 meters at nanosecond lifetimes must be highly relativistic, that is, traveling very close to the speed of light, which is about 3x10⁸ meters per second. t is time in the particle’s frame, and t’ is the time in the lab. The Lorentz factor is the gamma symbol.

\(t' = \frac{30m}{3\times 10^8 \,m/s}=10^{-7}\, \text{seconds}\)

\(t'=\gamma t\Longrightarrow\quad \gamma =\frac{t'}{t}=\frac{10^{-7}}{10^{-8}}=10\)

\(\gamma\approx 10\)

The Lorentz factor is approximately 10, meaning time for the particle runs about 1/10 the speed it does in the lab, and if the particle were a meter stick, it would look like a 10-centimeter stick as it whizzed by you.

I didn’t get everything correct on the final, but I did well enough, and I got those right. It’s been nice to work on songs again. They travel more slowly, and you can get your hands on them. I’m looking forward to singing some for you.

Your fan,

Jonathan Byrd