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Cross-topic physical model selection

Learning objectives

  • Outcome

    For a described phenomenon, list at least two candidate models and state the assumption by which one of them is rejected.

  • Outcome

    Compute a dimensionless ratio of characteristic quantities (L/rL/r, a/λa/\lambda, λ/L\lambda/L) and use its value to assign the problem to the geometric, wave, or quantum regime.

  • Outcome

    State at least two conditions under which the chosen model becomes inapplicable, and propose a measurement that distinguishes two competing models.

Understand the idea

Choosing a physical model for a phenomenon: a practicum

Text

Choosing a physical model is the first step of any solution: first we decide what may be neglected, and only then do we calculate. The candidates almost always come from topics already studied: point mass, rigid body, ideal gas, geometric optics, wave model, classical or quantum particle. Each model has a list of assumptions, and it works as long as those assumptions hold. The practical technique is to estimate characteristic quantities and form a dimensionless ratio: for instance L/rL/r for a point mass, a/λa/\lambda for geometric optics, λ/L\lambda/L for quantum behaviour. If the ratio is much greater or much less than one, the choice is unambiguous; if it is of order one, we are at the edge of the model and the description must be made more detailed. The chosen model is then tested against predictions: agreement with measurement within uncertainty supports it, a systematic discrepancy signals that the model must be changed.

Worked examples

Direct

Problem

A satellite of size L=5L = 5 m moves in a circular orbit of radius r=7.0×106r = 7.0\times10^{6} m. May its motion be described by the point-mass model? Carry out the check with a dimensionless ratio.

Solution

Step 1. The point-mass assumption: the size of the body is much smaller than the distances over which the motion changes appreciably, that is L≪rL \ll r.

Step 2. Compute the ratio:

Lr=57.0×106≈7.1×10−7.\frac{L}{r} = \frac{5}{7.0\times10^{6}} \approx 7.1\times10^{-7}.

Step 3. Since L/r≪1L/r \ll 1, all points of the satellite move essentially identically, and rotation about its own axis does not affect the orbit.

Answer: the point-mass model applies; the orbital period follows from T=2πr3/(GM)T = 2\pi\sqrt{r^{3}/(GM)}, which does not contain the satellite mass. The model fails if we care about the spacecraft's orientation — then a rigid-body model is needed.

Application

Problem

Sound of frequency f=1000f = 1000 Hz (v=343v = 343 m/s) passes through a doorway of width a=0.9a = 0.9 m, and light of wavelength λ=550\lambda = 550 nm passes through the same opening. For each case decide which model is required: geometric optics (straight rays) or the wave model (diffraction). Justify with the ratio a/λa/\lambda.

Solution

Step 1. Sound wavelength.

λsnd=vf=3431000≈0.343 m.\lambda_{\text{snd}} = \frac{v}{f} = \frac{343}{1000} \approx 0.343\ \text{m}.

Step 2. Ratio for sound.

aλsnd=0.90.343≈2.6.\frac{a}{\lambda_{\text{snd}}} = \frac{0.9}{0.343} \approx 2.6.

The ratio is of order one: the opening is comparable to the wavelength, so diffraction is strong — sound is heard around the corner and there is no sharp shadow. The wave model is required.

Step 3. Ratio for light.

aλlt=0.9550×10−9≈1.6×106.\frac{a}{\lambda_{\text{lt}}} = \frac{0.9}{550\times10^{-9}} \approx 1.6\times10^{6}.

Here a≫λa \gg \lambda, so the diffraction spread is of order λ/a∼10−6\lambda/a \sim 10^{-6} rad and is invisible. The geometric model applies (rays, sharp shadow).

Answer: sound — wave model, light — geometric model; the threshold is set not by the nature of the wave but by the ratio a/λa/\lambda.

Transfer

Problem

Decide whether to treat (a) a football m=0.43m = 0.43 kg flying at v=25v = 25 m/s and (b) an electron with kinetic energy E=10E = 10 eV as classical particles. The criterion is the ratio of the de Broglie wavelength λ=h/p\lambda = h/p to the object scale LL (L≈0.22L \approx 0.22 m for the ball, L≈0.3L \approx 0.3 nm for the atomic scale in a crystal).

Solution

Step 1. The ball.

p=mv=0.43⋅25=10.75 kg⋅m/s,p = mv = 0.43\cdot 25 = 10.75\ \text{kg·m/s},

λ=hp=6.63×10−3410.75≈6.2×10−35 m.\lambda = \frac{h}{p} = \frac{6.63\times10^{-34}}{10.75} \approx 6.2\times10^{-35}\ \text{m}.

λL=6.2×10−350.22≈2.8×10−34≪1.\frac{\lambda}{L} = \frac{6.2\times10^{-35}}{0.22} \approx 2.8\times10^{-34} \ll 1.

Wave effects are unobservable — a classical particle.

Step 2. The electron. Convert the energy: E=10E = 10 eV =1.60×10−18= 1.60\times10^{-18} J. From p=2mEp = \sqrt{2mE}:

p=2⋅9.11×10−31⋅1.60×10−18≈1.71×10−24 kg⋅m/s,p = \sqrt{2\cdot 9.11\times10^{-31}\cdot 1.60\times10^{-18}} \approx 1.71\times10^{-24}\ \text{kg·m/s},

λ=6.63×10−341.71×10−24≈3.9×10−10 m=0.39 nm.\lambda = \frac{6.63\times10^{-34}}{1.71\times10^{-24}} \approx 3.9\times10^{-10}\ \text{m} = 0.39\ \text{nm}.

Step 3. The ratio.

λL=0.390.3≈1.3∼1.\frac{\lambda}{L} = \frac{0.39}{0.3} \approx 1.3 \sim 1.

Answer: the ratio is of order one — the electron diffracts on the crystal and the quantum (wave) model is needed; for the ball λ/L≪1\lambda/L \ll 1 and the classical description is fully sufficient. The same criterion λ/L\lambda/L explains why the same formula λ=h/p\lambda = h/p leads to different conclusions at different scales.

Practice

Text

Goal: see for yourself where geometric optics ends and the wave regime begins, and obtain a numerical decision rule.

Where to work: in a simulation only — PhET "Wave Interference" (single-slit/diffraction mode) or an equivalent. Never point a real laser into an eye or at people.

  1. Set λ=500\lambda = 500 nm and slit width a=20λa = 20\lambda. Sketch the pattern behind the slit — this is the "geometric" regime (a narrow strip, almost a shadow).
  2. Reduce the slit: 10λ10\lambda, 5λ5\lambda, 2λ2\lambda, 1λ1\lambda, 0.5λ0.5\lambda. For each value record the ratio a/λa/\lambda and sketch the pattern.
  3. Note at which a/λa/\lambda the central maximum stops being a narrow strip and broadens noticeably.
  4. State your own rule: for example, "a/λ≳20a/\lambda \gtrsim 20 — geometric model, a/λ≲5a/\lambda \lesssim 5 — wave model".
  5. Test the rule at a second wavelength (λ=650\lambda = 650 nm) and record whether it holds. If not, adjust the thresholds.

Safety: work in the simulation only; if you additionally listen to a tone through a doorway, keep the volume moderate and do not disturb others.

Visual reference

Purpose

To show that model selection is not a matter of taste but a testable decision rule based on a dimensionless ratio of characteristic quantities.

Representation

A flowchart. At the top, a box "Phenomenon and question". Three candidate branches leave it: "point mass / rigid body", "wave / geometric", "classical / quantum". Each branch contains its own ratio test with a threshold: L/r≪1L/r \ll 1, a/λa/\lambda (threshold near 20), λ/L\lambda/L (threshold near 1). At the bottom there are two outcome boxes: "model works — compute" and "edge of the model — refine the description", with an arrow from the second back to the list of candidates. Alongside are three small sketches: a point on an orbit with the segment rr; a slit of width aa with wavefronts of wavelength λ\lambda; a particle with an attached sinusoid of wavelength λ\lambda.

Real-world examples

A football free kick: a point mass or a spinning sphere?

Scenario

A free kick from 25 m25\ \text{m}: a ball of mass m=0,43 kgm = 0{,}43\ \text{kg} flies at v=25 m/sv = 25\ \text{m/s} and stays in the air for about 1,0 s1{,}0\ \text{s}. The model 'point mass in a gravitational field' predicts a drop of 12gt2≈4,9 m\frac{1}{2} g t^2 \approx 4{,}9\ \text{m} over that time, with weight mg≈4,2 Nmg \approx 4{,}2\ \text{N}. An estimate of the drag force Fd≈12ρCdAv2F_d \approx \frac{1}{2} \rho C_d A v^2 with ρ=1,2 kg/m3\rho = 1{,}2\ \text{kg/m}^3, Cd≈0,25C_d \approx 0{,}25 and cross-section A≈0,038 m2A \approx 0{,}038\ \text{m}^2 gives Fd≈3,6 NF_d \approx 3{,}6\ \text{N} — the same order of magnitude as mgmg.

Connection

This is an exercise in model selection: by comparing FdF_d and mgmg we test the assumption of the ideal mechanical model 'a point mass with no medium resistance' and decide whether point mechanics suffices or aerodynamics with the Magnus force from the ball's spin is needed. The selection criterion is the ratio of the characteristic forces and the acceptable error in the predicted trajectory.

An electric kettle: heat capacity alone or losses as well?

Scenario

A kettle of power P=1,5 kWP = 1{,}5\ \text{kW} heats m=1,0 kgm = 1{,}0\ \text{kg} of water from 20 ∘C20\ ^\circ\text{C} to 100 ∘C100\ ^\circ\text{C} with c=4200 J/(kg⋅K)c = 4200\ \text{J/(kg}\cdot\text{K)}. An energy balance without losses gives t=mcΔTP=1,0⋅4200⋅801500≈224 st = \frac{m c \Delta T}{P} = \frac{1{,}0 \cdot 4200 \cdot 80}{1500} \approx 224\ \text{s}. A stopwatch shows about 250 s250\ \text{s}, roughly 10 %10\ \% more: the average absorbed power is 1344 W1344\ \text{W}, and the difference P−1344 W≈156 WP - 1344\ \text{W} \approx 156\ \text{W} goes to the surroundings.

Connection

A choice is made between two thermal models: a closed system with perfect insulation (Qel=QheatingQ_{\text{el}} = Q_{\text{heating}}) and a system with heat losses. The discrepancy between 224 s224\ \text{s} and 250 s250\ \text{s} is an experimental test of the no-loss assumption; the power difference estimates whether a model with losses is required.

Wi-Fi at home: a ray or a wave?

Scenario

A phone receives a Wi-Fi signal at frequency f=2,4 GHzf = 2{,}4\ \text{GHz}. The wavelength is λ=cf=3⋅1082,4⋅109=0,125 m\lambda = \frac{c}{f} = \frac{3\cdot 10^8}{2{,}4\cdot 10^9} = 0{,}125\ \text{m}. A room 5 m5\ \text{m} wide is 40λ40\lambda, so propagation inside the room is well described by the geometric (ray) approximation. A doorway of 0,9 m0{,}9\ \text{m} is only 7λ7\lambda, and diffraction at its edges is noticeable; a wall 0,25 m0{,}25\ \text{m} thick is 2λ2\lambda, so the transmitted wave is strongly attenuated.

Connection

One must choose between the ray model (straight-line propagation, reflection) and the wave model (diffraction, interference). The criterion is the ratio of the obstacle or aperture size aa to the wavelength λ\lambda: for a≫λa \gg \lambda the ray works, for a∼λa \sim \lambda the wave picture does. Here λ\lambda is obtained from v=λfv = \lambda f for an electromagnetic wave with speed cc.