# Classical Stern-Gerlach Experiment
> You cannot be shocked by two spots on a photographic plate until you know exactly what classical
> physics predicted instead — and earning that prediction honestly is harder than it looks. This
> lesson builds the 1922 apparatus from the ground up: dipoles, gradients, torques, and the one
> subtlety (precession) that makes the whole thing work as a measuring instrument at all.
## Learning Objectives
After this lesson you will be able to:
1. Compute the **force** and **torque** on a magnetic dipole in an arbitrary field, and explain why a *uniform* field can never separate a beam.
2. Show that a spinning rotor's magnetic moment satisfies $\boldsymbol{\mu} = \gamma\mathbf{L}$, and derive the **Larmor** precession frequency from the torque equation.
3. Contrast the motion of a dipole *without* angular momentum (libration) against one *with* it (precession), and explain why only the latter yields a constant deflecting force.
4. Derive the classical prediction for the detector pattern from an isotropic source — a **uniform band with sharp edges**, not a Gaussian blur.
5. Estimate the deflection of a silver-atom beam with realistic 1922 apparatus numbers.
6. State precisely which classical assumptions the experiment was designed to test, so that the quantum result in the next lesson lands as a genuine contradiction.
---
## Intuition
Everything we can learn about an atom, we learn by letting it interact with something we can see.
The Stern-Gerlach experiment is the archetype: it converts an invisible internal property of an
atom — its magnetism — into a visible position on a screen. The strategy is worth stating in the
abstract, because the whole program uses it: **to measure a quantity, arrange for it to control a
force, then let the force act long enough that the resulting displacement is macroscopic.**
Why magnetism? Because an atom containing moving charge is a tiny magnet, and magnets respond to
magnetic fields in a way we can compute exactly. But there is a catch that turns out to be the
entire engineering problem. A magnet in a *uniform* field feels a twist and no push: a compass
needle turns toward north, it does not fly north. To get a *push* you need the field to be
different at the two ends of the magnet — an inhomogeneous field, a gradient. That is why the
Stern-Gerlach magnet has one sharp pole piece and one grooved one: it is a machine for making the
field change rapidly with height.
There is a second catch, subtler and more interesting, and it is the reason this lesson is longer
than "put a dipole in a gradient." A dipole that is *just* a little bar magnet will swing back and
forth in a field like a pendulum, so the component of its moment along the field — the part that
sets the force — keeps changing sign as it swings. Such an object gets pushed up, then down, then
up, and the beam smears in a complicated way that encodes nothing clean. But a dipole that carries
**angular momentum** behaves completely differently: instead of swinging, it *precesses*, like a
gyroscope, and the component along the field stays exactly constant. Only then is the apparatus a
faithful meter. Atoms, fortunately, are the second kind [Fre, §1.1].
---
## Theory
### Force and torque on a magnetic dipole
A small current loop or bar magnet is characterized by a single vector, its **magnetic dipole
moment** $\boldsymbol{\mu}$, which points from the south pole to the north pole of the equivalent
needle and whose magnitude measures the strength of the magnet. For a planar loop of current $I$
enclosing area $A$ with unit normal $\hat{n}$ given by the right-hand rule, $\boldsymbol{\mu} = IA\,\hat{n}$.
Everything about how the object responds to a magnetic field $\mathbf{B}$ follows from one scalar,
the interaction energy
$$
U = -\boldsymbol{\mu}\cdot\mathbf{B} .
$$
The energy is lowest when the moment is aligned with the field, which is why compass needles point
north. From $U$ we get both of the mechanical responses at once. The **torque** is the derivative of
$U$ with respect to orientation, and the force is minus its gradient with respect to position:
$$
\boldsymbol{\tau} = \boldsymbol{\mu}\times\mathbf{B},
\qquad
\mathbf{F} = -\nabla U = \nabla(\boldsymbol{\mu}\cdot\mathbf{B}) .
$$
Read these two formulas carefully, because they say different things. The torque depends only on the
field *at* the dipole; it is nonzero even in a perfectly uniform field. The force depends on how the
field *changes* from place to place; it vanishes identically when $\mathbf{B}$ is uniform, no matter
how strong the field is. A uniform field reorients magnets; only a non-uniform field moves them.
### The projection that sets the force
Orient the apparatus so the field points mainly along $\hat{z}$ and varies mainly along $z$:
$\mathbf{B} \approx B_z(z)\hat{z}$. Then $\boldsymbol{\mu}\cdot\mathbf{B} = \mu_z B_z(z)$, and
$$
F_z = \mu_z\,\frac{\partial B_z}{\partial z} .
$$
The single number that controls the deflection is $\mu_z = \mu\cos\alpha$, the **projection** of the
moment onto the field axis, where $\alpha$ is the angle between $\boldsymbol{\mu}$ and $\mathbf{B}$.
A dipole tipped by $\alpha$ is pushed as though it were a weaker dipole pointing exactly along the
axis. Since $\cos\alpha$ runs continuously over $[-1, +1]$ as the dipole is tilted from parallel to
antiparallel, classical physics says the deflecting force runs continuously over
$[-\mu\,\partial_z B_z,\ +\mu\,\partial_z B_z]$. Hold on to that word *continuously* — it is the
prediction the experiment was built to confirm, and the one it destroyed.
> **Careful.** The direction of $\mathbf{B}$ and the direction in which $|\mathbf{B}|$ *increases*
> are independent choices. Below the north pole of a magnet the field points up and gets stronger as
> you go up; below its south pole the field points down and still gets stronger as you go up. The
> sign of $F_z$ depends on the gradient, so you must track both. Freericks makes this the point of a
> dedicated graphical construction [Fre, §1.1]; algebraically it is just the statement that $\mu_z$
> and $\partial_z B_z$ each carry their own sign.
### Two classical rotors: the needle and the loop
Now the physics that decides whether the apparatus works. Consider a dipole free to rotate, and ask
how $\mu_z$ evolves in time. There are two very different classical answers.
**Case 1 — a needle with no angular momentum.** Model the magnet as a rigid body of moment of
inertia $I_{\text{rot}}$ that is not spinning. The torque $\tau = -\mu B\sin\alpha$ drives the
orientation angle according to
$$
I_{\text{rot}}\,\ddot{\alpha} = -\mu B\sin\alpha ,
$$
which is *exactly the pendulum equation*. The needle swings toward alignment, overshoots (nothing
stops instantly), swings back, and — with no friction, as in vacuum — repeats the cycle forever in
a motion called **libration**. Its
projection $\mu_z = \mu\cos\alpha(t)$ oscillates, so $F_z$ oscillates too, in general changing sign.
An atom of this kind entering the magnet would be pushed up and down repeatedly, and where it landed
would depend on the phase of its swing when it entered. The screen pattern would be a mess that
encodes the initial *phase* as much as the initial *tilt*.
**Case 2 — a rotor that carries angular momentum.** Now suppose the magnetism comes from something
actually circulating, so that the moment is tied to an angular momentum $\mathbf{L}$ by a constant
of proportionality. For any rigid body whose charge and mass are distributed alike, the current and
the mass flow are the same motion, and the two vectors are strictly parallel:
$$
\boldsymbol{\mu} = \gamma\,\mathbf{L}, \qquad \gamma = \frac{q}{2m} ,
$$
where the constant $\gamma$ is called the **gyromagnetic ratio**. (Quick derivation for a circular
orbit: charge $q$ going around a loop of radius $r$ at speed $v$ has period $T = 2\pi r/v$, hence
current $I = q/T = qv/2\pi r$ and moment $\mu = I\pi r^2 = qvr/2 = (q/2m)(mvr) = \gamma L$.)
### Precession: why the loop is the honest model
With $\boldsymbol{\mu} = \gamma\mathbf{L}$, the rotational equation of motion becomes a closed
equation for $\mathbf{L}$ alone:
$$
\frac{d\mathbf{L}}{dt} = \boldsymbol{\tau} = \boldsymbol{\mu}\times\mathbf{B} = \gamma\,\mathbf{L}\times\mathbf{B} .
$$
Two conservation laws fall straight out. Dotting with $\mathbf{L}$ gives
$\tfrac{d}{dt}|\mathbf{L}|^2 = 2\gamma\,\mathbf{L}\cdot(\mathbf{L}\times\mathbf{B}) = 0$: the length
is fixed. Dotting with $\hat{z}$ for $\mathbf{B} = B\hat{z}$ gives
$$
\frac{dL_z}{dt} = \gamma\,(\mathbf{L}\times\mathbf{B})\cdot\hat{z} = 0 ,
$$
because $\mathbf{L}\times\mathbf{B}$ has no $z$-component when $\mathbf{B}\parallel\hat{z}$. **The
projection along the field is a constant of the motion.** The vector neither aligns nor swings: with
fixed length and fixed $z$-component, its tip is confined to a circle, and it sweeps that circle at
a steady rate — this motion is **precession**, the same motion you see in a spinning top or a
gyroscope whose axle traces a cone instead of falling over. Writing $\mathbf{L}$ in components shows
the sweep rate is independent of the tilt angle:
$$
\omega_L = |\gamma| B ,
$$
the **Larmor frequency**. Physically, precession is angular momentum conservation asserting itself:
the torque cannot change $|\mathbf{L}|$ or $L_z$, so all it can do is rotate $\mathbf{L}$ about the
field axis.
This is the result that rescues the experiment. Because $\mu_z$ is *constant* during precession, the
force $F_z = \mu_z\,\partial_z B_z$ is constant throughout the flight through the magnet. Each atom
receives a steady transverse impulse proportional to its own $\mu_z$, and the landing position on
the screen is a faithful, monotonic readout of that one number. A librating needle would give no
such clean map; a precessing rotor does. An inhomogeneous magnet is therefore a **$\mu_z$-meter**,
and only for objects that carry angular momentum.
### Why an atom should behave like a current loop
In 1922, the reigning picture of the atom was Rutherford's nucleus orbited by electrons, quantized
after Bohr and Sommerfeld. Whatever else was wrong with it, it made one prediction that is still
right: the orbiting electron is a current loop, so an atom should carry a magnetic moment tied to
its orbital angular momentum by $\gamma = -e/2m_e$ (negative, because the electron's charge is
negative, so $\boldsymbol{\mu}$ is antiparallel to $\mathbf{L}$). With angular momentum of order
$\hbar$, the natural unit of atomic magnetism is the **Bohr magneton**
$$
\mu_B = \frac{e\hbar}{2m_e} = 9.274\times10^{-24}\ \text{J/T} = 5.788\times10^{-5}\ \text{eV/T} .
$$
So the expected moment of a silver atom is "of order one Bohr magneton," and the expected motion in
the magnet is precession at the Larmor rate with a constant deflecting force. Everything up to here
is classical mechanics and classical electromagnetism, with one quantum input (the *scale* $\hbar$)
and no quantum weirdness at all.
### The apparatus
```mermaid
flowchart LR
OVEN["Oven
silver vapor, ~1300 K"] --> S1["Slit 1"]
S1 --> S2["Slit 2
(collimation)"]
S2 --> MAG["Inhomogeneous magnet
length ~5 cm
gradient ~10³ T/m"]
MAG --> DRIFT["Field-free drift"]
DRIFT --> PLATE["Glass plate
(detector)"]
```
Four stages, each doing one job. A hot oven of metallic silver emits atoms through a pinhole, giving
a thermal beam with mean speed of order several hundred metres per second. A second aperture
downstream performs **collimation**: it discards all but the atoms travelling nearly parallel to the
axis, so that the *undeflected* beam makes a narrow line on the plate. Everything the experiment
can resolve is set here — a deflection smaller than the collimated beam width is invisible. Next,
the beam crosses the field gradient, where each atom picks up a transverse velocity proportional to
its $\mu_z$. Finally a field-free drift region lets those tiny transverse velocities accumulate into
a visible separation before the atoms strike a glass plate and stick.
Silver was an excellent choice: it is heavy enough to travel in straight lines, evaporates at
accessible temperatures, condenses permanently where it lands, and — the part nobody could have
designed on purpose — a silver deposit far too thin to see develops into a visible dark trace when
exposed to sulfur, which the experimenters supplied by smoking cheap cigars over the plates
[Fre, §1.2].
### The classical prediction: a uniform band with sharp edges
Now assemble the prediction. The oven has no reason to prefer any direction, so the moments emerge
**isotropically**: the unit vector $\hat{\mu}$ is uniformly distributed over the sphere. The
deflection of an atom is proportional to $\mu_z = \mu\cos\alpha$, so we need the distribution of
$\cos\alpha$, not of $\alpha$. Solid angle on the sphere is
$$
d\Omega = \sin\alpha\,d\alpha\,d\phi = -\,d(\cos\alpha)\,d\phi ,
$$
so a uniform distribution over the sphere is a *uniform* distribution in $\cos\alpha$ on $[-1,+1]$.
(This is Archimedes' hat-box theorem: equal slices of a sphere by parallel planes have equal area.)
Therefore classical physics predicts something sharper than the "smear" of folklore. The pattern on
the plate is a **band of uniform intensity**, symmetric about the undeflected line, extending out to
$\pm z_{\max}$ where $z_{\max}$ corresponds to $|\mu_z| = \mu$, with **abrupt edges** and *no
concentration anywhere* — least of all at the extremes. Real effects soften this: the thermal spread
in speeds means slower atoms deflect more (the deflection scales as $1/v^2$), and the finite beam
width blurs the edges. But the qualitative signature is unmistakable: **one connected blob covering
every intermediate deflection, densest nowhere, with nothing special about the endpoints.**
Everything the next lesson does depends on your having internalized this prediction. It is not "we
had no idea what would happen." It is a specific, quantitative, edge-to-edge continuum.
### Worked example: how big is the deflection?
Take a silver atom, $M = 1.79\times10^{-25}$ kg, moving at $v = 660$ m/s through a magnet of length
$\ell = 5$ cm with gradient $\partial_z B_z = 10^3$ T/m, followed by a drift of $D = 20$ cm. Consider
the extreme case $\mu_z = \mu_B$. The acceleration inside the magnet is
$$
a = \frac{\mu_B\,\partial_z B_z}{M} = \frac{9.274\times10^{-24}\times10^{3}}{1.79\times10^{-25}}
= 5.18\times10^{4}\ \text{m/s}^2 .
$$
The time in the field is $t_1 = \ell/v = 7.58\times10^{-5}$ s, giving a displacement inside the
magnet of $z_1 = \tfrac12 a t_1^2 = 1.49\times10^{-4}$ m and an exit transverse velocity of
$v_z = a t_1 = 3.93$ m/s. The drift takes $t_2 = D/v = 3.03\times10^{-4}$ s and adds
$z_2 = v_z t_2 = 1.19\times10^{-3}$ m. The total is
$$
z_{\max} = z_1 + z_2 = \frac{a\ell}{v^2}\left(\frac{\ell}{2} + D\right) \approx 1.34\ \text{mm} ,
$$
so the classical band should be about $2z_{\max} \approx 2.7$ mm tall. Two lessons hide in this
arithmetic. First, the drift region contributes nearly nine tenths of the deflection — the magnet
supplies the impulse, the flight path supplies the magnification, exactly as a lever arm does.
Second, the whole effect is sub-millimetre-per-decimetre: with a beam only tens of microns wide and
a plate read under a microscope, this experiment sat right at the edge of what 1922 could do, which
is why it failed repeatedly before it succeeded.
### A Stern-Gerlach experiment that cannot work
Suppose a colleague proposes an improved apparatus using a beautifully *uniform* field, arguing that
a cleaner field must give a cleaner separation. The proposal is hopeless, and it is worth being able
to say exactly why [Fre, §1.1.5].
In a uniform field, $\nabla(\boldsymbol{\mu}\cdot\mathbf{B}) = 0$, so $\mathbf{F} = 0$ identically:
no atom is deflected, whatever its moment. The field is not inert — each atom's moment precesses at
$\omega_L = |\gamma|B$ about the field axis — but precession moves no centre of mass. Every atom
flies straight through and lands in the undeflected line, and the plate shows a single narrow trace
identical to the one you get with the magnet switched off. The apparatus sorts nothing, because
sorting requires the *energy* to depend on position, and in a uniform field it does not.
The comparison also sharpens what a gradient buys you. A source that emitted only one value of
$\mu_z$ would, in a real inhomogeneous magnet, produce a single displaced line; in a uniform field it
would produce a single undisplaced line. The two experiments differ observably — one line moves when
you reverse the gradient, the other does not — so gradient-off and single-valued-source are
distinguishable hypotheses, not the same null result.
---
## Hands-on (Python)
Simulate the classical plate and confirm the uniform-band prediction, including the thermal
velocity spread that softens its edges.
```python
import numpy as np
import matplotlib.pyplot as plt
rng = np.random.default_rng(0)
N = 200_000
mu_B, M = 9.274e-24, 1.79e-25 # J/T, kg
grad, ell, D = 1.0e3, 0.05, 0.20 # T/m, m, m
v0 = 660.0 # m/s
# Nominal band edge: |cos(alpha)| = 1 at the mean speed (the worked example).
z_edge = (mu_B * grad / M) * ell * (ell / 2 + D) / v0**2
# (a) Isotropic moments -> cos(alpha) uniform on [-1, 1] (Archimedes).
cos_alpha = rng.uniform(-1.0, 1.0, N)
# (b) Thermal speed spread (deflection scales as 1/v^2, so slow atoms
# overshoot the nominal edge and soften it into a tail).
v = np.clip(v0 * (1.0 + 0.15 * rng.normal(size=N)), 250.0, None)
# (c) Deflection: impulse in the magnet, magnified over the drift, plus the
# finite width of the collimated beam.
z = cos_alpha * z_edge * (v0 / v) ** 2
z += 0.02 * z_edge * rng.normal(size=N)
plt.hist(z / z_edge, bins=300, density=True)
plt.xlabel("deflection / nominal band edge")
plt.ylabel("intensity")
plt.title("Classical prediction: a filled band, not two spots")
plt.show()
print(f"nominal band edge = {z_edge * 1e3:.2f} mm")
print(f"within 10% of the centre = {np.mean(np.abs(z) < 0.1 * z_edge):.3f}")
print(f"outer 10% of the band = {np.mean(np.abs(z) > 0.9 * z_edge):.3f}")
# Expect a band edge of ~1.34 mm and ~0.10 of the atoms near the centre --
# the flat-density prediction of E3. The outer bin runs a little higher
# (~0.16) only because the 1/v^2 tail lets slow atoms overshoot the nominal
# edge. Nothing is special about the extremes; the middle is not empty.
```
The number to remember from the last line: **the classical centre is not empty.** In the next
lesson, it will be.
---
## Exercises
**E1 (easy).** A bar magnet is released in a perfectly uniform magnetic field. Describe its
subsequent motion, and state what happens to its centre of mass. Repeat for a spinning gyroscope-like
magnet with $\boldsymbol{\mu} = \gamma\mathbf{L}$.
Solution
In a uniform field the force is $\mathbf{F} = \nabla(\boldsymbol{\mu}\cdot\mathbf{B}) = 0$, so in
both cases the centre of mass moves in a straight line at constant velocity — no deflection
whatsoever. The *orientations* differ: the non-spinning bar magnet feels torque
$\boldsymbol{\tau} = \boldsymbol{\mu}\times\mathbf{B}$ with no angular momentum to stabilize it, so it
librates about alignment like a frictionless pendulum, overshooting forever. The spinning magnet
precesses about the field axis at $\omega_L = |\gamma|B$, keeping both $|\boldsymbol{\mu}|$ and
$\mu_z$ fixed. Torque changes orientation; only a gradient changes position.
Solution
$\omega_L = eB/2m_e = (1.602\times10^{-19} \times 0.1)/(2\times 9.109\times10^{-31})
= 8.79\times10^{9}$ rad/s, so $f_L = \omega_L/2\pi = 1.40$ GHz and the period is
$T_L = 7.1\times10^{-10}$ s. During $t_1 = 7.6\times10^{-5}$ s the moment precesses about
$t_1/T_L \approx 1.1\times10^{5}$ times — roughly a hundred thousand revolutions. This is why the
transverse components of $\boldsymbol{\mu}$ contribute nothing: they average to zero over the flight
many times over, leaving only the constant $\mu_z$ to produce a net force.
Solution
The deflection is $z \propto \mu_z = \mu\cos\alpha$. For an isotropic direction, the measure is
$d\Omega = \sin\alpha\,d\alpha\,d\phi = -d(\cos\alpha)\,d\phi$, so $u \equiv \cos\alpha$ is uniform on
$[-1,1]$ with density $p(u) = 1/2$. Since $z = z_{\max}u$ is a linear map, $z$ is uniform on
$[-z_{\max}, z_{\max}]$ with density $1/(2z_{\max})$ — a flat band.
"Outer 10% on either side" means $|z| > 0.9\,z_{\max}$, i.e. $|u| > 0.9$, with probability
$2\times(0.1)/2 = 0.10$. So only one atom in ten lands near an edge, and exactly the same fraction
lands within 10% of the *centre*. Classically the edges are in no way preferred — which is precisely
what makes the observed result impossible to accommodate.
Solution
Inside the magnet, $z_1 = \tfrac12 a t_1^2$ with $t_1 = \ell/v$; afterwards the transverse velocity
$v_z = a t_1$ is constant, so $z_2 = v_z t_2 = a t_1 D/v$. Hence
$$
\frac{z_2}{z_1} = \frac{a t_1 D/v}{\tfrac12 a t_1^2} = \frac{2D}{\ell} ,
$$
which for $D = 0.20$ m, $\ell = 0.05$ m gives $8$ — matching $1.19/0.149 \approx 8$. The total
$z_{\max} = (a\ell/v^2)(\ell/2 + D)$ is *linear* in $D$ but only linear in the gradient too, and
gradients are limited by pole-piece geometry, magnet saturation, and the requirement that the field
stay uniform across the beam width. Drift length is free: it costs vacuum pipe. The kinematic reason
is that the magnet's job is to deliver a transverse *impulse*, and once delivered, time does the
amplifying at no further cost.
Solution
*Multipole correction.* The dipole force is the first term of an expansion of
$\mathbf{F} = \nabla\!\int \mathbf{M}\cdot\mathbf{B}$ in powers of $r\,\partial/\partial z$. The
relative size of the next term is $\sim r\,(\partial_z B_z)/B \approx
10^{-10}\times10^{3}/0.1 = 10^{-6}$. Utterly negligible: the field varies by one part in $10^{6}$
across the atom, so the atom is a point dipole for every practical purpose.
*Transverse gradients.* In the field-free bore, $\nabla\cdot\mathbf{B} = 0$ forces
$\partial_x B_x + \partial_y B_y = -\partial_z B_z \ne 0$, so a gradient along $z$ *requires*
transverse gradients of comparable magnitude — there is no such thing as a purely one-dimensional
gradient. The corresponding force components are $F_x = \mu_x\partial_x B_x$ and similarly for $y$.
They do not spoil the measurement because $\mu_x$ and $\mu_y$ precess at the Larmor frequency,
completing $\sim10^{5}$ revolutions during the flight (E2), so their time-averaged contribution
vanishes to a part in $10^{5}$, while $\mu_z$ is strictly constant and accumulates coherently. This
is the deep reason the experiment needs precessing rotors rather than librating needles: precession
is what averages the unwanted force components away while protecting the wanted one.
Answers
1. The force is $\nabla(\boldsymbol{\mu}\cdot\mathbf{B})$, which vanishes for uniform $\mathbf{B}$ however large; only a gradient makes the energy position-dependent. A uniform field gives one undeflected line — the same picture as with the magnet off, except that the moments precess (invisibly) on the way through.
2. Without angular momentum the dipole librates like a pendulum, so $\mu_z$ oscillates and the deflecting force changes sign during the flight; with angular momentum it precesses, holding $\mu_z$ exactly constant and delivering a steady impulse proportional to that one number. Only the latter maps "landing position" faithfully onto "projection".
3. Dot with $\mathbf{L}$: $\tfrac{d}{dt}|\mathbf{L}|^2 = 2\gamma\mathbf{L}\cdot(\mathbf{L}\times\mathbf{B}) = 0$. Dot with $\hat{z}$ (with $\mathbf{B} = B\hat{z}$): $\dot{L_z} = \gamma(\mathbf{L}\times\mathbf{B})\cdot\hat{z} = 0$. Fixed length and fixed $z$-component confine the tip to a circle traversed at rate $\omega_L = |\gamma|B$ — precession on a cone.
4. A band of *uniform* intensity between $\pm z_{\max}$ with sharp edges (softened by thermal and beam-width effects), symmetric about the centre. The intensity is the same everywhere inside the band — no peak at the edges, and emphatically no gap in the middle.
5. The drift region: $z_2/z_1 = 2D/\ell = 8$. The magnet supplies a transverse impulse, and the field-free flight magnifies it linearly in the drift length.