Physics in Everyday Life: The Quantum Mechanics and Relativity Running Your Phone, GPS and MRI

0. Key points

• The semiconductors in your phone rest on a consequence of quantum mechanics: an electron in a crystal can occupy only certain discrete bands of energy. Fix the band gap $E_g$ and you have fixed, through a single exponential, both the color the material emits and the current that flows through it of its own accord at room temperature.
• Miniaturization stalled because of tunneling. Every $1\ \mathrm{nm}$ shaved off the gate insulator multiplies the leakage current by roughly $7 \times 10^{7}$. That number falls out of an exponential you can evaluate by hand.
• A GPS satellite clock runs fast relative to a clock on the ground by about $38\ \mu\mathrm{s}$ per day. This is the difference between a special-relativistic loss of $-7.2\ \mu\mathrm{s}$ and a general-relativistic gain of $+45.7\ \mu\mathrm{s}$. Left uncorrected, the positioning error accumulates to about $11\ \mathrm{km}$ in a single day.
• What MRI actually sees is a minute imbalance of orientation among the hydrogen nuclei in your body: five nuclei in every million. That is still enough, because $1\ \mathrm{mm}^3$ contains $3 \times 10^{14}$ of them.
• The three technologies share one equation, $E = hf$. A band gap becomes the color of light, the Zeeman splitting of a nucleus becomes a radio frequency, and the hyperfine structure of a cesium atom becomes the second itself.

1. Motivation: twentieth-century physics in your pocket

Tell a physicist of 1900 that the theory they are about to build will, a hundred years later, be carried around by the whole of humanity in a slab that fits in the palm of a hand, and they will probably not believe you. Quantum mechanics and relativity both began as small discrepancies in the corner of a laboratory: a black-body spectrum that would not fit, an ether that could not be found.

Yet today, in the few seconds it takes you to open a map on your phone and check where you are, the following happens.

1. Your finger touches the glass, a change in the capacitance of a transparent electrode is read out, and a silicon integrated circuit turns it into a number. Inside that circuit, electrons move through discrete energy bands.
2. Differences in the arrival times of radio signals from GPS satellites at an altitude of roughly $2\times 10^{4}\ \mathrm{km}$ are computed. The atomic clocks aboard those satellites have had special- and general-relativistic corrections built into them from the start.
3. And if you are unwell, you go to a hospital and lie down inside an MRI scanner. The hydrogen nuclei in your body emit radio waves by nuclear magnetic resonance.

In this article we take these three in turn and carry every calculation through to a number. Anyone can say “relativity is being used here”; being able to produce “how many microseconds per day” yourself makes the statement suddenly concrete. The only mathematics we need is exponentials, square roots, and the approximation $(1+x)^{1/2} \approx 1 + x/2$.

2. Preliminaries: one equation running through all three

The three technologies look like they belong to entirely different fields, but at their root lies the same relation.

Definition 2.1（The Planck relation）

A single photon of an electromagnetic wave of frequency $f$ (angular frequency $\omega = 2\pi f$) carries energy

$$E = hf = \hbar \omega ,$$

where $h = 6.62607\times 10^{-34}\ \mathrm{J\,s}$ is the Planck constant and $\hbar = h/(2\pi)$ the reduced Planck constant.

Conversely, a system whose two energy levels differ by $\Delta E$ absorbs and emits electromagnetic radiation at $f = \Delta E / h$.

This equation is a dictionary translating “energy difference” into “frequency”. As we shall see, $\Delta E$ is the band gap in a semiconductor (visible to infrared), the Zeeman splitting of a nuclear spin in MRI (radio), and the hyperfine structure of the cesium atom in an atomic clock (microwave).

The constants used below are collected here. Values are given to four or five significant figures, based on the CODATA 2018 recommended values.

Symbol	Value	Meaning
$c$	$2.99792\times 10^{8}\ \mathrm{m/s}$	speed of light in vacuum
$h$	$6.62607\times 10^{-34}\ \mathrm{J\,s}$	Planck constant
$hc$	$1239.84\ \mathrm{eV\,nm}$	conversion between wavelength and energy
$k_B$	$1.38065\times 10^{-23}\ \mathrm{J/K}$	Boltzmann constant
$k_B T$ ($T = 300\ \mathrm{K}$)	$0.02585\ \mathrm{eV}$	thermal energy at room temperature
$m_e$	$9.109\times 10^{-31}\ \mathrm{kg}$	electron mass
$GM_\oplus$	$3.986\times 10^{14}\ \mathrm{m^3/s^2}$	geocentric gravitational constant
$R_\oplus$	$6.371\times 10^{6}\ \mathrm{m}$	mean radius of the Earth

Tip

Commit $hc = 1239.84\ \mathrm{eV\,nm}$ to memory. In the form “divide $1240$ by the energy in eV to get the wavelength in nm”, it comes up every single time one discusses visible light.

3. The smartphone: discrete energies turned into electronics

3.1. Why “half a conductor”?

Metals conduct well; glass does not conduct at all. Semiconductors sit in between, but what makes them interesting is not that they conduct moderately — it is that whether they conduct can be switched from outside. That switch is the transistor, and ten billion transistors together make the heart of a phone.

Why can it be switched? Because the energies available to an electron in a crystal are not continuous but arranged in bands.

Definition 3.1（Band gap）

The energies available to an electron in a crystal split into several continuous regions (energy bands), separated by regions in which no electron state exists (forbidden gaps). The highest band completely filled with electrons at absolute zero is the valence band, the empty band above it the conduction band, and their energy difference

$$E_g = E_{\text{bottom of conduction band}} - E_{\text{top of valence band}}$$

is called the band gap. A material with $E_g = 0$ is a metal; one whose $E_g$ is well above $k_B T$ and of order a few eV is a semiconductor; one with a far larger gap is an insulator.

The reason bands exist is quantum mechanical. In an isolated atom the electron energies form discrete levels; line up $10^{23}$ atoms into a crystal and the wave functions of neighboring atoms overlap, splitting each level into a dense band. In a periodic crystal, an electron wave of a certain wavelength is Bragg-reflected and cannot propagate, so a gap opens there — that is the forbidden gap.

Band diagrams for an intrinsic, an n-type and a p-type semiconductor. The shaded lower band is the filled valence band; the outlined upper band is the conduction band.

3.2. What the band gap determines, part 1: the color of the light

Proposition 3.2（Band gap and emission wavelength）

In a direct-gap semiconductor of band gap $E_g$, when an electron at the bottom of the conduction band recombines with a hole at the top of the valence band and emits one photon, the vacuum wavelength $\lambda$ of that photon is

$$\lambda = \frac{hc}{E_g}, \qquad \text{numerically}\quad \lambda\,[\mathrm{nm}] = \frac{1239.84}{E_g\,[\mathrm{eV}]}.$$

Proof（Proposition 3.2）

The energy the electron loses in recombining is exactly $E_g$. By conservation of energy, the emitted photon carries energy $E_g$ as well. By Definition 2.1 the photon energy is $E = hf$, and in vacuum $f = c/\lambda$, so

$$E_g = \frac{hc}{\lambda} \iff \lambda = \frac{hc}{E_g}.$$

Substituting $hc = 1239.84\ \mathrm{eV\,nm}$ from the table below Definition 2.1 gives the relation for $E_g$ measured in eV and $\lambda$ in nm.

∎

Example 3.3（Silicon does not glow; gallium nitride glows blue）

Feeding the tabulated values into Proposition 3.2:

Material	$E_g$	$\lambda = 1239.84/E_g$	Appearance
germanium Ge	$0.66\ \mathrm{eV}$	$1879\ \mathrm{nm}$	infrared
silicon Si	$1.12\ \mathrm{eV}$	$1107\ \mathrm{nm}$	infrared
gallium arsenide GaAs	$1.42\ \mathrm{eV}$	$873\ \mathrm{nm}$	near infrared (remote controls)
gallium phosphide family	$\approx 2.0\ \mathrm{eV}$	$620\ \mathrm{nm}$	red
InGaN (blue LED)	$\approx 2.76\ \mathrm{eV}$	$449\ \mathrm{nm}$	blue
gallium nitride GaN	$3.4\ \mathrm{eV}$	$365\ \mathrm{nm}$	ultraviolet

One proviso matters here. Proposition 3.2 assumes a direct gap. Si and Ge are indirect-gap materials — the bottom of the conduction band and the top of the valence band lie at different points in momentum space — so a photon alone cannot conserve momentum and a phonon (a lattice vibration) must assist. The emission probability is therefore smaller by orders of magnitude, and silicon does not emit light. That is why LEDs and lasers are made of GaAs or GaN, while silicon is left to do the computing.

3.3. What the band gap determines, part 2: the current that flows by itself

At absolute zero a semiconductor is an insulator, but at room temperature thermal energy lifts valence electrons across the gap. The number of these spontaneously generated carriers fixes the character of the material.

Proposition 3.4（Intrinsic carrier density）

In an intrinsic (impurity-free) semiconductor of band gap $E_g$, the density $n_i$ of conduction electrons at temperature $T$ satisfies, for $E_g \gg k_B T$,

$$n_i = \sqrt{N_c N_v}\,\exp\!\left(-\frac{E_g}{2k_B T}\right),$$

where $N_c$ and $N_v$ are the effective densities of states of the conduction and valence bands respectively; in typical semiconductors these are of order $10^{19}\ \mathrm{cm^{-3}}$ at room temperature.

Remark 3.5

This formula follows by integrating the Fermi distribution $f(E) = [\exp((E-E_F)/k_BT) + 1]^{-1}$ against the density of states of each band, approximating it by a Maxwell–Boltzmann distribution where $E - E_F \gg k_B T$, and imposing the charge-neutrality condition $n = p$ of an intrinsic semiconductor. The derivation is in Chapter 1 of Sze–Ng, Physics of Semiconductor Devices. Here we use only the result, and we care about one feature of it: $E_g/2k_BT$ sits in the exponent.

Example 3.6（Si and Ge differ by a factor of 3600 in free electrons）

At room temperature $T = 300\ \mathrm{K}$ we have $k_B T = 0.02585\ \mathrm{eV}$, hence $2k_B T = 0.0517\ \mathrm{eV}$.

Silicon ($E_g = 1.12\ \mathrm{eV}$, $N_c = 2.8\times 10^{19}$, $N_v = 1.04\times 10^{19}\ \mathrm{cm^{-3}}$):

$$\begin{aligned} \frac{E_g}{2k_BT} &= \frac{1.12}{0.0517} = 21.66, \qquad e^{-21.66} = 3.9\times 10^{-10},\\ \sqrt{N_cN_v} &= \sqrt{2.8\times 10^{19}\cdot 1.04\times 10^{19}} = 1.71\times 10^{19}\ \mathrm{cm^{-3}},\\ n_i &= 1.71\times 10^{19}\cdot 3.9\times 10^{-10} = 6.7\times 10^{9}\ \mathrm{cm^{-3}}. \end{aligned}$$

The measured value is $1.0\times 10^{10}\ \mathrm{cm^{-3}}$, agreeing to within a factor of 1.5.

Germanium ($E_g = 0.66\ \mathrm{eV}$, $N_c = 1.04\times 10^{19}$, $N_v = 6.0\times 10^{18}\ \mathrm{cm^{-3}}$):

$$\begin{aligned} \frac{E_g}{2k_BT} &= \frac{0.66}{0.0517} = 12.77, \qquad e^{-12.77} = 2.9\times 10^{-6},\\ \sqrt{N_cN_v} &= \sqrt{1.04\times 10^{19}\cdot 6.0\times 10^{18}} = 7.9\times 10^{18}\ \mathrm{cm^{-3}},\\ n_i &= 7.9\times 10^{18}\cdot 2.9\times 10^{-6} = 2.3\times 10^{13}\ \mathrm{cm^{-3}}. \end{aligned}$$

The measured value is $2.4\times 10^{13}\ \mathrm{cm^{-3}}$; here the agreement is nearly exact.

A difference of merely $0.46\ \mathrm{eV}$ in $E_g$ produces about a factor of $3500$ in the number of free electrons. Such is the power of the exponential. And this difference is part of the answer to “why are modern integrated circuits made of Si rather than Ge?” The current that leaks through a transistor switched OFF is more than three orders of magnitude larger in Ge.

The atomic density of a silicon crystal is $5.0\times 10^{22}\ \mathrm{cm^{-3}}$. With $n_i = 10^{10}\ \mathrm{cm^{-3}}$, there is one free electron for every $5\times 10^{12}$ atoms. Mix in phosphorus at $10^{16}\ \mathrm{cm^{-3}}$ — one atom in five million — and the free-electron count jumps by a factor of $10^{6}$ at a stroke. This is doping, and it is why the semiconductor industry insists on “eleven nines” purity (99.999999999 %). If unintended impurities crept in at $10^{16}\ \mathrm{cm^{-3}}$, the entire design would be ruined.

3.4. What stopped miniaturization: tunneling

A transistor sandwiches a thin insulating film (silicon dioxide $\mathrm{SiO_2}$, say) between the gate electrode and the semiconductor, and a voltage on the gate switches the channel beneath it ON and OFF. The thinner the insulator, the better the control, so for forty years the industry made it thinner. Then, in the 2000s, it hit a wall: electrons slip straight through the insulator.

Proposition 3.7（Tunneling transmission through a rectangular barrier）

When a particle of mass $m$ and energy $E$ is incident on a rectangular potential barrier of height $V_0\ (> E)$ and thickness $d$, setting

$$\kappa = \frac{\sqrt{2m(V_0 - E)}}{\hbar},$$

the transmission probability $T$ is, in the regime $\kappa d \gg 1$, approximated by

$$T \approx 16\,\frac{E(V_0-E)}{V_0^{2}}\,e^{-2\kappa d} \sim e^{-2\kappa d}.$$

That is, the transmission falls off exponentially in the thickness $d$.

Remark 3.8

The formula is obtained by solving the one-dimensional Schrödinger equation inside and outside the barrier and matching the wave function and its derivative at the boundaries. The essential point is that inside the barrier the wave number becomes purely imaginary, $i\kappa$, so the wave function decays as $e^{-\kappa x}$. Derivations appear in Volume V of the Feynman Lectures on Physics and in the tunneling section of any standard quantum mechanics text. In classical mechanics the transmission probability for $E < V_0$ is exactly $0$, so this is a purely quantum-mechanical effect.

Example 3.9（One nanometer thinner, seventy million times more leakage）

The barrier height seen by an electron at a $\mathrm{Si/SiO_2}$ interface is $V_0 - E \approx 3.1\ \mathrm{eV}$. First compute $\kappa$:

$$\begin{aligned} 2m_e(V_0-E) &= 2 \cdot 9.109\times 10^{-31}\ \mathrm{kg} \cdot 3.1 \cdot 1.602\times 10^{-19}\ \mathrm{J} = 9.05\times 10^{-49},\\ \sqrt{2m_e(V_0-E)} &= 9.51\times 10^{-25}\ \mathrm{kg\,m/s},\\ \kappa &= \frac{9.51\times 10^{-25}}{1.0546\times 10^{-34}} = 9.02\times 10^{9}\ \mathrm{m^{-1}} = 9.02\ \mathrm{nm^{-1}}. \end{aligned}$$

Hence $2\kappa = 18.0\ \mathrm{nm^{-1}}$, and $e^{-2\kappa d}$ by thickness is

Oxide thickness $d$	$2\kappa d$	$e^{-2\kappa d}$
$3.0\ \mathrm{nm}$	$54.1$	$3\times 10^{-24}$
$2.0\ \mathrm{nm}$	$36.1$	$2\times 10^{-16}$
$1.5\ \mathrm{nm}$	$27.1$	$2\times 10^{-12}$
$1.0\ \mathrm{nm}$	$18.0$	$1.5\times 10^{-8}$

Each $1\ \mathrm{nm}$ removed multiplies the leakage by $e^{18.0} = 6.6\times 10^{7}$, roughly seventy million. Going from $3\ \mathrm{nm}$ to $1\ \mathrm{nm}$ multiplies it by $10^{16}$.

This is not a metaphor; it is what happened. At $1.2\ \mathrm{nm}$ — a mere five atomic layers of $\mathrm{SiO_2}$ — leakage began to dominate power consumption, and the industry gave up on “thinner still”. The move it made instead was to switch to a high-permittivity material (hafnium oxide $\mathrm{HfO_2}$, relative permittivity $\approx 25$, more than six times the $3.9$ of $\mathrm{SiO_2}$), producing a film that is physically thick but electrically thin. Intel put this into volume production at the 45 nm node in 2007. In other words, your phone contains an answer to a deadline set by the Schrödinger equation.

4. GPS: converting a clock error into meters

4.1. Positioning is subtraction of times

A GPS receiver measures distance from the difference between the transmission time encoded in a satellite’s signal and the time of reception. Radio waves travel at the speed of light, so a timing error $\Delta t$ becomes a distance error $c\,\Delta t$:

$$c \times 1\ \mathrm{ns} = 2.998\times 10^{8}\ \mathrm{m/s} \times 10^{-9}\ \mathrm{s} = 0.30\ \mathrm{m}.$$

One nanosecond is thirty centimeters. To position to within a few meters, the clocks must agree to about ten nanoseconds. But a satellite clock and a ground clock do not fundamentally run at the same rate: the slowing due to velocity (time dilation（Theorem 3.1）［Einstein and Feynman］) and the speeding up due to the difference in gravitational potential (gravitational shift in clock rate（Proposition 5.2）［Einstein and Feynman］) act at the same time. Below we treat both in a single formula.

Definition 4.1（Proper time）

The time actually recorded by a clock moving through spacetime is its proper time $\tau$. In terms of the line element $ds^2$ written with a coordinate time $t$, proper time is defined by

$$d\tau^2 = -\frac{ds^2}{c^2}$$

(with the sign convention in which $ds^2 < 0$ along a timelike world line). In relativity, “the time a clock records” is always proper time, never coordinate time.

4.2. Thirty-eight microseconds a day

Theorem 4.2（Difference in rate between an orbiting clock and a ground clock）

Approximate the Earth’s gravitational field by the Schwarzschild metric of a mass $M$, and neglect the Earth’s rotation and oblateness, the eccentricity of the orbit, and the influence of other bodies. Let $\tau_{\mathrm{sat}}$ be the proper time of a clock on a circular orbit of geocentric radius $r$, and $\tau_{\mathrm{gnd}}$ that of a clock at rest at geocentric radius $R$. Then, to first order in the small quantities $GM/(rc^2) \ll 1$ and $v^2/c^2 \ll 1$,

$$\frac{\tau_{\mathrm{sat}} - \tau_{\mathrm{gnd}}}{\tau_{\mathrm{gnd}}} \approx \frac{GM}{c^{2}}\left(\frac{1}{R} - \frac{3}{2r}\right).$$

In particular the right-hand side is positive (the satellite clock runs fast) when $r > \tfrac{3}{2}R$, and negative (it runs slow) when $r < \tfrac{3}{2}R$.

Proof（Theorem 4.2）

Write the Schwarzschild metric in spherical coordinates $(t, r, \theta, \varphi)$ and restrict to the equatorial plane $\theta = \pi/2$:

$$ds^{2} = -\left(1 - \frac{2GM}{rc^{2}}\right)c^{2}dt^{2} + \left(1 - \frac{2GM}{rc^{2}}\right)^{-1}dr^{2} + r^{2}d\varphi^{2}.$$

The satellite. On a circular orbit $dr = 0$, so by Definition 4.1

$$c^{2}d\tau_{\mathrm{sat}}^{2} = \left(1 - \frac{2GM}{rc^{2}}\right)c^{2}dt^{2} - r^{2}d\varphi^{2}.$$

Setting the coordinate velocity $v = r\,d\varphi/dt$ and factoring out $dt^2$,

$$\frac{d\tau_{\mathrm{sat}}}{dt} = \sqrt{1 - \frac{2GM}{rc^{2}} - \frac{v^{2}}{c^{2}}}.$$

By hypothesis the correction terms under the root are small compared with $1$, so with $\sqrt{1-x} \approx 1 - x/2$ (valid for $|x| \ll 1$),

$$\frac{d\tau_{\mathrm{sat}}}{dt} \approx 1 - \frac{GM}{rc^{2}} - \frac{v^{2}}{2c^{2}}.$$

The ground clock. It is at rest, so $dr = d\varphi = 0$ and

$$\frac{d\tau_{\mathrm{gnd}}}{dt} = \sqrt{1 - \frac{2GM}{Rc^{2}}} \approx 1 - \frac{GM}{Rc^{2}}.$$

Take the ratio. To first order in the small quantities,

$$\frac{d\tau_{\mathrm{sat}}}{d\tau_{\mathrm{gnd}}} \approx \left(1 - \frac{GM}{rc^{2}} - \frac{v^{2}}{2c^{2}}\right)\left(1 + \frac{GM}{Rc^{2}}\right) \approx 1 + \frac{GM}{c^{2}}\left(\frac{1}{R} - \frac{1}{r}\right) - \frac{v^{2}}{2c^{2}}.$$

(The second-order terms are of order $10^{-19}$ and have been dropped.)

Use the circular-orbit condition. For a circular orbit in the Newtonian approximation, gravity balances the centripetal acceleration, $GM/r^{2} = v^{2}/r$, that is $v^{2} = GM/r$. Substituting,

$$\frac{d\tau_{\mathrm{sat}}}{d\tau_{\mathrm{gnd}}} \approx 1 + \frac{GM}{c^{2}}\left(\frac{1}{R} - \frac{1}{r} - \frac{1}{2r}\right) = 1 + \frac{GM}{c^{2}}\left(\frac{1}{R} - \frac{3}{2r}\right).$$

Subtracting $1$ from both sides gives the stated formula. The sign statement follows from $1/R - 3/(2r) > 0 \iff r > \tfrac{3}{2}R$.

∎

Example 4.3（GPS satellites: +38.5 μs per day, 11 km per day uncorrected）

A GPS satellite orbits at radius $r = 2.656\times 10^{7}\ \mathrm{m}$ (altitude about $20{,}200\ \mathrm{km}$); take the ground clock at $R = R_\oplus = 6.371\times 10^{6}\ \mathrm{m}$. First check the orbital speed:

$$v = \sqrt{\frac{GM}{r}} = \sqrt{\frac{3.986\times 10^{14}}{2.656\times 10^{7}}} = \sqrt{1.501\times 10^{7}} = 3874\ \mathrm{m/s},$$

about $14{,}000\ \mathrm{km/h}$.

The special-relativistic contribution (slowing due to velocity):

$$-\frac{v^{2}}{2c^{2}} = -\frac{(3874)^{2}}{2(2.998\times 10^{8})^{2}} = -\frac{1.501\times 10^{7}}{1.798\times 10^{17}} = -8.35\times 10^{-11}.$$

Multiplying by one day $= 86400\ \mathrm{s}$ gives $-8.35\times 10^{-11} \times 86400 = -7.21\times 10^{-6}\ \mathrm{s}$: the clock loses $7.2\ \mu\mathrm{s}$ per day.

The general-relativistic contribution (gain from sitting in a shallower gravitational potential). Using $GM/c^{2} = 3.986\times 10^{14}/8.988\times 10^{16} = 4.435\times 10^{-3}\ \mathrm{m}$,

$$\frac{GM}{c^{2}}\left(\frac{1}{R} - \frac{1}{r}\right) = 4.435\times 10^{-3}\left(1.5696\times 10^{-7} - 3.765\times 10^{-8}\right) = 5.29\times 10^{-10}.$$

Over a day this is $5.29\times 10^{-10}\times 86400 = 4.57\times 10^{-5}\ \mathrm{s}$: the clock gains $45.7\ \mu\mathrm{s}$ per day.

The total (using Theorem 4.2 directly gives the same answer):

$$4.435\times 10^{-3}\left(1.5696\times 10^{-7} - \frac{1.5}{2.656\times 10^{7}}\right) = 4.435\times 10^{-3} \times 1.0049\times 10^{-7} = 4.457\times 10^{-10},$$$$4.457\times 10^{-10}\times 86400\ \mathrm{s} = 3.85\times 10^{-5}\ \mathrm{s} = 38.5\ \mu\mathrm{s}\ \text{(gained, per day)}.$$

Convert to distance. $c \times 38.5\ \mu\mathrm{s} = 2.998\times 10^{8}\times 3.85\times 10^{-5} = 1.15\times 10^{4}\ \mathrm{m}$. Neglect the correction and the positioning error piles up at about $11\ \mathrm{km}$ per day. It would take less than half a day for your car navigation to announce that you are in the next prefecture.

In practice this correction is made before launch. The reference frequency of the atomic clock carried by the satellite is offset slightly from its nominal ground value of $10.23\ \mathrm{MHz}$ and set to $10.22999999543\ \mathrm{MHz}$. The relative offset is

$$\frac{10.23 - 10.22999999543}{10.23} = 4.4647\times 10^{-10},$$

and $\times 86400\ \mathrm{s} = 38.6\ \mu\mathrm{s}$, agreeing with our hand-computed $38.5\ \mu\mathrm{s}$ to within 0.3 % (the difference comes from neglecting the Earth’s rotation and oblateness). A clock that becomes correct only once it reaches orbit is built on the ground and launched. The same $38\ \mu\mathrm{s}$, obtained by setting up the special-relativistic and gravitational effects separately, is computed in Example 5.3［Einstein and Feynman］.

flowchart TD
A["Satellite atomic clock (altitude 20,200 km)"] --> B["Special relativity: speed 3.87 km/s<br/>clock loses 7.2 μs/day"]
A --> C["General relativity: shallower gravitational potential<br/>clock gains 45.7 μs/day"]
B --> D["Net +38.5 μs/day"]
C --> D
D --> E["Uncorrected: positioning error about 11 km/day"]
D --> F["Fix: offset the reference frequency before launch<br/>10.23 MHz to 10.22999999543 MHz"]

The two effects on a GPS satellite clock, and their net balance

Remark 4.4

The sign condition $r = \tfrac{3}{2}R$ in Theorem 4.2 corresponds to an altitude of $\tfrac{3}{2}\times 6371 - 6371 = 3186\ \mathrm{km}$. Below that, the velocity effect beats the gravitational one and the clock runs slow. For the International Space Station (altitude about $400\ \mathrm{km}$, $r = 6.771\times 10^{6}\ \mathrm{m}$),

$$4.435\times 10^{-3}\left(1.5696\times 10^{-7} - \frac{1.5}{6.771\times 10^{6}}\right) = -2.86\times 10^{-10},$$

which is $-2.47\times 10^{-5}\ \mathrm{s} = -24.7\ \mu\mathrm{s}$ per day, about $0.009\ \mathrm{s}$ per year. Astronauts do come back marginally younger than the rest of us — by a hundredth of a second a year. Rip Van Winkle they are not.

5. MRI: making nuclei sing for the camera

5.1. A nucleus is a spinning top

The proton (the nucleus of hydrogen) has spin $1/2$ and with it a magnetic moment. Put a magnet in a magnetic field and it tries to align; for a quantum-mechanical spin, what appears is not an alignment of “direction” but a splitting of energy levels.

Definition 5.1（Zeeman splitting and the Larmor frequency）

Place a spin-$1/2$ nucleus of gyromagnetic ratio $\gamma$ in a static magnetic field $B_0$. Its energy level splits in two, with separation

$$\Delta E = \gamma \hbar B_0$$

(Zeeman splitting). By Definition 2.1, the frequency of the radiation that drives transitions between these two levels is

$$f_L = \frac{\Delta E}{h} = \frac{\gamma B_0}{2\pi},$$

called the Larmor frequency. For the proton, $\gamma/2\pi = 42.577\ \mathrm{MHz/T}$.

Proposition 5.2（Larmor frequencies in clinical MRI）

With the proton gyromagnetic ratio $\gamma/2\pi = 42.577\ \mathrm{MHz/T}$, the Larmor frequencies for static fields $B_0 = 1.5\ \mathrm{T}$ and $3.0\ \mathrm{T}$ are

$$f_L(1.5\ \mathrm{T}) = 63.87\ \mathrm{MHz}, \qquad f_L(3.0\ \mathrm{T}) = 127.7\ \mathrm{MHz}$$

respectively.

Proof（Proposition 5.2）

Substitute into $f_L = (\gamma/2\pi) B_0$ from Definition 5.1:

$$f_L(1.5\ \mathrm{T}) = 42.577\ \mathrm{MHz/T} \times 1.5\ \mathrm{T} = 63.866\ \mathrm{MHz},$$$$f_L(3.0\ \mathrm{T}) = 42.577\ \mathrm{MHz/T} \times 3.0\ \mathrm{T} = 127.73\ \mathrm{MHz}.$$∎

$63.9\ \mathrm{MHz}$ and $127.7\ \mathrm{MHz}$: FM radio in Japan broadcasts between $76$ and $95\ \mathrm{MHz}$, so MRI works with radio waves right next door to FM. That is why an MRI room is wrapped in a thick radio-frequency shield — to keep outside broadcasts from showing up in the image as noise.

5.2. What is visible is five parts in a million

Here is where MRI becomes strange. Compare the size of the Zeeman splitting with the thermal energy at room temperature.

Proposition 5.3（Thermal-equilibrium polarization of nuclear spins）

For an ensemble of spin-$1/2$ nuclei in thermal equilibrium at field $B_0$ and temperature $T$, the relative difference between the population $N_+$ of the lower level and the population $N_-$ of the upper level (the polarization) is

$$P = \frac{N_+ - N_-}{N_+ + N_-} = \tanh\!\left(\frac{\Delta E}{2k_B T}\right) \approx \frac{\Delta E}{2k_B T} = \frac{\gamma \hbar B_0}{2 k_B T} \qquad (\Delta E \ll k_B T).$$

Proof（Proposition 5.3）

In thermal equilibrium the populations follow the Boltzmann distribution, $N_-/N_+ = e^{-\Delta E/k_BT}$. Setting $x = \Delta E/(2k_BT)$ we may write $N_+ \propto e^{x}$ and $N_- \propto e^{-x}$, so

$$P = \frac{e^{x} - e^{-x}}{e^{x} + e^{-x}} = \tanh x.$$

For $|x| \ll 1$ we have $\tanh x = x - x^3/3 + \cdots \approx x$, hence $P \approx \Delta E/(2k_BT)$. Substituting $\Delta E = \gamma\hbar B_0$ from Definition 5.1 gives the claim.

∎

Example 5.4（At body temperature and 1.5 T, the imbalance is five per million）

Take $B_0 = 1.5\ \mathrm{T}$ and body temperature $T = 310\ \mathrm{K}$. Putting $f_L = 63.87\ \mathrm{MHz}$ from Proposition 5.2 into $\Delta E = h f_L$ from Definition 5.1,

$$\Delta E = 6.626\times 10^{-34}\ \mathrm{J\,s} \times 6.387\times 10^{7}\ \mathrm{Hz} = 4.232\times 10^{-26}\ \mathrm{J}.$$

The thermal energy at body temperature, meanwhile, is

$$k_B T = 1.3806\times 10^{-23} \times 310 = 4.280\times 10^{-21}\ \mathrm{J}.$$

By Proposition 5.3,

$$P \approx \frac{4.232\times 10^{-26}}{2\times 4.280\times 10^{-21}} = 4.94\times 10^{-6}.$$

So out of a million protons, only about five are polarized on balance. The remaining 99.9995 % cancel one another out and contribute nothing to the signal.

Imaging is nonetheless possible because the numbers are enormous. The number of protons in $1\ \mathrm{mm}^3$ of water (mass $1\ \mathrm{mg}$) is

$$\frac{1\times 10^{-3}\ \mathrm{g}}{18\ \mathrm{g/mol}} \times 2 \times 6.022\times 10^{23} = 6.7\times 10^{19}$$

(the factor $2$ because each $\mathrm{H_2O}$ molecule carries two hydrogens). Of these, the net polarized number is

$$6.7\times 10^{19} \times 4.94\times 10^{-6} = 3.3\times 10^{14}.$$

That is 330 trillion per $1\ \mathrm{mm}^3$ — quite enough for a strong signal.

Note also that $P \propto B_0$ in Proposition 5.3 means a $3\ \mathrm{T}$ machine has twice the polarization of a $1.5\ \mathrm{T}$ one. This is the reason for the push to higher fields (and since receiver sensitivity also rises with frequency, the actual improvement in signal-to-noise ratio is larger still).

5.3. How do we know where the signal came from?

We now know that hydrogen inside the body emits radio waves. But to make a picture we must distinguish which position a signal came from. The answer is to make the Larmor frequency itself depend on position.

Superimpose on the static field a gradient field $G$ (units $\mathrm{T/m}$) whose strength varies in proportion to position. By Definition 5.1, the resonance frequency at position $x$ becomes

$$f(x) = \frac{\gamma}{2\pi}\left(B_0 + G x\right),$$

so frequency and position are in one-to-one correspondence. Fourier transform the received signal into its frequency components and what you get is exactly the spatial distribution — the central idea of MRI, found independently by Lauterbur and Mansfield in 1973 and recognized by the 2003 Nobel Prize in Physiology or Medicine.

Example 5.5（With a 10 mT/m gradient, 1 mm is 426 Hz）

Take $G = 10\ \mathrm{mT/m} = 1.0\times 10^{-2}\ \mathrm{T/m}$. A displacement of $\Delta x = 1\ \mathrm{mm} = 1.0\times 10^{-3}\ \mathrm{m}$ shifts the resonance frequency by

$$\Delta f = \frac{\gamma}{2\pi} G \Delta x = 42.577\times 10^{6}\ \mathrm{Hz/T} \times 1.0\times 10^{-2}\ \mathrm{T/m} \times 1.0\times 10^{-3}\ \mathrm{m} = 425.8\ \mathrm{Hz}.$$

Conversely, measuring frequency to a resolution of $426\ \mathrm{Hz}$ fixes position to a resolution of $1\ \mathrm{mm}$. Relative to the $63.87\ \mathrm{MHz}$ carrier, $426\ \mathrm{Hz}$ is $6.7\times 10^{-6}$ — again a precision of one part in a million. This is why the static field of an MRI scanner is held uniform to a few ppm across the whole imaging volume.

Incidentally, with a receiver bandwidth of $\pm 32\ \mathrm{kHz}$ the field of view is

$$\frac{2\times 32000\ \mathrm{Hz}}{42.577\times 10^{6}\times 1.0\times 10^{-2}\ \mathrm{Hz/m}} = \frac{64000}{4.258\times 10^{5}} = 0.150\ \mathrm{m} = 15\ \mathrm{cm}.$$

Imaging a head therefore calls for a somewhat weaker gradient or a wider bandwidth — and so on: every setting of the machine is tied together by this one equation.

Image contrast — the fact that white and gray matter in the brain appear in different shades — comes less from the number of hydrogen nuclei than from how fast the excited spins return to equilibrium: the longitudinal relaxation time $T_1$ and the transverse relaxation time $T_2$ differ from tissue to tissue. Since these are set by the motional state of the water molecules and their interaction with the surrounding macromolecules, MRI sees not only “where the hydrogen is” but “what kind of environment the hydrogen is in”.

6. What the three have in common

Set them side by side once more.

	Phone (semiconductors)	GPS	MRI
Central theory	quantum mechanics (band theory, tunneling)	special and general relativity	quantum mechanics (nuclear spin) + statistical mechanics
Energy difference $\Delta E$	band gap $\approx 1\ \mathrm{eV}$	Cs-133 hyperfine structure $\approx 4\times 10^{-5}\ \mathrm{eV}$	Zeeman splitting $\approx 2.6\times 10^{-7}\ \mathrm{eV}$
Corresponding frequency	$\approx 10^{14}\ \mathrm{Hz}$ (infrared to visible)	$9.192631770\ \mathrm{GHz}$ (microwave)	$6.4\times 10^{7}\ \mathrm{Hz}$ (VHF)
Decisive number	$e^{-E_g/2k_BT}$, $e^{-2\kappa d}$	$38.5\ \mu\mathrm{s}$ per day	polarization $5\times 10^{-6}$
Theory to application	1928 (Bloch) → 1947 (transistor)	1915 (general relativity) → 1978 (first GPS satellite)	1946 (NMR) → 1977 (human imaging)

Running through all three columns is $E = hf$ from Definition 2.1. In a semiconductor the band gap sets the color of the light; in MRI the splitting of a nuclear spin sets the radio frequency; and in the atomic clocks of GPS — this is the mildly startling part — the statement that the hyperfine transition of the ground state of cesium-133 has frequency $9\,192\,631\,770\ \mathrm{Hz}$ has been the definition of the second itself since 1967 (a definition retained after the 2019 revision of the SI). Our unit of time is an atomic energy difference read through $E = hf$.

The other thing they share is exponentials and a feeling for orders of magnitude. Semiconductor performance is set by the contest between two exponentials, $e^{-E_g/2k_BT}$ and $e^{-2\kappa d}$; GPS cannot ignore a relative error of $10^{-10}$; MRI picks up an imbalance of $10^{-6}$. The “approximations” and “order-of-magnitude estimates” of a high-school textbook are working tools used daily in exactly these situations.

Note

One occasionally reads that “quantum mechanics is about a world unrelated to everyday life”. That is not quite right. The accurate statement is that the consequences of quantum mechanics are everywhere in daily life, while the characteristic strangeness of quantum mechanics (superposition（Definition 2.1）［Schrödinger's Cat］ and the measurement problem) is hard to see at everyday scales. On the latter, see Schrödinger’s cat (in particular correlation with the environment destroys interference（Proposition 6.1）［Schrödinger's Cat］), and on the relation to determinism, Laplace’s demon (why quantum effects are invisible in daily life（Example 5.4）［Laplace's Demon and Determinism］).

For all this practical success, physics is not finished. The performance limits of semiconductors, room-temperature superconductivity, error correction for quantum computers, and the unification of gravity with quantum theory (for the scale on which both matter, see Planck units（Definition 6.1）［Unsolved Problems in Physics］) — the open problems are collected in unsolved problems in physics. Whatever today’s phone would have been to a physicist of 1900, the same thing is surely happening again by 2120.

7. Exercises

Exercise 7.1Easy

A green LED emits at about $530\ \mathrm{nm}$. What is the band gap of its semiconductor, in eV? And can you light such a material by connecting a single $1.5\ \mathrm{V}$ dry cell directly across it? Give your reasoning.

Solution

By Proposition 3.2, $E_g = 1239.84/\lambda\,[\mathrm{nm}]$, so

$$E_g = \frac{1239.84}{530} = 2.34\ \mathrm{eV}.$$

You cannot light it. An electron crossing a potential difference $V$ gains energy $eV$, so a $1.5\ \mathrm{V}$ cell supplies only $1.5\ \mathrm{eV}$. Emitting one photon requires $E_g = 2.34\ \mathrm{eV}$, as used in the derivation of Proposition 3.2, so a forward voltage of at least $2.34\ \mathrm{V}$ is needed. Indeed the forward voltage of a green LED is around $2.0$ to $3.0\ \mathrm{V}$, higher than a red LED (about $1.8\ \mathrm{V}$) and lower than a blue or white one (about $3.0$ to $3.4\ \mathrm{V}$). LED forward voltages differ by color precisely because the band gaps differ.

Exercise 7.2Standard

An atomic clock is placed on a geostationary satellite (orbital radius $r = 4.2164\times 10^{7}\ \mathrm{m}$). By how many microseconds per day does it drift relative to a clock on the ground, and does it run fast or slow? Use $GM/c^{2} = 4.435\times 10^{-3}\ \mathrm{m}$ and $R_\oplus = 6.371\times 10^{6}\ \mathrm{m}$.

Solution

Apply Theorem 4.2 directly (a geostationary orbit is circular with essentially zero eccentricity, so the hypotheses hold). First the bracket:

$$\frac{1}{R} = \frac{1}{6.371\times 10^{6}} = 1.5696\times 10^{-7}\ \mathrm{m^{-1}},$$$$\frac{3}{2r} = \frac{1.5}{4.2164\times 10^{7}} = 3.5578\times 10^{-8}\ \mathrm{m^{-1}},$$$$\frac{1}{R} - \frac{3}{2r} = 1.5696\times 10^{-7} - 3.5578\times 10^{-8} = 1.2138\times 10^{-7}\ \mathrm{m^{-1}}.$$

Multiplying by $GM/c^{2}$,

$$\frac{GM}{c^{2}}\left(\frac{1}{R} - \frac{3}{2r}\right) = 4.435\times 10^{-3} \times 1.2138\times 10^{-7} = 5.384\times 10^{-10}.$$

The value is positive, so the clock runs fast (consistent with the sign condition in Theorem 4.2: $r = 4.2164\times 10^{7} > \tfrac{3}{2}R = 9.56\times 10^{6}$). Per day,

$$5.384\times 10^{-10}\times 86400\ \mathrm{s} = 4.65\times 10^{-5}\ \mathrm{s} = 46.5\ \mu\mathrm{s}.$$

This exceeds the $38.5\ \mu\mathrm{s}$ of a GPS satellite, because a higher orbit means a larger gravitational effect (the gain) and a lower speed, hence a smaller special-relativistic effect (the loss).

Exercise 7.3Standard

For an MRI scanner with $B_0 = 3.0\ \mathrm{T}$, find the following.

(1) The proton polarization $P$ at body temperature $T = 310\ \mathrm{K}$. (2) The net polarized number of protons in $1\ \mathrm{mm}^{3}$ of water. (3) The frequency resolution required to achieve a spatial resolution of $0.5\ \mathrm{mm}$ with a gradient field $G = 20\ \mathrm{mT/m}$.

Solution

(1) By Proposition 5.2, $f_L(3.0\ \mathrm{T}) = 127.73\ \mathrm{MHz}$, so from $\Delta E = h f_L$ in Definition 5.1,

$$\Delta E = 6.626\times 10^{-34}\times 1.2773\times 10^{8} = 8.464\times 10^{-26}\ \mathrm{J}.$$

The thermal energy at body temperature is $k_BT = 1.3806\times 10^{-23}\times 310 = 4.280\times 10^{-21}\ \mathrm{J}$. Since $\Delta E \ll k_BT$ (the ratio is about $2\times 10^{-5}$), the approximation in Proposition 5.3 applies and

$$P \approx \frac{8.464\times 10^{-26}}{2\times 4.280\times 10^{-21}} = 9.89\times 10^{-6},$$

about $10^{-5}$, or roughly ten per million. This is exactly twice the $1.5\ \mathrm{T}$ value of Example 5.4, consistent with $P \propto B_0$ in Proposition 5.3.

(2) As in Example 5.4, $1\ \mathrm{mm}^{3}$ of water contains $6.7\times 10^{19}$ protons, so

$$6.7\times 10^{19}\times 9.89\times 10^{-6} = 6.6\times 10^{14}.$$

(3) The same computation as Example 5.5:

$$\Delta f = \frac{\gamma}{2\pi}G\,\Delta x = 42.577\times 10^{6}\times 2.0\times 10^{-2}\times 0.5\times 10^{-3} = 425.8\ \mathrm{Hz}.$$

Doubling the gradient makes half the distance correspond to the same frequency difference. Stronger gradients buy finer resolution: that is the basic design principle of MRI. (In practice there is a limit, because rapidly switched gradients stimulate peripheral nerves.)

References

• S. M. Sze and K. K. Ng, Physics of Semiconductor Devices, 3rd ed., Wiley, 2007 — Chapter 1 (crystal structure and band theory) and Chapter 4 (the MOS capacitor). The numerical values for intrinsic carrier densities and effective densities of states follow the tables in this book.
• C. Kittel, Kotai Butsurigaku Nyumon (Introduction to Solid State Physics, 8th ed.), Maruzen, 2005 (in Japanese) — Chapters 7 and 8 (energy bands and semiconductor crystals).
• R. P. Feynman, R. B. Leighton and M. Sands, Feynman Butsurigaku V: Ryoshi Rikigaku (The Feynman Lectures on Physics, Vol. V: Quantum Mechanics), Iwanami Shoten (in Japanese) — the chapters on tunneling and on two-level systems.
• N. Ashby, “Relativity in the Global Positioning System”, Living Reviews in Relativity 6 (2003), article 1. doi:10.12942/lrr-2003-1 — the standard review of relativistic effects in GPS, including the discussion of the reference-frequency offset.
• E. M. Haacke, R. W. Brown, M. R. Thompson and R. Venkatesan, Magnetic Resonance Imaging: Physical Principles and Sequence Design, Wiley, 1999 — the basics of nuclear magnetic resonance, spatial encoding by gradient fields, and relaxation times.
• BIPM, The International System of Units (SI), 9th edition, 2019 — https://www.bipm.org/en/publications/si-brochure — the definition of the second (the cesium-133 hyperfine transition frequency $9\,192\,631\,770\ \mathrm{Hz}$).
• NIST, CODATA Internationally Recommended Values of the Fundamental Physical Constants — https://physics.nist.gov/cuu/Constants/ — the source of the values of $h$, $k_B$, $m_e$ and the proton gyromagnetic ratio used above.

Appendix: Where the numbers come from, and what to watch for

What the GPS calculation ignores. In Theorem 4.2 we neglected the Earth’s rotation. The surface at the equator moves at about $465\ \mathrm{m/s}$, so the ground clock too is slowed, by $-v^2/2c^2 = -1.2\times 10^{-12}$ (that is $-0.10\ \mu\mathrm{s}$ per day). Including this increases the satellite’s relative gain by about $0.1\ \mu\mathrm{s}$. Furthermore, since the Earth is not a perfect sphere, operational practice expands the gravitational potential in spherical harmonics and uses a coordinate system fixed to the rotating Earth (defined so that clocks on the geoid all run at the same rate). The rigorous treatment is in Ashby’s review. Periodic variations from orbital eccentricity (for GPS, $e \approx 0.02$ or less) are handled at the receiver as a correction term of the form $-2\sqrt{GMa}\,e\sin E/c^2$.

On the semiconductor numbers. The effective densities of states $N_c, N_v$ are computed from density-of-states effective masses and differ by a few percent between sources. The factor-of-1.5 discrepancy for silicon in Example 3.6 comes from this uncertainty together with our neglect of the slight shrinking of the band gap with temperature (about $-2.7\times 10^{-4}\ \mathrm{eV/K}$ for Si near $300\ \mathrm{K}$). For order-of-magnitude estimates the accuracy is ample. As for the tunneling transmission, Proposition 3.7 idealizes the barrier as rectangular; a real gate insulator requires accounting for the tilt of the barrier under the applied field (Fowler–Nordheim tunneling) and for interface states. The conclusion “exponential in the thickness” is unaffected.

On the MRI polarization. Proposition 5.3 assumes thermal equilibrium. Recent hyperpolarization techniques polarize nuclear spins strongly by artificial means before introducing them into the body, reaching a few percent polarization — more than ten thousand times the thermal-equilibrium value — with $^{129}$Xe or $^{13}$C. Hyperpolarized xenon MRI, which images the airways of the lung directly, is entering clinical use. The figure $10^{-6}$ should be understood as conditional on thermal equilibrium.