van der Waals forces — London dispersion, dipole-dipole, dipole-induced dipole: Visual walkthrough

Step 1 — Two charges: the only rule we start with

WHAT. Everything below is built from one fact: two electric charges push or pull along the line joining them. If the charges have the same sign they repel; opposite signs, they attract.

WHY this tool. We need some law that turns "charge + distance" into "force". That law is Coulomb's law. It is the only ingredient we are allowed to use — every van der Waals force is secretly Coulomb's law wearing a disguise.

PICTURE. In the figure, a red charge $+ q$ and a violet charge $- q$ sit a distance $r$ apart. The arrow is the pull between them.

Figure — van der Waals forces — London dispersion, dipole-dipole, dipole-induced dipole

The problem this creates. A molecule is neutral — it has equal $+$ and $-$ . So $q_{total} = 0$ and naïvely $U = 0$ . If Coulomb's law were the whole story, nothing would ever stick. The rest of the derivation is about the small attraction that survives neutrality.

Step 2 — A dipole: charge that is separated, not cancelled

WHAT. A dipole is a $+ q$ and a $- q$ held a small distance $d$ apart. Its strength is one number, the dipole moment: $μ = q d$

WHY. A neutral molecule is not featureless: its $+$ and $-$ live in slightly different places (think $H^{δ +} - C l^{δ -}$ ). So although the total charge is zero, the arrangement is lopsided. $μ$ measures that lopsidedness. Bigger $q$ or bigger separation $d$ → bigger $μ$ . (More on this in Dipole moment and molecular polarity.)

PICTURE. The figure shows the $+ q$ / $- q$ pair with the gap $d$ marked, and the dipole moment drawn as an orange arrow pointing from $-$ to $+$ .

Step 3 — Why a dipole's reach dies as $1/ r^{3}$ (the master fact)

WHAT. We now ask: how strong is the electric field of a dipole a distance $r$ away? A single charge's field falls off as $1/ r^{2}$ . A dipole's field falls off faster — as $1/ r^{3}$ .

WHY it falls faster. From far away you see both charges almost on top of each other. The $+ q$ pull and the $- q$ pull nearly cancel — you only feel the tiny leftover because one charge is a hair closer than the other. That "hair of difference" is what shrinks the reach.

HOW (the cancellation, shown as subtraction). Put a test point a distance $r$ from the dipole's centre. The $+$ charge is at distance $r - \frac{d}{2}$ , the $-$ charge at $r + \frac{d}{2}$ . The field is the difference of two nearly-equal potentials: $E \propto from + q \frac{1}{r - \frac{d}{2}} - from - q \frac{1}{r + \frac{d}{2}}$

For a small gap ( $d ≪ r$ ) these two terms nearly kill each other; what is left is $E \propto \frac{d}{r ^{2} - \frac{d ^{2}}{4}} \approx \frac{d}{r ^{2}} \cdot \frac{1}{r} = \frac{d}{r ^{3}} \propto \frac{μ}{r ^{3}} .$

PICTURE. The figure overlays two curves: a single charge's field ( $\sim 1/ r^{2}$ , the higher curve) and the dipole's ( $\sim 1/ r^{3}$ , the lower, faster-dying curve). The gap between them is the cancellation.

Step 4 — Dipole–dipole: two permanent dipoles talk

WHAT. Molecule 1 has a permanent dipole $μ_{1}$ ; molecule 2 has $μ_{2}$ . Molecule 1 makes a field (one factor $1/ r^{3}$ ); molecule 2's dipole sits in it. The energy of a dipole $μ_{2}$ in a field $E$ is $U = - μ_{2} E$ .

WHY the minus sign. When $μ_{2}$ points the way the field pulls it, energy drops (it "wants" that alignment) — that is an attraction, hence negative $U$ .

HOW. Plug $E_{1} \propto μ_{1} / r^{3}$ into $U = - μ_{2} E_{1}$ : $U_{dd, aligned} \propto - \frac{μ _{1} μ _{2}}{r ^{3}} .$ Dipole counter: molecule 1 spoke once → one factor $1/ r^{3}$ . ✔

PICTURE. Two orange dipole arrows, head-to-tail (the favourable alignment), with the field lines of the left dipole threading the right one.

Step 5 — Tumbling turns $r^{3}$ into $r^{6}$ (the Keesom twist)

WHAT. In a real gas or liquid the molecules spin and tumble. Sometimes two dipoles line up head-to-tail (attract, $U < 0$ ); sometimes head-to-head (repel, $U > 0$ ). We must average over all orientations.

WHY it is not just zero. If every orientation were equally likely, attraction and repulsion would cancel to exactly zero. But attractive orientations are slightly more likely — they are lower in energy, and low-energy states are favoured (this is the Boltzmann bias, one factor of $1/ k_{B} T$ ). So a tiny net attraction survives.

HOW the power doubles. The surviving average is proportional to the square of the interaction energy (because the bias itself scales with that energy): $U_{dd} \propto - \frac{1}{k _{B} T} (Step 4 \frac{μ _{1} μ _{2}}{r ^{3}})^{2} = - \frac{μ _{1}^{2} μ _{2}^{2}}{k _{B} T r ^{6}} .$ Squaring the $r^{3}$ gives $r^{6}$ . This is the first appearance of $6$ .

PICTURE. A tumbling pair of dipoles with a tilted balance: the "attract" pan sits a hair lower than the "repel" pan — that small tilt is the whole effect.

Step 6 — Dipole–induced dipole: one permanent, one squishy (Debye)

WHAT. Now one molecule is polar ( $μ$ ) and its neighbour is non-polar but squishy — its electron cloud can be pushed out of shape. The polar molecule's field distorts the neighbour, creating a dipole in it.

WHY the field appears twice. Two jobs need doing:

Induce the dipole: $μ_{ind} = α E$ — one factor of $E \propto 1/ r^{3}$ .
Interact with it: $U \propto - μ_{ind} E$ — a second factor of $E \propto 1/ r^{3}$ .

Two dipole-fields multiplied → $1/ r^{6}$ . Second appearance of $6$ — and no temperature this time, because the induced dipole always lines up favourably (it is created by the field, so it never fights it).

HOW. $U_{di} \propto - μ_{ind} E = - (α E) E = - α E^{2} \propto - \frac{α μ ^{2}}{r ^{6}} .$ Here $α$ is the polarisability — "how much dipole per unit field" (see Polarisability and Fajans' rules).

PICTURE. A polar dipole on the left; on the right a round, initially symmetric cloud gets stretched into a small induced dipole that points cooperatively.

Step 7 — London dispersion: attraction from nothing permanent

WHAT. Take two atoms with no permanent dipole at all — argon, say. Their electrons are still moving. At any frozen instant the cloud is a little lopsided → an instantaneous dipole. That flicker makes a field, which induces a matching dipole in the neighbour, and the two flickers correlate — they dance in step — giving a net pull.

WHY it still gives $1/ r^{6}$ . Same double-duty as Debye: one factor of field to induce, one to interact. Both instantaneous dipoles are proportional to how squishy the clouds are ( $α$ ), so: $U_{London} \propto - \frac{α _{1} α _{2}}{r ^{6}} .$ Third appearance of $6$ . The quantum treatment adds the ionisation energy $I$ (how tightly electrons are held), giving London's full form.

PICTURE. Three snapshots of the same two atoms in time: the left cloud flickers lopsided, the right cloud responds in step each time — the correlation is the attraction.

Step 8 — Degenerate & limiting cases (never skipped)

WHAT / WHY / PICTURE — walk every corner so nothing surprises you:

$μ = 0$ (non-polar molecule). Keesom ( $μ^{2} μ^{2}$ ) → $0$ ; Debye ( $μ^{2}$ ) → $0$ . Only London survives. This is why noble gases have only dispersion.
$r \to \infty$ (far apart). Every term $\to 0$ ; $1/ r^{6}$ collapses very fast, so vdW forces are short-range.
$r \to$ small (clouds overlap). The $- 1/ r^{6}$ attraction is not the whole story: at very short range electron clouds repel hard ( $+ 1/ r^{12}$ ). Adding the two gives the Lennard-Jones potential with its minimum — the equilibrium separation.
$T \to \infty$ (very hot). The Keesom term $\propto 1/ T \to 0$ : tumbling washes out the permanent-dipole alignment. Debye and London do not carry $T$ , so they persist.
$α \to 0$ (perfectly rigid cloud, an idealisation). Debye and London vanish; only rigid permanent dipoles could still interact.

PICTURE. The full $U (r)$ curve: steep repulsive $+ 1/ r^{12}$ wall on the left, the gentle $- 1/ r^{6}$ attractive tail on the right, and the well minimum where they balance.

The one-picture summary

The figure lines up all three mechanisms side by side. Read left to right and count the dipole-fields: Keesom (two permanent dipoles, averaged, $μ^{2} μ^{2} / T$ ), Debye ( $μ^{2} α$ ), London ( $α_{1} α_{2}$ ). Three different starting stories — but every one uses a dipole field twice, and $(1/ r^{3})^{2} = 1/ r^{6}$ each time. That shared $1/ r^{6}$ is the attractive tail of the Lennard-Jones curve, and it is why bigger, squishier molecules (more $α$ , more electrons) always stick harder — the key to Boiling point and intermolecular forces.

Recall Feynman retelling — say it in plain words

Coulomb's law says charges pull on charges. Molecules are neutral, so the $+$ and $-$ nearly cancel — but not perfectly, because they sit slightly apart: that leftover is a dipole. A dipole's reach dies off as $1/ r^{3}$ because from far away you only see the tiny difference between its two charges. Now the three forces are just how many dipoles are talking and whether they are permanent. Two permanent dipoles tumbling? Attraction wins by a whisker, and averaging squares the $1/ r^{3}$ into $1/ r^{6}$ (that's Keesom, weakened by heat). One permanent dipole squeezing a squishy neighbour? The field acts twice — once to make the dipole, once to feel it — so $1/ r^{6}$ again (Debye, no heat term). No permanent dipoles at all? The electron clouds flicker and dance in step, still using a field twice → $1/ r^{6}$ (London). Add a hard $1/ r^{12}$ wall for when clouds bump, and you get the Lennard-Jones well. Three stories, one number: six.

Recall Predict then check

Q: Which term disappears for a non-polar molecule like $C H_{4}$ ? ::: Keesom and Debye (both need $μ$ ); only London remains. Q: Why does the Keesom term shrink as temperature rises but London does not? ::: Keesom relies on dipoles staying aligned; heat randomises orientations ( $\propto 1/ T$ ). London's flicker-correlation is not orientation-locked, so it carries no $T$ . Q: Where does the power $6$ come from, in one sentence? ::: A dipole field goes as $1/ r^{3}$ , and every mechanism uses it twice, so $(1/ r^{3})^{2} = 1/ r^{6}$ .

Links: parent van der Waals forces · Coulomb's law · Dipole moment and molecular polarity · Polarisability and Fajans' rules · Lennard-Jones potential · Hydrogen bonding · Boiling point and intermolecular forces · States of matter and condensation