Einstein relation between mobility and diffusion: Visual walkthrough

Step 0 — The two ways a carrier moves (the whole stage)

WHAT. Picture a thin bar of semiconductor. Inside it are tiny charged particles (call them carriers). We track how many sit at each position. The count-per-volume at position $x$ is the carrier density $n (x)$ — literally "how crowded is it here?"

WHY. Before any maths, we must name the two reasons a carrier ever leaves a spot:

A push from a field (an invisible slope that shoves charges one way) → this is drift.
Random jiggling — carriers bounce around from heat and, purely by accident, wander from crowded corners to empty ones → this is diffusion.

PICTURE. In the figure, the left panel shows carriers all pushed the same way by a field (drift). The right panel shows carriers with no field, spreading from a crowded lump into empty space (diffusion). Same little dots, two behaviours.

Figure — Einstein relation between mobility and diffusion

Step 1 — Write the drift current

WHAT. Turn on a field $E$ . Every carrier acquires a steady drift speed $v_{d r i f t} = μ E$ . If there are $n$ carriers per volume each carrying charge $- q$ moving at speed $μ E$ , the electric current they make is $J_{n, drift} = q n μ_{n} E$

WHY this form. Current = (charge per carrier) × (carriers per volume) × (speed). The subscript $n$ just marks "electrons." We use $q$ (positive) because the two sign flips — negative charge $- q$ moving against the field direction — cancel to give a current in the same direction as $E$ .

PICTURE. The arrows all point one way; their length is the drift speed $μ_{n} E$ . More dots (bigger $n$ ) or a stronger field (bigger $E$ ) → longer green current arrow.

Step 2 — Write the diffusion current

WHAT. Now the spreading current. A crowd flows from high density to low density. The steepness of the crowd is measured by the derivative $\frac{d n}{d x}$ — read as "how fast does $n$ change as I step in $x$ ." The diffusion flux (carriers per area per second) is $F_{diff} = - D_{n} \frac{d n}{d x}$

WHY the minus sign, and why a derivative? We need a tool that answers "which way, and how strongly, is the crowd leaning?" A single number $n$ can't say that — only its slope can. That is exactly what a derivative is: the slope of $n$ versus $x$ . The minus sign says carriers flow down the slope (from many toward few). See Diffusion current and Fick's law.

Turning flux into current. Multiply flux by charge. Electrons carry $- q$ : $J_{n, diff} = (- q) (- D_{n} \frac{d n}{d x}) = + q D_{n} \frac{d n}{d x}$ Two minus signs make a plus.

PICTURE. The density curve slopes downhill to the right; the red flux arrows point downhill (toward fewer carriers). Steeper slope → longer arrows.

Step 3 — Demand zero net current (equilibrium)

WHAT. Add the two pieces. The total electron current is $J_{n} = q n μ_{n} E + q D_{n} \frac{d n}{d x}$ Now impose thermal equilibrium: no battery, no wires, nothing flowing in or out. Then the net current must be zero everywhere: $J_{n} = 0$ .

WHY this is the key trick. With no external drive, the field-push and the random-spread must exactly cancel at every point — otherwise charge would pile up and current would flow, contradicting "nothing is connected." Setting the sum to zero forces drift and diffusion to speak the same language, which is what lets us extract the ratio $D / μ$ .

PICTURE. Green drift arrows (pushing right) and red diffusion arrows (spreading left) at the same point, equal in length, tip-to-tip — a perfect standoff. Net arrow = zero.

Setting $J_{n} = 0$ : $q n μ_{n} E = - q D_{n} \frac{d n}{d x}$

Step 4 — Trade the field for a potential slope

WHAT. A field is the downhill slope of an electric potential $ψ (x)$ (think: height of a landscape). Formally $E = - \frac{d ψ}{d x}$

WHY. We want everything written in one language: slopes of $x$ -dependent quantities. The field $E$ and the density slope $d n / d x$ are both "slopes," but they live in different variables. Rewriting $E$ as a slope of $ψ$ lets us later connect $ψ$ to $n$ (Step 5). The minus sign: the field points downhill, from high potential to low.

PICTURE. A gentle potential hill $ψ (x)$ ; the field arrow $E$ points down the slope. Cancel the $q$ on both sides of the equilibrium equation and substitute $E = - d ψ / d x$ : $n μ_{n} \frac{d ψ}{d x} = D_{n} \frac{d n}{d x}$

The two remaining minus signs (from $E$ and from the RHS of Step 3) cancelled, leaving both sides positive.

Step 5 — Plug in how carriers pile up on the hill (the physics input)

WHAT. Here is the one physical fact we import. At temperature $T$ , carriers are less likely to sit where the potential energy is high. For an electron the potential energy is $- q ψ$ (higher $ψ$ = lower energy for a negative charge), and the fraction sitting there follows the Boltzmann distribution: $n (x) = n_{0} exp (\frac{q ψ ( x )}{k _{B} T})$

WHY an exponential — why this tool? We need a rule for "how does crowding respond to a potential hill?" Statistical physics answers: the count drops off exponentially with energy over $k_{B} T$ . The exponential is exactly the function whose slope is proportional to itself — which is why differentiating it gives back $n$ times a simple factor. That self-similarity is what makes the next step collapse. See Boltzmann distribution in semiconductors.

PICTURE. Density high in the valleys, low on the peaks — an upside-down copy of the potential hill from Step 4, plotted on the same axis.

Differentiate (slope of an exponential = itself times the slope of the inside): $\frac{d n}{d x} = n_{0} exp (\frac{q ψ}{k _{B} T}) \cdot \frac{q}{k _{B} T} \frac{d ψ}{d x} = n \cdot \frac{q}{k _{B} T} \frac{d ψ}{d x}$ The first factor is just $n$ again — that is the exponential's magic.

Step 6 — Substitute and watch everything cancel

WHAT. Put the boxed $d n / d x$ from Step 5 into the balance from Step 4: $n μ_{n} \frac{d ψ}{d x} = D_{n} \cdot n \frac{q}{k _{B} T} \frac{d ψ}{d x}$

WHY it finishes here. Look at what appears on both sides: the density $n$ and the potential slope $\frac{d ψ}{d x}$ . Neither depends on $D$ or $μ$ — they are just "wherever we happen to be." Cancel them: $μ_{n} = D_{n} \frac{q}{k _{B} T} ⟹ \frac{D _{n}}{μ _{n}} = \frac{k _{B} T}{q}$

PICTURE. Terms $n$ and $d ψ / d x$ struck out on both sides; what survives is a bare constant — the same value everywhere in the bar, set only by temperature.

Step 7 — The degenerate case: when the picture breaks

WHAT. Step 5 assumed carriers are rare enough that the simple Boltzmann exponential holds. In a heavily doped crystal, carriers get crowded and start blocking each other's states (Pauli exclusion). The correct count is then Fermi–Dirac, not Boltzmann.

WHY it changes the answer. The self-similar exponential of Step 5 no longer applies, so the clean cancellation gives a bigger ratio: $\frac{D}{μ} = \frac{k _{B} T}{q} \cdot \frac{F _{1/2} ( η )}{F _{- 1/2} ( η )} > \frac{k _{B} T}{q}$ The factor is $> 1$ for degenerate material and $\to 1$ in the dilute limit.

PICTURE. Two curves: at low doping $D / μ$ sits flat at $k_{B} T / q$ ; as doping climbs into the degenerate zone it bends upward.

The one-picture summary

Everything above, compressed: two currents → set net to zero → swap field for potential slope → feed in the Boltzmann exponential → cancel → out drops $k_{B} T / q$ .

Recall Feynman retelling in plain words

Imagine a room with a slope on the floor (that's the potential) and warm air making dust jiggle (that's the temperature). Two things move the dust: gravity sliding it downhill (drift, controlled by how slippery the floor is — mobility) and random jiggling spreading it out (diffusion). Seal the room so nothing enters or leaves. Now nothing can be net flowing — so at every spot, "sliding down" and "jiggling apart" must exactly tie. But how the dust piles up on the slope is a known law: fewer specks the higher you go, dropping off exponentially with height over the warmth $k_{B} T$ . When you write down "sliding = jiggling" and plug in that pile-up law, every messy detail about the floor and the dust cancels. What's left is a pure number — warmth over charge, $k_{B} T / q$ . That's the Einstein relation: how easily you're pushed and how fast you spread are two faces of the same warmth.

Active recall

Connections

Drift current and mobility
Diffusion current and Fick's law
Boltzmann distribution in semiconductors
Thermal voltage V_T
Continuity equation and carrier transport
Degenerate semiconductors & generalized Einstein relation
Fermi-Dirac vs Maxwell-Boltzmann statistics

Visual walkthrough — Einstein relation between mobility and diffusion