Microscopic origin of resistivity

Why does a copper wire resist the flow of current at all? Not because "resistance" is a fixed label stamped on the metal, but because free electrons drifting under an electric field keep colliding with the vibrating ions of the lattice. Between collisions an electron accelerates; at each collision it forgets its gained velocity and starts again. The free-electron theory turns this simple collision picture into an exact formula, $\rho = m/(ne^2\tau)$ , that derives Ohm's law from microscopic physics rather than treating resistivity as a number to look up in a table. This article builds that formula from the drift-velocity expression you already know, reads physical meaning into each symbol, and uses it to explain two facts every student meets in the next topic: why a metal's resistivity climbs with temperature, and why the same rule flips for semiconductors.

The core insight

Resistivity is not arbitrary — the free-electron theory shows $\rho = m/(ne^2\tau)$ , so it is fixed by electron mass $m$ , free-electron density $n$ , and the relaxation time $\tau$ between lattice collisions.

The core idea

From drift velocity to current density. When a field $E$ is switched on, each free electron accelerates between collisions and reaches an average drift velocity

v_d = \frac{eE\tau}{m}

where: $e$ is the electron charge, $m$ its mass, and $\tau$ the relaxation time — the average time between successive collisions with the lattice.

From current density to conductivity. The current density carried by $n$ electrons per unit volume drifting at $v_d$ is $J = nev_d$ . Substituting the expression above:

J = ne\left(\frac{eE\tau}{m}\right) = \frac{ne^2\tau}{m}\,E

Comparing with the microscopic Ohm's law $J = \sigma E$ identifies the conductivity, and its reciprocal gives resistivity:

\sigma = \frac{ne^2\tau}{m}, \qquad \rho = \frac{1}{\sigma} = \frac{m}{ne^2\tau}

where: $\sigma$ is conductivity, $\rho$ is resistivity, and $n$ , $e$ , $m$ , $\tau$ are as above. This is the key result of the whole topic: resistivity is not a mysterious material constant but the direct outcome of three microscopic quantities.

Reading $\tau$ physically. A larger $\tau$ means electrons travel farther, on average, before scattering off a lattice ion or an impurity — so conductivity is high and resistivity low. A smaller $\tau$ means frequent scattering, more resistivity. Since $\tau$ sits in the denominator of $\rho$ , anything that shortens the free flight between collisions raises resistivity.

Why metals get more resistive as they heat up. Raising temperature makes lattice ions vibrate with larger amplitude, presenting a bigger effective target and so a higher collision frequency. That shortens $\tau$ , and because $\rho \propto 1/\tau$ , resistivity rises — the microscopic root of the familiar $\rho_T = \rho_0[1+\alpha(T-T_0)]$ rule for metals.

Why semiconductors behave oppositely. In a semiconductor, heating releases many more charge carriers, so $n$ grows faster than $\tau$ shrinks. Since $\rho = m/(ne^2\tau)$ , a rising $n$ dominates over a falling $\tau$ , and resistivity falls with temperature — same formula, opposite dominant variable.

Why resistivity depends on the material. Copper, silver, nichrome and manganin are all metallic conductors, yet their resistivities differ by orders of magnitude. The free-electron model attributes this entirely to differences in $n$ and $\tau$ : pure, well-ordered metals like copper and silver have both a high free-electron density and a long relaxation time (few scattering centres), giving low resistivity, while alloys like nichrome have an irregular lattice packed with foreign atoms that scatter electrons frequently, sharply shortening $\tau$ even though $n$ is comparable. This is exactly why nichrome, not copper, is chosen for heating coils — a short $\tau$ is a feature, not a defect, when the goal is to dissipate electrical energy as heat.

Figure

Two metal rods at different temperatures showing electrons colliding more often with more strongly vibrating lattice ions in the hot rod, giving a shorter relaxation time and higher resistivity

The core formula, derived

Derivation ADeriving resistivity from the free-electron model

The goal is to obtain the microscopic expression $\rho = m/(ne^2\tau)$ for the resistivity of a metal purely from the drift-velocity picture of conduction electrons and the macroscopic definition $J = \sigma E$ , without invoking any empirical resistivity table — this is the derivation that converts resistivity from a measured number into a predicted one.

Step 1. Start from the drift velocity an electron reaches under a field $E$ . Between collisions it accelerates at $a = eE/m$ ; if $\tau$ is the average time it survives before the next collision, its average drift speed picked up from the field is $v_d = a\tau = eE\tau/m$ .

Step 2. Convert drift velocity into current density. With $n$ free electrons per unit volume, each carrying charge $e$ and drifting at $v_d$ , the current density is $J = nev_d$ . Substituting Step 1's expression for $v_d$ gives $J = ne\left(\dfrac{eE\tau}{m}\right) = \dfrac{ne^2\tau}{m}E$ .

Step 3. Match against the macroscopic form of Ohm's law. Ohm's law in field form states $J = \sigma E$ . Comparing this directly with Step 2's result, the coefficient of $E$ must be the conductivity: $\sigma = \dfrac{ne^2\tau}{m}$ .

Step 4. Invert for resistivity. Since $\rho = 1/\sigma$ by definition, $\rho = \dfrac{m}{ne^2\tau}$ . This is the target result: resistivity is built entirely from the electron mass $m$ , the free-electron density $n$ , the electron charge $e$ , and the relaxation time $\tau$ — no separate empirical input is needed once these microscopic quantities are known.

Worked example

For copper, $n = 8.5\times10^{28}\ \text{m}^{-3}$ and the measured resistivity at room temperature is $\rho = 1.7\times10^{-8}\ \Omega\cdot\text{m}$ . Find the relaxation time $\tau$ . (Take $e = 1.6\times10^{-19}\ \text{C}$ , $m = 9.1\times10^{-31}\ \text{kg}$ .)

Step 1. Rearrange $\rho = m/(ne^2\tau)$ for $\tau$ : $\tau = \dfrac{m}{ne^2\rho}$ .

Step 2. Compute the denominator piece by piece: $e^2 = (1.6\times10^{-19})^2 = 2.56\times10^{-38}$ , so $ne^2 = (8.5\times10^{28})(2.56\times10^{-38}) = 2.176\times10^{-9}$ , and $ne^2\rho = (2.176\times10^{-9})(1.7\times10^{-8}) = 3.70\times10^{-17}$ .

Step 3. Divide: $\tau = \dfrac{9.1\times10^{-31}}{3.70\times10^{-17}} = 2.46\times10^{-14}\ \text{s}$ .

The relaxation time is about $2.5\times10^{-14}\ \text{s}$ — an electron in copper collides roughly $4\times10^{13}$ times every second, the characteristic scale of scattering in a metal.

Competitive stretch

Two wires A and B, same length and area, same $n$ , but $\tau_A = 3\times10^{-14}\ \text{s}$ and $\tau_B = 1\times10^{-14}\ \text{s}$ . Since $\rho \propto 1/\tau$ with $m$ , $n$ , $e$ fixed, $\rho_B/\rho_A = \tau_A/\tau_B = 3$ . Wire B, whose electrons scatter three times as often, has three times the resistivity of wire A — more frequent collisions demand a larger field to sustain the same current.

Common traps

Treating resistivity as something only measurable, not derivable — the free-electron model gives a genuine physical derivation, $\rho = m/(ne^2\tau)$ .
Assuming a higher electron density $n$ must raise resistivity; $n$ sits in the denominator, so more carriers mean lower resistivity, not higher.
Applying the metal temperature trend to semiconductors — in metals $\tau$ falling with temperature dominates ( $\rho$ up), but in semiconductors rising $n$ dominates ( $\rho$ down).
Confusing resistivity $\rho$ (a material property from this derivation) with resistance $R$ (a property of a specific wire, $R = \rho l/A$ , which also depends on geometry).

Ask the tutor: Want to see how this relaxation-time picture connects to the actual resistance-versus-temperature graphs for conductors and semiconductors?

Why does more electron density n lower resistivity instead of raising it?

It seems intuitive that "more electrons" should mean more obstruction, but that confuses carriers with obstacles. In $\rho = m/(ne^2\tau)$ , $n$ measures how many charge carriers are available to carry current, not how many scattering centres exist. More carriers sharing the same driving field means each one needs to drift less to supply a given current, so less field is needed overall — lower resistivity. The obstruction to flow comes from $\tau$ (how often carriers collide), not from $n$ .

Is tau the same for every electron in the metal, or an average?

It is an average. Individual electrons have wildly different free times between collisions — some travel briefly, some longer — but $\tau$ in the formula is the statistical mean relaxation time across all conduction electrons at a given temperature. This average is what enters the macroscopic $\rho = m/(ne^2\tau)$ ; the free-electron model deliberately smooths over individual electron histories to get a bulk material property.

Does this formula also explain why alloys like nichrome have much higher resistivity than pure copper?

Yes, qualitatively. Alloys and impurity-laden metals have irregular lattice structures that scatter electrons far more often than a clean, periodic lattice of a pure metal — so their effective $\tau$ is much shorter than copper's. With $n$ and $m$ roughly comparable, a smaller $\tau$ directly gives a larger $\rho$ through $\rho = m/(ne^2\tau)$ , which is exactly why nichrome and manganin are chosen for heating elements and standard resistors.