Lesson 00

How to use this course

Every student learns \(V=IR\) in the first week of a circuits course, and it can seem too simple to have needed a discoverer. In the 1820s it was not simple at all. Physicists had no agreed meaning for current, tension or resistance, no reliable meters and no steady sources. Georg Simon Ohm, a schoolteacher in Cologne, built his own instruments, found the law by measurement, and then derived it from a theory modelled on Joseph Fourier’s mathematics of heat. German physics largely ignored him for a decade.

Prerequisites for Part II

First-year electricity and magnetism: potential, current, the magnetic field of a wire.
Calculus of one variable and simple linear ODEs.
For Lessons 07 and 09, partial derivatives and the heat equation at the level of a first PDE course.
For Lesson 10, Fourier series.

Each technical lesson opens with a paragraph for the general reader, then gives precise statements, derivations and exercises with folded solutions. Ohm wrote of “electroscopic force”, “reduced length” and the “galvanic chain”; each lesson translates these into modern terms and says what he actually measured.

Reference

Timeline

1789	Born 16 March in Erlangen, son of the locksmith Johann Wolfgang Ohm and Maria Elisabeth Beck.
1792	Birth of his brother Martin Ohm, later a mathematician in Berlin.
1805	Enters the University of Erlangen; leaves after three semesters.
1806	September: mathematics teacher at a school in Gottstatt, near Nidau, Switzerland.
1809	March: private tutor in Neuchâtel.
1811	Returns to Erlangen; doctorate on 25 October; lectures as Privatdozent for three semesters.
1813	Teacher of mathematics and physics at a school in Bamberg.
1817	Teacher of mathematics and physics at the Jesuit Gymnasium in Cologne.
1820	Ørsted discovers electromagnetism; Ohm begins electrical experiments.
1825	First paper, on the decrease of the magnetic action of a current with wire length.
1826	January: thermoelectric experiments yield \(X=a/(b+x)\). Two papers on conduction. From August, a year’s leave in Berlin.
1827	Die galvanische Kette, mathematisch bearbeitet, Berlin.
1828	March: resigns from Cologne. Teaches in Berlin for several years.
1833	Professor at the Polytechnic School in Nuremberg; director from 1839.
1841	Copley Medal of the Royal Society.
1842	Foreign Member of the Royal Society.
1843	Formulates the acoustic law; dispute with August Seebeck follows.
1845	Full member of the Bavarian Academy of Sciences.
1849	Moves to Munich: curator of the Academy’s physical collection and extraordinary professor at the university.
1852	Chair of experimental physics at the University of Munich.
1854	Dies in Munich on 6 July, aged 65.
1881	The International Electrical Congress in Paris names the unit of resistance the ohm.

Part I · Formation

Lesson 01

Erlangen, 1789–1806

Georg Simon Ohm was born in Erlangen, then in the Protestant principality of Brandenburg-Bayreuth, on 16 March 1789. His father, Johann Wolfgang Ohm, was a master locksmith. His mother, Maria Elisabeth Beck, a tailor’s daughter, died when he was ten. Of seven children, three survived to adulthood.

The father had no formal education beyond his trade, but he had taught himself mathematics and philosophy, and he taught his sons at home. The lessons were demanding. Georg and his younger brother Martin, born in 1792, learned mathematics far beyond what the town Gymnasium offered, which gave them little science. They also learned the locksmith’s craft. That skill with metal and tools shaped Ohm’s later experiments, in which he made his own wires and instruments.

At fifteen Ohm was examined by Karl Christian von Langsdorf, professor of mathematics at Erlangen, who was struck by the brothers’ ability and compared them to the Bernoulli family. In 1805 Ohm entered the University of Erlangen. His father, unhappy with his progress and short of money, made him leave after three semesters.

Lesson 02

The wandering teacher, 1806–1817

In September 1806 Ohm took a post teaching mathematics at a school in Gottstatt, a former monastery near Nidau in Switzerland. In March 1809 he moved to Neuchâtel as a private tutor. Langsdorf, who had left Erlangen for Heidelberg, advised him by letter to study mathematics on his own from the classics: Leonhard Euler, Pierre-Simon Laplace and Sylvestre François Lacroix. Ohm did so.

He returned to Erlangen in April 1811 and received his doctorate there on 25 October 1811. He lectured as a Privatdozent, unpaid apart from student fees, for three semesters, then gave up for lack of money and prospects. In 1813 the Bavarian government sent him to teach mathematics and physics at a school in Bamberg. He found it poor and overcrowded. When it closed in 1816 he helped out at another Bamberg school.

On 11 September 1817 Ohm accepted a post at the Jesuit Gymnasium in Cologne, then in Prussia’s new Rhineland province.

Lesson 03

Cologne and the laboratory, 1817–1826

The Cologne Gymnasium had a well-equipped physics collection, and Ohm taught there for ten years. Among his pupils was the young Peter Gustav Lejeune Dirichlet. Ohm read the French mathematical physicists, above all Joseph Fourier’s Théorie analytique de la chaleur (1822), which showed how a single partial differential equation could describe the flow of heat through any body.

In 1820 Hans Christian Ørsted found that an electric current deflects a compass needle. Within months André-Marie Ampère, Jean-Baptiste Biot and Félix Savart had turned the discovery into a science of forces between currents. For Ohm the deflected needle was something else: a meter. The force of a current on a magnet could measure how much current flowed, something the electrometers of Alessandro Volta’s generation could not do.

Ohm decided that research was his route to a university post. Working in the school laboratory, he built a torsion balance in which a magnetized needle, hung from a fine wire, sat above a straight section of the circuit. He published his first results in 1825 and his decisive experiments in 1826. In August 1826 he obtained a year’s leave on half pay and went to Berlin, where his brother Martin taught, to write the theory. The result was Die galvanische Kette, mathematisch bearbeitet (The galvanic circuit investigated mathematically), published in Berlin in 1827.

Part II · The physics

Lesson 04

The experiments of 1825–1826

For the general reader. Ohm wanted to know how the length of a wire affects the current through it. His first attempt failed because his batteries kept fading. On the advice of the editor Johann Christian Poggendorff he switched to a thermocouple, a junction of two metals held at different temperatures, which gives a small but very steady push. With it he found a simple formula that fitted every wire he tried.

Goals

Understand Ohm’s apparatus and why the thermocouple mattered.
Interpret \(X=a/(b+x)\) as \(I=\mathcal E/(R_{\text{int}}+R)\).
Extract constants from data by linearization.

1825: the fading pile

Ohm’s first paper, in Poggendorff’s Annalen der Physik und Chemie in 1825, used a zinc–copper pile with dilute acid. He measured the magnetic force on his needle with wires of several lengths. The pile polarized and weakened during each run, so readings drifted. He fitted the data to a logarithmic formula, which was wrong. Poggendorff, in an editorial note, suggested a thermoelectric source instead. Thomas Johann Seebeck had discovered in 1821 that a circuit of two metals carries a current when its junctions are at different temperatures.

January 1826: the thermocouple

Ohm made a circuit of bismuth and copper. One junction stood in boiling water, the other in ice. The temperature difference, and with it the driving force, stayed constant for half an hour or more. The circuit was closed through two cups of mercury, into which he dipped the ends of test wires. He used eight copper wires of the same thickness, of lengths 2, 4, 6, 10, 18, 34, 66 and 130 inches. The torsion needed to bring the needle back to its rest position measured the current.

Ohm’s empirical law (1826)

For a fixed source and wires of one material and cross-section, the magnetic force \(X\) varies with the wire length \(x\) as

\[ X=\frac{a}{b+x}, \]

where \(a\) depends only on the temperature difference of the thermocouple and \(b\) only on the rest of the circuit. In modern terms \(X\propto I\), \(a\propto\mathcal E\), \(b\) is the internal resistance expressed as an equivalent wire length, and \(x\) is the external resistance: \(I=\mathcal E/(R_{\text{int}}+R)\).

He then tested brass wire and wires of other thicknesses. Each could be replaced by a copper wire of some “reduced length” giving the same current. The reduced length was proportional to the real length and inversely proportional to the cross-section, so the formula covered them all. This is the first clear experimental definition of resistance as a property of a conductor.

Exercise 4.1Show that if \(X=a/(b+x)\), then \(1/X\) is a linear function of \(x\). How would you find \(a\) and \(b\) from a plot?

Show solution

\(1/X=b/a+x/a\). Plot \(1/X\) against \(x\): the slope is \(1/a\) and the intercept is \(b/a\). So \(a=1/\text{slope}\) and \(b=\text{intercept}/\text{slope}\); \(b\) is minus the \(x\)-intercept.

Exercise 4.2Suppose the needle reads \(X=300\) with the 2-inch wire and \(X=60\) with the 130-inch wire. Find \(a\) and \(b\), and predict the reading for the 34-inch wire.

Show solution

\(300(b+2)=60(b+130)\), so \(300b+600=60b+7800\), \(b=30\) inches and \(a=300\times32=9600\). For \(x=34\): \(X=9600/64=150\). The internal resistance of the circuit equals that of 30 inches of test wire.

Exercise 4.3Why did the thermocouple’s low emf not harm the experiment, while the pile’s higher emf did?

Show solution

The law concerns ratios at a fixed source. What matters is that \(\mathcal E\) and \(R_{\text{int}}\) stay constant during a run. A thermocouple’s emf depends only on its junction temperatures, which ice and boiling water fix. A pile’s emf and internal resistance change as gas collects on its plates. Because the bismuth–copper circuit had very low total resistance, even its few millivolts drove a current large enough to deflect the needle.

Lesson 05

Die galvanische Kette: the Fourier analogy

For the general reader. Heat flows from hot to cold, faster when the temperature drops steeply. Fourier had made this into mathematics. Ohm proposed that electricity flows the same way, driven by differences of what he called electroscopic force, the quantity an electroscope measures. From this one assumption he derived his experimental law and predicted how the electroscopic force must vary along a wire.

Goals

State Ohm’s flux law and derive the circuit law from it.
Find the potential profile along a circuit with a source.
Relate Volta’s contact tensions to Ohm’s sources.

The flux law

Ohm’s electroscopic force \(u\) is the electric potential. He assumed that the flow of electricity through a cross-section of a wire is proportional to the conductivity, the area and the gradient of \(u\), exactly as Fourier’s heat flux is proportional to the temperature gradient.

Ohm’s law (1827, modern notation)

In a uniform wire of conductivity \(\sigma\) and cross-section \(\omega\), the steady current is

\[ I=-\sigma\,\omega\,\frac{du}{dx}. \]

In a steady state \(I\) is the same at every cross-section, so \(u\) is linear along each uniform segment. For a closed circuit with total tension (emf) \(\mathcal E\) concentrated at the junctions,

\[ I=\frac{\mathcal E}{\sum_k \ell_k/(\sigma_k\omega_k)}=\frac{\mathcal E}{\sum_k R_k}. \]

Derivation. Steady flow with no accumulation of charge gives \(dI/dx=0\), so \(I\) is constant around the loop. On segment \(k\), \(du/dx=-I/(\sigma_k\omega_k)\), so the fall of \(u\) across it is \(IR_k\) with \(R_k=\ell_k/(\sigma_k\omega_k)\). Going once around the circuit, \(u\) returns to its starting value. The drops \(IR_k\) must therefore be balanced by the jumps at the sources, whose sum is \(\mathcal E\). Hence \(\mathcal E=I\sum R_k\).

Ohm took the source jumps to be Volta’s contact tensions at the junctions between unlike conductors. The steady state is a sawtooth: \(u\) jumps up at each source and falls linearly along each conductor.

A second analogy

The Fourier analogy is now so familiar that it runs in reverse: engineers speak of thermal resistance and solve heat problems with circuit diagrams. Fourier’s law \(\mathbf q=-k\nabla T\), Ohm’s law \(\mathbf J=-\sigma\nabla u\), and Fick’s law of diffusion \(\mathbf j=-D\nabla c\) all have the same form.

Exercise 5.1A cell of emf 2 V and internal resistance 1 Ω drives current through an external wire of resistance 3 Ω. Take the potential at the negative terminal as 0. Sketch \(u\) around the circuit and give the terminal voltage.

Show solution

\(I=2/4=0.5\ \mathrm A\). Inside the cell, \(u\) jumps by 2 V at the source and falls by \(0.5\times1=0.5\) V across the internal resistance, so the positive terminal is at 1.5 V. It then falls linearly by 1.5 V along the external wire, back to 0. The terminal voltage is \(1.5\ \mathrm V=\mathcal E-Ir\).

Exercise 5.2A circuit is a copper wire of length 1 m joined to an iron wire of length 1 m, same cross-section, with a thermoelectric source of emf \(\mathcal E\) at one junction and negligible internal resistance. Iron’s resistivity is about 6 times copper’s. What fraction of \(\mathcal E\) is dropped along each wire?

Show solution

Resistances in ratio 1 : 6, current common, so the drops are \(\mathcal E/7\) along copper and \(6\mathcal E/7\) along iron. The potential gradient is six times steeper in the iron, as Ohm’s sawtooth predicts.

Lesson 06

Resistance, resistivity and combinations

For the general reader. Ohm found that a wire’s resistance grows with its length and shrinks with its thickness, and that every material has its own characteristic resistance. From these facts follow the rules for connecting wires end to end or side by side, the arithmetic of every electrical circuit.

Goals

Use \(R=\rho\ell/A\) and its temperature dependence.
Derive the series and parallel rules from Ohm’s law and conservation of charge.
Compute Joule heating.

Resistance of a uniform conductor

A uniform conductor of length \(\ell\), cross-sectional area \(A\) and resistivity \(\rho=1/\sigma\) has resistance

\[ R=\frac{\rho\,\ell}{A}. \]

For metals near room temperature, \(\rho(T)\approx\rho_0\,[1+\alpha(T-T_0)]\), with \(\alpha\approx3.9\times10^{-3}\ \mathrm K^{-1}\) for copper.

Series and parallel \[ R_{\text{series}}=R_1+R_2+\cdots,\qquad \frac{1}{R_{\text{parallel}}}=\frac1{R_1}+\frac1{R_2}+\cdots. \]

In series the current is common and the voltage drops add. In parallel the voltage is common and the currents add, because charge is conserved at the junction. Ohm’s “reduced length” is \(R\) expressed in units of a standard wire; his combination rules for reduced lengths are these rules.

Power dissipated in a resistor is \(P=IV=I^2R=V^2/R\), established by James Prescott Joule in 1841 by calorimetry, and used by him as evidence that Ohm’s law correctly describes energy flow.

Exercise 6.1Find the resistance of 100 m of copper wire of diameter 1.0 mm (\(\rho=1.7\times10^{-8}\ \Omega\,\mathrm m\)). By how much does it change between 20 °C and 60 °C?

Show solution

\(A=\pi(0.5\times10^{-3})^2=7.85\times10^{-7}\ \mathrm{m^2}\). \(R=1.7\times10^{-8}\times100/7.85\times10^{-7}=2.17\ \Omega\). The increase is \(\alpha\Delta T=0.0039\times40=15.6\%\), to about 2.5 Ω. Temperature control mattered in Ohm’s experiments for exactly this reason.

Exercise 6.2A wire of resistance \(R\) is cut into \(n\) equal pieces, which are then connected in parallel. Find the new resistance. Then stretch the original wire to twice its length at constant volume: what is its resistance?

Show solution

Each piece has \(R/n\); \(n\) in parallel give \(R/n^2\). Stretching doubles \(\ell\) and halves \(A\) (volume fixed), so \(R\to\rho(2\ell)/(A/2)=4R\).

Exercise 6.3Show that among all ways of splitting a fixed current \(I\) between two parallel resistors \(R_1,R_2\), the actual split minimizes the total dissipated power.

Show solution

Let \(I_1+I_2=I\). \(P=I_1^2R_1+(I-I_1)^2R_2\); \(dP/dI_1=2I_1R_1-2(I-I_1)R_2=0\) gives \(I_1R_1=I_2R_2\): equal voltage across both, the Ohm’s-law split. This is the simplest case of a general principle: steady currents in a resistive network minimize dissipation subject to the current constraints.

Lesson 07

Ohm’s law in a continuum and the Drude model

For the general reader. Ohm wrote his law for wires. Gustav Kirchhoff rewrote it for any shape of conductor, relating the current at each point to the electric field there. Seventy years after Ohm, once the electron had been discovered, Paul Drude explained why the law holds: electrons accelerate in the field, collide with the lattice, and settle into an average drift proportional to the field.

Goals

State the local form \(\mathbf J=\sigma\mathbf E\) and derive \(R=\rho\ell/A\) from it.
Solve a steady conduction problem with Laplace’s equation.
Derive \(\sigma=ne^2\tau/m\) in the Drude model.

Local Ohm’s law

In an isotropic ohmic conductor the current density is proportional to the electric field,

\[ \mathbf J=\sigma\mathbf E=-\sigma\nabla u. \]

In steady state \(\nabla\cdot\mathbf J=0\), so in a uniform conductor \(\nabla^2u=0\), with \(\mathbf J\cdot\mathbf n=0\) on insulated surfaces and \(u\) prescribed on electrodes.

Kirchhoff formulated conduction in this form in 1845–49 and showed that Ohm’s electroscopic force must be identified with the electrostatic potential. This settled a question Ohm had left open: whether the quantity driving current is the same one that acts on electroscopes.

Drude’s model (1900)

Treat conduction electrons, density \(n\), charge \(-e\), mass \(m\), as free particles that undergo random collisions, losing their drift momentum with mean time \(\tau\) between collisions. The average momentum obeys

\[ \frac{d\langle\mathbf p\rangle}{dt}=-e\mathbf E-\frac{\langle\mathbf p\rangle}{\tau}. \]

Drude conductivity

In steady state \(\langle\mathbf p\rangle=-e\tau\mathbf E\), the drift velocity is \(\mathbf v_d=-e\tau\mathbf E/m\), and \(\mathbf J=-ne\mathbf v_d\) gives

\[ \sigma=\frac{ne^2\tau}{m}. \]

The law is linear because the drift velocity is tiny compared with electron thermal speeds, so collisions are hardly affected by the field. Quantum theory (Sommerfeld, Bloch) keeps the formula with \(m\) replaced by an effective mass and \(\tau\) interpreted at the Fermi surface. Ohm’s law fails in semiconductor junctions, gas discharges, superconductors and at very high fields, and it is a constitutive law, not a fundamental one.

Exercise 7.1For copper, \(n=8.5\times10^{28}\ \mathrm{m^{-3}}\) and \(\rho=1.7\times10^{-8}\ \Omega\,\mathrm m\). Find \(\tau\), and the drift speed in a 1 mm² wire carrying 1 A.

Show solution

\(\tau=m/(ne^2\rho)=9.11\times10^{-31}/(8.5\times10^{28}\times(1.6\times10^{-19})^2\times1.7\times10^{-8})\approx2.5\times10^{-14}\ \mathrm s\). \(v_d=I/(neA)=1/(8.5\times10^{28}\times1.6\times10^{-19}\times10^{-6})\approx7\times10^{-5}\ \mathrm{m/s}\), about a quarter of a metre per hour.

Exercise 7.2A cylindrical shell conductor fills \(a\lt r\lt b\), length \(L\), resistivity \(\rho\), with current flowing radially from the inner to the outer electrode. Find its resistance.

Show solution

Through any cylinder of radius \(r\), \(I=J(r)\,2\pi rL\), so \(E=\rho I/(2\pi rL)\). Integrating, \(V=\int_a^bE\,dr=\dfrac{\rho I}{2\pi L}\ln\dfrac ba\). Hence \(R=\dfrac{\rho}{2\pi L}\ln\dfrac ba\). The same answer follows from solving \(\nabla^2u=0\) with \(u=u(r)\).

Exercise 7.3Add an oscillating field \(\mathbf E e^{-i\omega t}\) to the Drude equation. Show that \(\sigma(\omega)=\sigma_0/(1-i\omega\tau)\). Up to what frequency does Ohm’s DC law hold for copper?

Show solution

Try \(\langle\mathbf p\rangle=\mathbf p_0e^{-i\omega t}\): \(-i\omega\mathbf p_0=-e\mathbf E-\mathbf p_0/\tau\), so \(\mathbf p_0=-e\tau\mathbf E/(1-i\omega\tau)\), giving \(\sigma(\omega)=\sigma_0/(1-i\omega\tau)\). The DC law holds while \(\omega\tau\ll1\), i.e. \(\omega\ll4\times10^{13}\ \mathrm{s^{-1}}\): through radio and microwave frequencies, into the far infrared.

Lesson 08

Networks: Kirchhoff and Wheatstone

For the general reader. Ohm’s book treated a single loop. In 1845 Gustav Kirchhoff, then a student in Königsberg, extended the law to networks of any complexity. In 1843 Charles Wheatstone popularized a balanced circuit, now called the Wheatstone bridge, that measures resistance by comparison. Both rest directly on Ohm.

Goals

State and apply Kirchhoff’s two laws.
Derive the balance condition of the Wheatstone bridge.
Find equivalent resistances by symmetry.

Kirchhoff’s laws (1845)

In a network of resistors and sources in a steady state:

at every node, the currents flowing in sum to zero, \(\sum I=0\);
around every closed loop, the sum of emfs equals the sum of \(IR\) drops, \(\sum\mathcal E=\sum IR\).

For a connected network with \(N\) nodes and \(B\) branches, there are \(N-1\) independent node equations and \(B-N+1\) independent loop equations, and the currents are uniquely determined.

The first law is conservation of charge. The second is Ohm’s sawtooth argument from Lesson 05, applied loop by loop. The count \(B-N+1\) of independent loops is the first Betti number of the network graph; Kirchhoff’s paper is an early result in graph theory, and it contains the matrix-tree theorem.

The Wheatstone bridge

Four resistors \(R_1,R_2\) (one arm) and \(R_3,R_x\) (the other) are connected across a source, with a galvanometer bridging the midpoints. The galvanometer reads zero when the midpoints are at equal potential:

\[ \frac{R_1}{R_2}=\frac{R_3}{R_x}\quad\Longleftrightarrow\quad R_x=\frac{R_2R_3}{R_1}. \]

A null measurement is independent of the source’s emf and of the galvanometer’s calibration. Samuel Hunter Christie described the circuit in 1833; Wheatstone credited him in his 1843 Bakerian Lecture, which also brought Ohm’s work before English readers.

Exercise 8.1Twelve equal resistors \(R\) form the edges of a cube. Find the resistance between opposite corners.

Show solution

By symmetry, the current \(I\) entering a corner splits equally into 3 edges (\(I/3\) each), then each of the 3 next nodes sends current into 2 edges (\(I/6\) each, 6 edges), then recombines through 3 edges (\(I/3\)). Along any path, \(V=\tfrac I3R+\tfrac I6R+\tfrac I3R=\tfrac56IR\). So \(R_{\text{eq}}=\tfrac56R\).

Exercise 8.2In a bridge with \(R_1=100\ \Omega\), \(R_2=250\ \Omega\), balance is found at \(R_3=48\ \Omega\). Find \(R_x\). Why does the balance not depend on the battery?

Show solution

\(R_x=R_2R_3/R_1=250\times48/100=120\ \Omega\). Each arm is a voltage divider; the midpoint potentials are \(\mathcal E\,R_2/(R_1+R_2)\) and \(\mathcal E\,R_x/(R_3+R_x)\) above the negative rail, both proportional to \(\mathcal E\). Their equality is a condition on ratios alone.

Exercise 8.3A network has 6 nodes and 9 branches. How many loop equations are independent? Check with the cube of Exercise 8.1.

Show solution

\(B-N+1=9-6+1=4\). For the cube, \(N=8\), \(B=12\), so 5 independent loops: the six faces, of which any five are independent (the sixth face is the sum of the others with orientation).

Lesson 09

Diffusion of charge and the telegraph

For the general reader. Ohm’s book also treated what happens before a circuit settles down. Following Fourier, he wrote an equation for how electroscopic force spreads along a conductor over time. Three decades later, long submarine telegraph cables showed exactly this kind of slow, smeared-out signal, and William Thomson’s theory of the cable used an equation of the same type.

Goals

Derive the diffusion equation for a resistive line with capacitance.
Find the diffusion time and Thomson’s \(L^2\) law.
Compare with Fourier’s heat equation.

The cable equation

Consider a long conductor with resistance \(r\) and capacitance to its surroundings \(c\), both per unit length, and negligible inductance and leakage. Let \(u(x,t)\) be the potential and \(I(x,t)\) the current. Ohm’s law on an element gives \(\partial u/\partial x=-rI\). Charge conservation, with charge \(cu\) per unit length, gives \(c\,\partial u/\partial t=-\partial I/\partial x\).

Diffusion equation for a resistive line \[ \frac{\partial u}{\partial t}=\frac{1}{rc}\,\frac{\partial^2u}{\partial x^2}. \]

This is Fourier’s heat equation with diffusivity \(D=1/(rc)\). A signal applied at one end of a line of length \(L\) reaches the far end on the time scale \(t\sim rcL^2\).

Ohm’s 1827 equation was of this Fourier type, though his capacitance term was a vaguely defined capacity of the conductor itself, not of a cable to the sea. William Thomson (later Lord Kelvin) derived the cable equation in 1855. Its \(L^2\) law showed why signalling across the Atlantic would be slow, and it shaped the design of the cables of 1858 and 1866. Oliver Heaviside later added inductance and leakage, producing the telegrapher’s equations.

Exercise 9.1A cable has \(r=3\ \Omega/\mathrm{km}\) and \(c=0.2\ \mu\mathrm F/\mathrm{km}\). Estimate \(rcL^2\) for \(L=4000\ \mathrm{km}\). What does doubling the length do?

Show solution

\(rc=3\times0.2\times10^{-6}=6\times10^{-7}\ \mathrm{s/km^2}\). \(rcL^2=6\times10^{-7}\times1.6\times10^{7}=9.6\ \mathrm s\). Doubling \(L\) multiplies the time by 4. Signals needed several seconds to build up, limiting transatlantic rates to a few words a minute in the first cables.

Exercise 9.2Show that \(u(x,t)=\dfrac{Q}{c\sqrt{4\pi Dt}}\,e^{-x^2/(4Dt)}\) solves the diffusion equation, and that the total charge \(\int cu\,dx\) is constant.

Show solution

Write \(u=At^{-1/2}e^{-x^2/(4Dt)}\). Then \(u_t=u\left(-\frac1{2t}+\frac{x^2}{4Dt^2}\right)\) and \(u_{xx}=u\left(\frac{x^2}{4D^2t^2}-\frac1{2Dt}\right)\), so \(u_t=Du_{xx}\). The Gaussian integral gives \(\int cu\,dx=Q\) for every \(t\).

Lesson 10

Ohm’s acoustic law

For the general reader. In 1843 Ohm turned to sound. He proposed that the ear hears a tone as a sum of pure sine tones, the components of its Fourier series, and that it can pick out each one. August Seebeck objected with experiments on sirens. The dispute was bitter, and Ohm withdrew from it, but Hermann von Helmholtz later made Ohm’s idea the foundation of his theory of hearing.

Goals

State Ohm’s acoustic law and its basis in Fourier analysis.
Understand Seebeck’s objection and the missing fundamental.
Compute Fourier components of simple waveforms.

Fourier’s theorem (for periodic signals)

A square-integrable function \(p(t)\) of period \(T\) can be written in the mean-square sense as

\[ p(t)=\frac{a_0}2+\sum_{n=1}^{\infty}\big(a_n\cos n\omega t+b_n\sin n\omega t\big),\qquad \omega=\frac{2\pi}T, \]

with \(a_n=\frac2T\int_0^Tp(t)\cos n\omega t\,dt\) and \(b_n=\frac2T\int_0^Tp(t)\sin n\omega t\,dt\).

Ohm’s acoustic law (1843)

The ear perceives a simple tone of frequency \(n\omega\) in a periodic sound only if the sound’s Fourier series contains a component \(a_n\cos n\omega t+b_n\sin n\omega t\) with non-zero amplitude. A complex tone is analysed by the ear into its sinusoidal components.

The dispute

Seebeck, son of the discoverer of thermoelectricity, used sirens that produced trains of pulses. He found that listeners heard a strong pitch at the repetition frequency even when the Fourier component at that frequency was weak or absent. He concluded that the ear does not simply analyse sounds into sines. Ohm’s replies relied on assumptions he could not fully justify, and he eventually conceded. In 1863 Helmholtz’s Die Lehre von den Tonempfindungen took Ohm’s law as its starting point and explained timbre by the strengths of harmonics. The pitch of the “missing fundamental” that Seebeck had heard was later understood as the brain’s inference from the pattern of harmonics, so both men had seen part of the truth.

Exercise 10.1Find the Fourier series of the square wave \(p(t)=1\) for \(0\lt t\lt T/2\), \(p(t)=-1\) for \(T/2\lt t\lt T\). Which harmonics would Ohm say the ear hears?

Show solution

\(a_n=0\) by oddness about \(t=0\) after extension; \(b_n=\frac{2}{T}\left(\int_0^{T/2}-\int_{T/2}^T\right)\sin n\omega t\,dt=\frac{2}{n\pi}(1-\cos n\pi)\). So \(b_n=4/(n\pi)\) for odd \(n\), 0 for even \(n\): \(p=\frac4\pi\left(\sin\omega t+\frac13\sin3\omega t+\frac15\sin5\omega t+\cdots\right)\). Ohm: the ear hears the fundamental and the odd harmonics, with amplitudes falling as \(1/n\), which is why a square wave sounds hollow, like a clarinet.

Exercise 10.2A signal is \(p(t)=\cos(2\pi\cdot400t)+\cos(2\pi\cdot600t)+\cos(2\pi\cdot800t)\). What is its period, and is there a Fourier component at the fundamental? What pitch do listeners report?

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The frequencies share greatest common divisor 200 Hz, so the period is 5 ms and the fundamental is 200 Hz. Its Fourier coefficient is zero. Listeners nevertheless report a pitch of 200 Hz: the missing fundamental, the effect that challenged a literal reading of Ohm’s law.

Part III · Life and legacy

Lesson 11

Rejection and recognition, 1827–1849

Ohm had hoped Die galvanische Kette would bring him a university chair. It did not. German physics in the 1820s was dominated by experimenters suspicious of mathematical theory, and by followers of the speculative Naturphilosophie, who preferred grand principles to measurement. One reviewer dismissed the book as “a web of naked fancies”. Ohm also quarrelled with the physicist Georg Friedrich Pohl and with the Prussian education official Johannes Schulze, which damaged his standing in Berlin.

When his leave ended in 1827 no university offer had come. In March 1828 he resigned from Cologne, and he spent the next five years teaching in Berlin schools, including the military school, on a modest income. In 1833 he accepted a professorship at the Royal Polytechnic School in Nuremberg, back in Bavaria. It carried the title of professor but was not a university. He became its director in 1839.

Recognition

Recognition came from abroad. Claude Pouillet in Paris confirmed the law with careful measurements, and Wheatstone promoted it in England. In 1841 the Royal Society awarded Ohm its Copley Medal; it elected him a Foreign Member in 1842. Wheatstone’s 1843 Bakerian Lecture built its measurements of resistance on Ohm’s law. At home, the Bavarian Academy of Sciences elected him a full member in 1845. In 1849 he published Beiträge zur Molecular-Physik, the first part of an intended larger work on molecular physics.

DiscussionOhm’s law was correct, carefully tested and clearly written, yet it took a decade to be accepted in Germany. Which factors mattered most: the style of argument, the author’s position, the instruments, or the state of the field? Compare with the reception of Fourier’s own theory of heat.

Lesson 12

Munich, death and legacy, 1849–1854

In 1849 King Maximilian II brought Ohm to Munich as curator of the Bavarian Academy’s physical collection and extraordinary professor at the university. In 1852, at 63, he received what he had wanted all his life: the chair of experimental physics at the University of Munich. He also advised the Bavarian telegraph administration. He wrote a physics textbook, Grundzüge der Physik (1854), and work on interference in crystal plates. He died in Munich on 6 July 1854, aged 65, and is buried in the Alter Südfriedhof there.

Legacy

The law: \(V=IR\) for circuits and \(\mathbf J=\sigma\mathbf E\) for continua remain the starting point of electrical engineering and of transport theory in condensed matter.
Method: Ohm brought Fourier’s style of mathematical physics into electricity, defining current, tension and resistance as measurable quantities related by an equation.
Descendants: Kirchhoff’s network laws, Wheatstone’s bridge, Thomson’s cable theory and the whole telegraph industry relied on the law.
Acoustics: his acoustic law, refined by Helmholtz, founded the analysis of timbre and auditory frequency analysis.
Units: in 1881 the International Electrical Congress in Paris named the unit of resistance the ohm. The unit of conductance was later called the mho, his name reversed, before the siemens replaced it.
Memorials: a monument in front of the Technical University of Munich, a bust in the Munich Hall of Fame, and the Technische Hochschule Nürnberg Georg Simon Ohm.

Final projectReproduce Ohm’s 1826 experiment with modern parts: a thermocouple (or a weak cell with a large series resistor to mimic it), wires of eight lengths, and a sensitive ammeter or a compass needle under the wire. Fit \(1/I\) against length, extract \(\mathcal E\) and the internal resistance, and discuss what limits your accuracy compared with his.

Reference

Sources and further reading

Biography

J. J. O’Connor and E. F. Robertson, Georg Simon Ohm, MacTutor History of Mathematics.
Georg Ohm, Wikipedia; and Georg Simon Ohm, Wikipedia (German).
K. L. Caneva, Ohm, Georg Simon, in Dictionary of Scientific Biography, vol. 10, Scribner, 1974.
K. Jäger, F. Heilbronner (eds.), Lexikon der Elektrotechniker, VDE Verlag, 2010.
Universität Flensburg, Histolab, Ohm’s torsion balance (replication of the 1826 apparatus).

Physics

G. S. Ohm, Vorläufige Anzeige des Gesetzes, nach welchem Metalle die Contaktelektricität leiten, Ann. Phys. Chem. 4 (1825) 79–88.
G. S. Ohm, Bestimmung des Gesetzes, nach welchem Metalle die Contaktelektricität leiten, Journal für Chemie und Physik (Schweigger), 1826.
G. S. Ohm, Die galvanische Kette, mathematisch bearbeitet, Berlin, 1827. English translation by W. Francis, The Galvanic Circuit Investigated Mathematically, Van Nostrand, 1891.
Ohm’s law, Wikipedia (history section, Kirchhoff and Drude forms).
G. Kirchhoff, Ueber die Auflösung der Gleichungen, auf welche man bei der Untersuchung der linearen Vertheilung galvanischer Ströme geführt wird, Ann. Phys. 72 (1847) 497–508.
W. Thomson, On the theory of the electric telegraph, Proc. R. Soc. 7 (1855) 382–399.
R. S. Turner, The Ohm–Seebeck dispute, Hermann von Helmholtz, and the origins of physiological acoustics, Brit. J. Hist. Sci. 10 (1977) 1–24. Journal page.
N. W. Ashcroft and N. D. Mermin, Solid State Physics, Saunders, 1976, ch. 1 (the Drude model).