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Isaac Barrow

Isaac Barrow, first Lucasian Professor of Mathematics

A course in two strands: the life of a Royalist classicist, traveller and preacher who became Cambridge’s first professor of mathematics, and the geometric lectures in which tangents and areas were first shown to be inverse problems.

Born October 1630, London Died 4 May 1677, London Lucasian Professor Cambridge, 1663–1669 Level advanced undergraduate
Lectiones geometricae, Lecture X, Prop. 11, redrawn. The ordinate of the upper curve measures the shaded area; its tangent at F meets the axis at T with DT : DF = R : DE.
Lesson 00

How to use this course

Isaac Barrow is remembered mostly as the teacher who stepped aside for Isaac Newton. That is a small part of his story. He was a professor of Greek before he was a professor of mathematics, a celebrated preacher, the editor of Euclid and Archimedes for English students, and the Master of Trinity College who commissioned its great library. In mathematics he wrote the most advanced geometric treatment of tangents and areas before the calculus, including a clear proof that finding tangents and finding areas are inverse problems.

Prerequisites for Part II
  • Single-variable calculus: derivatives, integrals, implicit differentiation.
  • Euclidean geometry of similar triangles and conic sections.
  • For Lesson 10, Snell’s law of refraction.

Each technical lesson opens with a paragraph for the general reader, then states Barrow’s results in modern notation, sketches his arguments, and ends with exercises whose solutions are folded. Barrow wrote in words and figures, never in formulas. Translating his propositions into symbols makes them easier to check but hides how hard they were to see. Keep that cost in mind as you read.

Reference

Timeline

1630Born in October in London, son of Thomas Barrow, a linen draper, and Ann Buggin.
c. 1634His mother dies; he is sent to his grandfather in Cambridgeshire.
1640sCharterhouse, then Felsted School in Essex under Martin Holbeach.
1646Enters Trinity College, Cambridge.
1649BA; elected fellow of Trinity.
1652MA, as a pupil of the classicist James Duport.
1655Refuses to sign the Engagement to the Commonwealth. Latin edition of Euclid’s Elements. Leaves for France.
1655–59Paris, Florence, Smyrna, Constantinople, Venice; returns through Germany and Holland.
1660Ordained; Regius Professor of Greek at Cambridge. English edition of Euclid.
1662Professor of Geometry at Gresham College, London.
1663Fellow of the Royal Society (20 May). First Lucasian Professor of Mathematics.
1664–66Lucasian lectures on the foundations of mathematics, later printed as Lectiones mathematicae.
1669Lectiones opticae. Sends Newton’s De analysi to John Collins. Resigns the Lucasian chair; Newton succeeds him.
1670Lectiones geometricae. Doctor of Divinity by royal mandate; chaplain to Charles II.
1673Appointed Master of Trinity College by the king (February 1672 in the old-style calendar).
1675Vice-Chancellor of the University. Editions of Archimedes, Apollonius’ Conics I–IV and Theodosius’ Spherics.
1676Building of the Wren Library begins, to a design by his friend Christopher Wren.
1677Dies 4 May in London of a fever, aged 46. Buried in Westminster Abbey.
1683Lectiones mathematicae published posthumously.
Part I · Formation
Lesson 01

London and the schools, 1630–1646

Isaac Barrow was born in London in October 1630. His father, Thomas Barrow, was a linen draper with connections at court; his mother, Ann Buggin, came from a Kent family. She died when Isaac was about four. His father remarried and sent the boy to live with his grandfather, also named Isaac, a justice of the peace at Spinney Abbey in Cambridgeshire.

He first went to Charterhouse in London. By the family’s later account he learned little there and was mainly known for fighting. His father moved him to Felsted School in Essex, where the Puritan headmaster Martin Holbeach gave him a solid grounding in Latin, Greek, Hebrew and logic. The change worked. Barrow became a strong classical scholar and remained one all his life.

The Civil War broke out when he was eleven. The Barrows were Royalists, and his father followed the king’s cause at considerable cost. An uncle who was a fellow of Peterhouse arranged a place for him there in 1643, but the uncle lost his fellowship in the parliamentary purge of the university and the place fell through. In 1646 Barrow entered Trinity College instead.

Lesson 02

Trinity in the Commonwealth, 1646–1655

Trinity under Parliament was not a comfortable place for an outspoken Royalist. Barrow survived through ability and through patrons. His tutor was James Duport, the Regius Professor of Greek, who protected him and directed his reading. Barrow took his BA and was elected a fellow of the college in 1649. He became MA in 1652.

His interests were broad. He read the new philosophy of René Descartes and Francis Bacon, studied anatomy and botany, and considered a career in medicine. Mathematics entered gradually, first as part of the study of astronomy and chronology, then for its own sake. In 1655 he published a compact Latin edition of all fifteen books of Euclid’s Elements, designed for students. Its economy made it the standard English school Euclid for roughly half a century; an English version followed in 1660 and an edition of Euclid’s Data in 1657.

In 1655 the Commonwealth required fellows to sign the Engagement, a pledge of loyalty to a government without king or House of Lords. Barrow refused. Instead of expulsion he received a university travelling award, which conveniently took him out of the country.

Lesson 03

The grand tour, 1655–1660

Barrow spent about four years abroad. In Paris, for some ten months, he found few mathematicians to talk to besides Gilles Personne de Roberval, whose kinematic method of tangents shaped Barrow’s later lectures (Lesson 05). From February 1656 he spent about eight months in Florence, where he met Vincenzo Viviani, the last pupil of Galileo, and studied the Medici collections. Plague prevented a visit to Rome.

He sailed for the East. On the voyage his ship was attacked by pirates, and Barrow is said to have helped defend it. He spent seven months at Smyrna and about a year and a half in Constantinople as a guest of the English ambassador, reading the Greek church fathers and writing a Latin poem on the religion of the Turks dated 1658. On the way home his ship caught fire in Venice and he lost most of his possessions. He returned through Germany and Holland and reached Cambridge in September 1659.

The Restoration of Charles II in 1660 transformed his position. He was ordained, and in 1660 he succeeded Duport as Regius Professor of Greek. Two years later, in 1662, he was also elected Professor of Geometry at Gresham College in London, a post with a salary of fifty pounds a year and a public audience. His career as a mathematician had begun.

Part II · The mathematics
Lesson 04

Euclid, magnitude and proportion

For the general reader. Before calculus, the language of rigorous mathematics in England was Euclid’s geometry. Barrow made Euclid accessible in a pocket edition, and in his first Lucasian lectures he argued that geometry is the foundation of all mathematics, including arithmetic. His argument depends on one subtle idea: how to compare quantities whose ratio is not a fraction.

Goals
  • Understand Barrow’s view of geometry as the basic mathematical science.
  • State and use Eudoxus’ definition of equal ratios (Elements V, Def. 5).
  • Connect that definition to the modern real number.

Continuous magnitude

In the lectures of 1664–1666, printed in 1683 as Lectiones mathematicae, Barrow treated the nature of mathematics, its objects and its methods. He held that number is an abstraction from continuous magnitude, so arithmetic and algebra are parts of geometry, not its rivals. Algebra, in his view, was a method of reasoning, closer to logic than to a separate science. This put him at odds with the algebraists of his day, including John Wallis, who saw arithmetic as prior. A large part of the lectures defends Euclid’s theory of ratio and proportion against its critics.

Definition (Euclid V, Def. 5, after Eudoxus)

Let \(a,b\) be magnitudes of one kind and \(c,d\) magnitudes of one kind. Then \(a:b=c:d\) if for all positive integers \(m,n\):

\[ ma>nb \iff mc>nd,\qquad ma=nb \iff mc=nd,\qquad ma<nb \iff mc<nd. \]

The definition never mentions what a ratio “is”. It says two ratios are equal when every rational number \(n/m\) lies on the same side of both. In 1872 Richard Dedekind turned this into a construction: a real number is the cut it makes in the rationals. Barrow’s defence of Euclid was a defence of exactly this idea.

Why the definition is needed

Proposition (Euclid VI.1)

Triangles with the same height are to one another as their bases.

Proof sketch. Let the triangles have bases \(b_1,b_2\) and areas \(A_1,A_2\). Lay off \(m\) copies of \(b_1\) and \(n\) copies of \(b_2\) along the base line, and join each division point to the apex. The \(m\) triangles on the first segment all have area \(A_1\), so \(mb_1\) carries area \(mA_1\); likewise \(nb_2\) carries \(nA_2\). Triangles of equal height with larger base have larger area, so \(mb_1\gtreqless nb_2\) exactly when \(mA_1\gtreqless nA_2\). By Definition 5, \(b_1:b_2=A_1:A_2\). No assumption that the bases are commensurable is needed.

Exercise 4.1Show that the diagonal and side of a square are incommensurable, that is, no segment measures both a whole number of times.

Show solution
If a unit measured the side \(s=q\) times and the diagonal \(d=p\) times, then \(d^2=2s^2\) gives \(p^2=2q^2\). Take \(p,q\) with no common factor. Then \(p^2\) is even, so \(p=2r\), and \(4r^2=2q^2\) gives \(q^2=2r^2\), so \(q\) is even too, a contradiction. Hence ratios such as \(d:s\) need Definition 5.

Exercise 4.2Using Definition 5, prove that if \(a:b=c:d\) and \(a>b\) then \(c>d\).

Show solution
Take \(m=n=1\). Since \(a>b\), the definition forces \(c>d\).

Exercise 4.3Let \(s,d\) be the side and diagonal of one square and \(s',d'\) those of another. Show \(d:s=d':s'\) directly from Definition 5.

Show solution
For both squares \(d^2=2s^2\) and \(d'^2=2s'^2\). For positive integers \(m,n\): \(md>ns \iff m^2d^2>n^2s^2 \iff 2m^2>n^2\), and the same chain holds for the primed square. Equality never occurs because \(2m^2\neq n^2\). So the three conditions agree and the ratios are equal.
Lesson 05

Curves generated by motion

For the general reader. Barrow opened his geometry lectures by describing curves as traced by moving points, with time as an independent, uniformly flowing quantity. This picture, learned partly from Roberval and Galileo, passed directly to Newton, whose “fluxions” are the speeds of such flowing quantities.

Goals
  • Find tangents as directions of a composed velocity.
  • Read the area under a velocity curve as distance travelled.
  • Apply the method to the parabola, cycloid and spiral.

Time as a line

The first lectures of Lectiones geometricae (1670) discuss how magnitudes are generated: lines by moving points, surfaces by moving lines. Barrow argued that time, though not itself a motion, can be represented by a straight line, since it has length, uniform succession and divisibility. If a point moves along one line while that line itself moves, the traced curve has at each moment a direction given by combining the two velocities.

Principle of composed motion

If a point \(P(t)=(x(t),y(t))\) is driven by two simultaneous motions with velocities \(\dot x\) and \(\dot y\) along perpendicular directions, the tangent to its path at \(P\) is the line through \(P\) in the direction \((\dot x,\dot y)\). Its subtangent, the segment of the axis between the foot of the ordinate and the tangent’s intercept, is \(t=y\,\dot x/\dot y\).

This is Roberval’s rule. The difficulty in using it is finding the right decomposition of motion, and Barrow’s lectures contain many ingenious ones. They also contain the converse use: if a speed is represented by an ordinate, the space traversed is represented by the area under the speed curve. Galileo had used this for uniformly accelerated fall. Barrow made it general, and it leads straight to Lesson 07.

Exercise 5.1A point moves along the parabola \(y^2=px\) so that \(y\) increases uniformly. Find the subtangent and give Apollonius’ classical description of it.

Show solution
Take \(y=ct\), so \(x=c^2t^2/p\), \(\dot x=2c^2t/p\), \(\dot y=c\). Then \(t_{\rm sub}=y\dot x/\dot y=ct\cdot 2ct/p=2x\). The tangent meets the axis at distance \(x\) behind the vertex: the vertex bisects the subtangent, as in Apollonius, Conics I.33.

Exercise 5.2A circle of radius \(r\) rolls along a line with centre speed \(v\). Show that the velocity of a point on the rim is perpendicular to the line joining it to the contact point, so the tangent to the cycloid passes through the top of the rolling circle.

Show solution
The velocity is the sum of translation \(v\) along the line and rotation with angular speed \(v/r\) about the centre. Equivalently, the circle rotates instantaneously about the contact point \(C\) with angular speed \(v/r\), so every point’s velocity is perpendicular to its radius from \(C\). The diameter through \(C\) ends at the top point \(A\); by Thales, the angle \(CPA\) is right, so the velocity at \(P\) points along \(PA\).

Exercise 5.3A body falls from rest with speed \(v=gt\). Represent time on a horizontal line and speed by ordinates. Show that the distance fallen is \(\tfrac12gt^2\) by Galileo’s area argument.

Show solution
The speed graph is a straight line through the origin. In each small time interval the distance is speed times time, an ordinate times a small width, so the total distance is the area of the triangle with base \(t\) and height \(gt\), namely \(\tfrac12gt^2\).
Lesson 06

The differential triangle

For the general reader. To find the slope of a curve at a point, take a second point very close to it. The two points and the lines joining them form a tiny triangle, nearly similar to the large triangle made by the tangent and the axis. Barrow turned this observation into a mechanical procedure for tangents, the closest thing in his lectures to differential calculus.

Goals
  • State Barrow’s rules for the method of tangents.
  • Apply them to algebraic curves, including the kappa curve.
  • Compare them with Fermat’s method and with implicit differentiation.

The rules of Lecture X

At the end of Lecture X, Barrow added a method he said he used on the advice of a friend, widely taken to be Newton. Let \(M\) be a point of the curve with abscissa \(x\) and ordinate \(y=MP\). Let \(N\) be a nearby point of the curve with abscissa \(x-e\) and ordinate \(y-a\). The small triangle with sides \(e\), \(a\) and the arc \(MN\) is nearly similar to the triangle \(TPM\) formed by the tangent, so \(a:e\approx MP:PT\).

Barrow’s rules (Lectiones geometricae, 1670)

Substitute \(x-e\) for \(x\) and \(y-a\) for \(y\) in the equation of the curve. Then:

  1. omit all terms containing a power of \(a\) or \(e\), or their product;
  2. omit all terms containing neither \(a\) nor \(e\), which cancel by the equation of the curve;
  3. replace \(a\) by the ordinate \(y\) and \(e\) by the subtangent \(t\), and solve for \(t\).

Rule 3 uses the similar triangles: \(a/e=y/t\). In modern terms rules 1 and 2 extract the linear part of \(F(x-e,y-a)-F(x,y)\), namely \(-eF_x-aF_y\), so \(a/e=-F_x/F_y=dy/dx\) and \(t=y/(dy/dx)\). Pierre de Fermat had used a similar device with one increment; Barrow’s use of two increments and the triangle makes the geometry explicit. The same small triangle, under the name “characteristic triangle”, became central for Gottfried Wilhelm Leibniz.

Exercise 6.1Use Barrow’s rules on \(x^3+y^3=r^3\).

Show solution
\((x-e)^3+(y-a)^3=r^3\). Rule 1 leaves \(x^3-3x^2e+y^3-3y^2a=r^3\); rule 2 leaves \(-3x^2e-3y^2a=0\). So \(a/e=-x^2/y^2\), and rule 3 gives \(y/t=-x^2/y^2\), \(t=-y^3/x^2\). The sign says the tangent meets the axis on the far side of the ordinate, since the curve is decreasing.

Exercise 6.2Barrow was the first to find tangents to the kappa curve \(x^2(x^2+y^2)=c^2y^2\). Find its subtangent.

Show solution
Write \(F=x^4+x^2y^2-c^2y^2\). The linear terms in \(e,a\) are \(-e(4x^3+2xy^2)-a(2x^2y-2c^2y)\). Setting them to zero, \(a/e=\dfrac{2x^3+xy^2}{y(c^2-x^2)}\), so \(t=y\,e/a=\dfrac{y^2(c^2-x^2)}{x(2x^2+y^2)}\).

Exercise 6.3Explain why rule 1 is legitimate, and why it is not a proof by modern standards.

Show solution
Dividing the remaining equation by \(e\), the omitted terms are of order \(e\) or \(a\) (with \(a/e\) bounded near a point where the tangent is not vertical), so they vanish as \(N\to M\). Barrow simply dropped them as “infinitely small” without a limit argument. The justification came with Cauchy’s limits in the 1820s.
Lesson 07

Tangents and areas are inverse

For the general reader. The fundamental theorem of calculus says that finding slopes and finding areas undo each other. Barrow proved a geometric version in 1670, before Newton or Leibniz published anything. He did not see it as the key to a new algorithm, and that is the difference between his work and theirs.

Goals
  • State Lecture X, Proposition 11 in Barrow’s terms and in modern form.
  • Follow his proof by comparison of areas.
  • Use the theorem to compute subtangents of area curves.

The statement

Let \(ZGE\) be a curve above an axis \(AD\), with ordinates increasing from \(A\). Fix a length \(R\). Build a second curve \(AIF\) whose ordinate at each point \(D\) is chosen so that the rectangle \(R\times DF\) equals the area under \(ZGE\) from \(A\) to \(D\). Barrow proves that if \(T\) on the axis is chosen with \(DE:DF=R:DT\), then the line \(TF\) touches the curve \(AIF\) at \(F\). The hero figure shows this configuration.

Theorem (Barrow, Lectiones geometricae X.11, modern form)

Let \(f\) be continuous and increasing on \([0,b]\), with \(f\ge0\), and let \(R>0\). Define \(z(x)=\frac1R\int_0^x f(s)\,ds\). Then at every \(x\in(0,b)\) the line through \((x,z(x))\) with slope \(f(x)/R\) is tangent to the graph of \(z\). Equivalently \(z'(x)=f(x)/R\), and the subtangent is \(Rz(x)/f(x)\).

Barrow’s proof

Let \(y=f(x)\) and let \(\ell(s)=z(x)+\frac{y}{R}(s-x)\) be the proposed tangent line. For \(s<x\), the area from \(s\) to \(x\) is at most \((x-s)y\) because \(f\) is increasing, so

\[ R\big(z(x)-z(s)\big)=\int_s^x f\le (x-s)\,y \quad\Longrightarrow\quad z(s)\ge \ell(s). \]

For \(s>x\), the area from \(x\) to \(s\) is at least \((s-x)y\), so again \(z(s)\ge\ell(s)\). The line meets the curve at \(F\) and lies on one side of it everywhere else, so it touches the curve at \(F\). This is precisely Barrow’s argument, phrased as a comparison of a curvilinear area with a rectangle. Monotonicity does the work that continuity and limits do in the modern proof.

Barrow presented this as one proposition among many. He also gave results in the opposite direction, deducing areas from known tangents. What he lacked was the idea that the inverse relation could be the basis of a general, symbolic method for computing areas. Newton and Leibniz supplied that.

Exercise 7.1Let \(f(x)=x^2\) and \(R=1\). Compute the subtangent of the area curve directly and by Barrow’s rule.

Show solution
\(z=x^3/3\), so directly \(t=z/z'=(x^3/3)/x^2=x/3\). Barrow’s rule gives \(Rz/y=(x^3/3)/x^2=x/3\).

Exercise 7.2Adapt the proof to a decreasing \(f\). On which side of the curve does the line now lie?

Show solution
For \(s<x\), \(\int_s^x f\ge(x-s)y\), so \(z(s)\le\ell(s)\); for \(s>x\), \(\int_x^s f\le (s-x)y\), so again \(z(s)\le\ell(s)\). The line lies above the area curve, which is now concave, and still touches it only at \(F\).

Exercise 7.3Use the theorem backwards: the curve \(z=x^{n+1}/(n+1)\) has subtangent \(x/(n+1)\). Deduce the area under \(y=x^n\).

Show solution
With \(R=1\), Barrow’s relation \(t=z/y\) holds for the area curve. For \(z=x^{n+1}/(n+1)\), \(z/t=x^n\), so this \(z\) has the property of the area curve for \(y=x^n\). Both vanish at \(0\), so the area from \(0\) to \(x\) is \(x^{n+1}/(n+1)\). (Uniqueness needs the fact that two curves with the same slope everywhere and the same starting value coincide.)
Lesson 08

Rules of the calculus in geometric dress

For the general reader. Lectures XI and XII contain dozens of propositions about tangents, areas and arc lengths. In 1916 the historian J. M. Child translated them and argued that many are geometric versions of standard rules of calculus. Not every historian accepts his strongest claims, but the correspondence is real and instructive.

Goals
  • Prove Barrow’s theorem on the area under the subnormal.
  • Read area transformations as substitution and integration by parts.
  • Judge what “anticipating the calculus” does and does not mean.

The subnormal theorem

At a point \((x,y)\) of a curve the normal meets the axis at a point whose distance from the foot of the ordinate is the subnormal \(n=y\,dy/dx\).

Theorem (Lectiones geometricae XI.1, modern form)

Let \(y=g(x)\) be increasing and differentiable on \([0,x]\) with \(g(0)=0\). If ordinates equal to the subnormals of \(g\) are erected along the axis, the area under the resulting curve from \(0\) to \(x\) equals \(\tfrac12 g(x)^2\).

In symbols, \(\int_0^x g\,g'\,ds=\tfrac12 g(x)^2\). Barrow’s proof divides the axis into small parts, uses the similarity of the differential triangle with the triangle formed by ordinate, normal and subnormal, and sums. Each small strip under the subnormal curve has area \(n\,\Delta x\approx y\,\Delta y\), and the sum of \(y\,\Delta y\) is the area of a right isosceles triangle with legs \(y\).

Transformations of area

Seventeenth-century geometers changed one area into another by moving strips. Slicing a region under a curve by vertical strips or by horizontal strips gives the same area. In symbols, for increasing \(g\) with \(g(0)=0\),

\[ \int_0^a g(x)\,dx+\int_0^{g(a)} g^{-1}(y)\,dy = a\,g(a), \]

which is integration by parts, \(\int g\,dx=xg-\int x\,dg\). Child identified propositions of this kind throughout Lectures XI and XII. Historians such as Michael Mahoney have stressed that Barrow framed them as theorems about particular configurations, without the general notation that turns them into rules.

Exercise 8.1Check the subnormal theorem for \(g(x)=x^2\).

Show solution
The subnormal is \(gg'=2x^3\). Its area from \(0\) to \(x\) is \(x^4/2=\tfrac12(x^2)^2\).

Exercise 8.2For the quarter circle \(y=\sqrt{r^2-x^2}\), \(0\le x\le r\), show the subnormal has length \(x\) and interpret the theorem for this decreasing curve.

Show solution
\(yy'=-x\), so the subnormal has length \(x\), pointing back toward the centre (the normal of a circle passes through the centre). The area under \(|n|=x\) from \(0\) to \(x\) is \(x^2/2=\tfrac12(r^2-y^2)\), which is \(\tfrac12(g(0)^2-g(x)^2)\): the theorem with a sign change for decreasing \(g\).

Exercise 8.3Use the strip identity to find \(\int_0^1\sqrt{x}\,dx\) from \(\int_0^1 y^2\,dy=\tfrac13\).

Show solution
With \(g(x)=\sqrt x\), \(g^{-1}(y)=y^2\), \(a=1\), \(g(a)=1\): \(\int_0^1\sqrt x\,dx=1-\tfrac13=\tfrac23\).
Lesson 09

The secant integral and Mercator’s chart

For the general reader. On Mercator’s map a straight line is a course of constant compass bearing, which is why sailors loved it. To draw the map one needs the integral of the secant function. Barrow gave the first proof of its formula that a modern reader can follow, and with it the first use of partial fractions in integration.

Goals
  • Derive the spacing of parallels on a Mercator chart.
  • Integrate the secant by partial fractions, following Barrow’s route.
  • Know the history: Wright, Bond, Gregory, Barrow.

Why the secant

On a sphere of radius \(R\), a parallel at latitude \(\varphi\) has length \(2\pi R\cos\varphi\). A cylindrical map stretches it to the equator’s length, a factor \(\sec\varphi\). To keep angles true, north–south distances must be stretched by the same factor, so the map coordinate of latitude \(\varphi\) is

\[ Y(\varphi)=R\int_0^{\varphi}\sec\theta\,d\theta . \]

Edward Wright tabulated this by summing secants in 1599. In the 1640s the navigation teacher Henry Bond noticed that Wright’s table agreed with a table of \(\log\tan(45^\circ+\varphi/2)\). James Gregory proved the conjecture in 1668 in an argument that is very hard to read. Barrow’s proof in Lectiones geometricae (1670) is geometric but transparent.

Theorem

For \(|\varphi|<\pi/2\),

\[ \int_0^{\varphi}\sec\theta\,d\theta=\tfrac12\ln\frac{1+\sin\varphi}{1-\sin\varphi}=\ln\tan\Big(\frac{\pi}{4}+\frac{\varphi}{2}\Big)=\ln(\sec\varphi+\tan\varphi). \]

Barrow’s route in modern notation

Write \(\sec\theta=\cos\theta/\cos^2\theta=\cos\theta/(1-\sin^2\theta)\) and put \(u=\sin\theta\), \(du=\cos\theta\,d\theta\). Then split the integrand into partial fractions:

\[ \int_0^{\sin\varphi}\frac{du}{1-u^2}=\frac12\int_0^{\sin\varphi}\Big(\frac1{1+u}+\frac1{1-u}\Big)du=\frac12\ln\frac{1+\sin\varphi}{1-\sin\varphi}. \]

Barrow expressed the logarithms as areas under a hyperbola, the standard representation since Grégoire de Saint-Vincent and Alphonse Antonio de Sarasa.

Exercise 9.1Show that the three closed forms in the theorem agree.

Show solution
Multiply numerator and denominator by \(1+\sin\varphi\): \(\frac{1+\sin\varphi}{1-\sin\varphi}=\frac{(1+\sin\varphi)^2}{\cos^2\varphi}=(\sec\varphi+\tan\varphi)^2\), giving the third form. For the second, with \(h=\varphi/2\), \(\tan(\pi/4+h)=\frac{1+\tan h}{1-\tan h}=\frac{\cos h+\sin h}{\cos h-\sin h}\); squaring gives \(\frac{1+\sin\varphi}{1-\sin\varphi}\).

Exercise 9.2On a Mercator chart with \(R=1\), how far from the equator is the parallel at \(60^\circ\)? Compare with its true arc distance.

Show solution
\(Y=\ln(\sec60^\circ+\tan60^\circ)=\ln(2+\sqrt3)\approx1.317\). The true distance is \(\pi/3\approx1.047\), so the chart stretches it by about 26%.
Lesson 10

Lectures on optics

For the general reader. Barrow’s optical lectures of 1669 were the most complete treatment of geometrical optics in England at the time. He made precise where rays reflected or refracted by curved surfaces meet, worked on lenses, and gave a simpler account of René Descartes’ explanation of the rainbow. Newton read the proofs, and his own optical work began where Barrow’s ended.

Goals
  • Locate the focus of rays reflected from a spherical mirror.
  • Use the paraxial lens formula.
  • Derive the angle of the primary rainbow from minimum deviation.

Foci and images

Lectiones opticae (1669, revised 1674) start from rectilinear propagation, the law of reflection and the sine law of refraction. Barrow defined the geometrical focus as the point where a pencil of rays from one point meets again after reflection or refraction, and he traced how that point moves as the rays depart from the axis. For narrow pencils near the axis his constructions agree with the modern paraxial formulas.

Paraxial results

A spherical mirror of radius \(r\) brings rays parallel to its axis to a focus at distance \(r/2\) from the mirror. A thin lens of focal length \(f\) forms the image of a point at distance \(u\) at distance \(v\) on the other side, with \(\frac1u+\frac1v=\frac1f\).

The rainbow

Minimum deviation (Descartes, simplified by Barrow)

A ray entering a spherical drop of refractive index \(n\) at incidence \(i\), refracted at angle \(r\) with \(\sin i=n\sin r\), and reflected once inside, is turned through \(D(i)=\pi+2i-4r\). \(D\) is stationary when \(\cos^2 i=(n^2-1)/3\). Rays pile up at this angle, producing the bright bow at \(\pi-D_{\min}\) from the antisolar point.

Derivation. \(dD/di=2-4\,dr/di\), and differentiating Snell’s law, \(\cos i=n\cos r\,dr/di\). Setting \(dr/di=\tfrac12\) gives \(\cos i=\tfrac n2\cos r\), so \(4\cos^2 i=n^2-n^2\sin^2r=n^2-\sin^2 i=n^2-1+\cos^2 i\), hence \(\cos^2 i=(n^2-1)/3\).

Exercise 10.1Compute the rainbow angle for water, \(n=4/3\).

Show solution
\(\cos^2i=(16/9-1)/3=7/27\), so \(i\approx59.4^\circ\). Then \(\sin r=\sin i/n\approx0.646\), \(r\approx40.2^\circ\). \(D=180^\circ+118.8^\circ-160.8^\circ\approx138.0^\circ\), so the bow appears about \(42^\circ\) from the antisolar point.

Exercise 10.2A ray parallel to the axis at height \(h=r\sin\theta\) reflects from a concave spherical mirror of radius \(r\). Show it crosses the axis at distance \(r/(2\cos\theta)\) from the centre. What happens as \(\theta\to0\)?

Show solution
The normal at the point of incidence passes through the centre \(C\) and makes angle \(\theta\) with the axis. The reflected ray makes angle \(\theta\) with the normal on the other side, so the triangle formed by \(C\), the point of incidence and the crossing point \(F\) is isosceles with base \(r\) and base angles \(\theta\). Hence \(CF=r/(2\cos\theta)\). As \(\theta\to0\), \(CF\to r/2\), the paraxial focus; for larger \(\theta\) the rays cross nearer the mirror (spherical aberration), which is why Barrow distinguished the focus of a narrow pencil.

Exercise 10.3An object stands 30 cm from a thin converging lens of focal length 10 cm. Where is the image, and what is the magnification?

Show solution
\(1/v=1/10-1/30=1/15\), so \(v=15\) cm on the far side. Magnification \(-v/u=-\tfrac12\): real, inverted, half size.
Part III · Life and legacy
Lesson 11

The Lucasian chair and Newton, 1663–1669

In 1663 Henry Lucas, a former member of Parliament for the university, left money to found a professorship of mathematics at Cambridge. Barrow was elected its first holder and gave up the Greek chair. He was elected a Fellow of the newly chartered Royal Society on 20 May 1663. His Lucasian lectures of 1664–1666 dealt with the foundations of mathematics (Lesson 04); those that followed covered optics and geometry, and one series of 1667 analysed the methods behind the results of Archimedes.

Isaac Newton had entered Trinity in 1661. How much he learned in Barrow’s lecture room is uncertain, but the two came to know each other well. Newton helped prepare the Lectiones opticae for the press. In 1669 Barrow sent Newton’s manuscript De analysi, on infinite series and the calculus of fluxions, to John Collins in London, the mathematical correspondent who circulated new work among English mathematicians. It was the first time Newton’s name became known outside Cambridge.

That autumn Barrow resigned the Lucasian chair, and Newton was elected to it. The traditional story says Barrow stepped aside because he saw that his pupil was the greater mathematician. Historians such as Mordechai Feingold have argued that Barrow also wanted to devote himself to divinity and was looking for preferment at court, and that the resignation served both ends. Within a year he was a royal chaplain and a Doctor of Divinity by royal mandate. The Lectiones geometricae appeared in 1670, edited with Collins’ help.

DiscussionBarrow’s geometric lectures contain the inverse relation of tangents and areas, yet Newton and Leibniz are credited with inventing the calculus. What do you think is required, beyond particular theorems, for a body of results to count as a new method?

Lesson 12

Master of Trinity, death and legacy, 1670–1677

In February 1673 Charles II appointed Barrow Master of Trinity College. He served as Vice-Chancellor of the university in 1675. His most visible act as Master was the new library. He persuaded his friend Christopher Wren to design it without fee, and the college raised money by public appeal. Building began in 1676 and took nearly twenty years; the Wren Library still houses the college’s manuscripts and early printed books. In 1675 Barrow also published editions of Archimedes, the first four books of Apollonius’ Conics and Theodosius’ Spherics.

Barrow was famous in his lifetime above all as a preacher. His sermons were long, carefully argued and widely read for more than a century; they were collected by John Tillotson and published after his death, along with A Treatise of the Pope’s Supremacy (1680). In April 1677, while in London, he caught a fever. He tried to treat it by fasting and with opium. He died on 4 May 1677, aged 46 and unmarried, and was buried in Westminster Abbey.

Legacy

  • Calculus: the clearest pre-Newtonian proof that tangents and areas are inverse, and a method of tangents using the differential triangle. His priority claims were revived in 1916 by J. M. Child and remain debated.
  • Teaching: his Euclid served English students for decades, and his lectures set the style of the Lucasian chair.
  • Optics: a rigorous geometrical optics on which Newton built.
  • Institutions: the first Lucasian professorship, held after him by Newton and, much later, by Dirac and Hawking, and the Wren Library at Trinity.

Final projectTranslate one proposition of Lectiones geometricae (Child’s 1916 English edition is a good starting point) into modern notation. State it precisely, prove it in modern terms, and write a paragraph on what Barrow’s geometric form makes easier or harder to see.

Reference

Sources and further reading

Biography

  1. J. J. O’Connor and E. F. Robertson, Isaac Barrow, MacTutor History of Mathematics.
  2. Isaac Barrow, Wikipedia.
  3. Trinity College Library, About the Wren Library.
  4. M. Feingold (ed.), Before Newton: The Life and Times of Isaac Barrow, Cambridge University Press, 1990.
  5. List of Vice-Chancellors of the University of Cambridge, Wikipedia.

Mathematics

  1. I. Barrow, Lectiones geometricae, London, 1670. English translation: J. M. Child, The Geometrical Lectures of Isaac Barrow, Open Court, 1916.
  2. I. Barrow, Lectiones opticae, London, 1669.
  3. I. Barrow, Lectiones mathematicae, London, 1683; English translation The Usefulness of Mathematical Learning, 1734.
  4. Barrow and Leibniz on the fundamental theorem of the calculus, arXiv:1111.6145.
  5. M. S. Mahoney, Barrow’s mathematics: between ancients and moderns, in Feingold (ed.), Before Newton, 1990.
  6. Integral of the secant function, Wikipedia (history of Wright, Bond, Gregory and Barrow).
  7. C. H. Edwards, The Historical Development of the Calculus, Springer, 1979.