courseshub.world · Lives in Mathematics and Physics

Caroline Herschel

Caroline Lucretia Herschel

A course in two strands: the life of the woman who left a Hanover kitchen to become a concert singer, a discoverer of comets and nebulae and the first woman in England paid for scientific work, and the astronomy she practised, from sweeping the sky to reducing thousands of observations by spherical trigonometry.

Born 16 March 1750, Hanover Died 9 January 1848, Hanover Comets eight, 1786–1797 Level advanced undergraduate
Two of her comets, drawn in one plane for clarity. Encke’s comet (dots at 30-day steps, crowded near aphelion) returns every 3.3 years; 35P/Herschel–Rigollet, found on 21 December 1788, every 155.
Lesson 00

How to use this course

Caroline Herschel is often presented as her brother’s helper, and for much of her life she described herself that way. The record shows more. She discovered eight comets and about a dozen nebulae and star clusters. She recorded and reduced nearly all the observations of William Herschel’s twenty-year survey of the nebulae. She produced, on her own, an index and supplement to John Flamsteed’s star catalogue, and in old age a reduced catalogue of some 2,500 nebulae that became a foundation of the New General Catalogue still in use.

Much of that work was mathematical. To turn a night’s notes into positions on the sky she used sidereal time, spherical trigonometry and the theory of precession. Part II teaches these methods, together with the orbital mechanics of the comets she found and the optics of the small telescope she found them with.

Prerequisites for Part II
  • Trigonometry and vectors in three dimensions.
  • Newtonian gravitation and Kepler’s laws.
  • Basic integration and power series.
  • For Lessons 04 and 10, the idea of aperture and magnification in a telescope (see Lesson 04 of the William Herschel module).

Each technical lesson opens with an italic paragraph for the general reader, then gives precise statements, derivations and exercises with folded solutions. Companion modules cover her brother William Herschel and her nephew John Herschel.

Reference

Timeline

1750Born 16 March in Hanover, eighth child and fourth daughter of Isaac Herschel, a military oboist, and Anna Ilse Moritzen.
c. 1760Typhus stunts her growth and damages the sight of one eye.
1767Death of her father.
1772Leaves Hanover on 15 August with William; settles in Bath.
1778First soloist in Handel’s Messiah in Bath, April; declines an engagement at the Birmingham festival.
1782Moves with William to Datchet in August; begins searching the sky.
1783First deep-sky discoveries, including NGC 2360 (26 February), M110 and NGC 253. Becomes William’s recorder for his sweeps from October.
1786Moves to Slough in April. 1 August: discovers her first comet.
1787George III grants her a salary of £50 a year as William’s assistant.
1788William marries Mary Pitt, 8 May; Caroline moves to lodgings. 21 December: discovers the comet now called 35P/Herschel–Rigollet.
1790–93Comets of 9 January 1790, 17 April 1790, 15 December 1791 and 7 October 1793.
17957 November: finds the comet later identified as Encke’s.
17976 August: eighth and last comet, found with the naked eye.
1798The Royal Society publishes her catalogue of 561 stars from Flamsteed’s observations, with a general index.
1822William dies 25 August; she returns to Hanover in October.
1828Gold Medal of the Astronomical Society of London for her reduced zone catalogue of William’s nebulae.
1835Honorary member of the Royal Astronomical Society, with Mary Somerville: the first women so elected.
1838Member of the Royal Irish Academy.
1846Gold medal for science from King Frederick William IV of Prussia, conveyed by Alexander von Humboldt.
1848Dies 9 January in Hanover, aged 97.
Part I · Formation
Lesson 01

Hanover, 1750–1772

Caroline Lucretia Herschel was born in Hanover on 16 March 1750, the eighth child and fourth daughter of Isaac Herschel and Anna Ilse Moritzen. Isaac was an oboist in the band of the Hanoverian Foot Guards, self-taught and keen that his children should learn. On a frosty night he took the small Caroline outside to see the constellations and a comet. Her mother took a different view. Daughters were to be useful in the house, and any education beyond reading, writing and needlework would only unfit them for it.

At about the age of ten Caroline caught typhus. The illness stunted her growth, leaving her about four feet three inches tall as an adult, and damaged the sight of one eye. Her family assumed she would never marry and would serve as the household’s unpaid help. In 1757, during the Seven Years’ War, her brothers William and Jacob left for England. Her father came back from the campaigns in poor health and died in 1767. For the next five years she ran the house for her mother and her brother Jacob, with little prospect of anything else. She later recalled years spent at knitting and housework.

William, by then an established musician in Bath, proposed that she join him and train as a singer for his concerts. On 15 August 1772 Caroline left Hanover with William, aged 22.

Lesson 02

Bath: singer and apprentice, 1772–1782

In Bath she kept William’s house and took daily singing lessons from him. She learned English, accounts and keyboard. Her voice developed, and she became the principal soprano in the oratorios William directed. In April 1778 she was first soloist in a performance of Handel’s Messiah in Bath and was offered an engagement at the Birmingham music festival. She declined it. She would sing only under her brother’s direction, and as his interests changed, her musical career faded.

From 1773 William’s spare hours went into grinding telescope mirrors, and the house became a workshop. Caroline was drawn in. She read to him and fed him by hand when he could not leave the polisher, copied catalogues and tables, and learned to make the paper tubes and eyepiece fittings. William also taught her mathematics: algebra, geometry and plane and spherical trigonometry, with the practical aim of reducing observations. She kept notebooks of rules and worked examples, aimed at the calculations she needed rather than at theory.

When William found Uranus in March 1781 and became, the following year, the King’s Astronomer, the household’s centre moved from music to astronomy. In August 1782 they left Bath for Datchet, near Windsor.

Lesson 03

Datchet and Slough: an astronomer, 1782–1788

At Datchet William set Caroline to sweep the sky for comets and interesting objects in strips, first with a small refractor and, from the summer of 1783, with a small Newtonian reflector he built for the purpose. According to Michael Hoskin’s reconstruction from her notebooks, her first discovery, on 26 February 1783, was the open cluster now called NGC 2360 in Canis Major. That year she also found the dwarf galaxy M110, companion of the Andromeda galaxy (independently of Charles Messier, who had seen it but not published it), the bright spiral galaxy NGC 253 in Sculptor, and several clusters including NGC 7789 in Cassiopeia.

Her finds showed William how many objects remained undiscovered, and in October 1783 he began his own systematic sweeps with the large 20-foot reflector. Caroline’s role changed. Night after night she sat at a desk by an open window, wrote down the descriptions he called out, read the clock and the telescope’s index, and the next day reduced the observations to positions on the sky. This work, described in Lessons 07–09, left her less time at her own telescope. On clear nights when William did not need her, she swept alone.

On 3 April 1786 the household moved to Slough. On 1 August 1786, with William away, she found a comet. She is often called the first woman to discover a comet, though Maria Kirch in Berlin had found one in the early 1700s. William showed it to the royal family at Windsor, and the discovery made Caroline known across Europe. In 1787 George III granted her an annual salary of £50 as her brother’s assistant, making her the first woman in England paid for scientific work.

Part II · The astronomy
Lesson 04

Sweeping the sky

For the general reader. A comet far from the Sun is a faint fuzzy patch, easy to miss among thousands of stars. To find one you must examine the whole sky systematically, field by field, again and again. Caroline Herschel’s method was to move a wide-field telescope steadily up and down while the turning Earth carried the sky past it. Her telescope was designed for this: low magnification, wide field, and as much light as the eye could use.

Goals
  • Compute how fast the sky drifts through a fixed telescope.
  • Find the conditions for a sweep to leave no gaps.
  • Understand the “richest-field” design through the exit pupil.

The sweeper

Her comet sweeper of 1783 was a Newtonian reflector of about 4.2 inches (107 mm) aperture, mounted so she could move it easily in altitude. Her low-power eyepiece gave a field of about 2°15′ at a magnification of roughly 15 to 24, according to her own and William’s estimates. From 1791 she also used a larger, 9-inch instrument.

Drift

Proposition (diurnal drift)

The sky turns once per sidereal day, 86,164 s, so a star of declination \(\delta\) drifts across a fixed telescope at angular speed

\[ \omega(\delta)=15.04''\cos\delta\ \text{ per second of time}. \]

A field of diameter \(f\) is crossed in time \(t=f/\omega(\delta)\).

The star moves on a small circle of radius \(\cos\delta\) (in units of the sphere’s radius), at \(360^\circ\) per 86,164 s \(=15.04''\) per second along the equator. Near the pole the drift is slow; on the equator it is fastest.

Proposition (no-gap condition)

Suppose the observer moves the telescope up and down through a band of declination of height \(H\), taking time \(T\) for each traverse, with a field of diameter \(f\). The band is covered without gaps if, during one traverse, the sky drifts by no more than the field width:

\[ \omega(\delta)\,T\le f \quad\Longleftrightarrow\quad \frac{H}{T}\ge\frac{H\,\omega(\delta)}{f}. \]

Between successive passes through a given declination the sky moves \(\omega T\) in right ascension, and the field covers a strip \(f\) wide, so the strips overlap when \(\omega T\le f\). In practice an observer wants a margin, since an object at the edge of the field is easily missed.

The richest-field telescope

The beam leaving the eyepiece has diameter \(D/M\), the exit pupil, for aperture \(D\) and magnification \(M\). If it is wider than the eye’s pupil \(p\), light is wasted; if narrower, the field is smaller than necessary. The lowest useful magnification is therefore \(M_{\min}=D/p\), and at that magnification the eye sees the widest field the telescope can show at full brightness. A telescope used this way is called a richest-field telescope, because it shows the most stars per field.

Exercise 4.1For a field of \(2^\circ15'\), how long does a star take to cross the field at declination \(0^\circ\) and at \(+60^\circ\)?

Show solution
\(f=8{,}100''\). At \(\delta=0\): \(t=8{,}100/15.04\approx539\) s, about 9 minutes. At \(+60^\circ\), \(\cos\delta=0.5\), so about 18 minutes.

Exercise 4.2She sweeps a band \(10^\circ\) high on the celestial equator with this field. What is the maximum time per traverse, and the minimum vertical speed, for no gaps? How long does it take to cover one hour of right ascension?

Show solution
\(T\le539\) s. The vertical speed must be at least \(10^\circ/539\ \text{s}\approx1.1^\circ\) per minute. One hour of right ascension passes in one sidereal hour (about 59.8 minutes of mean time) regardless of the sweeping speed, so a band \(10^\circ\times15^\circ\) takes about an hour, with about 6.7 traverses. Comet hunting is slow work.

Exercise 4.3Take a dark-adapted pupil of 7 mm. What is the richest-field magnification of a 107 mm aperture? Compare with her eyepieces.

Show solution
\(M_{\min}=107/7\approx15\). Her own estimate for the finder eyepiece was 15, exactly the richest-field value; William’s estimate of 24 gives an exit pupil of 4.5 mm, still a good match for a less dark-adapted eye.
Lesson 05

Comets on parabolas

For the general reader. Most comets come from far beyond the planets, swing once around the Sun and leave again. Near the Sun their orbits are so stretched that a parabola fits them well. Edmond Halley and Isaac Newton showed how to compute such orbits. With a parabola, a few positions measured over a few weeks are enough to predict where a comet will be, how close it will come to the Sun and how fast it will move.

Goals
  • Write the parabolic orbit in polar form.
  • Derive Barker’s equation relating time to position.
  • Compute speeds and timescales of a comet’s passage.

The orbit

A body moving under the Sun’s gravity with exactly zero total energy follows a parabola with the Sun at the focus. With perihelion distance \(q\) and true anomaly \(\nu\) (the angle from perihelion, seen from the Sun),

\[ r=\frac{2q}{1+\cos\nu}=q\sec^2\frac{\nu}{2}. \]

Energy zero gives \(v^2=2GM/r\), the escape speed at each distance. The angular momentum per unit mass is \(h=r^2\dot\nu=\sqrt{2GMq}\), since at perihelion \(v=\sqrt{2GM/q}\) is perpendicular to the radius.

Theorem (Barker’s equation)

Let \(T\) be the time of perihelion and \(u=\tan(\nu/2)\). Then

\[ u+\frac{u^3}{3}=\sqrt{\frac{GM}{2q^3}}\;(t-T). \]

With the Gaussian constant \(k=0.01720209895\), \(q\) in AU and time in days, the right side is \(\dfrac{k}{\sqrt2\,q^{3/2}}(t-T)\approx\dfrac{0.012163}{q^{3/2}}(t-T)\).

Proof. From \(r^2\,d\nu=h\,dt\) and \(r=q(1+u^2)\), with \(d\nu=2\,du/(1+u^2)\):

\[ q^2(1+u^2)^2\cdot\frac{2\,du}{1+u^2}=\sqrt{2GMq}\,dt \quad\Longrightarrow\quad 2q^2\Big(u+\frac{u^3}{3}\Big)=\sqrt{2GMq}\,(t-T). \]

Dividing by \(2q^2\) gives the result. \(\square\)

The cubic in \(u\) has exactly one real root, since its left side is strictly increasing, so the position at any time is determined. For practical work it was solved from tables. Comet orbits in Caroline Herschel’s day were usually computed in this parabolic approximation, and only later, for comets observed at more than one return, as ellipses.

Exercise 5.1A comet on a parabola has \(q=1\) AU. How many days does it take to go from \(\nu=-90^\circ\) to \(\nu=+90^\circ\), and how far from the Sun is it at those points?

Show solution
At \(\nu=\pm90^\circ\), \(u=\pm1\) and \(r=2q=2\) AU. Then \(u+u^3/3=\pm4/3\), so \(t-T=\pm(4/3)/0.012163\approx\pm109.6\) days. The passage takes about 219 days.

Exercise 5.2Show that \(v=42.1\ \text{km/s}/\sqrt{r}\) for a parabolic comet with \(r\) in AU, and find its speed at perihelion when \(q=0.5\) AU.

Show solution
\(v=\sqrt{2GM_\odot/r}\). With \(GM_\odot=1.327\times10^{20}\ \mathrm{m^3s^{-2}}\) and \(1\) AU \(=1.496\times10^{11}\) m, \(\sqrt{2GM_\odot/1\ \text{AU}}=42.1\) km/s, Earth’s orbital speed times \(\sqrt2\). At \(q=0.5\) AU, \(v=42.1/\sqrt{0.5}\approx59.6\) km/s.

Exercise 5.3For small \(t-T\), show that \(\nu\approx\sqrt{2GM/q^3}\,(t-T)\), and interpret the result.

Show solution
For small \(u\), \(u\approx\nu/2\) and \(u^3\) is negligible, so \(\nu/2\approx\sqrt{GM/2q^3}(t-T)\), i.e. \(\nu\approx\sqrt{2GM/q^3}(t-T)=(v_q/q)(t-T)\). Near perihelion the comet moves uniformly along a circle of radius \(q\) at the perihelion speed.
Lesson 06

Periodic comets: Encke and 35P

For the general reader. Two of Caroline Herschel’s comets are now known to return. The comet of 7 November 1795 had already been seen by Pierre Méchain in 1786, and was seen again in 1805 and 1818. Johann Franz Encke showed in 1819 that all four sightings were one comet on a 3.3-year orbit, the shortest known for a bright comet. Her comet of 21 December 1788 was found again by Roger Rigollet in 1939, after a journey of about 155 years, and is called 35P/Herschel–Rigollet.

Goals
  • Use Kepler’s third law and the vis-viva equation for elliptical orbits.
  • Compute perihelion, aphelion and speeds from orbital elements.
  • See how repeated apparitions identify a periodic comet.
Theorem (ellipse elements and vis-viva)

For an elliptical orbit about the Sun with semi-major axis \(a\) (AU) and eccentricity \(e\): perihelion \(q=a(1-e)\), aphelion \(Q=a(1+e)\), period \(P=a^{3/2}\) years, and at distance \(r\) the speed satisfies

\[ v^2=GM_\odot\Big(\frac2r-\frac1a\Big). \]

Proof of vis-viva. Energy and angular momentum per unit mass are conserved: \(\tfrac12v^2-GM/r=E\) and \(h=r v_\perp\). At perihelion and aphelion the velocity is perpendicular to the radius, so \(v_qq=v_QQ\). Substituting into the energy equation at both points and eliminating \(v_q,v_Q\) gives \(E=-GM/(q+Q)=-GM/2a\). Then \(v^2=2(E+GM/r)=GM(2/r-1/a)\). \(\square\)

The orbital elements below are modern values.

2P/Encke: \(a=2.218\) AU, \(e=0.847\), \(P=3.30\) yr. 35P/Herschel–Rigollet: \(a=28.8\) AU, \(e=0.974\), inclination \(64.2^\circ\), \(P\approx155\) yr; next perihelion expected in 2092.

Exercise 6.1Compute \(q\), \(Q\) and the perihelion speed of Encke’s comet.

Show solution
\(q=2.218\times0.153=0.339\) AU, inside Mercury’s orbit; \(Q=2.218\times1.847=4.10\) AU, inside Jupiter’s. With \(29.8\) km/s as Earth’s speed at 1 AU, \(v_q=29.8\sqrt{2/0.339-1/2.218}=29.8\sqrt{5.448}\approx69.5\) km/s.

Exercise 6.2For 35P, check that \(P=155\) years agrees with \(a=28.8\) AU. Find \(q\), \(Q\) and the ratio of perihelion to aphelion speed.

Show solution
\(155^{2/3}\approx28.9\) AU, consistent. \(q=28.84\times0.026=0.748\) AU and \(Q=28.84\times1.974=56.9\) AU, well beyond Neptune. By conservation of angular momentum, \(v_q/v_Q=Q/q\approx76\).

Exercise 6.3Encke’s comet was seen in January 1786 and November 1795. How many orbits fit the interval, and what period results? Why was the link not obvious at the time?

Show solution
The interval is about 9.8 years, so three orbits give \(P\approx3.27\) years, close to 3.30. The comet is faint and well placed only at some returns, so most returns went unseen, and observers could not tell from two apparitions whether the interval held one, two or three orbits. Encke needed four apparitions and a full elliptical solution, including planetary perturbations, to settle it.
Lesson 07

Clocks, sidereal time and the meridian

For the general reader. During William Herschel’s sweeps, the telescope pointed roughly south and moved only up and down. The position of each nebula along the sky came from a clock: the moment it crossed the field. Caroline Herschel read that clock and wrote down the times, and then converted them into right ascension, the celestial equivalent of longitude. The key is that the stars keep their own time, sidereal time, which runs about four minutes a day faster than the Sun’s.

Goals
  • Relate sidereal and solar days.
  • Convert transit times to right ascension.
  • Use a reference star to calibrate a sweep.
Proposition (sidereal day)

The Earth rotates prograde with sidereal period \(P\) and orbits the Sun with period \(T=365.2422\) mean solar days. The mean solar day \(S=1\) day satisfies \(\dfrac1P=\dfrac1S+\dfrac1T\), so

\[ P=\frac{T}{T+1}\ \text{days}\approx23^{\mathrm h}\,56^{\mathrm m}\,4.09^{\mathrm s}. \]

In one sidereal day the Earth turns once relative to the stars. In one solar day it must turn a little more, because it has moved along its orbit and the Sun has shifted east among the stars. Over a year the Earth makes one more turn relative to the stars than relative to the Sun, which gives the formula. Mean solar time therefore runs slow against sidereal time by the factor \(1.0027379\).

Transit rule

A star crosses the local meridian when the local sidereal time equals its right ascension: \(\mathrm{LST}=\alpha\). If a sidereal clock reads \(t_1\) when a reference star of right ascension \(\alpha_1\) transits, then an object transiting at clock time \(t\) has \(\alpha=\alpha_1+(t-t_1)\). If the clock keeps mean solar time, the interval must first be multiplied by \(1.0027379\).

Herschel’s sweeps used the telescope near, not exactly on, the meridian, so the reduction also needed corrections for the telescope’s offset, found from known stars observed in the same sweep. Caroline’s reduction notebooks consist largely of such corrections.

Exercise 7.1Verify that \(P=T/(T+1)\) days gives \(23^{\mathrm h}56^{\mathrm m}4^{\mathrm s}\). By how much does a star’s transit time advance each night by a solar clock?

Show solution
\(365.2422/366.2422=0.997270\) days \(=86{,}164.1\) s \(=23^{\mathrm h}56^{\mathrm m}4.1^{\mathrm s}\). Each star transits 3 m 55.9 s earlier each night by mean solar time, about two hours per month.

Exercise 7.2A reference star with \(\alpha=5^{\mathrm h}14^{\mathrm m}32^{\mathrm s}\) transits at a mean-time clock reading of 22:03:10. A nebula transits at 23:41:40. Find its right ascension.

Show solution
The mean-time interval is 1 h 38 m 30 s \(=5{,}910\) s. In sidereal time this is \(5{,}910\times1.0027379=5{,}926.2\) s \(=1^{\mathrm h}38^{\mathrm m}46.2^{\mathrm s}\). So \(\alpha=6^{\mathrm h}53^{\mathrm m}18^{\mathrm s}\). Ignoring the correction would err by 16 s of time, about \(4'\) of arc on the equator, more than the size of many nebulae.
Lesson 08

Spherical trigonometry and reductions

For the general reader. The sky is a sphere, and distances and angles on a sphere obey rules different from those on a flat page. To get the second coordinate of each nebula, its distance from the celestial pole, Caroline Herschel converted the height of the telescope into declination. To check catalogues, find stars and compare positions, she used the formulas of spherical trigonometry, which William had taught her in Bath.

Goals
  • Convert meridian altitude into declination and polar distance.
  • Prove and use the spherical law of cosines.
  • Convert between equatorial and ecliptic coordinates.

On the meridian

An observer at latitude \(\varphi\) sees the celestial equator crossing the meridian at altitude \(90^\circ-\varphi\) in the south. An object transiting south of the zenith at altitude \(h\) therefore has declination \(\delta=h+\varphi-90^\circ\), and north polar distance \(\mathrm{NPD}=90^\circ-\delta=180^\circ-\varphi-h\). Herschel’s catalogues, and Caroline’s zone catalogue, were arranged by polar distance.

Theorem (spherical law of cosines)

In a spherical triangle on the unit sphere with sides \(a,b,c\) (arcs of great circles) and angle \(A\) opposite \(a\),

\[ \cos a=\cos b\cos c+\sin b\sin c\cos A. \]

Proof. Place the vertex at \(A\) at the north pole, \(\mathbf A=(0,0,1)\). The other vertices are \(\mathbf B=(\sin c,0,\cos c)\) and \(\mathbf C=(\sin b\cos A,\sin b\sin A,\cos b)\). The side \(a\) is the angle between \(\mathbf B\) and \(\mathbf C\), so \(\cos a=\mathbf B\cdot\mathbf C=\sin c\sin b\cos A+\cos c\cos b\). \(\square\)

With the pole as one vertex, \(b,c\) become polar distances and \(A\) the difference in right ascension, so the angular separation \(\theta\) of two objects satisfies

\[ \cos\theta=\sin\delta_1\sin\delta_2+\cos\delta_1\cos\delta_2\cos(\alpha_1-\alpha_2). \]

Equatorial and ecliptic coordinates

Older catalogues, and the motions of comets and planets, often used ecliptic longitude \(\lambda\) and latitude \(\beta\). The ecliptic is tilted by \(\varepsilon\approx23.44^\circ\) to the equator about the line to the vernal equinox. Rotating about that axis gives

\[ \begin{aligned} \sin\delta&=\sin\beta\cos\varepsilon+\cos\beta\sin\varepsilon\sin\lambda,\\ \cos\delta\cos\alpha&=\cos\beta\cos\lambda,\\ \cos\delta\sin\alpha&=\cos\beta\cos\varepsilon\sin\lambda-\sin\beta\sin\varepsilon. \end{aligned} \]

Exercise 8.1Slough is at latitude about \(51.5^\circ\) N. An object transits at altitude \(30^\circ\) south of the zenith. Find its declination and north polar distance. What is the most southerly declination visible on the meridian?

Show solution
\(\delta=30+51.5-90=-8.5^\circ\), so NPD \(=98.5^\circ\). At altitude \(0^\circ\), \(\delta=-38.5^\circ\); in practice haze limits sweeps to a few degrees higher. NGC 253, at about \(-25^\circ\), transits at only about \(13.5^\circ\), which is why it is hard to see from England.

Exercise 8.2Find the separation of two stars at \((\alpha,\delta)=(0^\circ,+40^\circ)\) and \((10^\circ,+42^\circ)\). Compare with the flat approximation \(\sqrt{(\Delta\alpha\cos\bar\delta)^2+(\Delta\delta)^2}\).

Show solution
\(\cos\theta=\sin40\sin42+\cos40\cos42\cos10=0.6428\times0.6691+0.7660\times0.7431\times0.9848=0.4301+0.5606=0.9907\), so \(\theta\approx7.8^\circ\). Flat: \(\bar\delta=41^\circ\), \(\Delta\alpha\cos\bar\delta=7.55^\circ\), \(\sqrt{7.55^2+2^2}\approx7.81^\circ\). Close agreement at this scale.

Exercise 8.3Find the declination of a point on the ecliptic (\(\beta=0\)) at longitude \(\lambda=90^\circ\), and interpret it.

Show solution
\(\sin\delta=\sin\varepsilon\), so \(\delta=+23.44^\circ\) and \(\alpha=90^\circ\). This is the Sun’s position at the June solstice.
Lesson 09

Precession and the epoch 1800

For the general reader. The coordinate grid on the sky slowly turns. The Earth’s axis wobbles like a spinning top, tracing a circle in about 26,000 years, so the celestial pole and equator drift among the stars. A position measured in 1783 and one measured in 1800 cannot be compared without a correction. Caroline Herschel’s great catalogue of 1825–28 reduced every one of William’s 2,500 nebulae to a common date, 1 January 1800. The Astronomical Society gave her its Gold Medal for this work.

Goals
  • Derive the first-order precession formulas.
  • Apply them to reduce a position to a standard epoch.
  • Estimate the size and period of the effect.

Derivation

Ignoring the small motion of the ecliptic itself, precession rotates the equatorial frame about the ecliptic pole at rate \(p\approx50.3''\) per year. Relative to that frame, the unit vector \(\mathbf u\) of a star rotates the other way about the ecliptic pole \(\mathbf K\), and its longitude increases by \(p\) per year. In equatorial coordinates with the x-axis towards the equinox,

\[ \mathbf K=(0,-\sin\varepsilon,\cos\varepsilon),\qquad \mathbf u=(\cos\delta\cos\alpha,\ \cos\delta\sin\alpha,\ \sin\delta),\qquad \dot{\mathbf u}=p\,\mathbf K\times\mathbf u. \]
Proposition (annual precession)

To first order,

\[ \frac{d\alpha}{dt}=m+n\sin\alpha\tan\delta,\qquad \frac{d\delta}{dt}=n\cos\alpha, \]

with \(m=p\cos\varepsilon\approx46.1''=3.07^{\mathrm s}\) and \(n=p\sin\varepsilon\approx20.0''=1.34^{\mathrm s}\) per year.

Proof. Compute \(\mathbf K\times\mathbf u=\big(-\sin\varepsilon\sin\delta-\cos\varepsilon\cos\delta\sin\alpha,\ \cos\varepsilon\cos\delta\cos\alpha,\ \sin\varepsilon\cos\delta\cos\alpha\big)\). The third component gives \(\cos\delta\,\dot\delta=p\sin\varepsilon\cos\delta\cos\alpha\). For \(\alpha\), use \(\dot\alpha=(x\dot y-y\dot x)/(x^2+y^2)\) with \(x^2+y^2=\cos^2\delta\):

\[ \dot\alpha=\frac{p\big[\cos\varepsilon\cos^2\delta+\sin\varepsilon\sin\delta\cos\delta\sin\alpha\big]}{\cos^2\delta}=p(\cos\varepsilon+\sin\varepsilon\sin\alpha\tan\delta).\qquad\square \]

The \(\tan\delta\) term grows without bound near the pole, where right ascension changes rapidly; near the pole the reductions need rigorous formulas instead. For sweeps reduced over a few decades and away from the pole, the first-order formulas suffice.

Exercise 9.1M110 lies near \(\alpha=0^{\mathrm h}40^{\mathrm m}\) (\(10^\circ\)), \(\delta=+41.7^\circ\). Reduce its position from 1783.0 to 1800.0.

Show solution
\(\dot\alpha=3.075+1.336\times\sin10^\circ\times\tan41.7^\circ=3.075+1.336\times0.1736\times0.891\approx3.282\) s/yr. Over 17 years, \(\Delta\alpha\approx+55.8\) s. \(\dot\delta=20.04''\cos10^\circ=19.74''\)/yr, so \(\Delta\delta\approx+336''=+5.6'\). Both shifts exceed the object’s visible size in a small telescope.

Exercise 9.2From \(p=50.29''\) per year, find the period of precession. By how much has the vernal equinox moved since Hipparchus (c. 130 BC) and 1800?

Show solution
\(1{,}296{,}000''/50.29''\approx25{,}770\) years. Over about 1,930 years the shift is \(1{,}930\times50.29''\approx97{,}000''\approx27^\circ\), nearly a full zodiacal sign.
Lesson 10

Her nebulae: seeing faint extended light

For the general reader. Among Caroline Herschel’s discoveries of 1783 are two objects now known to be galaxies: M110, a small companion of the Andromeda galaxy, and NGC 253, a large spiral in Sculptor. Neither was understood as a galaxy until the twentieth century. A faint galaxy is hard to see because its light is spread out, and no telescope can make that light look brighter per unit area. Her small sweeper, used at its lowest power, came close to the best possible view of such objects.

Goals
  • Prove that a telescope cannot raise the surface brightness of an extended object.
  • Understand why low power reveals faint nebulae.
  • Use the distance modulus to compare her galaxies.
Theorem (image surface brightness)

Let an extended object have surface brightness \(I\) (flux per unit solid angle). Seen through a telescope of aperture \(D\), transmission \(\tau\) and magnification \(M\), by an eye with pupil \(p\), the retinal surface brightness relative to the naked eye is

\[ \frac{I_{\text{tel}}}{I_{\text{eye}}}=\tau\Big(\frac{D}{Mp}\Big)^2 \quad\text{when } D/M\le p,\qquad\text{and } \tau \text{ when } D/M\ge p. \]

Proof. A patch of solid angle \(\Omega\) delivers flux \(\tau I\Omega\cdot\pi D^2/4\) to the telescope. It appears to the eye with solid angle \(M^2\Omega\). If the exit pupil \(D/M\) fits within the eye’s pupil, all this flux enters the eye, so the apparent surface brightness is \(\tau I\pi D^2/(4M^2)\), compared with \(I\pi p^2/4\) for the naked eye. If the exit pupil is larger, the eye admits only the fraction \((Mp/D)^2\) of the beam, and the ratio is \(\tau\). \(\square\)

So the image of a nebula is at best as bright per unit area as the naked-eye view (times \(\tau\)), and that best is reached at the richest-field magnification of Lesson 04. A telescope helps because it enlarges the object: the eye detects faint light more easily over a larger area.

Distance modulus

An object of absolute magnitude \(M_{\text{abs}}\) at distance \(d\) parsecs has apparent magnitude \(m=M_{\text{abs}}+5\log_{10}(d/10\,\mathrm{pc})\), ignoring absorption.

Exercise 10.1Her 107 mm sweeper with two speculum reflections (\(\tau\approx0.36\)) is used at magnification 15 and at 30, with \(p=7\) mm. Compare the surface brightness of a nebula in each case with the naked-eye value.

Show solution
At 15×, the exit pupil is 7.1 mm \(\ge p\), so the ratio is \(\tau=0.36\). At 30×, the exit pupil is 3.6 mm, and the ratio is \(0.36\times(3.57/7)^2\approx0.094\). Doubling the power makes the nebula larger but nearly four times dimmer per unit area, which is why she searched at low power and switched to 30× only to examine a suspect.

Exercise 10.2Take, as round values, M110 at \(m\approx8.5\) and \(d\approx0.82\) Mpc, and NGC 253 at \(m\approx7.1\) and \(d\approx3.5\) Mpc. Find their absolute magnitudes and the ratio of their luminosities.

Show solution
M110: \(5\log_{10}(8.2\times10^4)=24.6\), so \(M_{\text{abs}}\approx-16.1\). NGC 253: \(5\log_{10}(3.5\times10^5)=27.7\), so \(M_{\text{abs}}\approx-20.6\). The difference of 4.5 magnitudes is a luminosity ratio of \(10^{0.4\times4.5}\approx63\). The fainter-looking of her two galaxies is the dwarf; the other is a giant spiral about four times further away.
Part III · Life and legacy
Lesson 11

Comets and Flamsteed, 1788–1822

On 8 May 1788 William married Mary Pitt, a wealthy widow. Caroline lost the management of the household and moved into lodgings nearby, though she continued to work at the observatory. Biographers disagree about how much the change hurt her. In later life she corresponded warmly with Mary, and she was devoted to her nephew John, born in 1792, whom she helped to raise.

Her comet discoveries continued: on 21 December 1788 (now 35P/Herschel–Rigollet), 9 January and 17 April 1790, 15 December 1791, 7 October 1793, 7 November 1795 (Encke’s comet) and 6 August 1797, the last seen with the naked eye. She corresponded about them with the leading astronomers, among them Nevil Maskelyne, the Astronomer Royal, and Joseph Banks, president of the Royal Society. Her claim to the first discovery of five of them was recorded in the Philosophical Transactions.

Flamsteed’s catalogue

William’s work depended on John Flamsteed’s British Catalogue of about 3,000 stars, published in 1725. It contained many errors, and William needed to know which stars it listed and where they came from. Caroline went back to Flamsteed’s original observations in the second volume of his Historia Coelestis and compared them with the catalogue, one by one. She found 561 stars that Flamsteed had observed but left out of the catalogue, and she compiled a general index to every observation of every catalogued star, with a list of errata. The Royal Society published the work in 1798 under her own name. It became a standard tool for anyone using Flamsteed.

After 1797 she discovered no more comets. Her time went to William’s reductions and to the double-star and nebula records. William’s health declined after 1807. When he died, on 25 August 1822, she decided to return to Hanover.

DiscussionCaroline Herschel consistently played down her own role, yet her comets, nebulae and Flamsteed index were original work. How should historians assign credit in a scientific partnership where one person records and reduces what the other observes? Does the answer change when one partner is denied any independent position?

Lesson 12

Hanover, honours and legacy, 1822–1848

She arrived in Hanover in October 1822, after fifty years in England. The city’s buildings made observing difficult. She threw herself instead into a last great reduction. For her nephew John, who was re-examining William’s nebulae, she arranged all of them, about 2,500 objects, into zones by north polar distance, with positions reduced to the epoch 1800. In 1828 the Astronomical Society of London awarded her its Gold Medal for this zone catalogue. John used it in re-examining the nebulae, and it fed directly into his General Catalogue of 1864 and so into J. L. E. Dreyer’s New General Catalogue of 1888. Her last entry in an observing book, in January 1824, concerns the great comet of 1823.

Honours came late and steadily. In 1835 the Royal Astronomical Society elected her and Mary Somerville honorary members, the first women it admitted. In 1838 the Royal Irish Academy elected her a member. On her 96th birthday in 1846 the King of Prussia, Frederick William IV, awarded her a gold medal for science, conveyed by Alexander von Humboldt. She wrote memoirs and letters, later edited by John’s wife and published in 1876.

She died in Hanover on 9 January 1848, aged 97, and was buried in the Gartengemeinde cemetery beside her parents, with a lock of William’s hair. Her epitaph says that her eyes in life were turned to the starry heavens.

Legacy

  • Discoveries: eight comets, five with unquestioned priority, including 35P/Herschel–Rigollet; an early sighting of Encke’s comet; about a dozen clusters and nebulae, among them NGC 2360, NGC 7789, M110 and NGC 253.
  • Catalogues: the 1798 index and supplement to Flamsteed; the zone catalogue of William’s nebulae, a direct ancestor of the NGC.
  • Firsts: the first woman in England paid for scientific work; with Somerville, the first female honorary member of the Royal Astronomical Society.
  • Names: the asteroid 281 Lucretia, the lunar crater C. Herschel, and the informal names Caroline’s Cluster and Caroline’s Rose for NGC 2360 and NGC 7789.

Final projectChoose one of her comets or nebulae. Using modern data, write a two-page account for a strong undergraduate that explains how she found it, computes one quantity from Part II (an orbital speed, a transit time, a precession correction or a surface brightness), and explains what the object is now known to be.

Reference

Sources and further reading

Biography

  1. J. J. O’Connor and E. F. Robertson, Caroline Lucretia Herschel, MacTutor History of Mathematics.
  2. Caroline Herschel, Wikipedia (with extensive references).
  3. Obituary of Caroline Herschel, Monthly Notices of the Royal Astronomical Society 8 (1848) 64–66.
  4. Royal Astronomical Society, The Herschel Archive: Caroline Herschel.
  5. Mrs J. Herschel (ed.), Memoir and Correspondence of Caroline Herschel, John Murray, 1876.
  6. M. Hoskin, Discoverers of the Universe: William and Caroline Herschel, Princeton University Press, 2011.
  7. M. Hoskin (ed.), Caroline Herschel’s Autobiographies, Science History Publications, 2003.
  8. R. Holmes, The Age of Wonder, HarperPress, 2008.

Astronomy

  1. C. L. Herschel, Catalogue of Stars, taken from Mr. Flamsteed’s Observations contained in the Second Volume of the Historia Coelestis, and not inserted in the British Catalogue, Royal Society, 1798.
  2. In Caroline Herschel’s footsteps, Sky & Telescope (her deep-sky discoveries, after Hoskin).
  3. 35P/Herschel–Rigollet and Comet Encke, Wikipedia (orbital elements and discovery history).
  4. W. M. Smart, Textbook on Spherical Astronomy, 6th ed., Cambridge University Press, 1977 (sidereal time, precession, coordinate transformations).
  5. J. M. A. Danby, Fundamentals of Celestial Mechanics, 2nd ed., Willmann-Bell, 1988 (parabolic orbits and Barker’s equation).