Part A · the equation of motion · spine

Flux, Gauss, and gravity stops being a force between pairs

Integration over regions and over surfaces, the divergence theorem, and the moment gravity becomes a field with a local law rather than a sum over pairs.

needs 04~2½ h slides + 5 h problemsproblems: ./mr problems 05

Unit 04 integrated along curves and got work and potential energy. This unit integrates over regions and over surfaces, and the payoff is larger than a second kind of integral. Newton's law of gravitation is a statement about a pair of point masses. Real bodies are not points, so every prediction needs a sum over pairs — an integral, and in general an ugly one. Gauss's law replaces the sum by a local statement, $\operatorname{div}\vec g = -4\pi G\rho$, which says the same thing about the field at each point and says it without reference to anything far away.

Two things follow, and they are the reason this unit sits where it does. First, the shell theorem — that a sphere pulls exactly as a point at its centre — becomes three lines instead of a page, where Newton needed a page of geometry and a reputation for having been delayed by it for years. Second, the derivative of the field, the tidal tensor $\partial_i\partial_j\Phi$, is a symmetric trace-free object in vacuum, and unit 33 will show it is the Newtonian limit of the Riemann curvature tensor. Everything difficult in this course's last part has a rehearsal here.

How to read this deck

One object runs through it: the Earth idealised as a uniform ball, and a straight tunnel bored through it. Section 1 gives the integrals to compute its field the hard way; section 2 the flux that makes that unnecessary; section 3 the field equation; section 4 what empty space forces on the potential; section 5 what the field does to a body of finite size. The deck closes on $84.3$ minutes and on two tidal bulges.

Terms with a dotted underline are glossary entries — click one, or press G for the whole list. A solid underline is a piece of the running example; hover it. N opens these notes, where the longer proofs live under Full proof; R stacks the deck into one readable page.

05 · overview

One ball, five sections

A uniform ball of mass $M$ and radius $R$ — the Earth, idealised. Newton's law gives the force between two points, so its field — the force per unit mass, $\vec g$ — is an integral over the ball, done afresh for every field point. Two questions, and the routes this deck takes to them:

route 1 · the tunnel: 84.3 minutes

1. The flux of $\vec g$ out of a closed surface — how much of it crosses the surface — counts only the mass inside: $-4\pi G$ times it. The $1/r^2$ makes this so.

2. The ball is the same in every direction, so $\vec g$ is radial and equal all over each sphere about the centre. One sphere then gives $g(r)$: inside, $\vec g = -(GM/R^3)\,\vec r$.

3. Along a straight tunnel only the component along it acts: $\ddot x = -(GM/R^3)\,x$, simple harmonic. Period $2\pi\sqrt{R^3/GM} = 84.3$ min, for every tunnel.

route 2 · tides: two bulges

1. The Earth falls freely towards the Moon, so its centre feels nothing. A point displaced by $\vec\xi$ feels the difference of the Moon's field across $\vec\xi$.

2. To first order that difference is a matrix times $\vec\xi$: the second derivatives of the Moon's potential $\Phi$ — potential energy per unit mass, $\vec g = -\nabla\Phi$ — a symmetric matrix with trace zero in empty space.

3. For a point mass $M$ at distance $d$ its eigenvalues are $(-2,1,1)\,GM/d^3$: stretched along the Moon line, both ways, and squeezed across it.

The plan. 1 integrals over regions, and one shell by brute force · 2 flux and the divergence theorem · 3 the field equation, the shell theorem in three lines, the tunnel · 4 empty space: why gravity cannot trap you · 5 tides.

Where each step is done. Route 1, step 1 is Gauss's law for gravity in section 3, resting on the solid angle and the divergence theorem of section 2; step 2 is What "by symmetry" actually buys and Worked example 2 — the shell theorem in three lines; step 3 is Worked example 3 — the tunnel, and 84 minutes. Route 2, step 1 is Two free-fallers, and what is left over; step 2 is The tidal tensor, where the trace is shown to be $4\pi G\rho$ and so zero where there is no mass; step 3 is Worked example 4 — the tidal tensor of a point mass, with the bulges read off on the slide after it.

The problem sheet follows the same order: P1–P3 are the integration machinery, P4–P6 flux and solid angle, P7–P9 Gauss's law and what symmetry does and does not buy, P10–P12 harmonic functions, the tidal tensor and the tunnel once more. Four of the twelve are proofs (P6, P8, P9, P10), and six end in a number.

A remark on the idealisation. The Earth is not uniform — it is centrally condensed, with a core about twice the mean density — and it is not a sphere. Both corrections are computable and neither changes any structural claim here. Unit 10 does the departure from sphericity properly, with Legendre polynomials, and gets the precession of a real satellite's orbit out of it. Problem P12 does a centrally condensed planet and finds the tunnel period changes by about ten per cent.

section 1

Integrating over regions

The same Riemann sum, with length replaced by volume — and then the two theorems that make it computable.

05 · regions

The integral over a box

definition · the multiple integral

Let $B = [a_1,b_1]\times\cdots\times[a_n,b_n]$ and let $f : B \to \mathbb R$ be bounded. A partition $P$ cuts each edge, splitting $B$ into subboxes $B_k$ of volume $|B_k|$. Put $$\begin{aligned} L(f,P) &= \textstyle\sum_k \big(\inf_{B_k} f\big)\,|B_k|,\\ U(f,P) &= \textstyle\sum_k \big(\sup_{B_k} f\big)\,|B_k| .\end{aligned}$$ $f$ is integrable if $\sup_P L = \inf_P U$, and then $\int_B f$ is that common value.

Word for word the one-variable definition, with length replaced by volume. Nothing new has been assumed.

measure zero

$N \subseteq \mathbb R^n$ has measure zero if for every $\varepsilon > 0$ it can be covered by countably many boxes of total volume less than $\varepsilon$. A point, a line segment, a smooth surface in $\mathbb R^3$: all measure zero.

Lebesgue's criterion · cited

A bounded $f$ on a box is integrable iff its set of discontinuities has measure zero.

Proved in Spivak, Calculus on Manifolds, Thm 3-8, for boxes in $\mathbb R^n$; Rudin, Principles, Thm 11.33, is the one-variable case. We use it only as a licence: every integrand in this course is continuous off a surface.

Why state the criterion at all, rather than "continuous functions are integrable"? Because the integrands of this unit are not continuous. The density of the ball is $\rho$ inside and $0$ outside, with a jump across the sphere $r = R$; the indicator function of any region is discontinuous exactly on that region's boundary. Lebesgue's criterion is what says those are still integrable, and it says it by measuring the boundary rather than by inspecting the function.

Measure zero is not the same as small. The rationals in $[0,1]$ have measure zero and are dense: enumerate them as $q_1, q_2, \dots$ and cover $q_k$ by an interval of length $\varepsilon 2^{-k}$, total $\varepsilon$. Countability is doing the work. A set of measure zero can also be uncountable — the Cantor set — but no such set will appear here.

A warning about the upper and lower sums. $\inf_{B_k} f$ requires $f$ bounded, which is why the definition says so. Unbounded integrands do occur in this unit — $1/|x - y|$ as $y$ approaches $x$ — and they are handled as improper integrals, exactly as in unit 00: integrate over the region with a small ball removed, and let the ball shrink. The slide "The field of a continuous distribution" shows that this particular limit exists.

05 · regions

Regions that are not boxes

definition

For bounded $D \subseteq \mathbb R^n$ and $f$ bounded on $D$, extend by zero: $$f_D(x) = \begin{cases} f(x) & x \in D\\ 0 & x \notin D\end{cases}$$ and set $\int_D f := \int_B f_D$ for any box $B \supseteq D$. The value does not depend on which $B$.

$D$ is admissible if $\partial D$ has measure zero. Then $f_D$ is discontinuous only on $\partial D$ together with the discontinuities of $f$, so Lebesgue's criterion applies. Its volume is $|D| = \int_D 1$.

the hypothesis is not free

$D = \mathbb Q^3 \cap [0,1]^3$ has $\partial D = [0,1]^3$, which does not have measure zero, and $\int_D 1$ does not exist: every subbox contains both rational and irrational points, so $L = 0$ and $U = 1$ for every partition.

Every region in this unit is a ball, a shell, a box, or a region with a piecewise-smooth boundary, and all of those are admissible.

Independence of the box. If $B, B' \supseteq D$ are two boxes, take a partition of $B \cup B'$ refining the faces of both. Outside $D$ the integrand is zero, so only the subboxes meeting $D$ contribute, and those are common to both computations. This is the same argument that shows the one-variable integral does not care about the interval's endpoints.

Why a smooth surface has measure zero. A $C^1$ surface in $\mathbb R^3$ is locally a graph $z = h(x,y)$ over a compact square. $h$ is uniformly continuous, so cutting the square into $N^2$ pieces makes the graph fit inside $N^2$ boxes of height $\varepsilon_N \to 0$ and base $1/N^2$: total volume $\varepsilon_N \to 0$. A sphere is a finite union of six such graphs.

"Admissible" is also called Jordan measurable, and $|D|$ the Jordan content of $D$. The distinction between Jordan content and Lebesgue measure is exactly the rational-cube example: Lebesgue measure copes, Jordan content does not, and Riemann integration is built on Jordan content. Nothing in this course needs the Lebesgue theory, and it will not be assumed.

05 · regions

Fubini: one integral at a time

Fubini's theorem

Let $B = R \times S$ with $R \subseteq \mathbb R^p$, $S \subseteq \mathbb R^q$ boxes, and let $f = f(x,y)$ be continuous on $B$. Then $$\begin{aligned}\int_B f &= \int_R\!\Big(\int_S f\,\mathrm{d}y\Big)\mathrm{d}x\\ &= \int_S\!\Big(\int_R f\,\mathrm{d}x\Big)\mathrm{d}y .\end{aligned}$$

full proof in the notes (N)

An $n$-fold integral is $n$ ordinary ones. Every explicit computation in this unit is Fubini plus a substitution.

two traps

On a non-box region the limits change when the order does; they are not carried along — the next slide shows how. And integrability is a hypothesis, not a formality: for a non-integrable $f$ the two iterated integrals can both exist and differ (the notes give one).

On a region that is not a box, Fubini is applied to $f_D$, the extension by zero of the previous slide — the proof in the notes covers it, since $f_D$ is integrable and, for every region in this unit, its inner integrals exist. The zero part simply cuts each inner integral off at the boundary of $D$.

Full proof (Fubini for an integrable $f$ whose inner integrals exist)

Let $P = P_R \times P_S$ be a partition of $B$ into subboxes $R_i \times S_j$. Write $g(x) = \int_S f(x,y)\,\mathrm dy$. For $x \in R_i$, $$\sum_j \Big(\inf_{R_i\times S_j} f\Big)|S_j| \;\le\; g(x) \;\le\; \sum_j \Big(\sup_{R_i\times S_j} f\Big)|S_j|,$$ because on each $S_j$ the integrand lies between those two constants. Multiplying by $|R_i|$ and summing over $i$ gives $L(f,P) \le L(g,P_R) \le U(g,P_R) \le U(f,P)$. Since $f$ is integrable, $L(f,P)$ and $U(f,P)$ can be brought within $\varepsilon$ of $\int_B f$, so $g$ is integrable on $R$ with $\int_R g = \int_B f$. The other order is the same argument with the roles swapped. For continuous $f$ on a compact box, both $f$ and every $g$ are continuous, so all the hypotheses hold automatically.

The counterexample behind the second warning. On $(0,1]^2$ let $f(x,y) = (x^2-y^2)/(x^2+y^2)^2$, which is unbounded near the origin and so not integrable. Since $\partial_y\big(y/(x^2+y^2)\big) = f$, for each $x > 0$ $$\int_0^1 f\,\mathrm dy = \frac{1}{1+x^2},\qquad \int_0^1\!\!\int_0^1 f\,\mathrm dy\,\mathrm dx = \arctan 1 = \frac\pi4 ,$$ and since $f(y,x) = -f(x,y)$ the other order gives $-\pi/4$. Both iterated integrals exist and they differ. The moral is that Fubini is a theorem about integrable functions, and integrability has to be checked (Lebesgue's criterion is how).

Tonelli's theorem is the companion for nonnegative measurable functions, where no integrability hypothesis is needed because the value $+\infty$ is allowed. It belongs to the Lebesgue theory and this course will not use it, but it is what a physicist is implicitly invoking when swapping orders in an infinite sum of positive terms.

05 · regions · example

One triangle, sliced two ways

the order is a choice, and it matters

$\displaystyle\int_0^1\!\!\int_x^1 e^{y^2}\,\mathrm{d}y\,\mathrm{d}x$ — the inner integral has no elementary antiderivative. The region is $\{0 \le x \le y \le 1\}$; describe it the other way round:

$$\begin{aligned}\int_0^1\!\!\int_0^y e^{y^2}\,\mathrm{d}x\,\mathrm{d}y &= \int_0^1\! y\,e^{y^2}\,\mathrm{d}y\\ &= \tfrac12(e-1) = 0.8591 .\end{aligned}$$

The same triangle twice. Left: fix $x$ and the slice runs from the diagonal up to $1$, so the inner integral is of $e^{y^2}$ alone. Right: fix $y$ and the slice runs from $0$ to the diagonal; the integrand does not depend on $x$, so the inner integral is just the slice's length $y$ — and $y$ is the factor that makes $e^{y^2}$ integrable.

Reading the limits off the picture. Each Fubini order asks a different question of the region: "for this value of the outer variable, which values of the inner one lie in $D$?" With $x$ outer, the vertical line at $x$ meets the triangle in $x \le y \le 1$; with $y$ outer, the horizontal line at $y$ meets it in $0 \le x \le y$. Swapping the order of the integral signs without redrawing the region is the first trap of the previous slide, and the drawing is the cheapest way to avoid it.

Why the swap helped. The integrand $e^{y^2}$ depends on $y$ only, so with $x$ inside it is a constant and the inner integral is a length. That length, $y$, is exactly the derivative of $y^2$ up to a factor $2$, which is what makes the outer integral elementary: $\int_0^1 y\,e^{y^2}\,\mathrm dy = \tfrac12\big[e^{y^2}\big]_0^1$.

05 · regions

Change of variables, and what a determinant is for

change of variables · cited

Let $\Psi : U \to V$ be a $C^1$ diffeomorphism between open sets of $\mathbb R^n$ and let $f$ be integrable on $V$. Then $$\int_V f = \int_U (f\circ\Psi)\,\big|\det D\Psi\big| .$$

Rudin, Principles, Thm 10.9; Spivak, Calculus on Manifolds, Thm 3-13. The proof is an induction on dimension using the inverse function theorem of unit 04, and it is long.

It stays true if $\Psi$ fails to be a diffeomorphism on a set of measure zero — which is exactly the case for every coordinate system we use.

Read the formula as bookkeeping. A small box of volume $\mathrm dV_u$ around $u$ is carried by $\Psi$ to a small curved region around $\Psi(u)$; the integrand is evaluated there, and the box's volume has to be converted into the image's.

A $C^1$ map is linear to first order at each point (unit 03), and $D\Psi(u)$ is that linear map. So the conversion factor is whatever a linear map does to volume — and the next slide shows that is $|\det|$. The theorem says the volume distortion is the local one, integrated.

Why the measure-zero exception is allowed. Suppose $\Psi$ is a diffeomorphism on $U \setminus N$ with $N$ of measure zero, and $\Psi(N)$ also of measure zero. Removing a closed $\varepsilon$-neighbourhood of $N$ leaves a region where the theorem applies; both sides change by at most (sup of the integrand) times the volume removed, which tends to zero. This is the licence to use spherical coordinates on the whole ball even though the polar axis is a genuine failure of the chart.

Historical note. The $n = 2$ case is Euler's (1769), for the specific substitutions he needed; Lagrange did $n = 3$; the general statement with the Jacobian determinant is Jacobi's, 1841, and the determinant is named after him for it. That the determinant is the volume factor was not obvious in advance — it was found by computing examples.

05 · regions

What a linear map does to volume

why $|\det|$, and not something else

For a linear $T : \mathbb R^n \to \mathbb R^n$ and a box $B$, $$|T(B)| = |\det T|\cdot|B| .$$

Both sides, read as functions of the $n$ columns of $T$, are multilinear, vanish when two columns agree, and equal $|B|$ at $T = I$. The determinant is the unique such function up to scale (linear algebra), so they agree.

A curved chart is linear only cell by cell, so it scales each small cell by its own factor $|\det D\Psi|$ — for polar coordinates, $r$.

Left: the columns of $T = \begin{pmatrix}1.5 & 0.5\\ 0.5 & 1.5\end{pmatrix}$ are the images of the unit vectors, and they span $T(B)$; its area is $\det T = 2.25 - 0.25 = 2$. Right: under the polar chart a cell $\mathrm dr\times\mathrm d\varphi$ becomes a near-rectangle of sides $\mathrm dr$ and $r\,\mathrm d\varphi$, so its area is $r\,\mathrm dr\,\mathrm d\varphi$ and $|\det D\Psi| = r$: cells far out are bigger.

The alternating-multilinear argument, carefully. Strictly, $|T(B)|$ is multilinear only up to sign, so the clean statement is: the signed volume $\det$ and the unsigned $|T(B)|/|B|$ agree in absolute value. Fix a box $B$ and let $V(c_1,\dots,c_n) = |T(B)|$ where $c_i$ are $T$'s columns. Shearing one column by a multiple of another maps $T(B)$ to a region of equal volume (Cavalieri: slice parallel to the sheared direction, each slice is translated), scaling a column by $\lambda$ scales the volume by $|\lambda|$, and two equal columns collapse the image into a hyperplane, volume zero. Those are the defining properties of $|\det|$ up to the normalisation $T = I$, which is $|B| = |B|$.

The polar cell, computed. $\Psi(r,\varphi) = (r\cos\varphi, r\sin\varphi)$ has columns $\partial_r\Psi = (\cos\varphi, \sin\varphi)$, of length $1$, and $\partial_\varphi\Psi = (-r\sin\varphi, r\cos\varphi)$, of length $r$, at right angles; $\det D\Psi = r\cos^2\varphi + r\sin^2\varphi = r$. The figure is the two-dimensional rehearsal for the next slide, where the same picture in three dimensions gives $r^2\sin\theta$: one more column, of length $r\sin\theta$.

05 · regions

Spherical coordinates, and where the chart dies

spherical coordinates

$\Psi(r,\theta,\varphi) = (r\sin\theta\cos\varphi,\ r\sin\theta\sin\varphi,\ r\cos\theta)$ on $r > 0$, $\theta \in (0,\pi)$, $\varphi \in (0,2\pi)$. $\theta$ is the polar angle from the $z$-axis, $\varphi$ the azimuth.

the coordinate singularity

On the polar axis $\theta \in \{0,\pi\}$, and at $r = 0$, $\varphi$ is not defined at all and $\Psi$ is not injective: the whole circle of $\varphi$ values maps to one point.

Nothing is wrong with space there: the chart is at fault, not the geometry — a distinction that returns at the Schwarzschild radius in unit 36. The bad set — the axis, plus the half-plane where $\varphi$ jumps from $2\pi$ to $0$ — has measure zero, so the change of variables formula still applies to the whole ball.

$P$ is reached by tilting $\theta$ down from the $z$-axis and turning $\varphi$ round it from the $x$-axis; the dashed drop to the $xy$-plane shows where $\varphi$ is measured. Near the pole the circles of constant $\theta$ shrink, and on the axis itself a whole circle of $\varphi$ values lands on one point.

The convention warning. Physics writes $(r,\theta,\varphi)$ with $\theta$ the polar angle; much of the mathematics literature writes $(r,\theta,\varphi)$ with $\theta$ the azimuth. A formula copied across that divide is wrong in a way that survives dimensional analysis, because both angles are dimensionless. Check which convention a source uses by looking at where its $\sin$ sits in the volume element.

What to look for in the figure. The point $P$'s distance from the $z$-axis is $r\sin\theta$ — the length of the dashed segment from $O$ to the foot of the drop. As $\theta \to 0$ that distance goes to zero, and so does the circle that $\varphi$ sweeps out: the small dashed circle near the pole is on its way to becoming a point. That shrinking is exactly the factor $\sin\theta$ in the volume element on the next slide, and its vanishing on the axis is the chart's failure.

05 · regions

The volume element $r^2\sin\theta$

The three columns of $D\Psi$ are $$\partial_r\Psi = \hat r,\quad \partial_\theta\Psi = r\,\hat\theta,\quad \partial_\varphi\Psi = r\sin\theta\,\hat\varphi,$$ with $\hat r, \hat\theta, \hat\varphi$ orthonormal. For a matrix with orthogonal columns $A^{\!\top}\!A$ is diagonal, so $(\det A)^2 = \prod_i\lVert c_i\rVert^2$ and

the Jacobian determinant $$\big|\det D\Psi\big| = r^2\sin\theta, \qquad \mathrm{d}V = r^2\sin\theta\,\mathrm{d}r\,\mathrm{d}\theta\,\mathrm{d}\varphi .$$

It vanishes at $r = 0$ and on the polar axis — the coordinate singularity of the previous slide, seen as cells squeezed to nothing.

check

$\displaystyle |B_R| = \int_0^R\!\!\int_0^\pi\!\!\int_0^{2\pi}\! r^2\sin\theta\,\mathrm{d}\varphi\,\mathrm{d}\theta\,\mathrm{d}r = 2\pi\cdot2\cdot\tfrac{R^3}{3} = \tfrac43\pi R^3$ ✓, and it has dimensions $L^3$ ✓.

The cell's three edges, drawn in the plane through the $z$-axis: $\mathrm dr$ outward, $r\,\mathrm d\theta$ along the meridian, and $r\sin\theta\,\mathrm d\varphi$ round the circle of latitude, whose radius is $r\sin\theta$. They are at right angles, so the volume is their product.

The columns, computed. $\partial_r\Psi = (\sin\theta\cos\varphi, \sin\theta\sin\varphi, \cos\theta)$, which is the unit radial vector $\hat r$. $\partial_\theta\Psi = r(\cos\theta\cos\varphi, \cos\theta\sin\varphi, -\sin\theta)$, of length $r$; call the unit vector $\hat\theta$. $\partial_\varphi\Psi = r\sin\theta(-\sin\varphi, \cos\varphi, 0)$, of length $r\sin\theta$; call it $\hat\varphi$. Their pairwise dot products are zero by direct computation, and $(\hat r, \hat\theta, \hat\varphi)$ is right-handed.

The figure is the computation. Each column times its coordinate step is one edge of the image cell: $\partial_r\Psi\,\mathrm dr$ has length $\mathrm dr$, $\partial_\theta\Psi\,\mathrm d\theta$ has length $r\,\mathrm d\theta$ (an arc of the circle of radius $r$), and $\partial_\varphi\Psi\,\mathrm d\varphi$ has length $r\sin\theta\,\mathrm d\varphi$ (an arc of the circle of latitude, radius $r\sin\theta$). Orthogonal edges, so the volume is the product; the orthogonal-columns identity is exactly that statement for the parallelepiped the columns span.

Cylindrical coordinates, for completeness and for problems P2 and P5: $\Psi(s,\varphi,z) = (s\cos\varphi, s\sin\varphi, z)$ with $s > 0$, giving $|\det D\Psi| = s$ and $\mathrm dV = s\,\mathrm ds\,\mathrm d\varphi\,\mathrm dz$. The singularity is the whole $z$-axis, again a set of measure zero, and again the azimuth is undefined there.

The determinant can of course also be expanded by brute force; the orthogonal-columns trick is worth keeping because it works for every orthogonal coordinate system in one line, and because it makes clear that $r^2\sin\theta$ is a product of three lengths, not a coincidence of algebra.

05 · regions · example

Worked example 1 — a shell, by brute force (setting up)

Take a thin shell of radius $a$, mass $m$. Model it as a solid shell of thickness $\delta$ and density $\rho = \sigma/\delta$, and let $\delta \to 0$ at the end. Put the field point $P$ on the $z$-axis at distance $r$; by symmetry that costs nothing.

Unit 04 gave the potential energy of one pair, $-Gm_1m_2/s$. Divide by the test mass at $P$ to get the potential $\Phi$, potential energy per unit mass, and superpose: $$\Phi(r) = -G\!\!\int_{\text{shell}}\frac{\mathrm{d}m}{s},\qquad s = \lVert \vec r - \vec a\rVert .$$

The law of cosines gives $s$ in the coordinates we have: with $\theta$ the polar angle of the source point, $$s = \sqrt{r^2 + a^2 - 2ar\cos\theta} .$$

$s$ depends on the source point only through $\theta$: the whole ring at that angle is at one distance from $P$. That is why the azimuth $\varphi$ will integrate out for free.

Why putting $P$ on the axis is free. The shell's density is invariant under every rotation about the centre (unit 03's $SO(3)$), so $\Phi$ is a function of $r$ alone. Choosing the axis through $P$ is a choice of basis, and the integral's value does not depend on it. The slide "What 'by symmetry' actually buys" makes this precise for the field as well as the potential.

Why $\Phi$ and not the force. $\Phi$ is a scalar, so the superposition is a sum of numbers; the force would be a sum of vectors pointing in different directions, and three integrals instead of one. $\Phi$ here is the potential energy per unit mass of a test body at $P$ — unit 04's potential energy divided by the test mass — and the field, the force per unit mass, is recovered at the end as $-\nabla\Phi$.

05 · regions · example

Worked example 1 — down to one integral

the integral

With $\mathrm dm = \rho\,r'^2\sin\theta\,\mathrm dr'\,\mathrm d\theta\,\mathrm d\varphi$ and nothing depending on $\varphi$, Fubini does the azimuth for free: $$\Phi(r) = -2\pi G\sigma a^2\!\!\int_0^\pi\!\frac{\sin\theta\,\mathrm{d}\theta}{\sqrt{r^2+a^2-2ar\cos\theta}} .$$

Substitute $u = \cos\theta$, $\mathrm du = -\sin\theta\,\mathrm d\theta$: $$\Phi(r) = -2\pi G\sigma a^2\!\!\int_{-1}^{1}\!\frac{\mathrm{d}u}{\sqrt{r^2+a^2-2aru}} .$$

One ordinary integral, no vectors left. The next slide evaluates it.

Why $u = \cos\theta$ is the natural variable: $u$ is height along the axis in units of $a$, and slices of equal height cut the shell into bands of equal area — a cap near the pole is wide in $\theta$ but small in radius, a band at the equator the reverse. So the mass is spread uniformly in $u$ over $[-1,1]$.

Where the $2\pi a^2$ comes from. The radial integral over $r' \in [a, a+\delta]$ of $\rho r'^2$ is $\rho a^2\delta + O(\delta^2) = \sigma a^2 + O(\delta)$, and the $\varphi$ integral gives $2\pi$. Letting $\delta \to 0$ leaves exactly $2\pi\sigma a^2$ multiplying the $\theta$ integral. Nothing has been assumed about a "surface element": the whole computation is a volume integral in spherical coordinates, which is all that section 1 has licensed. The area element of a sphere is derived properly in section 2, and agrees.

A by-product worth naming. The same computation with the integrand $1$ instead of $1/s$ gives the area of the band between $\theta_1$ and $\theta_2$ as $2\pi a^2(\cos\theta_1 - \cos\theta_2)$ — it depends only on the band's height $a(\cos\theta_1 - \cos\theta_2)$ along the axis, not on where it sits. This is Archimedes' theorem that a sphere and its circumscribing cylinder have equal lateral area, which he had engraved on his tomb. It is also why a uniform distribution of points on a sphere is generated by taking $\cos\theta$ uniform on $[-1,1]$. The figure is that theorem: equal heights, equal areas.

05 · regions · example

Worked example 1 — the cancellation

evaluate

$\dfrac{\mathrm d}{\mathrm du}\Big[-\dfrac{\sqrt{r^2+a^2-2aru}}{ar}\Big] = \dfrac{1}{\sqrt{r^2+a^2-2aru}}$, so

$$\begin{aligned}\int_{-1}^1 &= \frac{\sqrt{(r+a)^2} - \sqrt{(r-a)^2}}{ar}\\[-2pt] &= \frac{(r+a) - |r-a|}{ar} .\end{aligned}$$

Outside ($r > a$): $|r - a| = r - a$, so the bracket is $2a$ and the integral is $2/r$.
Inside ($r < a$): $|r - a| = a - r$, the bracket is $2r$, the integral is $2/a$.

With $m = 4\pi a^2\sigma$: outside, $\Phi = -Gm/r$; inside, $\Phi = -Gm/a$, constant. The field $\vec g$, the force per unit mass on a test body, is $-\nabla\Phi$ (unit 04: force is minus the gradient of potential energy). So outside $\vec g = -(Gm/r^2)\,\hat r$, and inside $\vec g = \vec 0$.

the shell theorem, and where it hides

Outside, the shell pulls exactly as a point of the same mass at its centre. Inside, the field is exactly zero — not small, zero.

The entire theorem is the step $\sqrt{(r-a)^2} = |r-a|$. Write $r - a$ there, as the algebra invites you to, and both halves are wrong.

The next slide plots both halves and checks them.

Newton's difficulty. Book I, Propositions LXX and LXXI of the Principia (1687) are this theorem — the interior half and the exterior half — proved by geometric arguments; the interior one pairs opposite cones. The claim that Newton delayed the Principia for years until he could prove it is a nineteenth-century conjecture, repeated everywhere and probably overstated. What is certainly true is that without it, "the Earth attracts as if its mass were at its centre" is an assumption, and the whole quantitative comparison with the Moon's orbit that makes the theory testable rests on it.

The inside result is stronger than the outside one. Outside, dimensional analysis and symmetry already tell you $\Phi = -Gm f(r/a)/r$ for some $f$; the theorem is that $f \equiv 1$. Inside, the answer is constant, so there is no force anywhere in the cavity, at any distance from the centre, however close to the wall. A test mass just inside the shell has almost all the mass far away and a little of it very near, and these cancel exactly. That exact cancellation is a property of the exponent $2$ alone, which section 2 makes precise.

05 · regions · example

The shell's potential and field, plotted

checks

$[Gm/r] = \mathrm{L^2T^{-2}}$, an energy per mass ✓. $a \to 0$: a point mass ✓. $r \to \infty$: $\Phi \to 0$ ✓. At $r = a$ both branches give $-Gm/a$ ✓, though $\vec g$ jumps. Numerically, with $G = m = a = 1$, quadrature gives $\Phi(2) = -0.500000$ and $\Phi(0.5) = -1.000000$ ✓ — the two dots.

$\Phi$ is continuous with a kink at the shell; $g = -\Phi'$ is the slope, so the kink is a jump in $g$, of size $Gm/a^2 = 4\pi G\sigma$.

Inside, $\Phi$ is flat, so its slope — the field — is zero everywhere in the cavity. Outside, both curves are the point mass's. The field jumps at the shell because the shell has finite mass in zero thickness.

The jump in $\vec g$ is real and is not a mistake. A surface density is an idealisation: an infinitely thin layer of finite mass. Crossing it, the normal component of $\vec g$ jumps by $-4\pi G\sigma$ — check: $0$ to $-Gm/a^2 = -4\pi G\sigma$. Section 3 will read that off Gauss's law with a pillbox, and the same statement in electrostatics is the discontinuity of $E_\perp$ across a charged sheet (unit 23). For a genuine shell of thickness $\delta$ the field falls smoothly through the layer.

What to look for. The radial component $g(r)$ is plotted, so negative means pointing inward. The potential's curve has no jump — a jump in $\Phi$ would need an infinite field — but its slope does, from $0$ to $Gm/a^2$, and the field curve shows the same fact as a vertical drop. The two quadrature values are where the formula was checked against a direct numerical evaluation of example 1's $u$-integral.

section 2

Flux

How much of a vector field crosses a surface — and the one theorem that turns that into a volume integral.

05 · flux

Surfaces, and the vector area element

parametrised surface

A $C^1$ map $\Sigma : D \subseteq \mathbb R^2 \to \mathbb R^3$ on an admissible $D$, with $D\Sigma$ of rank $2$ and $\Sigma$ injective on the interior of $D$. Its two tangent vectors at $(u,v)$ are $\partial_u\Sigma$ and $\partial_v\Sigma$.

the vector area element $$\mathrm{d}\vec A = (\partial_u\Sigma \times \partial_v\Sigma)\,\mathrm{d}u\,\mathrm{d}v .$$

Direction: normal to both tangents, so normal to the surface. Length: $\lVert\partial_u\Sigma\times\partial_v\Sigma\rVert$, the area of the parallelogram they span, which is the local area distortion of $\Sigma$.

A rectangle $\mathrm du\times\mathrm dv$ in the domain lands on the surface as a small curved patch. To first order it is the parallelogram spanned by $\partial_u\Sigma\,\mathrm du$ and $\partial_v\Sigma\,\mathrm dv$; $\mathrm d\vec A$ is their cross product, normal to the patch, with length its area.

The cross product, and an apology for it. The natural object being integrated here is not a vector field at all but a 2-form — an antisymmetric bilinear function that eats the two tangent vectors and returns a number. In three dimensions, and only there, an inner product and an orientation convert a 2-form into a vector, and that conversion is the cross product. Unit 34 does this properly and proves the divergence theorem, Green's theorem and Stokes' theorem as one statement in any dimension. Until then the cross product is a convenience with an expiry date.

The figure as a computation. This is the change-of-variables picture one dimension down: a $C^1$ map is linear to first order, so a small rectangle goes to a small parallelogram, and the area of a parallelogram spanned by two vectors in $\mathbb R^3$ is the length of their cross product. The rank-$2$ condition in the definition is what keeps that parallelogram from collapsing to a segment.

05 · flux

Flux, and why the parametrisation does not matter

the flux of a field $$\int_\Sigma \vec F\cdot\mathrm{d}\vec A = \iint_D \vec F(\Sigma)\cdot(\partial_u\Sigma\times\partial_v\Sigma)\,\mathrm{d}u\,\mathrm{d}v$$

Read: at each patch, how much of $\vec F$ points through it, times the patch's area. Writing $\hat n$ for the unit normal and $\mathrm dA = \lVert\mathrm d\vec A\rVert$, the integrand is $\vec F\cdot\hat n\,\mathrm dA$: only the component of $\vec F$ across the surface counts.

independence of the parametrisation

Reparametrising by a $C^1$ diffeomorphism changes the flux by the factor $\operatorname{sgn}\det$ of that reparametrisation, and by nothing else — unit 04's parametrisation invariance of the line integral, one dimension up.

full proof in the notes (N)

So flux is a property of the surface and the field, plus one binary choice — which side the normal points to. The next slide is about that choice.

Full proof of parametrisation-independence

Let $\Sigma' = \Sigma \circ h$ with $h : D' \to D$ a $C^1$ diffeomorphism. The chain rule (unit 03) gives $D\Sigma' = D\Sigma \circ Dh$, so with $h = (h_1, h_2)$, $$\partial_{u'}\Sigma' \times \partial_{v'}\Sigma' = \big(\partial_{u'}h_1\,\partial_u\Sigma + \partial_{u'}h_2\,\partial_v\Sigma\big)\times\big(\partial_{v'}h_1\,\partial_u\Sigma + \partial_{v'}h_2\,\partial_v\Sigma\big) = \det(Dh)\,(\partial_u\Sigma\times\partial_v\Sigma),$$ where the cross terms $\partial_u\Sigma\times\partial_u\Sigma$ vanish by antisymmetry and the two surviving terms combine into the determinant. Integrating and applying the change of variables theorem, which supplies $|\det Dh|$, leaves exactly $\operatorname{sgn}\det(Dh)$. A reparametrisation with $\det Dh > 0$ preserves the flux; one with $\det Dh < 0$ reverses its sign.

The scalar area element is $\mathrm dA = \lVert\partial_u\Sigma\times\partial_v\Sigma\rVert\,\mathrm du\,\mathrm dv$, used for integrating scalars (mass, charge, the mean value property of section 4). With the unit normal $\hat n = (\partial_u\Sigma\times\partial_v\Sigma)/\lVert\cdot\rVert$, the flux is $\int_\Sigma \vec F\cdot\hat n\,\mathrm dA$, the form it will usually be written in.

05 · flux

Orientation, and the sphere's area element

orientation

A choice of unit normal field $\hat n$, continuous over the surface. A closed surface is by convention oriented outward.

two ways this goes wrong

Sign. Flip $\hat n$ and every flux changes sign. Gauss's law is a statement about the outward one; half the sign errors in this subject are here.

Existence. Not every surface admits a continuous $\hat n$: on a Möbius band, transport $\hat n$ once round and it comes back reversed. Flux is undefined there. Every surface in this unit is the boundary of a solid region, and those are always orientable.

the sphere of radius $a$

$\Sigma(\theta,\varphi) = a(\sin\theta\cos\varphi, \sin\theta\sin\varphi, \cos\theta)$. From the spherical-coordinate columns, $\partial_\theta\Sigma = a\hat\theta$ and $\partial_\varphi\Sigma = a\sin\theta\,\hat\varphi$, and $\hat\theta\times\hat\varphi = \hat r$, so

$$\mathrm{d}\vec A = a^2\sin\theta\,\hat r\,\mathrm{d}\theta\,\mathrm{d}\varphi ,$$

outward, as wanted. Total area $\displaystyle a^2\!\!\int_0^\pi\!\!\int_0^{2\pi}\!\sin\theta\,\mathrm d\varphi\,\mathrm d\theta = 4\pi a^2$ ✓.

And the band between $\theta_1$ and $\theta_2$ has area $2\pi a^2(\cos\theta_1 - \cos\theta_2)$ — the formula example 1 obtained as a limit of volumes, now derived directly.

Why a boundary is orientable. If $\Sigma = \partial D$ for a solid $D$, define $\hat n$ at each boundary point as the unit vector pointing out of $D$. It is well defined and continuous wherever the boundary is $C^1$, with no choice to make. The orientability of boundaries is the reason the divergence theorem never has to worry about Möbius bands, while Stokes' theorem (unit 04) does have to be told which way the boundary curve runs.

Swapping $\theta$ and $\varphi$ in the parametrisation reverses the sign, by the theorem on the previous slide, and gives the inward normal. There is no way to tell from the formula alone which one you have; the only reliable check is to evaluate $\hat n$ at one explicit point and look. At $\theta = \pi/2$, $\varphi = 0$ the formula above gives $\hat n = (1,0,0)$, which does point out of the ball.

Notation. $\oint$ will be written for an integral over a closed surface, as unit 04 wrote it for a closed curve. It carries no mathematical content beyond the reminder that $\partial\Sigma = \emptyset$, which is exactly the hypothesis the divergence theorem needs.

05 · flux

Solid angle

solid angle

For an oriented surface $\Sigma$ not meeting the origin, $$\Omega(\Sigma) = \int_\Sigma \frac{\hat r\cdot\mathrm{d}\vec A}{r^2} = \int_\Sigma \frac{\vec r\cdot\mathrm{d}\vec A}{r^3} .$$ It is the signed area of the radial projection of $\Sigma$ onto the unit sphere: the fraction of the sky that $\Sigma$ covers, times $4\pi$. Dimensionless; the unit is called the steradian.

on a sphere about the origin

$\mathrm d\vec A = a^2\sin\theta\,\hat r\,\mathrm d\theta\,\mathrm d\varphi$ and $r = a$, so the $a^2$ cancels exactly: $$\Omega = \int_0^\pi\!\!\int_0^{2\pi}\!\sin\theta\,\mathrm d\varphi\,\mathrm d\theta = 4\pi$$ for every radius. That cancellation is the whole of section 3.

Both intercepts of the cone have $\hat r\cdot\mathrm d\vec A/r^2$ equal: a patch twice as far and twice as wide, or tilted, projects to the same piece of sky.

The Moon subtends $6.418\times10^{-5}$ sr and the Sun $6.794\times10^{-5}$ sr — a ratio of $0.945$, which is why total eclipses are only just total.

Why the tilt does not matter. The integrand is $\hat r\cdot\hat n\,\mathrm dA/r^2$. Tilting a patch away from the line of sight multiplies its area by $1/\cos\alpha$ and multiplies $\hat r\cdot\hat n$ by $\cos\alpha$; the product is unchanged. Moving it from $r$ to $\lambda r$ along the rays multiplies the area by $\lambda^2$ and divides by $\lambda^2$. Solid angle is a property of the set of directions a surface occupies, and of nothing else about it — which is why it is exactly what a flux of an inverse-square field measures.

The numbers. Angular radius $\alpha$ gives $\Omega = 2\pi(1 - \cos\alpha)$ (problem P4). For the Moon, radius $1737.4$ km at mean distance $3.844\times10^5$ km, $\alpha = 0.2590^\circ$ and $\Omega = 6.418\times10^{-5}$ sr, about $5.11\times10^{-6}$ of the whole sky. For the Sun, $\alpha = 0.2666^\circ$ and $\Omega = 6.794\times10^{-5}$ sr. The near-equality is a coincidence of the present epoch: the Moon is receding at about $3.8$ cm per year, and total eclipses will end in something like $600$ million years.

Signed. If $\Sigma$ folds so that some rays cross it twice, the two crossings enter with opposite signs of $\hat r\cdot\hat n$ and cancel. That is exactly what makes Gauss's law's "$0$ if the mass is outside" true (section 3), and it is why the definition is stated with the vector area element rather than with areas.

05 · flux

Why $1/r^2$, and no other power

Consider the radial field $\vec F_p = \hat r/r^p$ and ask for its flux through a sphere of radius $R$ about the origin:

$$\oint_{S_R}\vec F_p\cdot\mathrm{d}\vec A = \frac{1}{R^p}\cdot 4\pi R^2 = 4\pi R^{2-p} .$$

theorem

In $\mathbb R^3$, the flux of $\hat r/r^p$ through a sphere about the origin is independent of the radius iff $p = 2$. In $\mathbb R^n$ the unique exponent is $n-1$.

The same exponent shows up a second way, through the divergence $\operatorname{div}\vec F = \sum_i\partial_iF_i$ (in Cartesian components; Divergence, defined without coordinates defines it properly, as a trace). Writing $\vec F_p = \vec x\,f(r)$ with $f = r^{-p-1}$, $$\operatorname{div}(\vec x f) = 3f + r f'(r) = (2-p)\,r^{-p-1},$$ which vanishes on $r>0$ exactly when $p = 2$.

Flux of $\hat r/r^p$ through spheres of growing radius $R$: the field weakens as $R^{-p}$ while the area grows as $R^2$. Only for $p = 2$ do the two cancel, at every radius.

The divergence computation, in full. For $\vec F = \vec x f(r)$ with $r = \lVert\vec x\rVert$: $\partial_i x_i = 3$ (sum over $i$), and $\partial_i r = x_i/r$, so $\partial_i(x_i f(r)) = 3f + x_i f'(r) x_i/r = 3f + rf'$. With $f = r^{-p-1}$, $rf' = -(p+1)r^{-p-1}$, giving $(3 - p - 1)r^{-p-1} = (2-p)r^{-p-1}$. Note the identity $\partial_i r = x_i/r$ is used constantly from here on; it is the chain rule applied to $r = (x_jx_j)^{1/2}$.

In $n$ dimensions. $\operatorname{div}(\vec x f) = nf + rf'$, so $f = r^{-p-1}$ gives $(n - 1 - p)r^{-p-1}$, zero iff $p = n-1$; and the area of the sphere of radius $R$ grows as $R^{n-1}$, so the flux $R^{n-1}/R^p$ is constant under the same condition. Two routes, one answer, which is the content of the divergence theorem (The divergence theorem, proved for a box).

05 · flux

What the exponent means

what this means physically

A field whose flux through every enclosing sphere is the same is a field with no sources except at the origin. The inverse square is not an empirical accident about the exponent; it is the statement that gravity is produced by mass and by nothing else, in three dimensions.

checks

Dimensions: $[\vec g]\,\mathrm L^2 = \mathrm{L^3T^{-2}}$, and $[GM] = \mathrm{L^3T^{-2}}$ ✓ — the two sides of Gauss's law will match.
$p = 2$, $n = 3$: flux $4\pi$, the solid angle of the whole sky ✓.
$p = 0$ (a constant radial field): flux $4\pi R^2$, growing — sources everywhere, and indeed $\operatorname{div} = 2/r \ne 0$ ✓.

Experimentally the exponent is $2$ to the precision of lunar laser ranging; departures at sub-millimetre range are sought as signs of extra dimensions.

The field-line picture of $p = 2$: lines start only at the mass and never end, so every sphere about it is crossed by all of them. On a sphere of area $4\pi R^2$ that makes their density, the field strength, fall as $1/R^2$.

What the field-line picture does and does not prove. Drawing lines that neither begin nor end in empty space is the statement that the divergence vanishes there; it does not derive it. The derivation is the previous slide. The picture is worth keeping because it makes Gauss's law obvious in advance: count the lines leaving a closed surface, and you have counted the mass inside. In $n$ dimensions the sphere's area grows as $R^{n-1}$, so the same picture gives a field falling as $R^{-(n-1)}$.

The empirical status. For gravity, the exponent is checked by torsion-balance experiments at short range and by planetary ephemerides and lunar laser ranging at long range: a deviation $r^{-(2+\epsilon)}$ would make every orbit precess (unit 08 shows why), and none beyond general relativity's is seen. Cited: Adelberger, Heckel and Nelson, Annu. Rev. Nucl. Part. Sci. 53 (2003) 77, for the bounds at each range. For electrostatics the classic test is the shell theorem itself: charge a closed conducting shell and look for any field inside. Williams, Faller and Hill (Phys. Rev. Lett. 26 (1971) 721) found none, and bounded $|\epsilon|$ at about $3\times10^{-16}$. Neither is a test of the mathematics here; both are tests of whether the physical law has the form assumed.

05 · flux

Divergence, defined without coordinates

divergence

For a $C^1$ field $\vec F : U \subseteq \mathbb R^n \to \mathbb R^n$, the total derivative $D\vec F(a)$ is a linear map $\mathbb R^n\to\mathbb R^n$ (unit 03). Define $$\operatorname{div}\vec F(a) := \operatorname{tr} D\vec F(a).$$ A number attached to a point, with no basis involved.

components

In any linear coordinates, $\operatorname{div}\vec F = \sum_i \partial_i F_i$.

Proof. The matrix of $D\vec F(a)$ in a basis has entries $\partial_jF_i$; the trace of a linear map is the sum of its matrix's diagonal entries in any basis, since $\operatorname{tr}(P^{-1}AP) = \operatorname{tr}(A)$. $\square$

House rule: the object is the trace of a linear map. The formula $\sum\partial_iF_i$ is a theorem about its matrix, not the definition.

"the field spreads out" is not the definition

$\hat r/r^2$ points away from the origin everywhere and has divergence $0$ everywhere it is defined. Spreading direction and divergence are different questions: the field spreads, but it also weakens at exactly the rate that keeps the flux constant.

Conversely $\vec F = (y, 0, 0)$ has $\operatorname{div} = 0$ and does not look radial at all.

The honest physical reading comes from the next slide: $$\operatorname{div}\vec F(a) = \lim_{|D|\to0}\frac1{|D|}\oint_{\partial D}\vec F\cdot\mathrm{d}\vec A,$$ the flux out of a shrinking region per unit volume — a local source strength.

Why the trace, and not some other invariant. $D\vec F(a)$ is a linear map, so it has $n$ eigenvalues and several basis-independent scalars — the trace, the determinant, the other coefficients of the characteristic polynomial. The trace is the one that appears, and the reason is that it is the first-order rate of change of volume under the flow of $\vec F$: the flow for time $t$ is approximately $I + tD\vec F$, whose determinant is $1 + t\operatorname{tr}D\vec F + O(t^2)$. Unit 14 uses exactly this to prove Liouville's theorem, that Hamiltonian flows preserve phase volume, and the calculation there is this sentence written out.

An honest caveat about "any coordinates". The proof above shows the formula holds in any linear coordinates. In curvilinear coordinates — spherical, say — the components of $\vec F$ are taken against a basis that varies from point to point, and $\sum\partial_iF_i$ is then not the divergence; the correct expression acquires the Jacobian, $\operatorname{div}\vec F = \frac{1}{r^2}\partial_r(r^2F_r) + \cdots$. That extra structure is the connection, and unit 32 builds it properly. Until then, compute divergences in Cartesian components.

The limit formula as an alternative definition. Some texts define the divergence by that limit and then prove it equals $\sum\partial_iF_i$. It is a defensible choice — it is manifestly geometric — but it has to prove the limit exists and is independent of the shape of the shrinking region, which is more work than taking a trace.

05 · flux

The divergence theorem, proved for a box

the divergence theorem

Let $D \subset \mathbb R^3$ be compact with piecewise-$C^1$ boundary, oriented outward, and let $\vec F$ be $C^1$ on an open set containing $D$. Then $$\int_D \operatorname{div}\vec F\,\mathrm{d}V = \oint_{\partial D}\vec F\cdot\mathrm{d}\vec A .$$

proof, for a box $B = [a_1,b_1]\times[a_2,b_2]\times[a_3,b_3]$

By linearity it is enough to take $\vec F = F_3\hat e_3$. Fubini, then the fundamental theorem of calculus on the inner integral:

$$\int_B \partial_3F_3\,\mathrm{d}V = \iint\Big[\int_{a_3}^{b_3}\!\partial_3F_3\,\mathrm{d}x_3\Big]\,\mathrm{d}x_1\mathrm{d}x_2$$ $$= \iint\big(F_3(\cdot,b_3) - F_3(\cdot,a_3)\big)\,\mathrm{d}x_1\mathrm{d}x_2 .$$

On the top face $\hat n = +\hat e_3$ and on the bottom $\hat n = -\hat e_3$, so that is exactly the flux through top and bottom; the four side faces have $\hat n\cdot\hat e_3 = 0$ and contribute nothing. Sum the three components. $\square$

what the theorem is made of

Fubini and the fundamental theorem of calculus, once per axis. Nothing else. The content is $\int_a^b f' = f(b)-f(a)$ with "boundary" now meaning a surface.

Unit 04's fundamental theorem of line integrals was the same statement for a curve; unit 04's Stokes' theorem for a surface with a boundary curve. Unit 34 proves all four at once, in every dimension.

check on a known case

$\vec F = \vec x$ has $\operatorname{div} = 3$, so the left side is $3|D|$. On a sphere of radius $R$, $\vec x\cdot\hat n = R$, so the right side is $R\cdot4\pi R^2 = 3\cdot\tfrac43\pi R^3$ ✓.

Where each hypothesis is used. "$C^1$ on an open set containing $D$" is used twice: to have $\operatorname{div}\vec F$ continuous, hence integrable, and to apply the fundamental theorem of calculus along each line, which needs $\partial_3F_3$ continuous along that line. Dropping smoothness at one interior point is not a technicality — the field $\hat r/r^2$ fails it at the origin and the theorem's conclusion is false for it there, with $0$ on the left and $4\pi$ on the right. That failure is not a defect; it is Gauss's law.

Piecewise-$C^1$ covers a box, whose boundary is six flat faces meeting along edges. Edges and vertices are sets of measure zero in the boundary surface and carry no flux, which is why they can be ignored rather than handled.

Names. The theorem is due to Lagrange (1762) in a special case, Gauss (1813) for the inverse-square field specifically, and Ostrogradsky (1826) in general — which is why continental texts call it Ostrogradsky's theorem. "Gauss's theorem" and "Gauss's law" are different statements in this course: the theorem is the mathematical identity above, the law is the physical statement about $\vec g$ in section 3.

05 · flux

From a box to any region

Step 1 · additivity. Cut $D$ into two pieces along a common face. Each interior face is traversed twice, once with $\hat n$ and once with $-\hat n$, and the two contributions cancel exactly. So both sides of the theorem are additive, and the theorem holds for any finite union of boxes.

Step 2 · exhaustion. Approximate $D$ from inside and outside by unions of small boxes. Both sides converge, because $\operatorname{div}\vec F$ is bounded on the compact $D$ and the boundary has finite area.

Step 3 · curved boundaries. Near each boundary point, change variables to flatten the boundary — this is the inverse function theorem of unit 04 — prove it there, and patch the local statements with a partition of unity.

Step 3 is where the work is. Stated as cited: Spivak, Calculus on Manifolds, Thm 5-8, which is this statement, and his Thm 5-5 for the general $n$; or Rudin, Principles, Thm 10.51. Unit 34 proves it in general.

The shared face carries opposite normals, so its two flux contributions cancel and only $\partial(D_1\cup D_2)$ survives.

the hypothesis that will bite

$\vec F$ must be $C^1$ on all of $D$. Every interesting application in section 3 has a field that is singular at a point, and the theorem is applied to a region that excludes it.

Why the cancellation argument is honest. It is sometimes presented as waving at a picture, but it is exact: the shared face is a single surface, and the flux integral over it appears in $\oint_{\partial D_1}$ with one orientation and in $\oint_{\partial D_2}$ with the reverse. By the parametrisation theorem of "Flux, and why the parametrisation does not matter", reversing the orientation reverses the sign, so the sum of the two is zero, term by term, for any $\vec F$ whatsoever. No limit and no estimate is involved.

What "partition of unity" means, in one sentence, since it will be used again in unit 34: a finite collection of smooth bump functions $\chi_k \ge 0$, each supported in one of the patches, with $\sum_k\chi_k = 1$ on $D$ — so any field splits as $\vec F = \sum_k \chi_k\vec F$ with each piece living in a single patch, where the flattened local statement applies. Constructing them needs only the existence of a smooth function that is $1$ on a ball and $0$ outside a larger one, which unit 11 will build for the fundamental lemma of the calculus of variations.

For the physics that follows, steps 1 and 2 are enough: every region used is a ball, a spherical shell, a box, a cylinder, or a region between two spheres, and all of those are limits of unions of boxes in the required sense.

section 3

Gravity becomes a field

From a sum over pairs to a differential equation satisfied at every point of space.

05 · the field

The field of a continuous distribution

Newton's law of gravitation · constitutive input

Two point masses attract along the line joining them with $\lVert\vec F\rVert = Gm_1m_2/s^2$, with $G = 6.67430\times10^{-11}\ \mathrm{m^3\,kg^{-1}\,s^{-2}}$. This is a force law, not a theorem: it is measured, as unit 00 insisted.

the gravitational field and its potential

For a density $\rho$ of compact support, $$\vec g(\vec x) = -G\!\!\int\!\rho(\vec y)\frac{\vec x - \vec y}{\lVert\vec x-\vec y\rVert^3}\,\mathrm{d}V(\vec y),$$ $$\Phi(\vec x) = -G\!\!\int\!\frac{\rho(\vec y)}{\lVert\vec x-\vec y\rVert}\,\mathrm{d}V(\vec y),$$ and $\vec g = -\nabla\Phi$. $\vec g$ is force per unit mass; $\Phi$ is potential energy per unit mass.

superposition

Both integrals are linear in $\rho$. That is the whole content of "gravitational forces add", and it is a physical assumption in exactly the same sense as the force law — one that general relativity will withdraw, since its field equations are nonlinear.

the integral converges inside the matter

Near $\vec y = \vec x$ the integrand grows as $s^{-2}$, while $\mathrm dV = s^2\sin\theta\,\mathrm ds\,\mathrm d\theta\,\mathrm d\varphi$ vanishes as $s^2$. The product is bounded, so the improper integral converges — the $s^2$ of the volume element beats the $s^{-2}$ of the force. In one or two dimensions it would not.

Dimensions: $[G] = \mathrm{L^3M^{-1}T^{-2}}$, $[\vec g] = \mathrm{LT^{-2}}$, $[\Phi] = \mathrm{L^2T^{-2}}$ ✓.

Why a field at all. Newton's law is a relation between two bodies with the distance between them in it, and it says nothing about what is at the points in between. The field $\vec g$ is a different sort of object: a function on space, defined at every point whether or not there is anything there to feel it. Newton found the idea of action at a distance intolerable — "so great an Absurdity, that I believe no Man who has in philosophical Matters a competent Faculty of thinking, can ever fall into it", in the 1693 letter to Bentley — and declined to speculate about the mechanism.

The field's promotion from bookkeeping device to physical object happens in unit 24, when it acquires a propagation delay and therefore an independent existence: if the Sun vanished, the Earth would keep orbiting for eight minutes, and the momentum missing during those minutes is carried by the field. In unit 29 the field acquires energy and momentum densities of its own, and in unit 35 it acquires the right to gravitate.

The convergence argument, in full. Fix $\vec x$ and split the integral into the ball $s = \lVert\vec y - \vec x\rVert \le \epsilon$ and the rest. On the ball, in spherical coordinates centred at $\vec x$, $\big|\int_{s\le\epsilon}\rho\,s^{-2}\,\mathrm dV\big| \le \rho_{\max}\!\int_0^\epsilon\!\!\int\! s^{-2}s^2\sin\theta\,\mathrm d\theta\,\mathrm d\varphi\,\mathrm ds = 4\pi\rho_{\max}\epsilon \to 0$. So the integral over the punctured region converges as $\epsilon \to 0$, uniformly in $\vec x$, and $\vec g$ is in fact continuous everywhere, including inside the matter. $\Phi$ is better still: its integrand goes as $s^{-1}$ and the same estimate gives $2\pi\rho_{\max}\epsilon^2$.

05 · the field

Gauss's law for gravity

For any closed surface $S$ bounding a region $D$, with no mass sitting exactly on $S$, $$\oint_S \vec g\cdot\mathrm{d}\vec A = -4\pi G\,M_{\text{enc}}, \qquad M_{\text{enc}} = \int_D\rho\,\mathrm{d}V .$$

proof

By superposition it suffices to take a single point mass $m$ at $\vec y$. Then $\vec g = -Gm(\vec x - \vec y)/\lVert\vec x-\vec y\rVert^3$, so $$\oint_S\vec g\cdot\mathrm{d}\vec A = -Gm\,\Omega_{\vec y}(S),$$ the solid angle of $S$ seen from $\vec y$. If $\vec y \in D$ that is $4\pi$; if $\vec y \notin D$ it is $0$. Integrate over $\rho$, swapping the two integrals by Fubini. $\square$

the two solid-angle values: full proof in the notes (N), and problem P6

Why the solid angle is $4\pi$ or $0$. From inside, each ray from $\vec y$ crosses $S$ once more outward than inward, so $S$ covers the whole sky. From outside, each ray that meets $S$ enters ($\hat r\cdot\hat n < 0$) and leaves ($\hat r\cdot\hat n > 0$) through patches of the same solid angle, which cancel.

Mass on the surface. The hypothesis excludes it because the solid angle of $S$ seen from a point of $S$ is not $4\pi$ or $0$ but something in between — $2\pi$ for a smooth boundary point. For a continuous $\rho$ this is a measure-zero worry and the law holds anyway; for a surface density, as in example 1's shell, it matters, and the right statement is that the flux through a surface lying exactly on the shell is $-4\pi G$ times half the shell's mass. Avoid the situation rather than handle it.

Full proof of the two solid-angle values

Put $\vec y$ at the origin and let $\vec F = \vec x/\lVert\vec x\rVert^3$, which is $C^1$ off the origin with $\operatorname{div}\vec F = 0$ there (the $p = 2$ case of "Why $1/r^2$"). If $0 \notin D$, $\vec F$ is $C^1$ on an open set containing the compact $D$, and the divergence theorem gives $\oint_S\vec F\cdot\mathrm d\vec A = 0$. If $0 \in \operatorname{int}D$, choose $\varepsilon$ with the closed ball $\bar B_\varepsilon \subset \operatorname{int}D$ and apply the theorem to $D' = D\setminus B_\varepsilon$, where $\vec F$ is $C^1$. The boundary of $D'$ is $S$ together with $S_\varepsilon$, and on $S_\varepsilon$ the normal out of $D'$ is $-\hat r$. So $0 = \oint_S\vec F\cdot\mathrm d\vec A - \varepsilon^{-2}\cdot4\pi\varepsilon^2$, and the flux through $S$ is $4\pi$. $\square$

The figure is the elementary version of that proof. It assumes each ray meets $S$ finitely often, which a general surface need not do; the divergence-theorem proof above does not need it, and that is why it is the proof of record.

05 · the field

Gauss's law: the sign, and a check

signs, once and for all

$\vec g$ points inward and $\hat n$ outward, so the flux is negative. Gravity has one sign of source and attracts; electrostatics has two and like charges repel, which is why its Gauss law reads $\oint\vec E\cdot\mathrm d\vec A = Q/\varepsilon_0$ with a plus. The dictionary is $-G \leftrightarrow 1/4\pi\varepsilon_0$, used in both directions in unit 23.

check

A sphere of radius $R$ around a point mass: $\vec g\cdot\hat n = -Gm/R^2$, area $4\pi R^2$, flux $-4\pi Gm$ — and $R$ has cancelled, as the $1/r^2$ slide promised. Dimensions: $\mathrm{LT^{-2}}\cdot\mathrm L^2 = \mathrm{L^3T^{-2}} = [GM]$ ✓.

One number per closed surface. The law constrains the flux, not the field; turning it into the field takes the symmetry argument two slides on.

What the law is and is not. It is an exact consequence of the inverse-square law plus superposition, so it contains no new physics — and it is worth being clear that it is also strictly weaker than what it came from: it constrains one number per surface, and there are not enough surfaces to recover $\vec g$ from it without help. The help is symmetry, and What "by symmetry" actually buys says exactly how much of it is needed.

History. Gauss proved the theorem in 1813, in a paper on the attraction of homogeneous ellipsoids; the general statement, for any force varying as the inverse square — electrical attraction and repulsion included — is in his memoir of 1840 on such forces. The gravitational case came first, which is the reverse of the order it is now taught in.

05 · the field

The field equation

the local form

On an open set where $\rho$ is $C^1$, $$\operatorname{div}\vec g = -4\pi G\rho, \qquad\text{equivalently}\qquad \nabla^2\Phi = 4\pi G\rho ,$$ with $\nabla^2\Phi := \operatorname{div}(\nabla\Phi)$, the Laplacian of section 4. The second is Poisson's equation.

proof

For every admissible $D$, the divergence theorem and Gauss's law give $$\int_D\operatorname{div}\vec g\,\mathrm{d}V = \oint_{\partial D}\vec g\cdot\mathrm{d}\vec A = -4\pi G\!\int_D\rho\,\mathrm{d}V .$$ So $\int_D h = 0$ for every ball $D$, where $h = \operatorname{div}\vec g + 4\pi G\rho$ is continuous. If $h(a) \ne 0$, continuity gives a ball on which $h$ keeps its sign, and the integral over it is not zero. Hence $h \equiv 0$. $\square$

what has just happened

The left-hand statement mentions only the field and the density at one point. An integral over every other mass in the universe has been replaced by a differential equation, local in space.

This is the form in which gravity will be generalised. Unit 35's Einstein field equations are the same kind of object — a local differential relation between a field and its source — and $\nabla^2\Phi = 4\pi G\rho$ is what they must reduce to when the field is weak and the motion slow.

checks

Dimensions: $[\operatorname{div}\vec g] = \mathrm{T^{-2}}$, $[G\rho] = \mathrm{L^3M^{-1}T^{-2}}\cdot\mathrm{ML^{-3}} = \mathrm{T^{-2}}$ ✓.
In vacuum $\rho = 0$: $\nabla^2\Phi = 0$, and indeed $\nabla^2(1/r) = 0$ away from the origin — section 4 uses nothing else ✓.

The localisation lemma is worth keeping. If $h$ is continuous on an open set and $\int_Dh = 0$ for every ball $D$ in it, then $h \equiv 0$. Proof as on the slide. It is the same move as unit 11's fundamental lemma of the calculus of variations, where the test objects are bump functions instead of balls, and it is how every integral conservation law in this course is converted into a differential one — unit 22 does it for mass and momentum and gets the continuity equation and the stress tensor.

Why $C^1$ and not merely continuous. The proof applies the divergence theorem to $\vec g$, so it needs $\vec g$ to be $C^1$. When $\rho$ is $C^1$ it is: write $\vec g(\vec x) = -G\int\rho(\vec x - \vec u)\,\vec u/\lVert\vec u\rVert^3\,\mathrm dV(\vec u)$ and differentiate under the integral, the derivative landing on $\rho$. For a density that is only continuous, $\Phi$ can fail to be twice differentiable and Poisson's equation can fail pointwise; the sharp condition is Hölder continuity of $\rho$. Cited: Kellogg, Foundations of Potential Theory (1929), ch. VI.

Where the hypothesis is needed. At a surface where $\rho$ jumps — the boundary of the ball, or example 1's shell — $\operatorname{div}\vec g$ does not exist in the classical sense, and the differential form has to be replaced by the integral form or by a jump condition. Unit 20 introduces distributions and the statement becomes clean again: $\nabla^2(-1/4\pi r) = \delta$, which retroactively legitimises writing a point mass as $\rho = m\delta$.

Read backwards, Poisson's equation is a recipe. Given $\rho$, solve for $\Phi$. That is a boundary value problem, and unit 23 solves it with a Green's function — the potential of a point source — recovering exactly the integral this section started from. The circle closes: the integral formula and the differential equation are the same statement, and which one is useful depends entirely on what you are given.

05 · the field

What "by symmetry" actually buys

equivariance of the field

Let $G \subseteq O(3)$ and suppose $\rho\circ A = \rho$ for every $A \in G$. Then $$\vec g(A\vec x) = A\,\vec g(\vec x) \quad\text{for all } A \in G .$$ Consequently $\vec g(\vec x)$ is fixed by every element of the stabiliser $G_{\vec x} = \{A \in G : A\vec x = \vec x\}$, so it lies in the subspace $\operatorname{Fix}(G_{\vec x})$.

Proof. Substitute $\vec y = A\vec z$ in the defining integral; $|\det A| = 1$ and $\lVert A\vec u\rVert = \lVert\vec u\rVert$, so $\vec g(A\vec x) = -G\!\int\rho(A\vec z)\,A(\vec x-\vec z)/\lVert\vec x-\vec z\rVert^3\,\mathrm dV = A\vec g(\vec x)$, using $\rho(A\vec z) = \rho(\vec z)$. $\square$

Read the consequence with $A\vec x = \vec x$: then $\vec g(\vec x) = A\,\vec g(\vec x)$. Any symmetry that leaves the point where it is must leave the field there unchanged too.

For a spherically symmetric $\rho$, the rotations about the axis through $\vec x$ fix $\vec x$. A field vector with any sideways part would be turned round a cone by them, yet must stay put: so the sideways part is zero and $\vec g(\vec x)$ points along the axis.

The stabiliser computation. For $\vec x \ne 0$ and $G = O(3)$, an element $A$ with $A\vec x = \vec x$ preserves the line through $\vec x$ pointwise and preserves the orthogonal plane, acting on it by any element of $O(2)$. A vector $\vec v$ fixed by all of them must have its component in that plane fixed by every rotation of the plane, hence zero. So $\operatorname{Fix}(G_{\vec x}) = \mathbb R\hat r$, which is the statement "$\vec g$ is radial", proved rather than asserted. The figure is this paragraph drawn.

Why $\rho$ invariant gives $\Phi$ invariant too. The same substitution in the potential integral gives $\Phi(A\vec x) = \Phi(\vec x)$: a scalar is invariant where a vector is equivariant. The difference is exactly that $\vec g$ has an index to be rotated and $\Phi$ has none, which is the first appearance of the distinction that unit 16 turns into the definition of a tensor of type $(p,q)$.

05 · the field

The spherical case: one unknown function

the spherical case

$\rho$ a function of $r$ alone: take $G = O(3)$. The stabiliser of $\vec x \ne 0$ is the copy of $O(2)$ fixing the axis through $\vec x$, whose fixed subspace is $\operatorname{span}(\hat r)$ — one dimension. So $$\vec g(\vec x) = g(r)\,\hat r ,$$ one unknown function of one variable. Gauss's law then determines it with a single equation per radius.

And $g(r)$ really is one number per sphere: a rotation carries any point of the sphere $S_r$ to any other and preserves $\rho$, so by equivariance it carries the field at one to the field at the other, same length.

That is the exact trade: symmetry reduces the unknown from a vector field on $\mathbb R^3$ to a scalar function on $(0,\infty)$, and Gauss's law does the rest. Neither alone is enough.

The same theorem gives $\vec g = g(s)\hat s$ for an infinite cylinder and $\vec g = g(z)\hat z$ for an infinite slab, once translations join the group.

05 · the field · example

Worked example 2 — the shell theorem in three lines

the whole argument

1. $\rho$ depends on $r$ alone, so $\vec g = g(r)\hat r$ by the equivariance theorem.

2. On the sphere $S_r$, $\vec g\cdot\hat n = g(r)$ is constant, so the flux is $4\pi r^2 g(r)$.

3. Gauss: $4\pi r^2g(r) = -4\pi G M(r)$, so $$\boxed{\,g(r) = -\frac{G\,M(r)}{r^2}\,}$$ with $M(r)$ the mass strictly inside radius $r$.

For the shell: $M(r) = 0$ for $r < a$ and $m$ for $r > a$. Both halves of the shell theorem, and the page of algebra in example 1 was a page of algebra reproving a cancellation that symmetry had already guaranteed.

the uniform ball

$M(r) = M r^3/R^3$ inside, $M$ outside, so

$$g(r) = \begin{cases} -\dfrac{GM}{R^3}\,r, & r \le R\\[6pt] -\dfrac{GM}{r^2}, & r \ge R .\end{cases}$$

Linear inside, inverse-square outside. Only the mass below you pulls; the shell above you does nothing.

At the surface $g_0 = GM/R^2 = 9.8203\ \mathrm{m\,s^{-2}}$, with $GM = 3.986004418\times10^{14}\ \mathrm{m^3s^{-2}}$, $R = 6371$ km.

checks

Continuous at $r = R$ ✓. $r\to0$: $g\to0$, as symmetry demands — no preferred direction at the centre ✓. $r\to\infty$: a point mass ✓. Dimensions $\mathrm{LT^{-2}}$ throughout ✓.

Why $M(r)$ and not $M$. Step 3 is Gauss's law applied to the sphere $S_r$, and the enclosed mass is by definition the integral of $\rho$ over the ball of radius $r$. For the uniform ball, $M(r) = \rho\cdot\frac43\pi r^3 = M(r/R)^3$. The statement "only the mass below you pulls" is the shell theorem applied to every shell above you, so the two results on this slide are one result used twice.

The surface value, from measurement. $GM$ for the Earth is known far more accurately than $G$ or $M$ separately — to nine figures, from satellite tracking, while $G$ is known to about two parts in $10^5$ and is the worst measured of the fundamental constants. That is why $GM$ is quoted as a single number. The $9.8203\ \mathrm{m\,s^{-2}}$ here is the value for a spherical Earth of mean radius; the measured surface gravity varies from $9.7803$ at the equator to $9.8322$ at the poles, from the rotation (unit 03's centrifugal term) and the flattening (unit 10).

What symmetry did not buy. Nothing in this argument determines $\vec g$ for a non-symmetric $\rho$, and nothing in it is a substitute for example 1 if the question is about a body that is not spherically symmetric. The slide "Where Gauss's law is true and useless" makes that precise.

05 · the field

The potential inside the ball

$\vec g = -\nabla\Phi$ and $\vec g$ is radial, so $\Phi$ depends on $r$ alone and $g(r) = -\Phi'(r)$. Integrating the two branches and matching at $r = R$:

the gravitational potential of a uniform ball $$\Phi(r) = \begin{cases} -\dfrac{GM}{2R^3}\big(3R^2 - r^2\big), & r \le R\\[8pt] -\dfrac{GM}{r}, & r \ge R \end{cases}$$

Inside it is a parabola in $r$ with its minimum at the centre — a harmonic-oscillator potential, which is the whole of the next example. Outside, the ball is a point.

checks

$r = R$: both branches give $-GM/R = -6.2565\times10^7\ \mathrm{J\,kg^{-1}}$ ✓.
$r = 0$: $\Phi(0) = -3GM/2R = -9.3847\times10^7\ \mathrm{J\,kg^{-1}}$, deeper than the surface value by exactly half of it ✓.
Poisson, inside: $\Phi = \text{const} + \frac{GM}{2R^3}r^2$ and $\nabla^2 r^2 = 6$, so $$\nabla^2\Phi = \frac{3GM}{R^3} = 3G\cdot\frac{\tfrac43\pi R^3\rho}{R^3} = 4\pi G\rho\ \checkmark$$ Poisson, outside: $\nabla^2(1/r) = 0$ for $r>0$ ✓, as it must be with no mass there.
Dimensions $\mathrm{L^2T^{-2}}$ ✓.

Two derivatives of $\Phi$, two checks: the first derivative reproduces $g$, the second reproduces $\rho$. Every result in this unit can be audited the same way.

05 · the field

The ball's field and potential, plotted

Inside, $g$ is linear in $r$: the mass below grows as $r^3$ and its pull falls as $1/r^2$. Zero at the centre, $-g_0$ at the surface.

$\Phi$ joins smoothly at $R$: same value and same slope, because $g = -\Phi'$ is continuous there.

The well's depth. $\Phi(0) - \Phi(R) = -GM/2R$, half the energy needed to escape from the surface. A body falling from rest at the surface reaches the centre at $\sqrt{GM/R} = 7.91\ \mathrm{km\,s^{-1}}$; escape takes $\sqrt{2GM/R} = 11.19\ \mathrm{km\,s^{-1}}$ — a ratio of exactly $\sqrt2$.

Both curves in units of the surface values, $g_0 = GM/R^2 = 9.8203\ \mathrm{m\,s^{-2}}$ and $GM/R$. At $r = R$ they meet at $-1$ only because of those units. A point mass would keep plunging (dotted); the ball's potential turns over into a parabola, the harmonic-oscillator well of the next example.

What to look for. The field's graph has a corner at the surface — a change of slope, not a jump, because the ball's density jumps from $\rho$ to $0$ there and the field is the integral of the density's effect. Compare the shell's plot, where the density was concentrated on a surface and the field itself jumped. The potential is one degree smoother again: no corner at all. Each integration smooths once.

An aside on the depth of the well. $\Phi(0) - \Phi(R) = -GM/2R$: getting from the surface to the centre is worth half as much energy as escaping from the surface to infinity. The escape speed from the surface is $\sqrt{2GM/R} = 11.19\ \mathrm{km\,s^{-1}}$ and the speed a body reaches at the centre, falling from rest at the surface through the tunnel, is $\sqrt{GM/R} = 7.91\ \mathrm{km\,s^{-1}}$ — a ratio of exactly $\sqrt2$, as the factor of a half requires.

05 · the field · example

Worked example 3 — the tunnel, and 84 minutes

the computation

Bore a straight tunnel, its line passing at perpendicular distance $b$ from the centre. With $\hat t$ along it and $x$ the signed distance from its midpoint, $\vec r = b\,\hat n_\perp + x\,\hat t$, $\hat n_\perp\cdot\hat t = 0$.

Inside, $\vec g = -(GM/R^3)\,\vec r$: proportional to the position vector itself. The frictionless walls push only normal to $\hat t$, so along the tunnel

$$\ddot x = \vec g\cdot\hat t = -\frac{GM}{R^3}\,\vec r\cdot\hat t = -\frac{GM}{R^3}\,x ,$$

and $b$ has vanished. With $\omega^2 = GM/R^3$ this is unit 00's third case; quadrature from $\dot x^2 = \omega^2(x_0^2-x^2)$ gives $T = 2\pi/\omega$, worked in the notes.

$\vec g = -\omega^2\vec r$ points at the centre. Split along and across the tunnel: along it, $-\omega^2x$, which is all the body feels; across it, $-\omega^2 b$, which the wall's normal force $N$ cancels. $b$ appears only in the part that does nothing.

The quadrature, in full. Multiplying $\ddot x = -\omega^2x$ by $\dot x$ and integrating (unit 00's third case) gives $\tfrac12\dot x^2 + \tfrac12\omega^2x^2 = \tfrac12\omega^2x_0^2$, where $x_0 = \sqrt{R^2-b^2}$ is the half-length of the tunnel and the body starts from rest at $x = x_0$. So $\dot x = -\omega\sqrt{x_0^2-x^2}$ on the first quarter cycle, and $$T = \frac{4}{\omega}\int_0^{x_0}\frac{\mathrm dx}{\sqrt{x_0^2-x^2}} = \frac{4}{\omega}\Big[\arcsin\frac{x}{x_0}\Big]_0^{x_0} = \frac{2\pi}{\omega},$$ with $x_0$ cancelling — the amplitude drops out, which is what makes the motion isochronous and is the one property of the linear restoring force that matters here.

Why $\vec r\cdot\hat t = x$. With $\vec r = b\hat n_\perp + x\hat t$ and the two unit vectors orthogonal, $\vec r\cdot\hat t = x$ exactly. Nothing is approximated. The perpendicular component of $\vec g$ is $-(GM/R^3)b\,\hat n_\perp$, which the tunnel wall supplies as a normal force — that is what a tunnel is for — and which does no work.

05 · the field · example

84 minutes, from anywhere to anywhere

the number

$\omega = 1.24153\times10^{-3}\ \mathrm{s^{-1}}$, so $T = 2\pi\sqrt{R^3/GM} = 5060.8\ \mathrm s = 84.3$ min, and the one-way trip is $42.2$ minutes — from anywhere to anywhere.

checks

Dimensions $\sqrt{\mathrm L^3/(\mathrm{L^3T^{-2}})} = \mathrm T$ ✓; independent of the falling body's mass ✓. Through the centre, top speed $\omega R = 7.91\ \mathrm{km\,s^{-1}}$, the circular-orbit speed at the surface — and $T$ is that orbit's period ✓. The tunnel motion is the orbit's shadow on a diameter.

Idealised away: uniform density — the real Earth is centrally condensed, and P12 gets $75.9$ min — and also air, rotation and rock at $5000\ \mathrm K$.

A satellite skimming the surface and a train in a tunnel along a diameter, released together: the train stays directly below the satellite, at $x = R\cos\omega t$, because the satellite's horizontal coordinate obeys the same $\ddot x = -\omega^2 x$. A chord of half-length $x_0$ is the shadow of a circle of radius $x_0$ at the same $\omega$.

The orbit connection, in detail. A circular orbit of radius $R$ has $\vec r(t) = R(\cos\omega t, \sin\omega t, 0)$ with $\omega^2 R = GM/R^2$, so $\omega = \sqrt{GM/R^3}$: the same $\omega$. Its first component obeys $\ddot x = -\omega^2x$. The tunnel through the centre and the skimming orbit are the same differential equation, and for the uniform ball the coincidence extends to every chord because the interior field is linear in $\vec r$. Any field linear in position gives a closed elliptical orbit centred on the origin — the isotropic oscillator — which is the other force law with closed orbits in Bertrand's theorem, met in unit 08.

The quoted number varies. $84.3$ min uses $GM = 3.986004418\times10^{14}$ and the mean radius $6371$ km. Textbooks usually quote $84.4$ min, which is $2\pi\sqrt{R/g}$ with the equatorial radius $6378$ km and $g = 9.81$, or $84.5$ min, which is $2\pi\sqrt{R^3/GM}$ with the equatorial radius $6378.137$ km. The spread is comparable to the flattening, and is smaller than the error made by assuming uniform density, so none of the digits after the first two mean anything about the real Earth.

The question is old. In their correspondence of 1679 Newton proposed dropping a body from a height to exhibit the Earth's rotation, and the exchange with Hooke that followed turned on what path it would take if the Earth let it through — the correspondence that set Newton back to work on orbits. The modern version, under the name gravity train, is Paul Cooper's analysis in the American Journal of Physics in 1966. The physics is right and the engineering is not: the deepest borehole ever drilled, the Kola superdeep, reached $12.3$ km, which is $0.2$ per cent of the way.

05 · the field

Where Gauss's law is true and useless

Gauss's law holds for every closed surface and every mass distribution. It determines $\vec g$ only when symmetry has already reduced $\vec g$ to one unknown number on the surface.

a uniform cube

Take $\rho$ constant on a cube of side $L$, and a concentric cubical surface outside it. Gauss gives $$\oint\vec g\cdot\mathrm{d}\vec A = -4\pi G\rho L^3$$ — one number. But the symmetry group is the $48$-element octahedral group, and the stabiliser of a generic face point is trivial, or on a mirror line is generated by one reflection: its fixed subspace has dimension $3$ or $2$. Only at face centres, edge midpoints and corners is it $1$. The direction of $\vec g$ is not fixed, its magnitude varies across the face, and one equation cannot produce two unknown functions of two variables.

The cube's actual field on its own surface, in the plane through its centre parallel to a face, computed from the defining integral. Normal and strongest at face centres, at $45^\circ$ at the edges, and turning in between: no single number $g$ can be pulled out of the flux integral.

What happens if you try anyway. Writing $\oint\vec g\cdot\mathrm d\vec A = g\cdot6L^2$ over the cube's own surface assumes both things that are false, and the resulting "$\lVert\vec g\rVert = 2\pi G\rho L/3 = 2.094\,G\rho L$ everywhere on the surface" is wrong by a factor that varies over it. Evaluating the defining integral numerically gives $2.597\,G\rho L$ at a face centre, $2.194\,G\rho L$ at the midpoint of an edge and $1.679\,G\rho L$ at a corner, where $\vec g$ points along the body diagonal and is normal to no face. The error is not small and is not a correction: it is the difference between an answer and a category mistake. The places symmetry does fix the direction are the face centres, edge midpoints and corners, whose stabilisers ($C_{4v}$, $C_{2v}$, $C_{3v}$) have one-dimensional fixed subspaces — and even there the group does not act transitively on the face, so the magnitude is still an unknown function.

How the figure was computed. For a uniform box the defining integral can be done in closed form: one integration of $(x_3 - y_3)/\lVert\vec x - \vec y\rVert^3$ along $y_3$ is elementary, and the remaining double integral of $1/\lVert\vec x-\vec y\rVert$ over each face has a primitive in logarithms and an arctangent (Nagy's formula, cited on the next slide). The values quoted above come from it and agree with a brute-force Riemann sum to the digits shown.

05 · the field

When Gauss's law does fix the field

the precise statement

Gauss's law determines $g$ on a surface $S$ when there is a group $G$ preserving $\rho$ and $S$ such that

$\operatorname{Fix}(G_{\vec x})$ is one-dimensional at each $\vec x \in S$, fixing the direction of $\vec g$; and
$G$ acts transitively on $S$, fixing its magnitude to one constant.

The three classic families: spheres about a spherically symmetric source, and — allowing translations, for idealised infinite sources — coaxial cylinders about a cylindrical one and planes parallel to a slab.

The cube of the previous slide fails both conditions on a generic face point. The sphere meets both everywhere, which is why example 2 took three lines.

Everything else needs the integral of example 1, or the multipole expansion of unit 10, or a numerical solution of Poisson's equation. The cube's field does have a closed form, in logarithms and arctangents — but it comes from doing the integral, not from Gauss (cited: Nagy, Geophysics 31 (1966) 362).

Transitivity is the second half, and it is easy to forget. Knowing $\vec g$ is radial is not enough to pull $g$ out of the flux integral: you also need $g$ to be the same at every point of the surface, which is what "some element of $G$ carries any point of $S$ to any other" gives. For the sphere both conditions hold; for the cube neither does. A useful test case is an ellipsoid, where the field is known in closed form (this is exactly Gauss's 1813 paper), but not by this method — it is found by a change of variables in the defining integral.

The honest role of Gauss's law in the non-symmetric case is as a constraint and a check, not a solution method: whatever $\vec g$ turns out to be, its flux through every closed surface must come out as $-4\pi GM_{\text{enc}}$. Problem P9 works the precise statement out.

section 4

Empty space

What Laplace's equation forces on a potential — and why gravity alone cannot hold anything still.

05 · empty space

The Laplacian is a trace

the Laplacian

Unit 04's Hessian of $f$ at $a$ is the symmetric bilinear form $\operatorname{Hess}f(a)$ with matrix $(\partial_i\partial_jf)$ — symmetric by Clairaut's theorem (unit 03). Define $$\nabla^2 f := \operatorname{tr}\operatorname{Hess}f = \operatorname{div}(\nabla f),$$ which in Cartesian coordinates is $\sum_i\partial_i^2f$.

Invariance: an orthogonal change of coordinates takes the matrix $H$ to $Q^{\!\top}HQ = Q^{-1}HQ$, whose trace is unchanged. The Laplacian is a geometric object, not an artefact of axes.

harmonic function

$f$ with $\nabla^2f = 0$ on an open set — a solution of Laplace's equation. By Poisson, $\Phi$ is harmonic exactly where there is no mass.

what the Laplacian measures

For $f$ of class $C^4$, the average of $f$ over the sphere of radius $r$ about $a$ is $$\langle f\rangle_r = f(a) + \frac{r^2}{6}\nabla^2f(a) + O(r^4).$$ So $\nabla^2f(a)$ is, up to a constant, how much $f$ at $a$ falls short of its own neighbourhood average.

full proof in the notes (N)

Harmonic therefore means: equal to its own average, to this order. Section 4's first theorem upgrades "to this order" to "exactly".

The factor is $r^2/2n$ in $\mathbb R^n$; the $6$ is $2\times3$.

Full proof of the averaging formula

Taylor in $\mathbb R^n$ (unit 04), to third order with a remainder $O(r^4)$ since $f$ is $C^4$: for $\hat n$ a unit vector, $f(a + r\hat n) = f(a) + r\,\nabla f(a)\cdot\hat n + \tfrac{r^2}{2}H_{ij}n_in_j + \tfrac{r^3}{6}C_{ijk}n_in_jn_k + O(r^4)$, with $C_{ijk} = \partial_i\partial_j\partial_kf(a)$. Average over the unit sphere. The linear term averages to zero because $\hat n$ and $-\hat n$ both occur. For the quadratic term, $\langle n_in_j\rangle = \delta_{ij}/3$: the average is a symmetric matrix invariant under all rotations, hence a multiple of $\delta_{ij}$, and taking the trace gives $\langle n_in_i\rangle = 1 = 3\lambda$. Hence $\tfrac{r^2}{2}H_{ij}\delta_{ij}/3 = \tfrac{r^2}{6}\nabla^2f(a)$. The cubic term averages to zero because it is odd in $\hat n$, leaving the $O(r^4)$ remainder. For $f$ only $C^2$ the same argument gives an error $o(r^2)$, which is all the slide's reading needs. $\square$

Why the trace needs an inner product, and why that is worth noticing now. The Hessian is a bilinear form — it eats two vectors — while a trace is naturally defined for a linear map, which eats one vector and returns one. Converting between them needs the inner product, exactly as converting the derivative (a covector) into the gradient (a vector) did in unit 04. In Cartesian coordinates with the standard inner product the conversion is invisible. Unit 32 makes it visible: there the Laplacian is $g^{ij}\nabla_i\nabla_j$, and the $g^{ij}$ is this inner product, no longer constant.

The averaging statement is also the reason the Laplacian governs diffusion (unit 18's parabolic case): $\partial_tu = k\nabla^2u$ says each point moves towards its neighbourhood average at a rate proportional to how far below it it is, which is a precise version of "heat flows from hot to cold".

05 · empty space

The mean value property

the mean value property

If $\Phi$ is harmonic on an open set containing the closed ball $\bar B(a,R)$, then for every $0 < r \le R$ $$\Phi(a) = \frac{1}{4\pi r^2}\oint_{S(a,r)}\Phi\,\mathrm{d}A .$$

proof

Put $A(r) = \dfrac{1}{4\pi}\displaystyle\int_{S^2}\Phi(a+r\hat n)\,\mathrm d\Omega$, the average over the unit sphere $S^2$ of directions $\hat n$, with $\mathrm d\Omega = \sin\theta\,\mathrm d\theta\,\mathrm d\varphi$. Differentiating under the integral sign (legitimate: see the notes), $$A'(r) = \frac{1}{4\pi}\int_{S^2}\!\nabla\Phi\cdot\hat n\,\mathrm{d}\Omega = \frac{1}{4\pi r^2}\oint_{S(a,r)}\!\nabla\Phi\cdot\mathrm{d}\vec A$$ $$= \frac{1}{4\pi r^2}\int_{B(a,r)}\!\nabla^2\Phi\,\mathrm{d}V = 0$$ by the divergence theorem. So $A$ is constant on $(0,R]$, and $A(r)\to\Phi(a)$ as $r\to0$ by continuity. $\square$

this is the shell theorem, read backwards

Apply it to $\Phi(\vec x) = -Gm/\lVert\vec x\rVert$, harmonic away from the origin, and to a sphere of radius $s$ centred at $\vec d$ with $s < \lVert\vec d\rVert$: $$\Big\langle \frac{1}{\lVert\vec x\rVert}\Big\rangle_{\text{sphere}} = \frac{1}{\lVert\vec d\rVert} .$$ That average is exactly example 1's integral, with the roles of shell and field point exchanged — and the answer, $1/\max$, is exactly what example 1 found.

Two computations that looked unrelated are one theorem. The inside case of example 1 is the same statement for a sphere enclosing the origin, where the mean value property does not apply, and the answer is $1/a$ rather than $1/\lVert\vec d\rVert$.

The solid-ball version. Averaging the sphere statement over $r$ with weight $4\pi r^2$ gives $\Phi(a) = \frac{1}{|B|}\int_B\Phi\,\mathrm dV$ as well: harmonic functions equal their solid averages too. Either version implies the other, and the solid one is often more convenient because it needs no surface measure.

The converse is true and is not obvious: a continuous function with the mean value property on every small sphere is harmonic, and in particular is automatically $C^\infty$. Cited: Evans, Partial Differential Equations, §2.2. For a $C^2$ function the converse is one line from the proof on the slide: $A'(r)$ is $r/3$ times the average of $\nabla^2 f$ over the ball, so if $\nabla^2 f(a) \ne 0$ then $A$ is not constant for small $r$. The continuous case needs the solution of Laplace's equation with given boundary values on a ball, which this course meets in unit 23. It is the reason harmonic functions are as rigid as they are — knowing one on the boundary of a region determines it inside, and knowing it on any open set determines it wherever it extends (real analyticity).

Differentiating under the integral sign is the Leibniz rule: if $F(r,\hat n)$ and $\partial_rF$ are continuous on $[0,R]\times S^2$, then $\frac{\mathrm d}{\mathrm dr}\int F\,\mathrm d\Omega = \int\partial_rF\,\mathrm d\Omega$. The proof is uniform continuity of $\partial_rF$ on the compact product plus the mean value theorem, and it is the single-variable result the learner already has, applied with a parameter.

05 · empty space

The maximum principle, and why gravity cannot trap you

the maximum principle

A harmonic function on a connected open $U$ that attains its maximum at an interior point is constant. Consequently, on a compact $D$, $\max_D\Phi = \max_{\partial D}\Phi$ — and the same for minima, applying this to $-\Phi$.

proof

Let $M = \max_U\Phi$ and $E = \{x\in U : \Phi(x) = M\}$, nonempty and closed in $U$ by continuity. $E$ is also open: if $\Phi(a) = M$ then for small $r$ the mean value property gives $M = \langle\Phi\rangle_r$ with $\Phi \le M$ pointwise, and a continuous function $\le M$ with average $M$ is identically $M$ — so a whole ball lies in $E$. A nonempty subset of a connected set that is open and closed is the whole set. $\square$

Earnshaw's theorem, now with its reason

Unit 04 met this as an arithmetic fact about one configuration: the Hessian's eigenvalues summed to zero, so they could not all be positive. That sum is $\nabla^2\Phi$, and it vanishes because there is no mass there.

So in vacuum $\Phi$ has no strict interior minimum anywhere, for any arrangement of sources: no point of empty space is a stable equilibrium under static gravity alone.

The same argument in electrostatics is why every charged-particle trap — Penning, Paul, magnetic, optical — uses motion, magnetism or time dependence to evade it.

The last step of the proof, spelled out. Suppose $h$ is continuous, $h \le M$ on a sphere, and its average is $M$. If $h(p) < M$ at one point, continuity gives a neighbourhood on which $h \le M - \epsilon$, of positive area, and the average is then strictly below $M$. So $h \equiv M$ on the sphere, for every small radius, hence on a ball.

Lagrange points are not a counterexample. $L_4$ and $L_5$ in the Sun–Earth system are stable, and they are points of empty space. The escape is that stability there is not a statement about a static potential: in the rotating frame the effective potential (gravitational plus centrifugal, unit 03) is not harmonic — the centrifugal term contributes $\nabla^2(-\tfrac12\omega^2s^2) = -2\omega^2$ — and even so $L_4$ is a maximum of it, stabilised only by the Coriolis term, which is velocity-dependent and does no work. Trojan asteroids at Jupiter's $L_4$ and $L_5$ are the observational confirmation, and the analysis is a standard application of unit 02's linearisation.

Earnshaw's theorem was proved in 1842 for the electrostatic case. Its loopholes are instructive because each is a hypothesis being broken: diamagnetic levitation works because $\chi < 0$ makes the energy, proportional to $-\chi B^2$, lowest where $B^2$ is smallest, and $B^2$ is not harmonic even though each component of $\vec B$ in empty space is: $\nabla^2(B_i^2) = 2\lVert\nabla B_i\rVert^2 \ge 0$, so $B^2$ can have a local minimum (though never a strict maximum) in empty space; a Paul trap works by making the saddle rotate faster than the particle can fall out of it; optical tweezers use radiation pressure, not a static conservative field.

05 · empty space

Two masses and a saddle

Two equal masses $m$ at distance $2a$. By symmetry the midpoint is an equilibrium: the pulls cancel. Is it stable?

Along the line joining them $\Phi$ has a maximum there — nudge the body towards either mass and it falls on. Across the line $\Phi$ has a minimum. The curvatures are $-4$, $+2$, $+2$ in units of $Gm/a^3$: they sum to $\nabla^2\Phi = 0$, as Earnshaw demands.

consistency check

$\Phi$ inside the ball does have a minimum, at the centre — no contradiction, since there $\rho \ne 0$ and $\Phi$ is not harmonic ✓.

$\Phi$ along two lines through the midpoint, computed: $-2Gm/(a(1-u^2))$ along the line of the masses, $-2Gm/(a\sqrt{1+u^2})$ across it, with $u$ the distance from the midpoint in units of $a$. One bends down, one up — a saddle, as every equilibrium in empty space must be.

The curvatures, computed. Each mass contributes the point-mass Hessian $(Gm/a^3)(\delta_{ij} - 3n_in_j)$ — unit 04 computed it, and worked example 4 redoes it — with $\hat n$ along the line of the masses, at distance $a$. Two of them give $(2Gm/a^3)(\delta_{ij} - 3n_in_j)$, eigenvalues $-4Gm/a^3$ along the line and $+2Gm/a^3$ twice across it. Trace zero, as the maximum principle requires of any point of empty space; one negative eigenvalue, so the midpoint is unstable along the line.

Reading the plot. Both curves pass through $-2Gm/a$ at the midpoint with zero slope. Differentiating twice at $u = 0$: $-2/(1-u^2)$ has second derivative $-4$, and $-2/\sqrt{1+u^2}$ has $+2$, matching the eigenvalues. The third direction, perpendicular to the page, is another copy of the second curve.

section 5

Tides

What a field does to a body that is not a point — and the object unit 33 will rename.

05 · tides

Two free-fallers, and what is left over

Two particles in free fall, at $\vec x(t)$ and $\vec x(t)+\vec\xi(t)$, both obeying $\ddot{\vec x} = -\nabla\Phi$. Subtract:

$$\begin{aligned}\ddot{\vec\xi} &= -\big[\nabla\Phi(\vec x + \vec\xi) - \nabla\Phi(\vec x)\big]\\ &= -D(\nabla\Phi)(\vec x)\,\vec\xi + O(\lVert\vec\xi\rVert^2),\end{aligned}$$

using the total derivative of unit 03 and Taylor of unit 04. In components,

the equation of tidal deviation $$\ddot\xi_i = -\big(\partial_i\partial_j\Phi\big)\,\xi_j + O(\lVert\vec\xi\rVert^2) .$$

The field itself has cancelled. An observer falling with the first particle cannot detect $\vec g$ at all; only its variation from place to place survives.

this is what a tide is

The Earth is in free fall towards the Moon. Its centre feels nothing. Its surface, displaced by $\vec\xi$ from the centre, feels $-\partial_i\partial_j\Phi_{\text{Moon}}\,\xi_j$ — and the ocean responds to that residue, not to the Moon's pull, which is thirty times larger and is entirely cancelled by the free fall.

and this is the equivalence principle in embryo

Unit 03's fictitious forces are proportional to $m$; so is gravity, so a uniform field can be abolished by falling. What cannot be abolished is $\partial_i\partial_j\Phi$ — which unit 30 makes the definition of a real gravitational field, and unit 33 renames the Riemann tensor.

The ratio that makes the point. The Moon's direct pull at the Earth's centre is $GM_{\mathrm m}/d^2 = 3.32\times10^{-5}\ \mathrm{m\,s^{-2}}$; the tidal residue across an Earth radius is $2GM_{\mathrm m}R/d^3 = 1.100\times10^{-6}\ \mathrm{m\,s^{-2}}$, smaller by the factor $2R/d = 3.31\times10^{-2}$. If the direct pull were what raised tides, they would be thirty times larger and there would be one bulge, not two. The observation that there are two, at antipodes, is the observation that the subtraction has happened.

The error term. $O(\lVert\vec\xi\rVert^2)$ is controlled by the third derivatives of $\Phi$, so the linear description is good when $\vec\xi$ is small compared to the scale over which $\nabla\Phi$ varies — here $d$, the Earth–Moon distance. With $R/d = 1.66\times10^{-2}$ the next term is smaller by that factor, which is why the near-side residual exceeds the far-side one by about five per cent — each departing from the linear value by $\tfrac32(R/d) = 2.5$ per cent, in opposite senses.

Naming. In general relativity this equation is the geodesic deviation equation, $\ddot\xi^\mu = -R^\mu{}_{\alpha\nu\beta}u^\alpha u^\beta\xi^\nu$, and its Newtonian limit is literally the boxed line with $R_{i0j0} \to \partial_i\partial_j\Phi$. Unit 33 derives it; the point of doing it here is that nothing in the derivation above needs anything beyond unit 04, so the reader arrives at unit 33 already knowing what the answer has to look like.

05 · tides

The tidal tensor

the tidal tensor

$T := \operatorname{Hess}\Phi$, the symmetric bilinear form $$T(\vec u,\vec v) = \sum_{i,j}\partial_i\partial_j\Phi\;u_iv_j$$ on each tangent space. It is a multilinear map. Its components in a basis are the numbers $T_{ij} = \partial_i\partial_j\Phi$; that they transform as $T' = Q^{\!\top}TQ$ under a change of basis is a theorem about the matrix of a bilinear form, not the definition of the object.

Unit 16 makes this the general definition of a $(0,2)$-tensor and shows every tensor in the rest of the course is the same kind of thing.

trace-free in vacuum

$\operatorname{tr}T = \nabla^2\Phi = 4\pi G\rho$. So in vacuum $T$ is trace-free, and by the spectral theorem it has an orthonormal eigenbasis — the principal axes — with real eigenvalues $$\lambda_1 + \lambda_2 + \lambda_3 = 0 .$$

the physical consequence, immediately

Unless all three vanish, they cannot all have the same sign. Every stretch is paid for by a squeeze. There is no such thing as a purely compressive tide in empty space — which is the eigenvalue form of Earnshaw's theorem, and the reason a tide always has an axis.

Dimensions: $[T] = \mathrm{T^{-2}}$, the square of a frequency — a positive eigenvalue $\lambda$ is an $\omega^2$ for the deviation equation, and a negative one a growth rate squared ✓.

Why insist on the bilinear-map definition. Three reasons, all of which are cashed later. (i) It is basis-free, so statements like "trace-free" are statements about the object and cannot be artefacts of coordinates. (ii) It generalises without modification to spaces with no preferred basis, which is every manifold in unit 31 onwards. (iii) The transformation law, taken as a definition, gives no way to decide which objects are tensors — it can only check a candidate, and a candidate has to come from somewhere. The Christoffel symbols of unit 32 are the standard cautionary tale: they carry indices, they look like a tensor, and they are not one.

The spectral theorem is linear algebra the reader already has: a real symmetric matrix is orthogonally diagonalisable, with real eigenvalues. Its use here is the whole of the next slide. Note that $T$ is symmetric because of Clairaut's theorem (unit 03) and for no other reason — for a $\Phi$ whose second partials were discontinuous the tidal tensor could be asymmetric, and the principal axes would not exist.

The decomposition that unit 33 will need. Any symmetric $T$ splits uniquely as $\frac13(\operatorname{tr}T)\,\delta + \big(T - \frac13(\operatorname{tr}T)\delta\big)$: a trace part fixed by the local density, and a trace-free part that is not. In general relativity the two pieces become the Ricci tensor and the Weyl tensor, and the field equations fix the first algebraically from the matter while leaving the second to propagate — which is why gravitational waves (unit 38) are a vacuum phenomenon and why tides exist at all where there is no matter.

05 · tides · example

Worked example 4 — the tidal tensor of a point mass

the computation

Unit 04 computed these second derivatives once; here they are the object, not a step. $\Phi = -GM/r$, so with $\partial_i r = x_i/r$, $$\partial_i\Phi = \frac{GM\,x_i}{r^3}, \qquad T_{ij} = \partial_j\partial_i\Phi = GM\Big(\frac{\delta_{ij}}{r^3} - \frac{3x_ix_j}{r^5}\Big)$$ $$= \frac{GM}{r^3}\big(\delta_{ij} - 3n_in_j\big)$$

with $\hat n = \vec x/r$. Trace: $\delta_{ii} = 3$ and $n_in_i = 1$, so $\operatorname{tr}T = (GM/r^3)(3-3) = 0$ ✓, as vacuum requires.

Eigenvectors. $(\delta - 3\hat n\hat n^{\!\top})\hat n = \hat n - 3\hat n = -2\hat n$, and for any $\vec v \perp \hat n$, $(\delta - 3\hat n\hat n^{\!\top})\vec v = \vec v$. So

$$\lambda = \frac{GM}{r^3}\,(-2,\ 1,\ 1),$$

the first along $\hat n$, the other two in the plane perpendicular to it. Sum $= 0$ ✓.

the numbers, for the Moon

$GM_{\mathrm m}/d^3 = 8.632\times10^{-14}\ \mathrm{s^{-2}}$. Across an Earth radius the stretch is $2(GM_{\mathrm m}/d^3)R = 1.100\times10^{-6}\ \mathrm{m\,s^{-2}}$, one part in $8.9\times10^{6}$ of $g_0$.

The Sun: $5.051\times10^{-7}\ \mathrm{m\,s^{-2}}$ — the Moon wins by $2.18$, although the Sun's direct pull is $179$ times larger. Tides go as $1/d^3$ and pulls as $1/d^2$.

The second derivative, step by step. $\partial_j(GMx_ir^{-3}) = GM\big(\delta_{ij}r^{-3} + x_i\cdot(-3)r^{-4}\partial_jr\big)$ and $\partial_jr = x_j/r$, giving $GM(\delta_{ij}r^{-3} - 3x_ix_jr^{-5})$. Symmetry in $i \leftrightarrow j$ is manifest, as Clairaut requires.

The equilibrium tide, as a number. The tidal potential relative to the Earth's centre is $\delta\Phi(\vec\xi) = \tfrac12T_{ij}\xi_i\xi_j = \frac{GM_{\mathrm m}}{2d^3}\big(\lVert\vec\xi\rVert^2 - 3(\hat n\cdot\vec\xi)^2\big)$. On the Earth's surface, $\lVert\vec\xi\rVert = R$, it ranges from $-GM_{\mathrm m}R^2/d^3$ under the Moon to $+GM_{\mathrm m}R^2/2d^3$ at the waist. The ocean surface is an equipotential of $g_0h + \delta\Phi$, so the height range is $$\Delta h = \frac{3GM_{\mathrm m}R^2}{2d^3g_0} = 0.535\ \mathrm m,$$ with the Sun adding $0.246$ m, for $0.78$ m when they line up at new and full moon — a spring tide. Observed ranges are metres to fifteen metres, because ocean basins have resonant periods near $12$ hours and the response is dynamic, not static; that is a forced-oscillator problem of the kind unit 06 solves and a shallow-water wave problem of the kind unit 18 sets up. The equilibrium number is the forcing, not the answer.

05 · tides · example

Reading the bulges off the eigenvectors

worked example 4, continued

$\ddot{\vec\xi} = -T\vec\xi$, so along $\hat n$: $\ddot\xi = +2GM\xi/r^3$ — stretch, outward, on both sides since $\xi$ and $-\xi$ are both eigenvectors. Perpendicular: $\ddot\xi = -GM\xi/r^3$ — squeeze.

Two bulges along one axis, and a waist around it. One eigenvector, one theorem.

For the Moon, $M$ is its mass and $r = d$ its distance. At a general point $\vec\xi$ of the Earth's surface the residual is $-T\vec\xi = (GM/d^3)\big(3(\hat n\cdot\vec\xi)\,\hat n - \vec\xi\big)$: the eigen-directions get pure stretch or pure squeeze, everything in between a mixture.

The residual acceleration $-T\vec\xi$ at sixteen points of the Earth's surface, computed, with the Moon off to the right at distance $d$. Outward and twice as long on the Moon line, at both ends; inward across it. The ocean settles into the dashed shape: two bulges and a waist (heights exaggerated by a factor of a few million).

What to look for. The arrows on the left are as long as those on the right and point the other way: the far side is pushed away from the Moon exactly as hard as the near side is pulled towards it, to this order. That symmetry under $\vec\xi \to -\vec\xi$ is automatic for anything linear in $\vec\xi$, and it is why the tensor gives two equal bulges where the "pulled harder, left behind" story gives no reason for them to match. The arrows at $45^\circ$ are tilted towards the Moon line: they are what drives water along the surface into the bulges.

Why two high tides a day. The bulges are fixed relative to the Moon while the Earth turns beneath them (unit 03's rotating frame), so a point on the surface passes through two of them per lunar day. The lunar day is $\big(1/86164\ \mathrm s - 1/2360592\ \mathrm s\big)^{-1} = 24$ h $50$ min, giving a semidiurnal period of $12$ h $25$ min — which is exactly the observed interval between successive high tides, and is the cleanest confirmation available that the eigenvector picture is right.

05 · tides

The explanation that is usually given, and why it is wrong

the familiar story

"The near side is pulled harder than the centre, so the water piles up there. The far side is pulled less, so it gets left behind. Hence two bulges."

The arithmetic of the differences is right, and the conclusion is right, but the reasoning smuggles in the frame. "Left behind" relative to what? Only relative to the freely falling centre — which is to say, only after the subtraction on the slide "Two free-fallers" has already been performed.

the statement that survives scrutiny

In the frame falling with the Earth's centre, the residual field on a test mass at $\vec\xi$ is exactly $-T\vec\xi$. Its eigenvector along the Moon line has eigenvalue $-2GM/d^3$, so the residual is outward at both ends of that axis, and inward around the waist. The two bulges and the sign of each come from one eigenvector, with no appeal to who is pulled harder.

The test of the difference: the familiar story gives no reason for the waist to be squeezed, and no reason for the two bulges to be of nearly equal size. The tensor gives both, and gives their five per cent asymmetry as the next term in $R/d$.

A second common error is to attribute the tides to the Earth's rotation about the Earth–Moon barycentre. That contributes a genuine centrifugal term — and it is already included, exactly, in the free-fall subtraction.

Why the barycentre story is a red herring. A frame falling freely with the Earth's centre is, over the timescale of a tide, the same frame as one rotating with the Earth–Moon system about the barycentre — the centrifugal acceleration at the Earth's centre in that frame is exactly the Moon's pull there, which is what "orbiting" means. So the centrifugal term is not an extra effect to be added; it is the very cancellation that leaves $-T\vec\xi$. Adding it a second time is double counting, and doing so is the origin of the persistent claim that the far bulge is centrifugal in nature and the near one gravitational. Both bulges have the same cause.

A sharper way to see that the frame is the whole question: tides would be identical if the Earth and Moon were falling straight towards each other on a radial trajectory with no orbital motion at all. Nothing in the derivation used the orbit.

The asymmetry, exactly. The near-side residual is $GM(1/(d-R)^2 - 1/d^2)$ and the far-side is $GM(1/d^2 - 1/(d+R)^2)$; expanding both to second order in $R/d$ gives $2GMR/d^3$ times $1 \pm \tfrac32 (R/d)$ respectively, with $\tfrac32\times1.66\times10^{-2} = 2.5$ per cent each way, so the near residual exceeds the far by about $5$ per cent. It is far too small to pick out of the ocean's dynamic response, but it is real, and it is the leading correction to the linear tidal tensor.

05 · closing

What this unit unlocks

The three results to keep.

$\displaystyle\int_D\operatorname{div}\vec F\,\mathrm dV = \oint_{\partial D}\vec F\cdot\mathrm d\vec A$ — Fubini and the fundamental theorem, once per axis.
$\nabla^2\Phi = 4\pi G\rho$ — a local law in place of a sum over pairs; with symmetry, $g(r) = -GM(r)/r^2$ in three lines.
$T_{ij} = \partial_i\partial_j\Phi$ — symmetric, trace-free in vacuum, eigenvalues $(-2,1,1)GM/r^3$ for a point mass.

And the habit underneath all three: define the object without coordinates — a trace, a bilinear form — and let the component formula be a theorem.

Where each one is used again.

10: what to do when the body is not spherical — the multipole expansion, and a real satellite's orbit.
22: the divergence theorem as the engine turning every integral conservation law into a local one, with the stress tensor supplying the flux.
23: the same vector calculus for electrostatics, with Green's functions solving Poisson's equation in general.
30: the tidal tensor is what an accelerating frame cannot imitate — which is why the equivalence principle is local.
33: $T_{ij} \to R_{i0j0}$, and $\nabla^2\Phi = 4\pi G\rho$ becomes the $00$ component of unit 35's field equations.

The line from here to unit 33 is worth stating once, plainly. In this unit the gravitational field is $\vec g$ and the thing that cannot be transformed away is its derivative $\partial_i\partial_j\Phi$. In general relativity the field is the metric, the "force" $\vec g$ becomes the connection — which can be transformed away at a point, exactly as $\vec g$ can be by falling — and the thing that cannot is the curvature. The statement "$T$ is trace-free in vacuum" becomes "the Ricci tensor vanishes in vacuum", and the trace-free part that survives becomes the Weyl tensor, which carries gravitational waves. Every structural feature is already visible here, in a unit that uses nothing beyond second partial derivatives.

If one thing survives from this unit, make it the audit habit on the slide "The potential inside the ball": differentiate your answer once to recover the field, twice to recover the source. It costs two lines and catches nearly everything.

05 · checkpoint

Done when

checkpoint

With a blank sheet and no notes, you can:

prove the shell theorem by Gauss's law in under half a page — stating why $\vec g$ is radial, computing the flux, and getting both the exterior and the interior half;
compute the tidal tensor $\partial_i\partial_j\Phi$ of a point mass, show $\operatorname{tr}T = 0$, find the eigenvalues $(-2,1,1)GM/r^3$, and read the two bulges off the eigenvectors.

Falsifiable both ways: a factor of $4\pi$ lost, $M$ used where $M(r)$ belongs, a trace that is not zero, or one bulge instead of two, each fails it. Record it with ./mr done 05.

Problems

./mr problems 05 — 12 problems, about 5 hours. Hints and worked solutions fold under each problem on the same page.

10 multipoles, when the body is not a sphere (needs 08 too) · 16 tensors properly · 22 the divergence theorem doing physics · 23 electrostatics · 30 and 33, where the tidal tensor gets its real name.

Try the checkpoint cold, on paper, before looking at anything. A good self-test for the first half: do it for a shell of non-uniform thickness and notice exactly where the argument breaks — it breaks at the equivariance step, not at Gauss's law, which remains true and stops being useful. For the second half: compute $\partial_i\partial_j\Phi$ for the interior of the uniform ball, where $\Phi = \text{const} + \frac{GM}{2R^3}r^2$, and find $T_{ij} = (GM/R^3)\delta_{ij}$ — isotropic, with trace $3GM/R^3 = 4\pi G\rho \ne 0$. A body inside matter is compressed in all three directions, which is the trace part, and is not tidally stretched at all. That contrast is the cleanest single illustration of why the trace matters.

If the derivation stalls, the usual place is the step from "$\rho$ is invariant" to "$\vec g$ is radial". Go back to "What 'by symmetry' actually buys" and redo the stabiliser computation; it is three lines and it is the only part of the argument that is not bookkeeping.

Flux, Gauss, and gravity stops being a force between pairs

One ball, five sections

Integrating over regions

The integral over a box

Regions that are not boxes

Fubini: one integral at a time

One triangle, sliced two ways

Change of variables, and what a determinant is for

What a linear map does to volume

Spherical coordinates, and where the chart dies

The volume element $r^2\sin\theta$

Worked example 1 — a shell, by brute force (setting up)

Worked example 1 — down to one integral

Worked example 1 — the cancellation

The shell's potential and field, plotted

Flux

Surfaces, and the vector area element

Flux, and why the parametrisation does not matter

Orientation, and the sphere's area element

Solid angle

Why $1/r^2$, and no other power

What the exponent means

Divergence, defined without coordinates

The divergence theorem, proved for a box

From a box to any region

Gravity becomes a field

The field of a continuous distribution

Gauss's law for gravity

Gauss's law: the sign, and a check

The field equation

What "by symmetry" actually buys

The spherical case: one unknown function

Worked example 2 — the shell theorem in three lines

The potential inside the ball

The ball's field and potential, plotted

Worked example 3 — the tunnel, and 84 minutes

84 minutes, from anywhere to anywhere

Where Gauss's law is true and useless

When Gauss's law does fix the field

Empty space

The Laplacian is a trace

The mean value property

The maximum principle, and why gravity cannot trap you

Two masses and a saddle

Tides

Two free-fallers, and what is left over

The tidal tensor

Worked example 4 — the tidal tensor of a point mass

Reading the bulges off the eigenvectors

The explanation that is usually given, and why it is wrong

What this unit unlocks

Done when

Problems

Next