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

Idea. Fubini says the double integral equals either iterated integral. The limits are not part of the theorem — they are a description of the region, and a region has two descriptions. Choosing the second is free, and sometimes it is the difference between an elementary answer and none.

1. The region is \(\{(x,y) : 0 \le x \le 1,\ x \le y \le 1\}\), the triangle with vertices \((0,0)\), \((0,1)\), \((1,1)\) — equivalently \(\{0 \le x \le y \le 1\}\). Read by \(y\) first: \[ \int_0^1\!\!\int_0^y e^{y^2}\,\mathrm dx\,\mathrm dy = \int_0^1 y\,e^{y^2}\,\mathrm dy = \Big[\tfrac12e^{y^2}\Big]_0^1 = \tfrac12(e-1) = 0.8591 . \] The \(x\) integral supplied the factor \(y\) that the substitution \(u = y^2\) needed.

2. \(D = \{0 \le x \le 1,\ x^2 \le y \le 1\} = \{0 \le y \le 1,\ 0 \le x \le \sqrt y\}\). \[ \int_0^1\!\!\int_{x^2}^1 xy\,\mathrm dy\,\mathrm dx = \int_0^1 x\,\frac{1-x^4}{2}\,\mathrm dx = \frac12\Big(\frac12 - \frac16\Big) = \frac16 , \] \[ \int_0^1\!\!\int_0^{\sqrt y} xy\,\mathrm dx\,\mathrm dy = \int_0^1 y\cdot\frac{y}{2}\,\mathrm dy = \frac16 . \]

3. Both equal \(\int_D xy\,\mathrm dA\), which is by definition \(\int_B (xy)_D\) over any box \(B \supseteq D\), with the integrand extended by zero. Fubini asserts that this single number equals each iterated integral; the differing limits are two ways of saying where the extension is nonzero.

Check. On \(D\) the integrand \(xy\) lies in \([0,1]\) and \(|D| = 1 - \tfrac13 = \tfrac23\), so the answer must lie in \([0, \tfrac23]\); \(\tfrac16\) does. For part 1, \(1 \le e^{y^2} \le e\) on a triangle of area \(\tfrac12\), so the answer lies in \([0.5, 1.359]\); \(0.8591\) does.

Idea. The change of variables theorem turns a hard region into an easy one at the cost of a determinant. For a linear map the determinant is the whole story; for spherical coordinates it is a product of three lengths, because the columns happen to be orthogonal.

1. The columns of \(D\Psi\) are \[ \partial_r\Psi = (\sin\theta\cos\varphi, \sin\theta\sin\varphi, \cos\theta),\quad \partial_\theta\Psi = r(\cos\theta\cos\varphi, \cos\theta\sin\varphi, -\sin\theta),\quad \partial_\varphi\Psi = r\sin\theta(-\sin\varphi, \cos\varphi, 0), \] of lengths \(1\), \(r\), \(r\sin\theta\). Their pairwise dot products vanish: the first two give \(\sin\theta\cos\theta(\cos^2\varphi + \sin^2\varphi) - \cos\theta\sin\theta = 0\), and the third is orthogonal to both because its third component is zero and its first two are \((-\sin\varphi,\cos\varphi) \perp (\cos\varphi,\sin\varphi)\). Hence \(A^{\!\top}A = \operatorname{diag}(1, r^2, r^2\sin^2\theta)\) and \((\det A)^2 = r^4\sin^2\theta\), so \(|\det D\Psi| = r^2\sin\theta\).

It degenerates where \(r = 0\) or \(\sin\theta = 0\): the origin and the polar axis, where the chart is also non-injective, since every \(\varphi\) gives the same point. That set is a line, of measure zero, and the change of variables theorem tolerates a failure on a null set — remove a thin closed neighbourhood of the axis, apply the theorem, and let its thickness go to zero; both sides move by at most (bound on the integrand) × (volume removed).

2. Let \(T = \operatorname{diag}(a,b,c)\), so \(T\) maps the unit ball \(B_1\) onto the ellipsoid \(E\). With \(f = 1\), \[ |E| = \int_E 1 = \int_{B_1}|\det T| = abc\,|B_1| = \tfrac43\pi abc . \]

3. \(M = \int_0^R 4\pi r^2\rho_0(1 - r/R)\,\mathrm dr = 4\pi\rho_0\big(\tfrac{R^3}{3} - \tfrac{R^3}{4}\big) = \tfrac13\pi\rho_0R^3\), and the mean density is \(M/(\tfrac43\pi R^3) = \rho_0/4\).

Check. Part 2 with \(a=b=c=R\) gives \(\tfrac43\pi R^3\) ✓. Part 3: the density runs from \(\rho_0\) at the centre to \(0\) at the surface, so the mean must lie strictly between; \(\rho_0/4\) does, and it is below \(\rho_0/2\) because most of the volume is near the surface, where the density is small ✓. Dimensions: a mass \(\rho_0R^3\) ✓.

Idea. The potential is linear in the mass, so decompose the ball into shells and add. Each shell's contribution at the centre is known from the shell theorem without any further integration — which is the entire point of proving that theorem first.

1. A shell of radius \(r\) and thickness \(\mathrm dr\) has mass \(\mathrm dm = 4\pi\rho r^2\,\mathrm dr\), and at any interior point — in particular at the centre — contributes \(-G\,\mathrm dm/r\). So \[ \Phi(0) = -G\!\int_0^R\frac{4\pi\rho r^2}{r}\,\mathrm dr = -4\pi G\rho\,\frac{R^2}{2} = -2\pi G\rho R^2 = -\frac{3GM}{2R}, \] using \(\rho = 3M/4\pi R^3\). This agrees with the slides' \(\Phi(0) = -3GM/2R\).

2. Build the ball outwards. When the shell of radius \(r\) arrives from infinity, the mass already present is \(M(r) = \tfrac43\pi\rho r^3\), a sphere, which by the shell theorem acts on the incoming shell as a point mass at the centre: the potential the shell falls through is \(-GM(r)/r\). So \[ U = -\int_0^R \frac{GM(r)}{r}\,\mathrm dm(r) = -G\!\int_0^R\frac{\tfrac43\pi\rho r^3}{r}\,4\pi\rho r^2\,\mathrm dr = -\frac{16\pi^2G\rho^2}{3}\!\int_0^R\! r^4\,\mathrm dr = -\frac{16\pi^2G\rho^2R^5}{15} . \] Substituting \(\rho^2 = 9M^2/16\pi^2R^6\) gives \(U = -\tfrac35 GM^2/R\). For the Earth, \(U = -0.6\times(3.986004418\times10^{14})\times(5.9722\times10^{24})/6.371\times10^6\), that is \[ U = -2.2419\times10^{32}\ \mathrm J . \]

3. They are different quantities. \(\Phi(0)\) is a potential per unit test mass at one point — the deepest point of the well. \(U\) is the self-energy of the whole distribution, \(U = \tfrac12\int\rho\Phi\,\mathrm dV\) — each pair of mass elements \(\mathrm dm, \mathrm dm'\) at separation \(s\) contributes \(-G\,\mathrm dm\,\mathrm dm'/s\) to \(U\) once, but to \(\int\Phi\,\mathrm dm\) twice, once from each end. It is an average of \(\Phi\) over all the mass, not just the centre, and halved because each pair of mass elements would otherwise be counted twice. Explicitly \(\langle\Phi\rangle = 2U/M = -\tfrac65 GM/R\), which is \(\tfrac45\) of \(\Phi(0)\) — the mass-weighted average is shallower than the centre — and the remaining factor \(\tfrac12\) is the double counting.

Check. Dimensions of \(U\): \(\mathrm{L^3M^{-1}T^{-2}}\cdot\mathrm M^2/\mathrm L = \mathrm{ML^2T^{-2}}\), an energy ✓. The sign is negative: work must be done to disperse the ball ✓. Comparison: a thin shell of the same mass and radius has \(U = -GM^2/2R\), coefficient \(\tfrac12 < \tfrac35\), so the ball is more tightly bound — as it should be, its mass being on average closer together ✓.

Idea. Solid angle is a flux, of the field \(\hat r/r^2\), and the disc is a flat surface with a constant normal — so the integral is elementary in plane polar coordinates.

1. Put the viewing point at the origin and the disc in the plane \(z = h\), parametrised by \(\Sigma(s,\varphi) = (s\cos\varphi, s\sin\varphi, h)\) with \(0 \le s \le a\). Then \(\mathrm d\vec A = \hat z\,s\,\mathrm ds\,\mathrm d\varphi\), \(r = \sqrt{s^2+h^2}\) and \(\hat r\cdot\hat z = h/r\). Hence \[ \Omega = \int_0^{2\pi}\!\!\int_0^a \frac{h}{(s^2+h^2)^{3/2}}\,s\,\mathrm ds\,\mathrm d\varphi = 2\pi h\Big[-\frac{1}{\sqrt{s^2+h^2}}\Big]_0^a = 2\pi\Big(1 - \frac{h}{\sqrt{h^2+a^2}}\Big). \]

2. \(h \to 0\): \(\Omega \to 2\pi\). Expected: from a point just above its centre, an infinite plane fills exactly half the sky, and \(4\pi/2 = 2\pi\) ✓. \(h \gg a\): \(1 - (1 + a^2/h^2)^{-1/2} = \tfrac{a^2}{2h^2} + O(a^4/h^4)\), so \(\Omega \approx \pi a^2/h^2\) — the area divided by the square of the distance, which is what a small distant object must give ✓.

3. \(a/h = 1737.4/384400 = 4.51978\times10^{-3}\), so \[ \Omega = 2\pi\Big(1 - \frac{1}{\sqrt{1 + (a/h)^2}}\Big) = 6.4177\times10^{-5}\ \mathrm{sr}, \] and the fraction of the sky is \(\Omega/4\pi = 5.107\times10^{-6}\), about one part in \(196\,000\).

The Moon is a sphere, not a disc. The correct cone is the tangent cone, with half-angle \(\alpha\) given by \(\sin\alpha = R/d\) rather than \(\tan\alpha = R/d\), and \(\Omega = 2\pi(1-\cos\alpha) = 6.41778\times10^{-5}\) sr. The disc value is \(6.41765\times10^{-5}\) sr: a relative error of \(2.0\times10^{-5}\), of order \((R/d)^2\), and irrelevant at any plausible precision.

Check. Dimensionless ✓. Both values agree with the small-object estimate \(\pi(R/d)^2 = 6.41775\times10^{-5}\) to four figures, as they must when \(R/d \ll 1\) ✓.

Idea. Two independent calculations — a volume one and a surface one — single out the same exponent, and the divergence theorem is the statement that this was never a coincidence.

1. With \(r = (x_jx_j)^{1/2}\), the chain rule gives \(\partial_ir = x_i/r\). Then, summing over \(i\), \[ \operatorname{div}(\vec x f(r)) = \partial_i\big(x_if(r)\big) = (\partial_ix_i)f + x_i f'(r)\frac{x_i}{r} = 3f(r) + rf'(r). \]

2. \(\hat r/r^p = \vec x\,r^{-p-1}\), so \(f = r^{-p-1}\), \(rf' = -(p+1)r^{-p-1}\), and \(\operatorname{div} = (3 - p - 1)r^{-p-1} = (2-p)r^{-p-1}\), zero on \(r>0\) iff \(p = 2\). The flux through the sphere of radius \(R\) is \(R^{-p}\cdot 4\pi R^2 = 4\pi R^{2-p}\), independent of \(R\) iff \(p = 2\). Same exponent.

3. In \(\mathbb R^n\), \(\partial_ix_i = n\), so \(\operatorname{div}(\vec xf) = nf + rf'\) and \(\operatorname{div}(\hat r/r^p) = (n-1-p)r^{-p-1}\). The sphere of radius \(R\) has area \(c_nR^{n-1}\), so the flux is \(c_nR^{n-1-p}\). Both single out \(p = n-1\): the inverse-square law is the three-dimensional member of a family.

4. The divergence theorem requires \(\vec F\) to be \(C^1\) on an open set containing the whole region. For \(p = 2\) and a ball containing the origin, \(\vec F\) is undefined at the centre, so the theorem does not apply and there is nothing to contradict. Applied where it is legitimate — the region between two concentric spheres, on which \(\vec F\) is smooth — it gives \(0 = (\text{flux through the outer sphere}) - (\text{flux through the inner one})\), that is, the flux is the same through every sphere. That is exactly the constancy found in part 2, now derived rather than computed.

Check. \(p = 0\), a field of constant length pointing outward: flux \(4\pi R^2\), growing with \(R\), and \(\operatorname{div} = 2/r > 0\) — sources everywhere, consistently ✓. \(n = 2\), \(p = 1\): the two-dimensional inverse-first-power field is the divergence-free one, which is why a line mass in three dimensions has a \(1/s\) field ✓.

Idea. Where the field is smooth, the divergence theorem gives zero immediately. Where it is not, excise the bad point with a small sphere: the theorem then applies to what is left, and the small sphere's flux is computable exactly because the integrand was built to be radius-independent.

Setting. \(\vec F(\vec x) = \vec x/\lVert\vec x\rVert^3\) is \(C^\infty\) on \(\mathbb R^3\setminus\{0\}\), with \(\operatorname{div}\vec F = 0\) there (P5 with \(p = 2\)).

Case \(0 \notin D\). \(D\) is compact and does not contain the origin, so \(\mathbb R^3\setminus\{0\}\) is an open set containing \(D\) on which \(\vec F\) is \(C^1\). The divergence theorem applies directly: \[ \oint_S\vec F\cdot\mathrm d\vec A = \int_D\operatorname{div}\vec F\,\mathrm dV = 0 . \]

Case \(0 \in \operatorname{int}D\). Choose \(\varepsilon > 0\) with \(\overline{B_\varepsilon} \subset \operatorname{int}D\), possible because the interior is open. Let \(D' = D\setminus B_\varepsilon\), compact with piecewise-\(C^1\) boundary \(S \cup S_\varepsilon\). On \(S_\varepsilon\) the normal pointing out of \(D'\) points towards the origin, so it is \(-\hat r\). \(\vec F\) is \(C^1\) on an open set containing \(D'\), so \[ 0 = \int_{D'}\operatorname{div}\vec F\,\mathrm dV = \oint_S\vec F\cdot\mathrm d\vec A + \oint_{S_\varepsilon}\vec F\cdot(-\hat r)\,\mathrm dA . \] On \(S_\varepsilon\), \(\vec F\cdot\hat r = 1/\varepsilon^2\) and the area is \(4\pi\varepsilon^2\), so the second term is \(-4\pi\). Hence \(\oint_S\vec F\cdot\mathrm d\vec A = 4\pi\), for every \(\varepsilon\) and so independently of it.

What goes wrong on \(D\) itself. \(\vec F\) is not defined at the origin, is unbounded near it, and \(\operatorname{div}\vec F\) does not exist there. The hypothesis "\(C^1\) on an open set containing \(D\)" fails at exactly one point — and that one point carries the entire \(4\pi\). Writing \(\int_D\operatorname{div}\vec F = 0\) would "prove" the flux is zero, which is false for a sphere about the origin by direct computation.

Check. For \(S\) a sphere of any radius about the origin, direct evaluation gives \(\Omega = 4\pi\), matching ✓. For \(S\) far from the origin, the projection reading says every ray either misses \(S\) or crosses it an even number of times with alternating signs of \(\hat r\cdot\hat n\), so the signed projected area is zero ✓. Dimensionless throughout ✓.

Idea. Spherical symmetry plus Gauss gives \(g(r) = -GM(r)/r^2\) with no computation; the only work is bookkeeping \(M(r)\) in three regions, and then integrating once with the constants fixed by continuity.

1. \(M(r) = 0\) for \(r \le a\); \(M(r) = \tfrac43\pi\rho(r^3-a^3)\) for \(a \le r \le b\); \(M(r) = M = \tfrac43\pi\rho(b^3-a^3)\) for \(r \ge b\). Hence \[ \vec g = \vec 0\ (r<a), \qquad \vec g = -\frac{4\pi G\rho}{3}\Big(r - \frac{a^3}{r^2}\Big)\hat r\ (a<r<b), \qquad \vec g = -\frac{GM}{r^2}\hat r\ (r>b). \]

2. Outside, \(\Phi = -GM/r\). In the shell, \(\Phi' = -g\) gives \(\Phi(r) = \tfrac{4\pi G\rho}{3}\big(\tfrac{r^2}{2} + \tfrac{a^3}{r}\big) + C\), and matching at \(r = b\) to \(-GM/b = -\tfrac{4\pi G\rho}{3}(b^3-a^3)/b\) gives \[ C = -\frac{4\pi G\rho}{3}\cdot\frac{3b^2}{2} = -2\pi G\rho b^2 . \] Inside the cavity \(\vec g = 0\), so \(\Phi\) is constant and equals its value at \(r = a\): \[ \Phi(r\le a) = \frac{4\pi G\rho}{3}\cdot\frac{3a^2}{2} - 2\pi G\rho b^2 = -2\pi G\rho\,(b^2 - a^2). \]

3. On \(a<r<b\), \(\tfrac{\mathrm d}{\mathrm dr}(r - a^3/r^2) = 1 + 2a^3/r^3 > 0\), so \(\lVert\vec g\rVert\) increases through the shell; outside it falls as \(1/r^2\). The maximum is at \(r = b\), where \(\lVert\vec g\rVert = GM/b^2 = \tfrac{4\pi G\rho}{3}(b^3-a^3)/b^2\).

4. \(a \to 0\): the shell formula becomes \(-\tfrac{4\pi G\rho}{3}r\,\hat r\) and \(\Phi = \tfrac{2\pi G\rho}{3}r^2 - 2\pi G\rho b^2\), which is exactly \(-GM(3b^2-r^2)/2b^3\) with \(M = \tfrac43\pi\rho b^3\) — the solid ball ✓. \(a \to b\) at fixed \(M\): \(\vec g = 0\) inside and \[ \Phi(r<a) = -2\pi G\rho(b^2-a^2) = -\frac{3GM}{2}\cdot\frac{b^2-a^2}{b^3-a^3} = -\frac{3GM}{2}\cdot\frac{b+a}{b^2+ab+a^2} \longrightarrow -\frac{3GM}{2}\cdot\frac{2b}{3b^2} = -\frac{GM}{b}, \] the thin shell ✓.

Check. \(\Phi\) is continuous everywhere and \(\vec g\) is continuous everywhere (no surface densities here), as they must be for a bounded, piecewise continuous \(\rho\) ✓. Dimensions: \(G\rho b^2\) has \(\mathrm{L^3M^{-1}T^{-2}}\cdot\mathrm{ML^{-3}}\cdot\mathrm L^2 = \mathrm{L^2T^{-2}}\) ✓.

Idea. A cavity is not a new problem. It is a full ball plus a ball of negative density in the hole, and superposition is exactly the statement that adding the two fields is legitimate.

1. Let \(\vec g_+\) be the field of the complete ball of density \(\rho\) and radius \(R\), and \(\vec g_-\) that of a ball of density \(-\rho\) and radius \(c\) centred at \(\vec d\). Their sum has density \(\rho\) everywhere in the body and \(0\) in the cavity, so by uniqueness of the defining integral it is the field of the hollowed body.

Take \(\vec x\) inside the cavity. Then \(\lVert\vec x\rVert < R\) and \(\lVert\vec x - \vec d\rVert < c\), so both interior formulas apply: \[ \vec g_+(\vec x) = -\frac{4\pi G\rho}{3}\vec x, \qquad \vec g_-(\vec x) = +\frac{4\pi G\rho}{3}(\vec x - \vec d), \] and adding, \[ \vec g(\vec x) = -\frac{4\pi G\rho}{3}\,\vec d , \] which does not depend on \(\vec x\): the field is uniform in the cavity, points from the cavity's centre towards the ball's centre, and has magnitude proportional to the offset.

2. \(\vec d = \vec 0\): \(\vec g = \vec 0\), which is the shell theorem — a concentric cavity has no field, since the mass outside it is a set of complete shells ✓. \(c \to 0\): the cavity shrinks to the single point \(\vec d\), where the intact ball's own field is \(-\tfrac{4\pi G\rho}{3}\vec d\) — the same answer ✓.

3. The two interior fields are each linear in position and their coefficients depend only on \(\rho\), not on the radius of the ball producing them. In the subtraction the position-dependent parts cancel identically, leaving only the offset. The radius \(c\) enters only through the requirement that \(\vec x\) be inside both spheres, so the result holds for any cavity contained in the ball — including one almost as large as the ball itself, where the remaining matter is a thin crescent and the field inside it is still exactly uniform.

Check. Dimensions: \(G\rho d\) is \(\mathrm{L^3M^{-1}T^{-2}}\cdot\mathrm{ML^{-3}} \cdot\mathrm L = \mathrm{LT^{-2}}\) ✓. Numerically, with \(G = \rho = R = 1\), \(c = 0.3\), \(\vec d = (0.4,0,0)\): at the point \(\vec x = (0.45, 0.05, 0)\), in spherical coordinates centred on \(\vec x\) the radial integral of the defining integral is trivial (the \(s^2\) of the volume element cancels the \(s^{-2}\)), leaving \(\vec g = G\rho\int \hat u\,\ell(\hat u)\,\mathrm d\Omega\) with \(\ell(\hat u)\) the length of matter along the ray from \(\vec x\) in direction \(\hat u\). Evaluating that angular integral numerically gives \((-1.67552, 0.00000, 0.00000)\), against the predicted \(-\tfrac{4\pi}{3}(0.4,0,0) = (-1.67552, 0, 0)\) ✓.

Idea. "By symmetry the field is radial" is a theorem with a hypothesis, and the hypothesis is about the stabiliser of a point. Gauss's law supplies one equation per surface; symmetry must reduce the unknown to one number before that equation can solve anything.

1. Substitute \(\vec y = A\vec z\) in the defining integral. The change of variables theorem contributes \(|\det A| = 1\), and \(\lVert A\vec u\rVert = \lVert\vec u\rVert\) since \(A\) is orthogonal: \[ \vec g(A\vec x) = -G\!\int\!\rho(\vec y)\frac{A\vec x-\vec y}{\lVert A\vec x-\vec y\rVert^3}\mathrm dV(\vec y) = -G\!\int\!\rho(A\vec z)\frac{A(\vec x-\vec z)}{\lVert\vec x-\vec z\rVert^3}\mathrm dV(\vec z) = A\,\vec g(\vec x), \] using \(\rho\circ A = \rho\) and the linearity of \(A\). If \(A \in G_{\vec x}\), that reads \(\vec g(\vec x) = A\vec g(\vec x)\), so \(\vec g(\vec x) \in \operatorname{Fix}(G_{\vec x})\).

2. \(G = O(3)\), \(\vec x \ne \vec 0\). An \(A\) fixing \(\vec x\) fixes the line \(\mathbb R\hat r\) pointwise and maps the orthogonal plane to itself, acting there as an arbitrary element of \(O(2)\). A vector fixed by all of them must have zero component in that plane, since no nonzero plane vector is fixed by every rotation of the plane. So \(\operatorname{Fix}(G_{\vec x}) = \mathbb R\hat r\) and \(\vec g = g(r)\hat r\).

The further property needed is transitivity: \(G\) must carry any point of the surface to any other, so that \(g\) takes the same value everywhere on it and can be taken outside \(\oint\vec g\cdot\mathrm d\vec A = g(r)\cdot4\pi r^2\). \(O(3)\) is transitive on each sphere, so this holds.

3. Gauss's law gives one number and nothing more: \[ \oint_{\text{cube }L'}\vec g\cdot\mathrm d\vec A = -4\pi G\rho L^3 . \] The surface has area \(6L'^2\), so the average of \(\vec g\cdot\hat n\) over it is \[ \langle\vec g\cdot\hat n\rangle = -\frac{4\pi G\rho L^3}{6L'^2} = -\frac{2\pi G\rho L^3}{3L'^2} . \] This is an average, not a value: it is attained at no particular point and cannot be used to predict \(\vec g\) anywhere.

4. Write \(G\) for the full symmetry group of the cube, of order \(48\).

• A generic point of a face is moved by every non-identity element, so \(G_{\vec x} = \{e\}\) and \(\operatorname{Fix} = \mathbb R^3\), dimension \(3\).
• A point on one of the face's four mirror lines (two joining midpoints of opposite edges, two along the diagonals) is fixed by the identity and by the reflection in the plane through that line normal to the face: \(\operatorname{Fix}\) is that plane, dimension \(2\).
• The face centre is fixed by the order-8 group \(C_{4v}\) — four rotations about the face normal and four reflections — whose fixed subspace is the normal axis, dimension \(1\).

So the direction of \(\vec g\) is determined by symmetry only at the face centre, and nowhere else on the face. Even there, the second condition fails: \(G\) does not act transitively on the face (the centre goes only to the centres of the six faces), so the magnitude of \(\vec g\) remains an unknown function of two variables on each face, and one scalar equation cannot determine it.

Check. The three fixed-subspace dimensions are \(3, 2, 1\) for stabilisers of orders \(1, 2, 8\) — larger stabiliser, smaller fixed subspace, as it must be ✓. The average in part 3 has dimensions \(\mathrm{LT^{-2}}\) ✓, and letting \(L' \to \infty\) it falls as \(1/L'^2\), matching the inverse-square field of a distant point mass ✓.

Idea. Everything here follows from one fact: a harmonic function equals its own spherical average. A strict interior extremum would make a function differ from its average, so there are none.

1. \(\partial_i(1/r) = -x_i/r^3\) using \(\partial_ir = x_i/r\), and then, summing over \(i\), \[ \nabla^2\frac1r = \partial_i\Big(-\frac{x_i}{r^3}\Big) = -\frac{3}{r^3} + 3\frac{x_ix_i}{r^5} = -\frac{3}{r^3} + \frac{3}{r^3} = 0 \qquad (r>0). \]

2. Let \(M = \sup_U\Phi\) be attained at an interior point, and let \(E = \{x \in U : \Phi(x) = M\}\). \(E\) is nonempty by hypothesis and closed in \(U\) because \(\Phi\) is continuous. \(E\) is open: take \(a \in E\) and \(r\) small enough that \(\overline{B(a,r)} \subset U\). The mean value property gives, for every \(s \le r\), \(M = \Phi(a) = \langle\Phi\rangle_{S(a,s)}\) with \(\Phi \le M\) pointwise. If \(\Phi(p) < M\) at some \(p\) on that sphere, continuity gives a patch of positive area on which \(\Phi \le M - \epsilon\), and then the average would be strictly below \(M\). So \(\Phi \equiv M\) on every such sphere, hence on \(B(a,r) \subseteq E\). A nonempty subset of a connected set that is both open and closed is the whole set, so \(E = U\) and \(\Phi \equiv M\).

3. Suppose \(\Phi\) is harmonic near \(a\) and \(a\) is a strict local minimum: \(\Phi(x) > \Phi(a)\) for all \(x \ne a\) in some ball. Then on any small sphere about \(a\) the average is strictly greater than \(\Phi(a)\), contradicting the mean value property. (Equivalently, apply part 2 to \(-\Phi\), which is also harmonic.) A test mass has potential energy \(m\Phi\), so a stable equilibrium requires a strict local minimum of \(\Phi\), and in vacuum there is none.

4. The hypothesis that fails is harmonicity. At the centre of the Earth \(\rho \ne 0\), so \[ \nabla^2\Phi = 4\pi G\rho = 4\pi(6.67430\times10^{-11})(5513.4) = 4.624\times10^{-6}\ \mathrm{s^{-2}} \ne 0 . \] Earnshaw's theorem says nothing about points inside matter, and the interior potential \(\Phi = \text{const} + GMr^2/2R^3\) indeed has a genuine minimum at \(r = 0\).

Check. The interior Laplacian can be computed a second way: \(\nabla^2(GMr^2/2R^3) = (GM/2R^3)\cdot6 = 3GM/R^3 = 3(1.5414\times10^{-6}) = 4.624\times10^{-6}\ \mathrm{s^{-2}}\), the same number ✓ — and \(GM/R^3 = \omega^2\) is exactly the tunnel frequency of worked example 3, so \(\nabla^2\Phi = 3\omega^2\) inside a uniform ball ✓.

Idea. Differentiate \(-GM/r\) twice, and then stop computing and start using the spectral theorem: the physics is in the eigenvalues, and the two bulges are one eigenvector.

1. \(\Phi = -GM/r\), so \(\partial_i\Phi = GM\,x_i/r^3\) and \[ T_{ij} = \partial_j\partial_i\Phi = GM\Big(\frac{\delta_{ij}}{r^3} + x_i\cdot\big(-3r^{-4}\big)\frac{x_j}{r}\Big) = GM\Big(\frac{\delta_{ij}}{r^3} - \frac{3x_ix_j}{r^5}\Big) = \frac{GM}{r^3}\big(\delta_{ij} - 3n_in_j\big), \] with \(\hat n = \vec x/r\). It is symmetric in \(i \leftrightarrow j\), as Clairaut's theorem requires.

2. \(\operatorname{tr}T = \frac{GM}{r^3}(\delta_{ii} - 3n_in_i) = \frac{GM}{r^3}(3-3) = 0\). For the eigenvectors, note \((\delta - 3\hat n\hat n^{\!\top})\hat n = \hat n - 3\hat n = -2\hat n\), and for any \(\vec v \perp \hat n\), \((\delta - 3\hat n\hat n^{\!\top})\vec v = \vec v\). So the eigenvalues are \[ \lambda_\parallel = -\frac{2GM}{r^3}\ \text{along } \hat n, \qquad \lambda_\perp = +\frac{GM}{r^3}\ \text{twice, in the plane} \perp \hat n, \] and \(-2 + 1 + 1 = 0\) ✓.

3. \(\ddot{\vec\xi} = -T\vec\xi\). Along \(\hat n\): \(\ddot\xi = +2GM\xi/r^3\), of the same sign as \(\xi\) — the separation grows, and it grows for \(\xi > 0\) and for \(\xi < 0\) alike, since both \(\pm\hat n\) are eigenvectors with the same eigenvalue. So the cloud is stretched outward at both ends of the line to the source: two bulges. Transversely, \(\ddot\xi = -GM\xi/r^3\), of the opposite sign to \(\xi\): the cloud is squeezed around a waist. The volume of the cloud is unchanged at leading order: released at rest, each separation moves as \(\vec\xi(t) \approx (I - \tfrac12t^2T)\vec\xi(0)\), so volumes scale by \(\det(I - \tfrac12t^2T) = 1 - \tfrac12t^2\operatorname{tr}T + O(t^4)\), and \(\operatorname{tr}T = 0\).

4. Moon: \(GM_{\mathrm m}/d^3 = 4.9029\times10^{12}/(3.844\times10^8)^3 = 8.632\times10^{-14}\ \mathrm{s^{-2}}\), so the stretch across an Earth radius is \[ 2\,\frac{GM_{\mathrm m}}{d^3}R_\oplus = 1.100\times10^{-6}\ \mathrm{m\,s^{-2}} . \] Sun: \(2GM_\odot R_\oplus/d_\odot^3 = 5.051\times10^{-7}\ \mathrm{m\,s^{-2}}\). The Moon dominates by a factor \(2.18\). The two direct pulls go the other way: \(GM_\odot/d_\odot^2 \big/ GM_{\mathrm m}/d^2 = 179\), the Sun winning by two orders of magnitude. The difference is one power of distance: a tide is a derivative of the field and so falls as \(M/d^3\), while the field itself falls as \(M/d^2\). The Sun is \(389\) times further away, and \(389\) is roughly the ratio \(179 \times 2.18\).

5. Inside a uniform ball, \(\Phi = \text{const} + GMr^2/2R^3\), so \[ T_{ij} = \frac{GM}{2R^3}\,\partial_i\partial_j(x_kx_k) = \frac{GM}{R^3}\,\delta_{ij} , \] isotropic, with \(\operatorname{tr}T = 3GM/R^3 = 4\pi G\rho \ne 0\). Every direction is an eigenvector with the same eigenvalue, so there is no preferred axis and no bulge: the deviation equation reads \(\ddot{\vec\xi} = -(GM/R^3)\vec\xi\), an isotropic compression in all three directions at once. That is the pure trace part of the tidal tensor, and it exists only where there is matter.

Check. Dimensions of \(T\): \(\mathrm{L^3T^{-2}}/\mathrm L^3 = \mathrm{T^{-2}}\), the square of a frequency ✓. The Moon's stretch is \(1.120\times10^{-7}\) of \(g_0\) — one part in \(8.9\) million, which is why the solid Earth barely notices and the ocean, free to flow, does ✓. And the equilibrium tide range this implies is \(\tfrac32GM_{\mathrm m}R^2/d^3g_0 = 0.535\) m for the Moon and \(0.246\) m for the Sun, in the right range for a real tide once the ocean's dynamic response is allowed for ✓.

Idea. Inside a uniform ball the field is linear in the position vector, so its component along any straight line is linear in the distance along that line — and simple harmonic motion follows with no reference to the line's position. For a different density profile the field is different and so is the answer; the method, reduction to quadrature, is unit 00's and does not change.

1. Write \(\vec r = b\,\hat n_\perp + x\,\hat t\) with \(\hat n_\perp \perp \hat t\). Inside, \(\vec g = -(GM/R^3)\vec r\), so the component along the tunnel is \(\vec g\cdot\hat t = -(GM/R^3)\,x\), and \[ \ddot x = -\omega^2x, \qquad \omega^2 = \frac{GM}{R^3}, \] with no \(b\). Hence \(T = 2\pi/\omega = 2\pi\sqrt{R^3/GM} = 84.3\) min for every tunnel. The tunnel runs from \(x = -x_0\) to \(x = +x_0\) with \(x_0 = \sqrt{R^2-b^2}\), so the first integral \(\dot x^2 = \omega^2(x_0^2-x^2)\) gives \[ v_{\max} = \omega\sqrt{R^2-b^2}. \] For \(b = 0\): \(v_{\max} = \omega R = 7909.8\ \mathrm{m\,s^{-1}}\). For \(b = R/2\): \(v_{\max} = \omega R\sqrt{3}/2 = 6850.1\ \mathrm{m\,s^{-1}}\).

2. \(M(r) = \int_0^r 4\pi r'^2\frac{A}{r'}\,\mathrm dr' = 2\pi Ar^2\), so \[ \lVert\vec g(r)\rVert = \frac{GM(r)}{r^2} = 2\pi GA, \] constant throughout the interior. The total mass is \(M = 2\pi AR^2\), so \(2\pi GA = GM/R^2 = g_0\): the constant interior value is the surface value.

3. Through the centre, \(\ddot x = -g_0\operatorname{sgn}(x)\), which is not linear in \(x\), so the motion is not simple harmonic. Multiply by \(\dot x\) and integrate (unit 00's third case): with release from rest at \(x = R\), \[ \tfrac12\dot x^2 + g_0|x| = g_0R \implies \dot x^2 = 2g_0(R - |x|), \] and a quarter period is the time from \(R\) to \(0\): \[ \frac T4 = \int_0^R\frac{\mathrm dx}{\sqrt{2g_0(R-x)}} = \Big[\frac{-2\sqrt{R-x}}{\sqrt{2g_0}}\Big]_0^R = \sqrt{\frac{2R}{g_0}}, \qquad T = 4\sqrt{\frac{2R}{g_0}} . \] With \(R = 6.371\times10^6\) m and \(g_0 = 9.8203\ \mathrm{m\,s^{-2}}\), \(T = 4556.4\ \mathrm s = 75.94\) min, against the uniform ball's \(84.3\) min.

4. Seismology says the Earth is strongly centrally condensed: the core is about twice the mean density, and the measured interior \(g\) stays near \(9.8\)–\(10.7\ \mathrm{m\,s^{-2}}\) through the whole mantle instead of falling linearly to zero. The constant-\(g\) model is therefore much the better guide over most of the depth, and since it gives a larger restoring force at every depth than the uniform model, the correction to \(84.3\) min is negative: the true period is shorter, and of order \(76\) min.

Check. Dimensions: \(\sqrt{\mathrm L/(\mathrm{LT^{-2}})} = \mathrm T\) ✓. The ratio of the two periods is \[ \frac{4\sqrt{2R/g_0}}{2\pi\sqrt{R/g_0}} = \frac{2\sqrt2}{\pi} = 0.9003, \] independent of \(R\) and \(g_0\) — and \(75.94/84.35 = 0.9003\) ✓. A last consistency check on part 1: \(v_{\max}\) at \(b = 0\) is \(\sqrt{GM/R} = 7.91\ \mathrm{km\,s^{-1}}\), the circular-orbit speed at the surface, as worked example 3 noted ✓.