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Let $s(t)$ be the length of the free straight segment from the current contact point to the mass, and let $\theta$ measure the contact point's rotation from first contact. For an inextensible taut rope, winding through angle $\theta$ uses length $R\theta$:The point mass follows an involute of the pole's circular cross-section. Its instantaneous velocity is perpendicular to the free rope. To verify the geometry explicitly, use planar polar unit vectors at the contact point:The rope's tension therefore does no work, and the speed stays at its initial value $v$. The geometry gives $v=s\dot{\theta }$. ConsequentlyThe distance from the pole's axis is $\sqrt{R^2+s^2}$, so a point mass first reaches the pole when $s=0$. The winding time around a circular pole is thusThis has units of time. The model assumes a light inextensible rope, no energy loss at contact and a point mass. The required tension $m{v}^2/s$ diverges at the ideal final contact, so finite strength and size matter very close to impact.

A spatially uniform external gravitational acceleration moves a body's centre of mass without stretching it. Tidal forces arise from the variation of that acceleration across the body. Let a distant mass $M$ lie at distance $D$ along unit vector $\mathbf{n}$, and let $|\mathbf{r}|\ll D$ denote position relative to the body's centre. Subtract the centre's acceleration. To first order,The tidal tensor is $T_{ij}=GM(3n_i{n}_j-{\delta }_{ij})/D^3$. It stretches along the line to the source with eigenvalue $2GM/D^3$ and compresses in the two perpendicular directions with eigenvalues $-GM/D^3$. Its trace vanishes outside the source, consistently with Laplace's equation for the gravitational potential.

After removal of the constant and uniform-acceleration terms, the tidal potential isAn ideal equilibrium ocean on a sphere of radius $R$ follows an equipotential of the body's gravity plus this perturbation. For surface gravity $g$, the leading displacement isup to a constant fixing total water volume. The quadrupolar form produces bulges on both the near and far sides; the far-side bulge follows from differential acceleration, without postulating a repulsive gravitational force.

The Moon and Sun both raise terrestrial tides, with relative strength set mainly by $M/D^3$, rather than by the total gravitational pull $M/D^2$. Their contributions reinforce near new and full Moon to produce spring tides, and partly cancel near quadrature to produce neap tides. Earth's rotation relative to the tidal pattern gives a strong semidiurnal component in the simplest geometry. Real tide heights and timing depend on ocean depth, coastlines, basin resonances, Coriolis acceleration, friction and the declination of the forcing bodies. Tidal dissipation produces phase lags and torques, exchanging rotational and orbital angular momentum and gradually altering the spin and orbit. The equilibrium bulge is therefore a useful starting model, not an exact local tide forecast.

Use local east, north and upward unit vectors at the equator, with Earth's rotation $\mathbf{\Omega }={\Omega }_E\hat{\mathbf{n}}$. The parcel starts at rest relative to the ground. To first order in ${\Omega }_E$, its unperturbed vertical velocity is $\dot{z}=-gt$ and its fall time from height $h$ isThe Coriolis acceleration is $-2\mathbf{\Omega }\times \mathbf{v}$. Since north crossed with downward points west, its negative points east:Integrating with $x_E(0)={\dot{x}}_E(0)=0$ gives the equatorial Coriolis deflection in free fall,For $h=200\mathrm{m}$, $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$ and ${\Omega }_E=2\pi /(86164\mathrm{s})=7.2921\times 10^{-5}{\mathrm{s}}^{-1}$, $t_f=6.39\mathrm{s}$ andThis neglects air resistance, variation of $g$ over the fall, and terms of order ${\Omega }_E^2$. The measured local $g$ already includes the background centrifugal correction. “Stationary” is interpreted as hovering above a fixed ground location, not being stationary in an inertial frame.

For the XXZ spin chain, write $\Delta =1+\alpha$ and replace each spin by a classical vector of length $S$. A bond is $J{S}^2{\mathbf{n}}_m^{\mathsf{T}}\mathrm{diag}(1,1,\Delta ){\mathbf{n}}_{m+1}$, whose minimum is $-J{S}^2\max (1,|\Delta |)$. For an even periodic chain all bonds can attain that bound simultaneously. The classical ground states of an XXZ spin chain therefore give $\alpha >0$ gives the two easy-axis Néel states ${\mathbf{S}}_m=(-1{)}^{m}S\hat{\mathbf{z}}$ and its reversal, with energy $-JN{S}^2(1+\alpha )$.

For $-2<\alpha <0$ the ground states are planar Néel configurations, ${\mathbf{S}}_m=(-1{)}^{m}S(\cos \phi ,\sin \phi ,0)$, with arbitrary $\phi$: their degeneracy is a circle and their energy is $-JN{S}^2$. This includes the easy-plane interval used later. Part (a), however, does not restrict negative $\alpha$ to that interval. For $\alpha <-2$, $\Delta <-1$ and the lowest states are instead the two uniform $z$-polarized ferromagnetic configurations, with energy $JN\Delta S^2$. At $\alpha =-2$, a rotation by $\pi$ about $z$ on alternating sites maps the classical energy to an isotropic ferromagnet: ${\mathbf{S}}_m=S((-1{)}^m{n}_x,(-1{)}^m{n}_y,n_z)$ for any unit vector $\mathbf{n}$ is a ground state. This special degeneracy is a sphere. At the other boundary, $\alpha =0$, the ordinary Néel direction is arbitrary on a sphere. These statements concern classical minimizing configurations, not the degeneracy of the exact finite-chain quantum ground state.

By the spectral theorem a symmetric positive-definite matrix $C$ has $C=Q\mathrm{diag}(c_i)Q^{\mathsf{T}}$, with $Q$ orthogonal and $c_i>0$. Define $S=Q\mathrm{diag}(\sqrt{c_i})Q^{\mathsf{T}}$. It is symmetric positive-definite and $S^2=C$. For uniqueness, any symmetric positive-definite square root $T$ commutes with $C=T^2$, hence preserves each eigenspace of $C$. On its $c$-eigenspace, $T$ is symmetric positive-definite and all its eigenvalues have square $c$, so all equal $\sqrt{c}$. Therefore $T=S$, including repeated eigenspaces.

For $\det F>0$, putThen $R^{\mathsf{T}}R=I$ and $\det R=\det F/\det U>0$, so $R\in SO(3)$; also $F=RU=VR$. Any other right polar decomposition has $F^{\mathsf{T}}F=U^2$, so uniqueness of the positive root fixes $U$ and then $R$. The left factor is consequently fixed too.

The principal stretches $v_i$ are the positive eigenvalues of $U$, equivalently the singular values of $F$. For a unit vector $N={\sum }_i{c}_i{N}_i$ in an orthonormal eigenbasis of $U$,Thus $\boxed{\underset{|N|=1}{\max }|FN|=v_{\max }}$, attained precisely by the unit vectors in the eigenspace of $U$ belonging to $v_{\max }$. With a simple largest eigenvalue these are its two opposite unit eigenvectors; degeneracy allows every unit combination within that eigenspace.

For the simple shear,A line with nonzero direction $N$ keeps its length exactly whenHence $\boxed{N_2=0\text{or}2N_1+\gamma N_2=0}$, with $N_3$ arbitrary. Every segment in the cube parallel to one of these directions preserves length; there is no restriction on its starting point other than remaining in the cube.

The two in-plane stretches are $v_{\pm }=(\sqrt{4+{\gamma }^2}\pm \gamma )/2$, and the third is one. A maximizing direction for $\gamma >0$ is $N\parallel (1,v_{+},0)$, since the $C$ eigenvalue equation gives $N_2/N_1=(v_{+}^2-1)/\gamma =v_{+}$. Therefore $\boxed{N\to \pm (1,1,0)/\sqrt{2}}$ as $\gamma \downarrow 0$. At exactly zero shear all directions tie; the limit refers to the selected maximizing direction for positive shear. These are the length-preserving directions and principal directions of simple shear.

Use separate moving orthonormal bases in the reference and current cylindrical coordinate systems. For pure torsion of an incompressible cylinder, the deformation map isThe unit vectors depend on the angle: ${\partial }_{\theta }{\mathbf{e}}_r={\mathbf{e}}_{\theta }$ and ${\partial }_{\theta }{\mathbf{e}}_{\theta }=-{\mathbf{e}}_r$. ConsequentlyIn particular, the last derivative includes the rotation of ${\mathbf{e}}_r$ along the cylinder. The material gradient has the reference coordinate scale factor $1/R$:The outer products specify the images of the three reference unit vectors. Thus, with current basis vectors indexing rows and reference basis vectors indexing columns,Its determinant is one, checking incompressibility. The radius is unchanged, so the current outer radius in part (d) is $a=A$. The formula at the coordinate axis is understood by continuity in Cartesian coordinates.

Let ${\mathbf{v}}_1=(c/2,0,0)$ and ${\mathbf{v}}_2=(c/(2\sqrt{2}),c/(2\sqrt{2}),0)$ in A. Each has Lorentz factor $g=2/\sqrt{3}$. The relative Lorentz factor of $S_2$ in the rest frame of $S_1$ follows from their four-velocity tensor contraction:Let $b=\sqrt{1-{\Gamma }^{-2}}$ be the relative speed divided by $c$. In the rest frame of $S_1$, take $\theta$ between the photon's propagation direction and $S_2$'s velocity. The observed angular frequency is the invariant $\omega =-k\cdot U$. Applying it to the emitter, whose rest angular frequency is ${\omega }_0$, gives ${\omega }_0=\Gamma {\omega }_1(1-b\cos \theta )$. Since ${\omega }_1={\omega }_0$,This is a relativistic Doppler effect calculation using both emitters' and receivers' rest clocks; the time delay of reception introduces no extra factor for constant velocities.

For the direction in A, write the wave four-vector as $k^{\mu }=({\omega }_A/c)(1,\mathbf{n})$, where $\mathbf{n}$ is a unit vector. Equality of the emitted and received angular frequencies gives $k\cdot (U_1-U_2)=0$. Since the two laboratory Lorentz factors are equal, their time components cancel:Thus the complete answer in three spatial dimensions isIf propagation is additionally assumed to lie in the stars' motion plane, this reduces to either direction along the internal angle bisector, $22.5^{\circ }$ or $202.5^{\circ }$ from $+x$. Reflection exchanging the stars' laboratory velocities explains the angle bisector by symmetry. The source does not specify coplanar propagation or the stars' positions, so it does not fix the sign or exclude an out-of-plane ray. For example $\mathbf{n}=(0,0,1)$ gives ${\omega }_1={\omega }_2=g{\omega }_A$ and satisfies the same no-shift condition. Initial positions can be chosen to make such a ray connect emission and reception. All these possibilities give the same rest-frame $\theta$ above.

The Pauli matrix multiplication law givesFor a unit vector $\hat{\mathbf{n}}$, it follows that $(\hat{\mathbf{n}}\cdot \mathbf{\sigma }{)}^2=I$. In the exponential series the even powers therefore multiply $I$ and the odd powers multiply $\hat{\mathbf{n}}\cdot \mathbf{\sigma }$. Summing them separately,Equivalently, the eigenvalues of $\hat{\mathbf{n}}\cdot \mathbf{\sigma }$ are $\pm 1$, so its two spectral projectors acquire phases $e^{\mp i\theta /2}$. The half-angle belongs to a spinor rotation, while the associated physical spin vector rotates through $\theta$.

Put $\alpha =\Delta \tau /\hslash$ and $\beta =E{x}_0\tau /\hslash$. The Pauli matrix multiplication law gives the pulse-and-precession spin one-half mapBecause $C^2+|\mathbf{v}{|}^2=1$, a suitable angle and unit vector for the rotation areIf the denominator vanishes, $U=\pm I$ and the axis is arbitrary. The Bloch vector $\mathbf{r}=⟨\mathbf{\sigma }⟩$ rotates through $\theta$ about $\mathbf{n}$ per period; after $N$ periods the angle is $N\theta$. The eigenphases equivalently give quasienergies $\pm \hslash \theta /(2\tau )$ modulo $2\pi \hslash /\tau$.

In the two-orbital subspace the position operator is $X=x_0{\sigma }_x$. The Heisenberg equation of motion between pulses gives $P/m=(i/\hslash )[H_0,X]=(2\Delta x_0/\hslash ){\sigma }_y$, henceThus two components of the Bloch vector directly describe the mean position and momentum; its third gives the orbital population imbalance. The electric kick rotates about the position axis, leaving $⟨X⟩$ unchanged at the kick, while free evolution mixes the position and momentum components. Starting in $|1s⟩$, one complete period giveswhich exhibits the induced motion explicitly. These are projected two-orbital observables.

The sphere $S^2$ is path-connected. For $u,v\in S^2$ which are not an antipodal pair, a continuous path isThe denominator never vanishes: a zero would make $u$ and $v$ antipodal. If $v=-u$, choose a unit vector $w$ perpendicular to $u$ and concatenate the paths from $u$ to $w$ and from $w$ to $v$ of the same form. Thus $S^2$ is connected. The quotient map $q_S$ is continuous and surjective, so the continuous image of a connected space theorem proves that ${\mathbb{RP}}^2$ is connected. In fact composing the paths above with $q_S$ also proves path connectedness.

Use the same $4\times 4$ square grid with four centre-based triangles per small square, but identify boundary points byThis gives the real projective plane. To see the model explicitly, a radial homeomorphism from the square, centred at $(1/2,1/2)$, to the closed disc carries the boundary involution to the antipodal map. The map from that disc to the projective planeis continuous and surjective: every one-dimensional vector subspace of ${\mathbb{R}}^3$ has a unit vector representative in the closed upper hemisphere. Its only identifications are antipodal boundary points. The compact-to-Hausdorff continuous bijection theorem therefore identifies this disc quotient space with ${\mathbb{RP}}^2$.

The square triangulation descends to an abstract simplicial complex $K_2$. Small-square centres remain distinct interior vertices; each triangle has one centre and two adjacent grid vertices. Boundary pairings identify whole grid edges, while preserving distinct vertices in each triangle, so this is a finite triangulation of the real projective plane. Its counts are $33$ vertices, $96$ edges and $64$ triangles, giving Euler characteristic one.

Choose $\sigma$ to be the interior triangle with vertices $(1/4,1/4)$, $(1/2,1/4)$ and $(3/8,3/8)$. Removing its relative interior and cutting from $(0,0)$ to $(1/4,1/4)$ gives the same nine-sided cut disc as before. Parametrize it by a convex polygon, transporting the boundary identifications. Its boundary word is nowThe boundary involution pairs the two $a$ sides and the two $b$ sides in matching traversal directions; the $c$ pair restores the slit. Leave $d,e,f$ unpaired. The quotient space isthe real projective plane with an open disc removed. In particular it is a compact connected surface with boundary consisting of one three-edge circle.

At an oriented internal surface with unit normal vector $\mathbf{n}$, the traction $\mathbf{t}(\mathbf{x},t;\mathbf{n})$ is the limiting contact force per unit surface area exerted by the material on the positive-$\mathbf{n}$ side on the material on the negative-$\mathbf{n}$ side:For a boundary of a chosen material volume, this convention gives the external traction when $\mathbf{n}$ is outward. The Cauchy stress tensor has componentsso the second index specifies the surface orientation and the first specifies the force component. The vectors ${\mathbf{e}}_i$ form the standard Cartesian coordinate system basis of unit vectors.

Cauchy's stress theorem states that the dependence of traction on the normal vector is linear:Repeated indices are summed according to Einstein notation. In particular, reversing the normal vector reverses the traction, and knowing the tractions on three mutually perpendicular coordinate planes determines the traction on every plane. Symmetry of the Cauchy stress tensor, for an ordinary fluid without body couples, follows separately from angular momentum conservation.

The Gram-Schmidt process constructs the vectors in order:The empty sum at $j=1$ is zero. Each subtraction removes the projection onto one previous unit vector, so $v_j$ is orthogonal to their span. Linear independence of the inputs ensures $v_j\ne 0$ in exact arithmetic, and normalization gives unit length.

In floating-point arithmetic, subtracting nearly equal projections can amplify roundoff and cause loss of orthogonality, especially for nearly dependent columns. Modified Gram-Schmidt, reorthogonalization, or Householder QR factorization can improve numerical behaviour; no implementation can remove the underlying sensitivity of nearly dependent input data.

For nonzero $x$, let $\varphi$ be its angle to the unit vector $q_1$. The scalar projection on the oriented $q_1$ axis is $\Vert x\Vert \cos \varphi$. Since $\Vert q_1\Vert =1$, the dot product identity gives $q_1^{T}x=\Vert x\Vert \cos \varphi$. Multiplying that signed scalar by $q_1$ places the foot at $p=(q_1^{T}x)q_1$.

The residual is perpendicular to the axis, because $q_1^T(x-p)=q_1^{T}x-(q_1^{T}x)(q_1^T{q}_1)=0$. An obtuse angle gives a negative scalar and a foot on the negative side of the line; $x=0$ has projection zero without needing an angle.

The first panel is a constructed geometric illustration. The second illustrates the directional stretch used later in Question 6.

A unit vector along the line is $d=(3,4{)}^{\top }/5$. Reflection preserves the orthogonal projection $d{d}^{\top }u$ onto that line and negates its perpendicular component. Therefore $Au=d{d}^{\top }u-(u-d{d}^{\top }u)$ andAs checks, $A(3,4{)}^{\top }=(3,4{)}^{\top }$, $A(-4,3{)}^{\top }=-(-4,3{)}^{\top }$, and $A^2=I$. Its determinant is minus one, as expected for a planar reflection matrix.

The Pauli matrices are the complex two-by-two matricesEach is a traceless matrix, a Hermitian matrix and a unitary matrix. Their eigenvalues are $+1,-1$, and direct matrix multiplication gives ${\sigma }_x{\sigma }_y=i{\sigma }_z$, with cyclic permutations and a minus sign for reversed order. This is the Pauli matrix multiplication law.

Together with the identity matrix, they form the Pauli operator basis of all complex two-by-two matrices. For spin one-half, the angular momentum operators are $S_j=\hslash {\sigma }_j/2$, so a spin measurement along a unit vector $n$ has outcomes $\pm \hslash /2$. For example, ${\sigma }_z$ has the standard basis as its eigenbasis, whereas ${\sigma }_x$ has normalized eigenvectors $(1,\pm 1{)}^T/\sqrt{2}$. This makes changing the measurement axis an explicit change of basis.^[ref]

For a source point mass $M$ at the origin, let $\mathbf{R}=R\mathbf{n}$, where $\mathbf{n}$ is an outward unit vector, and let $\mu =GM$ be the source's standard gravitational parameter. Differentiating $g_i=-\mu R_i/R^3$ gives the tidal tensorwhere ${\delta }_{ij}$ is the Kronecker delta and $I$ is the identity matrix. In an orthonormal basis with first axis radial, the tidal tensor isThe radial eigenvalue $2\mu /R^3$ separates nearby freely falling test particles; the two transverse eigenvalues $-\mu /R^3$ bring them together. The farther radial test particle falls inward less rapidly, so its acceleration relative to the nearer one points outward. Transverse gravitational accelerations converge toward the source. Thus an initially resting small cloud stretches radially and squeezes in both transverse directions. These are instantaneous relative accelerations; the subsequent shape also depends on initial velocities and the changing tidal tensor along the cloud's free fall.

Outside the source, the matrix trace is $2\mu /R^3-\mu /R^3-\mu /R^3=0$. This matches $\mathrm{tr}T=-{\nabla }^2\Phi =0$ in a region with zero mass density, by the Poisson equation for Newtonian gravity. Zero matrix trace means no leading change of a small initially resting cloud's volume at order $t^2$, not that each separation stays fixed. Inside matter, the same sign convention gives $\mathrm{tr}T=-4\pi G\rho$ instead.

Power iteration finds a dominant eigenvalue and its eigenvector by repeatedly applying a matrix and rescaling the resulting vector. The essential idea is selective amplification: components along different eigenvectors grow at different rates, so one direction can eventually dominate. It is the computational form of the power method and is useful when applying a large sparse matrix to a vector is cheap but computing its whole spectral decomposition is expensive.

For a square matrix over the real numbers or complex numbers $A$, choose a unit vector $x_0$ and repeatHere $\Vert \cdot {\Vert }_2$ is the Euclidean norm. If $y_{k+1}=0$, stop before division: the current vector is an eigenvector for the zero eigenvalue, and this iteration cannot continue. Otherwise, in exact arithmetic,Repeated normalization keeps the stored vectors at manageable size without changing their directions; it does not guarantee that an individual multiplication cannot exceed the range of floating-point arithmetic. The dominant eigenvalue means largest modulus, which need not mean largest signed real value.^[ref]

For a unit vector $u$ spanning the target eigenspace and a normalized iterate $x_k$, defineThis is the norm of the component in the orthogonal complement of $u$, equivalently the sine of the acute angle to its line. It ignores the arbitrary sign or complex unit factor of an eigenvector. Under power iteration convergence for a diagonalizable matrix, ${\epsilon }_k=O({\gamma }^k)$ in Big O notation. The orthogonal projection defines how we measure direction; it does not assume that the eigenbasis of $A$ is orthogonal. In computation, the projection form avoids subtracting nearly equal numbers inside the square root.

For each unit vector iterate compute the Rayleigh quotient ${\rho }_k=x_k^{\ast }A{x}_k$ and the eigenpair residual $r_k=A{x}_k-{\rho }_k{x}_k$. Use the same multiplication $A{x}_k$ for both diagnostics and the next normalized iterate. For any complex matrix, this Rayleigh quotient minimizes $\Vert A{x}_k-\mu x_k{\Vert }_2$ over complex scalars $\mu$, because $x_k^{\ast }{r}_k=0$. Under power iteration convergence for a diagonalizable matrix, ${\rho }_k\to {\lambda }_1$ and $\Vert r_k{\Vert }_2\to 0$, even when the vectors alternate signs or phases.^[ref]

A dimensionless stopping quantity iswhen the denominator is positive; $\Vert A{\Vert }_2$ is the operator norm, and a known upper bound may replace it if computing it is costly. A small eigenpair residual certifies a small backward error of an approximate eigenpair, but does not establish that the eigenvalue found is dominant.

For a Hermitian matrix with orthonormal eigenbasis $u_j$, write $x_k={\sum }_j{a}_j{u}_j$. Then the independently useful identityshows that some eigenvalue lies within $\Vert r_k{\Vert }_2$ of ${\rho }_k$. If a specific target eigenvalue ${\lambda }_1$ is separated from all the others by ${\delta }_k={\min }_{j>1}|{\lambda }_j-{\rho }_k|>0$, it also gives ${\epsilon }_k\le \Vert r_k{\Vert }_2/{\delta }_k$. This separation is measured from the computed Rayleigh quotient; it differs from the dominant-modulus spectral gap. For a nonnormal matrix, the same orthogonal expansion is unavailable and a small eigenpair residual alone need not imply a comparably small distance to the spectrum.