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An orthogonal change of coordinates relates two orthonormal bases of a real inner product space. With the new basis vectors as columns of an orthogonal matrix $Q$, old and new coordinate columns obey $x=Qy$. Then $x^{T}x=y^{T}y$, so Euclidean norm, Euclidean distance and angles are preserved. A quadratic form $x^{T}Ax$ becomes $y^T(Q^{T}AQ)y$. For a symmetric matrix, orthogonal diagonalization chooses $Q$ so that the transformed quadratic form is a sum of eigenvalues times squared coordinates. This explains the principal-axis descriptions of a quadric surface.

Under orthogonal projection from one plane onto another, a circle of radius $R$ becomes an ellipse with semiaxes $R$ and $R|\cos \alpha |$, where $\alpha$ is the angle between the planes' unit normal vectors. To prove this, choose one in-plane unit vector parallel to the planes' intersection and the other perpendicular to it. The first vector is unchanged by the projection; the second has projected length $|\cos \alpha |$ and remains perpendicular to the first. Parametrizing the circle in that orthonormal basis therefore gives projected coordinates $(R\cos t,R\cos \alpha \sin t)$. Parallel planes have a circular image, and perpendicular planes give the degenerate case of a line segment. An arbitrary center simply projects to the center of the image.

Two vectors are orthogonal when their inner product is zero.

For real Euclidean vectors, the inner product is the dot product, so $(3,4)$ and $(4,-3)$ are orthogonal: $3(4)+4(-3)=0$. For nonzero Euclidean vectors, this means their angle is $90^{\circ }$. The zero vector is orthogonal to every vector, although it has no direction or defined angle.^[ref]

In an abstract inner product space, orthogonality always refers to its chosen inner product. For example, the weighted inner product $⟨u,v{⟩}_w=4u_1{v}_1+u_2{v}_2$ makes $(1,1)$ and $(1,-4)$ orthogonal, since $4-4=0$. Their ordinary dot product is $-3$, so their arrows are not perpendicular in the usual Cartesian coordinate system. The algebraic definition also applies to functions, where arrows need not be available.

Parallax localization of a virtual image matches the apparent depth of a virtual image to a directly viewed physical marker. A sideways change of viewpoint changes their relative apparent positions if their depths differ. Adjust the marker until that relative motion vanishes, and use its coordinate to estimate image position. The marker must be viewed directly; viewing it through the same optical lens compares two images rather than an image and a known physical depth. ^[ref].

For two nearly aligned points at distances $L_i,L_m$ from an eye displaced sideways by $\delta$, the small-angle approximation gives a change of relative viewing angle proportional to $\delta (1/L_i-1/L_m)$. It vanishes when $L_i=L_m$, and changes sign across that match. When both distances are large, the difference is small, explaining the poor depth precision near the thin lens condition $u=f$.

Rotate an orthonormal basis counterclockwise by $\theta$ while keeping the vector $v$ fixed. The new basis vectors, written in the old basis, are the columns of $P=R_{\theta }$. The change-of-basis matrix convention therefore givesFor example, the fixed vector with old coordinate vector $(2,0{)}^T$ has new coordinate vector $(1,-\sqrt{3}{)}^T$ after the basis turns by $\pi /3$ radians. Its negative second coordinate means that the arrow lies below the new first axis. By contrast, an active planar rotation through that same angle moves the arrow to old coordinates $(1,\sqrt{3}{)}^T$. The sign difference follows from which object moves. The passive component formulas are $x^{\prime }=x\cos \theta +y\sin \theta$, $y^{\prime }=-x\sin \theta +y\cos \theta$.^[ref]

If coordinate vectors are instead written as rows, transposing the active equation gives $(v^{\prime }{)}^T=v^T{R}_{\theta }^T$. Switching to rows changes the side and the transpose; it does not by itself turn an active planar rotation into passive rotation of coordinates.

Use the unitary discrete Fourier transformThe transformed operators obey $[a_k,a_q^{†}]={\delta }_{kq}$. The hopping terms contribute $2A\cos k$ to the number coefficient. In the pairing sum, the odd sine part cancels between $k$ and $-k$. ThuswhereThe factor one half in the full momentum sum avoids counting an anomalous pair twice. For $H_f>0$, ${\alpha }_k>|{\Delta }_k|$ for every allowed momentum. SetReal even angles ensure consistency between opposite momenta, and ${\cosh }^2{\theta }_k-{\sinh }^2{\theta }_k=1$ preserves the canonical commutation relations. The anomalous coefficient becomes ${\Delta }_k\cosh 2{\theta }_k-{\alpha }_k\sinh 2{\theta }_k=0$. Normal ordering then giveswithThe field-polarized planar ferromagnetic spin wave has energy gap $ϵ(0)=S\sqrt{H_f(H_f+2J)}$. The momenta run over the whole Brillouin zone once, including the self-inverse momenta $0$ and, for even $L$, $\pi$; these use single-mode squeezing rather than introducing an extra independent partner.

The zero-field case has a zero-mode obstruction to bosonic Bogoliubov diagonalization. Its nonzero-momentum modes have $ϵ(k)=2\sqrt{2}JS|\sin (k/2)|$, but ${\alpha }_0={\Delta }_0=JS$, so no finite squeezing angle exists for the uniform mode. With $Q_0=(a_0+a_0^{†})/\sqrt{2}$, its Hamiltonian is $JS{Q}_0^2-JS/2$, not an identically zero oscillator. For example, its matrix element between the vacuum and the two-boson number state is $JS/\sqrt{2}$, whereas $ϵ(0)b_0^{†}{b}_0$ would vanish. Consequently the requested oscillator diagonalization is literal for nonzero field; at zero field treat the collective orientation separately or introduce a field regulator before taking a limit.

Each simple pendulum moves in the vertical plane containing the pivots. Ignoring contact between the ideal rods and point masses, each rod can turn through a full circle. Thus the configuration space is the torusChoose generalized coordinates ${\theta }_1,{\theta }_2$, measured from the downward vertical, positive towards the right and understood modulo $2\pi$. These angles are local coordinates on the torus, rather than a single globally one-to-one real coordinate chart. With upward Cartesian coordinate $y$, the bob positions areTheir velocities satisfy $|{\dot{\mathbf{r}}}_j{|}^2=l^2{\dot{\theta }}_j^2$, so kinetic energy from Cartesian position vectors givesThe spring length is the Euclidean distanceTaking gravitational potential energy zero at pivot height gives $V_g=-mgl(\cos {\theta }_1+\cos {\theta }_2)$. The spring potential energy is $V_s=k(d-r{)}^2/2$, since $r$ is its natural length. Consequently the Lagrangian isIn the paper's notation $t({\theta }_1,{\theta }_2)=V_s$; this function is distinct from the time variable. This is the exact Lagrangian of two spring-coupled pendulums.

One alternative Lagrangian isIt corresponds to setting gravitational potential energy zero at the downward position. The constant disappears from every Euler-Lagrange equation. More generally, total-time-derivative invariance of a Lagrangian allows $\tilde{L}=L+dF({\theta }_1,{\theta }_2,t)/dt$ for any smooth $F$.

Write $w=\sqrt{1+2k}$. The initial conditions on the normal mode coordinates are $u(0)=v(0)=0$ and $\dot{u}(0)=\dot{v}(0)=k$. Solving the two harmonic oscillator equations and recovering the original angles gives the exact solution within the harmonic approximation:Their angular velocities are ${\dot{\theta }}_1=k(\cos t+\cos wt)/2$ and ${\dot{\theta }}_2=k(\cos t-\cos wt)/2$, which verify all four initial conditions.

To expose the beat, define $\Delta =w-1$ and $\Omega =(w+1)/2$. Replacing $1/w$ by $1$ changes each angle by $O(k^2)$ uniformly in time. The sum-to-product identities then giveThe displayed angular velocities are exact in the linear system. The absolute and signed beat envelopes have leading amplitudes $k|\cos (\Delta t/2)|$ and $k|\sin (\Delta t/2)|$. Because $\Delta =k+O(k^2)$, at $\tau =\pi /k$ we have $\Delta \tau /2=\pi /2+O(k)$ and $\Omega \tau =\tau +\pi /2+O(k)$. HenceThe first simple pendulum has negligible oscillation amplitude. Its kinetic energy is $T_1(\tau )=O(k^4)$, while the second has $T_2(\tau )=k^2{\cos }^2\tau /2+O(k^3)$. Thus, to the retained order, all instantaneous kinetic energy belongs to the second simple pendulum. The fast phase can make that kinetic energy small: the whole initial energy $k^2/2$ is transferred as the second simple pendulum's combined kinetic energy and gravitational potential energy, sinceThis is energy transfer after an impulse in two weakly coupled oscillators. It does not require $\cos \tau =1$.

At $2\tau$, $\Delta (2\tau )/2=\pi +O(k)$, so the beat has returned the oscillation to the first simple pendulum:The energy localization returns, although the fast oscillation phase need not equal its initial phase. The exact first minimum of the leading beat envelope is at $\pi /\Delta$, whose leading value is $\pi /k$. Keeping the slow phase $\Delta t$ is essential: expanding $\sin (wt)$ at fixed $t$ and then using $t=O(1/k)$ would lose the energy exchange. Although angles are retained only to order $k$, their leading energies are of order $k^2$.

Now take zero torque and $\mathbf{\omega }(0)=(1,0,1)$. The axial angular velocity is ${\omega }_3=1$ and the complex transverse component obeys $\dot{z}=i\nu z$ with $z(0)=1$ and $\nu =(I_3-I_1)/I_1$. Hence the angular momentum components in the principal body frame areThe transverse projection therefore traces a circle with signed angular frequency $\nu$. Its magnitude is constant, $L=\sqrt{I_1^2+I_3^2}$.

In the inertial frame, zero torque means that $\mathbf{L}$ is fixed. Usingand the derivative of a body-fixed basis vector, the symmetry axis satisfiesThus there is uniform torque-free precession of a symmetric top about the fixed angular momentum, at a constant cone angle $\beta$ and space angular frequencyThe cone angle stays constant, with no nutation. The body-frame and space-frame frequencies of free symmetric-top precession differ: $\nu$ describes rotating components, whereas ${\Omega }_{\mathrm{space}}$ describes the symmetry axis in space. The formulas also cover $I_3=I_1$, where the angular momentum components in the principal body frame are constant.

Work on $I>0$, with a local choice of the periodic angle $\theta$. The inverse formulas areIn particular, ${\theta }_q=p/(2I)$, ${\theta }_p=-q/(2I)$, $I_q=\alpha q$ and $I_p=p/\alpha$. Their Poisson bracket isThe other fundamental Poisson brackets vanish by antisymmetry. Thus this is a canonical transformation for every $\alpha >0$, on local charts away from the origin. At $I=0$ the angle is undefined.

In the new coordinates the harmonic oscillator Hamiltonian isFor $I$ to be a conserved quantity throughout phase space, Hamilton's equations require $\dot{I}=-{\partial }_{\theta }{H}_0=0$ identically. The two coefficients must coincide. For the usual $m>0$ and $\omega >0$, positivity of $\alpha$ selectsThese are harmonic oscillator action-angle variables; the action variable is $I=H_0/\omega$.

Snell's law is $n_a\sin {\theta }_a=n_b\sin {\theta }_b$, with both angles measured from the interface normal vector. Let $\psi$ be the meridional ray's angle to the optical axis inside the core and let $n_0$ be the incident medium's refractive index. At the entrance face,At the cylindrical wall the angle of incidence is $i=\pi /2-\psi$. For $n_{\mathrm{inner}}>n_{\mathrm{outer}}$, the critical angle for total internal reflection satisfies $\sin i_c=n_{\mathrm{outer}}/n_{\mathrm{inner}}$. At the limiting accepted ray,Consequently the optical fiber acceptance angle obeysThe requested expression follows for incidence from air, $n_0\simeq 1$, provided the right-hand side is at most one. Strict total internal reflection occurs inside this limiting cone. The derivation uses the standard meridional ray model; skew launches need additional geometry.

If $\sqrt{n_{\mathrm{inner}}^2-n_{\mathrm{outer}}^2}>n_0$, every forward entrance angle allowed by this ideal model meets the side-wall condition, and the limiting half-angle is $\pi /2$, rather than an undefined inverse sine. For example, $n_{\mathrm{inner}}=1.6$, $n_{\mathrm{outer}}=1.1$, $n_0=1$ gives $\sqrt{1.35}>1$. Thus the unsaturated printed formula is not valid for every possible pair of refractive indices.

Write ${\theta }_i$ for the internal angle at $v_i$, using the numbering in part (b), and use the hyperbolic plane with Gaussian curvature $-1$. The necessary and sufficient conditions are equal lengths for each pair of equally labelled polygon sides, and total angle $2\pi$ at each vertex equivalence class. Explicitly, the two angle conditions areNecessity follows because the identification must preserve arc length along each seam, and a smooth interior point of a Riemannian surface has total angle $2\pi$. For sufficiency, equally long hyperbolic geodesic segments can be identified by isometries. Across their interiors, two half-disc neighbourhoods join to a hyperbolic disc. At an identified vertex, the incident sectors form a cyclic disc with total angle $2\pi$, so no cone singularity of a surface metric remains. These local isometries therefore extend the Riemannian metric smoothly everywhere. This is the smooth hyperbolic polygon gluing criterion.

The ten internal angles sum to $4\pi$. The area of a hyperbolic geodesic polygon consequently givesEquivalently, the Gauss-Bonnet theorem gives $-\mathrm{Area}(X)=2\pi \chi (X)=-4\pi$.

For a regular parametric surface $\mathbf{r}(u,v)$, its tangent vectors ${\mathbf{r}}_u,{\mathbf{r}}_v$ are linearly independent. The first fundamental form is the induced inner product on the tangent plane, writtenIt measures squared lengths and angles; its determinant satisfies $EG-F^2=|{\mathbf{r}}_u\times {\mathbf{r}}_v{|}^2>0$. Choose the unit normalThe second fundamental form, with shape operator $S=-dN$, isThe subscript distinguishes the coefficient $g_2$ from a genus or a Riemannian metric. In these coordinates the shape operator has matrix $I^{-1}II$, so the Gaussian curvature, the product of its two principal curvatures, isReversing $N$ reverses $II$ and both principal curvatures, leaving the Gaussian curvature unchanged.

Let $R_t$ be the horizontal rotation matrix through angle $t$. Define the screw motionIts linear part is an orthogonal matrix with determinant one, and its remaining part is a translation. Thus it is a proper Euclidean motion of Euclidean three-space, with ${\Phi }_s{\Phi }_t={\Phi }_{s+t}$.

In horizontal-vector notation the given parametric surface satisfies $\mathbf{r}(u,v)=(R_v(1+\cos u,\sin u),v)$. ThereforeThe defining circle in each horizontal plane is entirely parametrized by $u$, so $\mathbf{r}$ covers $X$. The displayed equality and ${\Phi }_{-t}={\Phi }_t^{-1}$ prove ${\Phi }_t(X)=X$. This one-parameter subgroup preserves the dot products and cross products defining $I$, $II$ and the chosen unit normal. Consequently their coordinate coefficients, and the Gaussian curvature, depend only on $u$.

For $0<b-a<2\pi$, let $D=\mathbf{r}([a,b]\times [c,d])$ and use the orientation of a surface determined by $N={\mathbf{r}}_u\times {\mathbf{r}}_v/|{\mathbf{r}}_u\times {\mathbf{r}}_v|$. Its positive boundary orientation follows the bottom side in increasing $u$, the right side in increasing $v$, the top side in decreasing $u$, and the left side in decreasing $v$. The Gauss-Bonnet theorem for this disc with four corners readswhere the ${\epsilon }_j$ are the signed exterior angles.

The bottom and top sides have cancelling geodesic curvature integrals: a screw motion takes one to the other with opposite boundary orientation. Explicitly, along increasing $u$ at fixed $v$, the speed is one and the signed geodesic curvature is $-\sin u/\sqrt{1+{\sin }^{2}u}$ on both sides. The right and left sides contribute, respectively, the integral along ${\gamma }_b$ and minus the integral along ${\gamma }_a$.

Although the coordinate sides are generally not perpendicular, the corner contributions still sum to $2\pi$. Indeed, let $\theta (u)$ be the angle between ${\mathbf{r}}_u$ and ${\mathbf{r}}_v$, soStarting at the bottom-left corner, the exterior angles are $\pi -\theta (a)$, $\theta (b)$, $\pi -\theta (b)$ and $\theta (a)$. Their sum is $2\pi$. Cancelling this from the Gauss-Bonnet theorem leaveswhich is the identity in part (e). For a rectangle covering more than one turn, apply the same argument on the parameter rectangle with its induced Riemannian metric, or partition it into narrower rectangles and cancel the artificial internal boundary contributions. This gives the version counting multiplicity, as required by the correction in part (e).

Order the conjugacy classes of the symmetric group $S_4$ by representatives $1,(12),(12)(34),(123),(1234)$; their sizes are $1,6,3,8,6$. For a three-dimensional rotation through an angle $\theta$, the eigenvalues are $1,e^{i\theta },e^{-i\theta }$, so the trace of a three-dimensional rotation is $1+2\cos \theta$.

Under the identification with the rotational symmetry group of a cube, elements of order two give half-turns, elements of order three give turns through $\pm 2\pi /3$, and elements of order four give turns through $\pm \pi /2$. A half-turn can have an axis through opposite edges or opposite faces, but both kinds have trace $-1$. ThusIts character inner product with itself isBy character orthogonality, a nonzero complex representation has character inner product one with itself exactly when it is an irreducible representation. Therefore $C$ is an irreducible representation of dimension three.

For comparison useful in part (e), the standard representation of the symmetric group has character equal to the number of fixed letters minus one, namely $(3,1,-1,0,-1)$ in this order. Multiplication by the sign representation changes this character to $(3,-1,-1,0,1)$. Thus $C$ is the tensor product of the standard representation of the symmetric group with the sign representation.

On Bob's polarization qubit, takeSuitable conditional unitary operators areIndeed $(I\otimes XZ)|{\Phi }^{+}⟩=|{\Psi }^{-}⟩$, $(I\otimes X)|{\Phi }^{-}⟩=|{\Psi }^{-}⟩$, and $(I\otimes Z)|{\Psi }^{+}⟩=-|{\Psi }^{-}⟩$. The last minus sign is a harmless global phase.

For an ideal half-wave plate whose fast axis makes angle $\theta$ with horizontal polarization, discard the common phase and use the Jones matrixIt implements $Z$ at $0^{\circ }$ and $X$ at $45^{\circ }$. Thus for outcome ${\Phi }^{+}$ use two half-wave plates, first $0^{\circ }$ then $45^{\circ }$ in the propagation direction; for ${\Phi }^{-}$ use one at $45^{\circ }$; for ${\Psi }^{+}$ one at $0^{\circ }$; for ${\Psi }^{-}$ use no plate. Ideal retardance is assumed at the photon wavelength, and the setting is chosen after receiving the detector's classical result.

In a real orthonormal basis, a matrix is a proper rotation matrix exactly when $R^{T}R=I$ and $\det R=1$. The first condition preserves dot products, and the second preserves orientation of a vector space.

Put $s=\sqrt{3}$, $p=(1+s)/3$, $q=(1-s)/3$ and $r=1/3$, so the rows of the given matrix are $(p,q,r),(r,p,q),(q,r,p)$. Their squared Euclidean norms and pairwise dot products areThus $R^{T}R=I$. For the sign of its determinant, use the circulant form:This proves that it is a rotation matrix. Its fixed axis is $(1,1,1{)}^T$, because each row sums to one. The rotation trace formula gives $\cos \theta =(\mathrm{tr}R-1)/2=\sqrt{3}/2$; the skew part selects a positive angle $\theta =\pi /6$ about the unit axis $(1,1,1{)}^T/\sqrt{3}$.

The original triangle has edge vectors $b-a=(-2,2,0{)}^T$ and $c-a=(-2,0,2{)}^T$. Its area is half the Euclidean norm of their cross product:The rotated vertices are twice the three columns of $R$:A rotation matrix preserves lengths and the angle between the two edge vectors, so it preserves their parallelogram area. More explicitly, $(Ru)\times (Rv)=R(u\times v)$ for a proper rotation matrix, and $R(1,1,1{)}^T=(1,1,1{)}^T$, so the rotated edge cross product is again $(4,4,4{)}^T$. Therefore

Use $s$ for cylindrical radius, to distinguish it from spherical radius $r$. In cylindrical coordinates $(s,\varphi ,z)$, the azimuthal angle $\varphi$ is measured from positive $x$ in the horizontal plane:Here $0\le \varphi <2\pi$ and $z\in \mathbb{R}$.

In spherical polar coordinates $(r,\theta ,\varphi )$, take $\theta$ as the polar angle from positive $z$ and use the same azimuthal angle $\varphi$:The ranges are $r\ge 0$, $0\le \theta \le \pi$ and $0\le \varphi <2\pi$. The radial direction has horizontal projection of length $r\sin \theta$, which explains the extra factor in $x,y$. The azimuth is undefined on the polar axis, and both angles are nonunique at the origin; these are coordinate degeneracies rather than geometric singularities.

The solid angle is the intercepted surface area on a sphere of radius $R$, divided by $R^2$. Take the cone axis as the polar axis of spherical coordinates, and call its half-opening angle $\theta$. With polar integration variable $\vartheta$,For small $\theta$ measured in radians, the Taylor series of the cosine givesThis is the solid angle of a circular cone, measured in steradians.

The supplied angular diameter is twice the half-opening angle, soHence either apparent disc subtendsThe fraction is about $0.0005409\%$ of the entire celestial sphere. Using the angular diameter directly in $\pi {\theta }^2$ would give four times the correct small-angle answer.