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For an isothermal sphere, put $\rho ={\rho }_c{e}^{-\Psi }$ and $r=a\xi$ with $a^2=K/(4\pi G{\rho }_c)$. The equilibrium equation becomes ${\xi }^{-2}({\xi }^2{\Psi }^{\prime }{)}^{\prime }=e^{-\Psi }$. Finite central density and smooth spherical symmetry require $\Psi (0)={\Psi }^{\prime }(0)=0$. The regular expansion is $\Psi ={\xi }^2/6-{\xi }^4/120+{\xi }^6/1890+O({\xi }^8)$. The enclosed mass is $M=4\pi {\rho }_c{a}^3{\xi }^2{\Psi }^{\prime }$, following by integrating the differential equation.

A pressure-confined isothermal sphere is a regular isothermal sphere truncated at a boundary where its internal pressure equals an imposed external pressure. The finite central density and the boundary density generally differ. Its radial stability depends on the dimensionless truncation radius as well as the fixed mass and temperature. The first fixed-mass turning point of an isothermal sphere occurs at dimensionless radius about $6.451$; the modestly concentrated branch is stable to the radial isothermal perturbations in this model.

An isothermal sphere with constant positive isothermal sound speed $c$ obeys the title hydrostatic equilibrium equation. Between two positive finite radii with bounded enclosed mass, its right-hand side has a finite integral. A positive starting density therefore remains positive at any proposed finite outer radius. It cannot meet a vacuum boundary with $p=c^2\rho =0$ there without an additional confining structure or a change of equation of state. The singular isothermal sphere makes the obstruction explicit: $\rho =c^2/(2\pi G{r}^2)$ and $p=c^4/(2\pi G{r}^2)$ never vanish at finite radius, and its mass grows as $2c^{2}r/G$ without bound at infinity.

On the escaping supersonic isothermal Parker wind branch, $w=u^2/c^2\to \infty$. Its integrated equation gives $w=4\ln x+\ln w-3+4/x$, so $w/(4\ln x)\to 1$ and the title result follows. A further iteration yields $w=4\ln x+\ln (4\ln x)-3+o(1)$. The leading logarithmic growth holds for any supersonic branch with a fixed finite integration constant, but does not describe a subsonic isothermal breeze. The unbounded speed is a formal consequence of indefinite isothermal heating in a prescribed central gravitational potential.

For a homogeneous extensive single-component thermodynamic phase, the Gibbs-Duhem equation at fixed temperature gives $dp=nd\mu$. With number density $n=N/V$, fixed $N$ gives $dn=-(n/V)dV$. The chain rule therefore givesThe derivatives are evaluated within the same smooth equation of state; an interface contribution or phase coexistence can invalidate this simple homogeneous derivation.

For $C_n=⟨g:g^n=e⟩$, the complex irreducible representations are precisely the $n$ one-dimensional maps $g\mapsto e^{2\pi ij/n}$, with $0\le j<n$. Their characters have the displayed values, forming the discrete Fourier character table. Indeed a representing matrix $A$ satisfies $A^n=I$. Applying the fundamental theorem of algebra to its characteristic polynomial gives an eigenvalue and an eigenvector over the complex numbers. Its line is an invariant subspace under every power of $A$; irreducibility therefore forces the entire space to be that line. Its eigenvalue is an $n$th root of unity. Different roots give inequivalent one-dimensional representations, and there are exactly $n$ possibilities.

All conjugacy classes are singletons because the group is Abelian. The number of the displayed irreducibles equals the group order, and their squared dimensions sum to $n$, in agreement with the sum of squares of irreducible representation dimensions. Over the real numbers, conjugate nonreal characters instead combine into the planes described by the irreducible real representations of a finite cyclic group.

A point mass attached to a light inextensible string winds around a fixed smooth circular pole of radius $R$. Let the initial straight free length be $L$, and assume the string remains taut without slipping over the part already laid on the pole. Since its tension is perpendicular to the point mass's velocity, the speed $v$ stays constant. If the contact point turns through $d\theta$, the free length $s$ obeys $ds=-Rd\theta$, while the involute geometry gives $v=s\dot{\theta }$. Thus $s\dot{s}=-Rv$, or $s^2=L^2-2Rvt$. First contact of the point mass with the pole occurs at $s=0$, after $L^2/(2Rv)$. In the ideal model, the required tension $m{v}^2/s$ diverges at contact; finite string strength or finite particle size limits that idealization.

For real $A,B$ with $AB<1$, the principal inverse tangent satisfiesTo prove it, multiply the complex numbers $1+iA$ and $1+iB$. Their product is $(1-AB)+i(A+B)$, whose positive real part places its complex argument in $(-\pi /2,\pi /2)$. The individual complex arguments are $\arctan A$ and $\arctan B$, and their sum is in $(-\pi ,\pi )$. Multiplication adds these complex arguments, so the positive real part forces the sum into the same principal range and gives the displayed formula.

If $AB>1$, both $A,B$ have the same sign. Add $\pi$ to the right-hand side when $A,B>0$, and subtract $\pi$ when $A,B<0$. If $AB=1$, the sum is $\pi /2$ or $-\pi /2$ according to the sign of $A+B$. These cases record the branch information that can be lost by applying the tangent addition formula alone.

Define the definite integral $F(k)={\int }_0^{\pi }{x}^k\sin xdx$. For real $k>1$, two applications of integration by parts giveThe boundary term $x^{k-1}\sin x$ vanishes at both endpoints. At $k=1$, direct integration gives $F(1)=\pi$, consistent with the displayed relation; $F(-1)$ is finite because $\sin x/x\to 1$ at zero. This recurrence and $F(0)=2$ evaluate every nonnegative integer moment, with separate even and odd chains.

If $F^{\prime }=f$ and $g$ is differentiable, the chain rule gives $(F\circ g{)}^{\prime }(x)=f(g(x))g^{\prime }(x)$. Taking antiderivatives gives the substitution ruleWriting $u=g(x)$ and $du=g^{\prime }(x)dx$ records this chain rule in reverse. For a continuous function $f$ and continuously differentiable $g$, the fundamental theorem of calculus gives the corresponding definite integral identityThe order of the transformed endpoints determines the sign. For example, ${\int }_0^{1}2x\cos (x^2)dx={\int }_0^1\cos udu=\sin 1$. If the substitution is used in the opposite direction by solving $x$ as a function of $u$, one must restrict to a suitable inverse branch; the tangent half-angle substitution illustrates why separate branches can be necessary even when the original integrand is smooth. The general change of variables formula extends substitution to higher dimensions.

The domain of integration is the set $D$ over which an integral is taken. For a multiple integral, inequalities or parameter ranges specify which points belong to $D$; a change of variables transforms both these bounds and the integration element through its Jacobian determinant. For example, a polar region $c\le r\le 1+\cos \varphi$ exists only at angles where $1+\cos \varphi \ge c$.

The set over which an integral is taken. For a double integral, boundary inequalities describe the allowed coordinate pairs, and slices of this set determine the limits in an iterated integral. The domain need not be a rectangle or connected. A change of variables transforms both the integrand and this set.

The displayed implicit ordinary differential equation has a singular initial point at the origin. On a nonconstant regular branch, treating $x$ as a function of $y$ gives $dx/dy+x\cot y=2\cos y$. The integrating factor $\sin y$ yields the first integral of an ordinary differential equationOne can verify this identity without dividing by $\sin y$: its derivative along a solution is $\sin y-(2\sin y-x)\cos y{y}^{\prime }=0$.

The initial point forces $C=0$, so $\sin y(x-\sin y)=0$. Near the origin this gives the two differentiable solutions $y=0$ and $y=\arcsin x$, with initial slopes zero and one. A division by $\sin y$ would lose the first solution. A piecewise switch between the two at zero would fail differentiability because their one-sided slopes differ.

There is no conflict with the Picard-Lindelöf theorem: solving explicitly for $y^{\prime }$ requires division by $2\sin y-x$, which vanishes at the initial point. A regular implicit equation or a local uniqueness theorem must be checked before claiming a unique solution through a singular point.

Linearize the Euler equation in a uniformly rotating frame about rest and impose incompressible flow, $\nabla \cdot \mathbf{u}=0$. Taking the curl eliminates the reduced pressure. For constant $\mathbf{\Omega }$, $\nabla \times (\mathbf{\Omega }\times \mathbf{u})=-(\mathbf{\Omega }\cdot \nabla )\mathbf{u}$, giving the displayed vorticity equation. Taking another curl and time derivative yields ${\partial }_t^2{\nabla }^2\mathbf{u}+4(\mathbf{\Omega }\cdot \nabla {)}^2\mathbf{u}=0$. These equations describe the restoring influence of background rotation on small disturbances without buoyancy or viscosity.

Insert a transverse plane wave $\mathbf{u}={\mathbf{u}}_0{e}^{i(\mathbf{k}\cdot \mathbf{r}-\omega t)}$ into the inertial-wave vorticity equation, with $\mathbf{k}\cdot {\mathbf{u}}_0=0$. It gives $\omega \mathbf{k}\times {\mathbf{u}}_0=2i(\mathbf{\Omega }\cdot \mathbf{k}){\mathbf{u}}_0$. Applying $\mathbf{k}\times$ again and using transversality gives the title dispersion relation. If $\theta$ is the angle to the rotation axis, $|\omega |=2|\Omega \cos \theta |$. At fixed direction the angular frequency does not depend on wavelength. Parallel wavevectors give the maximum $2|\Omega |$; perpendicular ones have zero frequency and belong to the nonoscillating balanced limit.

For a uniform planar lamina with $|x_1-s{x}_2|\le a$ and $|x_2|\le b$, put $u=x_1-s{x}_2$, $v=x_2$. The Jacobian determinant is one, the mass is $M=4ab\sigma$, and symmetry puts the centre of mass at zero. Uniform rectangular moments give $⟨u^2⟩=a^2/3$, $⟨v^2⟩=b^2/3$, $⟨uv⟩=0$. Substitution into $I_{ij}=\int (|x{|}^2{\delta }_{ij}-x_i{x}_j)dm$ yields the displayed inertia tensor. The in-plane principal moments of inertia are $M[a^2+(1+s^2)b^2\pm \sqrt{(a^2+(s^2-1)b^2{)}^2+4s^2{b}^4}]/6$; the perpendicular moment is their sum. This separates the geometric shear from the elementary rectangular integrals.

For a nonnegative mass distribution relative to an origin, the second-moment matrix is $\mathsf{S}=\int \mathbf{r}{\mathbf{r}}^{\mathsf{T}}dm$ and the inertia tensor is $\mathsf{I}=\mathrm{tr}(\mathsf{S})1-\mathsf{S}$. Taking the trace gives the displayed inverse formula, which must be positive semidefinite. Conversely, any such $\mathsf{S}$ can be realized by pairs of point masses along its orthogonal principal directions. Thus this is also a realizability test when total mass and dimensions are unrestricted. In principal coordinates it is equivalent to the moment-of-inertia triangle inequalities: $I_2+I_3-I_1=2S_{11}\ge 0$, with cyclic counterparts. A symmetric positive definite matrix alone need not be a physical inertia tensor.

For constant scalar response $\mathbf{m}={\alpha }_B{\mathbf{B}}_0$ and a body small compared with the applied field's spatial variation scale, the gradient of the induced magnetic dipole energy gives $\mathbf{F}={\alpha }_B\nabla B_0^2/2$. Along an axis this is $F_z=md{B}_0/dz$. For an ellipsoidal body, put ${\chi }_{\mathrm{eff}}=\chi /(1+N_d\chi )$ to obtain $F_z={\mu }_{0}V{\chi }_{\mathrm{eff}}{H}_0{H}_0^{\prime }$. Positive effective magnetic susceptibility attracts the body towards stronger field; a negative response repels it. Reversing the source current leaves the force unchanged because both the induced moment and the field gradient reverse. Shape and orientation enter through the demagnetizing factor, not merely through volume.

A grounded hemispherical boss of radius $a$ on the grounded plane $z=0$ in an otherwise uniform downward electric field has the displayed exterior electric potential, with $r^2=x^2+y^2+z^2$. It vanishes both on the hemisphere $r=a$ and on the surrounding plane, and tends to $E_{0}z$ at large distance. The outward normal field gives surface charge $-3{\epsilon }_0{E}_0\cos \theta$ on the hemisphere and $-{\epsilon }_0{E}_0(1-a^3/s^3)$ on the plane at cylindrical radius $s>a$. The apex field is $3E_0$, so a field-limited breakdown voltage is reduced by a factor three. For a capacitor with finite spacing $d\gg a$, this is a leading local approximation; it is not an exact solution on the distant second plate.

For a uniform on-axis plane wave reference, choose its constant wave phase to be zero, so $R=A_R>0$. The reconstructed cross terms are then proportional to $O$ and its complex conjugate $O^{\ast }$. This pair creates the twin image in holography. An outgoing curved wavefront and its conjugate have opposite curvature: under forward propagation, one can appear to diverge from a virtual image while the other converges to a real image. In in-line holography their depths differ, but a defocused member still overlaps the other. This is the same conjugation of spatial wave phase used in phase conjugation, although a transmission hologram here launches both components forward rather than automatically creating a returning phase-conjugate mirror.