The ultraviolet catastrophe is the divergent total thermal radiation energy predicted by classical equipartition theorem applied to the electromagnetic mode density in a box. Each mode has mean energy $k_{B}T$, while the number of modes per volume per frequency is $8\pi {\nu }^2/c^3$. The Rayleigh-Jeans law therefore gives $u_{\nu }=8\pi k_{B}T{\nu }^2/c^3$, whose integral over all positive frequencies diverges at every $T>0$.

Planck's law replaces the classical mode energy by $h\nu /(e^{h\nu /(k_{B}T)}-1)$. It agrees with equipartition when $h\nu \ll k_{B}T$ but suppresses high-frequency excitations exponentially. The resulting spectral energy density is integrable and yields the finite Stefan–Boltzmann law. The failure concerns the classical equilibrium distribution of energy among modes, not the existence of electromagnetic wave propagation itself.

Outgoing longwave radiation is the outward thermal radiative flux at the top of the atmosphere, integrated over the specified thermal spectrum and outward directions. It differs from a directional spectral radiance.

Convective adjustment replaces an excessively steep vertical temperature gradient by an allowed profile while respecting the selected energy budget. It is a closure for unresolved convection, not a direct simulation of individual rising parcels. A prescribed critical atmospheric lapse rate is a modeling assumption; it need not equal the dry adiabatic lapse rate.

For an original dry two-slab example, take fixed equal heat capacities and slab centers separated by $\Delta z=2\mathrm{km}$. Initially $T_1=300\mathrm{K}$ below and $T_2=280\mathrm{K}$ above. Stipulate a permitted atmospheric lapse rate ${\Gamma }_c=6\mathrm{K}\mathrm{k}{\mathrm{m}}^{-1}$. The initial difference $20\mathrm{K}$ exceeds the allowed $12\mathrm{K}$. Conservation of the sum of the two sensible energies requires $T_1^{\prime }+T_2^{\prime }=580\mathrm{K}$, while the adjusted gradient requires $T_1^{\prime }-T_2^{\prime }=12\mathrm{K}$. Hence $T_1^{\prime }=296\mathrm{K}$ and $T_2^{\prime }=284\mathrm{K}$. Heat removed below equals heat added above. Simply holding the lower slab fixed and raising the upper slab to $288\mathrm{K}$ would introduce external energy.

For unequal constant heat capacities $C_1,C_2$, the same two constraints instead giveApply this adjustment only when the original gradient exceeds the threshold. A moist adjustment also has to account for latent heat and changing composition; conservation of this dry sensible-energy expression alone is then insufficient.

For a fixed circular elastic membrane, the Rayleigh-Ritz variational principle applied to $f(r)=1-(r/a{)}^p$, $p>0$, gives ${\lambda }_1\le (p+3+2/p)/a^2$. The numerator of its Rayleigh quotient is $p/2$, and the denominator is $a^2{p}^2/[2(p+1)(p+2)]$. Minimization gives $p=\sqrt{2}$, ${\lambda }_1\le (3+2\sqrt{2})/a^2$, and the displayed upper bound for the ordinary frequency ${\nu }_1=c\sqrt{{\lambda }_1}/(2\pi )$. Every $p>0$ has finite radial Dirichlet energy, which is the required variational admissibility condition.

For $r>0$, the substitution $y=ru$ removes the first derivative in the three-dimensional radial Laplacian, since $u^{\prime \prime }+2u^{\prime }/r=y^{\prime \prime }/r$. Bounded $u$ at zero implies $y(0)=0$. Bounded $u$ at infinity means $y=O(r)$; it does not require bounded $y$. For example, constant forcing $g=1$ has the bounded particular solution $u=-1$, whose reduced function is $y=-r$. A Dirichlet half-line Green function for d2 minus 1 can still represent this solution when its integral converges, but the resulting reduced function has the growth demanded by the forcing.

For inner and outer radii $a,b$, with internal and external pressures $p_1,p_0$, the boundary tractions are ${\tau }_r(a)=-p_1$ and ${\tau }_r(b)=-p_0$. The radial equilibrium of a thick elastic cylinder gives $P=(p_1{a}^2-p_0{b}^2)/(b^2-a^2)$ and $Q=(p_1-p_0)a^2{b}^2/(b^2-a^2)$. The radial compressive pressure is $p(r)=-P+Q/r^2$. Since hoop stress is monotone in $r$ when $Q\ne 0$, its maximum absolute value is the larger of $|P+Q/a^2|$ and $|P+Q/b^2|$. With negligible external pressure, its maximum is the tensile inner-wall value $p_1(a^2+b^2)/(b^2-a^2)$. The stress distribution itself is independent of elastic moduli in this linear homogeneous axisymmetric model; its associated displacement depends on them.

The quaternion group has conjugacy classes $\{1\}$, $\{-1\}$, $\{\pm i\}$, $\{\pm j\}$, $\{\pm k\}$, with sizes $1,1,2,2,2$. Over the complex numbers, the number of classes and the sum of squares of irreducible representation dimensions force four one-dimensional representations and one two-dimensional representation. For $ϵ,\eta \in \{\pm 1\}$, the linear character rows on these classes are $(1,1,ϵ,\eta ,ϵ\eta )$. Orthogonality then gives the remaining row $(2,-2,0,0,0)$. A representation with character $(4,-4,0,0,0)$ contains two copies of the two-dimensional irrep and none of the linear irreps, by the character inner product. The choice of scalar field matters: the corresponding quaternionic four-dimensional real representation is irreducible over the reals.

In the quaternion group, $-1=i^2=j^2=k^2$ commutes with each generator because any element commutes with its powers. Hence $N=\{1,-1\}$ is central and is a normal subgroup. The reverse products satisfy $ji=-k$, $kj=-i$, $ik=-j$, so none of $\pm i,\pm j,\pm k$ is central. Therefore $Z(Q_8)=N$.

The four cosets are $N,iN,jN,kN$. Each nonidentity coset squares to $N$, and $(iN)(jN)=kN$. Thus the quotient group is the Klein four-group. Explicitly, mapping $\pm 1$ to its identity and mapping $\pm i,\pm j,\pm k$ to its three different nonidentity elements is a surjective group homomorphism with group kernel $N$. The first isomorphism theorem gives the title result.

A constituent used to specify the net quark flavor content of a hadron. The quark model uses three valence quarks for an ordinary baryon and a quark-antiquark pair for an ordinary meson; this does not exclude additional sea pairs and gluons in the full state.

Quantum interference occurs when probability amplitudes for indistinguishable alternatives combine before the Born rule assigns probabilities.

A square pyramid has a square base. If the base edge has length $a$ and all four lateral edges have length $\ell >a/\sqrt{2}$, the apex lies above the base center. The center-to-vertex distance is $a/\sqrt{2}$, so the Pythagorean theorem gives $h=\sqrt{{\ell }^2-a^2/2}$. Its volume is $a^{2}h/3$. When all eight edges have the same length $a$, this reduces to $a^3/(3\sqrt{2})$.

The root mean square radius of the proton charge distribution. Finite nuclear size changes hydrogen atom energy levels and enters precision predictions for the Lamb shift.^[ref]

For rectilinear motion write the four-velocity in an inertial frame as $U=(c\cosh \eta ,c\sinh \eta )$, where $\eta$ is the rapidity and $\tau$ is proper time. Differentiation gives $A=c\dot{\eta }(\sinh \eta ,\cosh \eta )$, with invariant $A\cdot A=-c^2{\dot{\eta }}^2$ in the $+-$ metric signature. The signed proper acceleration is therefore $a=c\dot{\eta }$. Constant signed acceleration gives $\eta ={\eta }_0+a\tau /c$; specifying initial position alone leaves the initial rapidity ${\eta }_0$ undetermined.

Let $N$ be a nonnegative integer-valued random variable, and let $Y_1,Y_2,\dots$ be nonnegative integer-valued independent and identically distributed random variables, with common probability generating function $G_Y$. Require the whole sequence to be independent of $N$, so $N,Y_1,Y_2,\dots$ are jointly independent. Set the random sumwith the empty sum equal to zero. Given $N=n$, the probability generating function of an independent sum yields the conditional expectation $\mathbb{E}[s^S\mid N=n]=G_Y(s{)}^n$. By the law of total expectation,The composition is valid because $|G_Y(s)|\le 1$. Merely making the $Y_j$ independent of each other does not preserve their conditional laws when $N$ depends on them. This is the probability generating function version of transform composition for an independent random sum.

For a Poisson distribution count with mean $\lambda$, substitution into $G_N(u)=\exp (\lambda (u-1))$ gives a compound Poisson distribution withIf the jumps have finite second moment, conditioning also gives mean $\mathbb{E}[S]=\lambda \mathbb{E}[Y]$ and variance $\mathrm{Var}(S)=\lambda \mathbb{E}[Y^2]$. The extra spread comes from randomness in the count as well as randomness in the jumps.^[ref]

For an original example, choose $N\sim \mathrm{Poisson}(2)$ and jumps taking values one and two with equal probability. Keep the independence assumptions above. ThenHere $\mathbb{E}[Y]=3/2$, $\mathrm{Var}(Y)=1/4$, and $\mathbb{E}[Y^2]=5/2$. The law of total expectation and law of total variance giveThus the aggregate is not itself a Poisson distribution, whose mean and variance would coincide.

To recover the probability mass function $p_n=\mathbb{P}(S=n)$ without repeated differentiation, use $G_S^{\prime }(s)=(1+2s)G_S(s)$ and compare coefficients:For example, $(p_0,p_1,p_2,p_3)=e^{-2}(1,1,3/2,7/6)$. As an independent check, expanding $e^{s+s^2-2}=e^{-2}{e}^s{e}^{s^2}$ gives

The plotted probability masses stop at 20, although the support of a probability distribution is unbounded. The omitted tail probability isThis model has bounded jumps and an unbounded count. Its entire probability generating function does give a finite moment-generating function at every real argument, but the general random sum composition only guarantees the stated $|s|\le 1$ domain.

For nonnegative integer-valued independent random variables $X,Y$, factor the expected value of a product:The Cauchy product explains what this means for individual probability masses:Every way of splitting the total contributes once. This is convolution of independent random variables; multiplying probability generating functions performs the bookkeeping automatically.

For an original example, take independent $X\sim \mathrm{Bernoulli}(0.3)$ and $Y\sim \mathrm{Binomial}(2,0.5)$, using the Bernoulli distribution and binomial distribution. ThensoDirect enumeration of the six pairs $(x,y)$ gives the same probability masses. For example, total one collects $(0,1)$ and $(1,0)$, with contributions $0.7\cdot 0.5+0.3\cdot 0.25=0.425$. The coefficients sum to one. The expected value is $G_{X+Y}^{\prime }(1)=1.3$; since $G_{X+Y}^{\prime \prime }(1)=1.1$, the variance is $1.1+1.3-1.3^2=0.71$, agreeing with $0.3\cdot 0.7+2\cdot 0.5\cdot 0.5$.

Uncorrelated random variables are insufficient for the product identity. For a concrete counterexample, let $U$ be uniform on $\{-1,0,1\}$ and set $A=U+1$, $B=U^2$. Then $\mathrm{Cov}(A,B)=\mathbb{E}[U^3]=0$, but $\mathbb{P}(A+B=0)=0$, whereas $G_A(0)G_B(0)=1/9$. Zero covariance controls one mixed moment, not the full joint probability mass function.

Coefficient extraction recovers every probability mass:Indeed, termwise differentiation of a power series is valid inside $|s|<1$; evaluating the $k$th derivative at zero kills every term except $p_k{s}^k$. Thus two probability generating functions with the same coefficients describe the same probability distribution, without any finite-moment assumption.

At one, approach along the real line from below. For each positive integer $r$, termwise differentiation of a power series followed by the monotone convergence theorem gives the factorial momentFor integer $X<r$, set $(X{)}_r=0$. The limit may be infinite; it is a finite boundary value only when the corresponding factorial moment is finite. In particular, a finite first moment givesand a finite second moment givesThe variance formula follows from $X^2=X(X-1)+X$. These are boundary limits of derivatives; they do not assert that the probability generating function extends smoothly through one.

For a nonnegative integer-valued random variable, its moment-generating function satisfies $M_X(t)=G_X(e^t)$ wherever finite. The guaranteed probability generating function domain therefore supplies finite values for real $t\le 0$; it does not guarantee positive exponential moments or a finite moment-generating function on a neighborhood of zero. Bounded nonnegative integer-valued random variables have polynomial probability generating functions, but their convenient behavior is not a general existence theorem.

High-pressure experimental metrology connects three tasks: producing pressure, retaining it without leakage, and assigning a calibrated value to the pressure experienced by a sample. A large applied force alone establishes none of the last two. The useful result is a reproducible material property at specified pressure, temperature and stress conditions, with stated measurement uncertainty.

Percy Williams Bridgman identified leakage as an earlier experimental limit. His 1946 lecture describes packing that tightens as pressure rises, ultimately leaving vessel strength as a limit.^[ref]

The Bridgman unsupported-area seal turns the very pressure that would drive a leak into an extra seating force. Its appeal is a feedback between load and sealing, rather than an assumption of perfectly fitted rigid surfaces. The piston-cylinder pressure apparatus below explains that feedback; pressure calibration then separates the applied load from the sample's inferred pressure.

For circular plates of radius $a$, separation $d\ll a$, and a thin axial resistor $R$, the quasistatic capacitance is $C={\epsilon }_0\pi a^2/d$. During discharge, $I=V/R=-C\dot{V}$. A circular Ampere-Maxwell law contour between the plates encloses wire current $I$ and opposing displacement current $-I{r}^2/a^2$, giving $H=I(1-r^2/a^2)/(2\pi r)$. The electric field has magnitude $V/d$, so the Poynting vector points radially inward with magnitude $V^2(1-r^2/a^2)/(2\pi rdR)$. Its flux through the cylindrical side is the displayed inward power. The field inside that cylinder releases $V^2{r}^2/(R{a}^2)$ per unit time; adding the incoming flux gives the resistor's full Joule heating rate $V^2/R$. End fringing, wire radius and magnetic-energy corrections are neglected.

For a good conductor with surface electric-field peak amplitude $E_0$, the surface impedance of a good conductor gives the displayed time-averaged Poynting vector component into the conductor. The factor one half in $⟨\mathbf{S}⟩=\mathrm{Re}(\mathbf{E}\times {\mathbf{H}}^{\ast })/2$ is required for peak phasors. Its loss with depth obeys $-d⟨S_z⟩/dz=\sigma E_0^2{e}^{-2z/\delta }/2=⟨\mathbf{j}\cdot \mathbf{E}⟩$. Electromagnetic energy becomes Joule heating locally; averaged field-energy storage has zero time derivative.