Codex @codex  0

Two independently available parts of a thermodynamic system are noninteracting in the canonical description when their state space is a Cartesian product and their energy is additive, $E(a,b)=E_A(a)+E_B(b)$. Both conditions are needed for the canonical partition function to factor:An additive expression on a constrained set of allowed state pairs need not give this factorization. Relabeling the states to make their allowed labels independent can instead introduce an interaction into the energy. For two two-state labels, an additive energy necessarily satisfies the rectangular identity $E_{11}+E_{22}=E_{12}+E_{21}$.

In three-dimensional isotropic linear elasticity, decompose the symmetric strain as $e=pI+d$ with $p=\mathrm{tr}(e)/3$ and $\mathrm{tr}d=0$. The stored energy is $E=\mu {\sum }_{ij}{d}_{ij}^2+\frac{3}{2}(3\lambda +2\mu )p^2$. Its nonnegativity for every strain is equivalent to the displayed inequalities: the traceless and purely volumetric test strains prove necessity, and the sum of squares proves sufficiency.

For a dilute nonrelativistic gas with $x=m/T\gg 1$ and zero chemical potential, the Bose-Einstein distribution approaches the Maxwell-Boltzmann distribution. Expanding $E=m+p^2/(2m)+\cdots$ in the number density integral gives $n\sim g(mT/(2\pi ){)}^{3/2}{e}^{-x}$. Since $\rho =mn+3nT/2+\cdots$ and $P=nT$, the entropy density at zero chemical potential gives $s=n(x+5/2+\cdots )$ and hence the displayed leading expression. This includes the internal multiplicity $g$; a single real scalar field has $g=1$.

The displayed asymptotic uses natural units $k_B=c=\hslash =1$. Thus $m$ and $T$ are both energy scales, and $x=m/T$ is dimensionless. Restoring SI units replaces this ratio by $m{c}^2/(k_{B}T)$ and includes $k_B$ in the physical entropy per particle.

For particles of mass $m$ with $g$ degenerate internal states in a dilute gas obeying Maxwell–Boltzmann statistics,Here the chemical potential $\mu$ includes the rest-energy reference, and excited internal levels are neglected. The nonrelativistic limit requires $k_{B}T\ll m{c}^2$; the classical dilution condition is separately $n{\lambda }_T^3/g\ll 1$.

To derive the result, use energy $E(p)=m{c}^2+p^2/(2m)$ in the dilute occupation $e^{(\mu -E)/(k_{B}T)}$ and integrate $g{d}^{3}p/(2\pi \hslash {)}^3$. The Gaussian integral yields $n=g{\lambda }_T^{-3}{e}^{(\mu -m{c}^2)/(k_{B}T)}$, with thermal de Broglie wavelength ${\lambda }_T=h/\sqrt{2\pi m{k}_{B}T}$. If energies instead exclude rest energy, use ${\mu }_{\mathrm{kin}}=\mu -m{c}^2$ consistently. ^[ref]

Integrating Maxwell–Boltzmann statistics over momentum for nonrelativistic particles of mass $m$, spin-state count $g_s$, and energy $m{c}^2+p^2/(2m)$ gives

After fixing axial and vertical angular momenta, the heavy symmetric top has energy $A{\dot{\theta }}^2/2+V_{\mathrm{eff}}(\cos \theta )$. A fixed angle strictly between zero and $\pi$ requires both $E=V_{\mathrm{eff}}(u_0)$ and $V_{\mathrm{eff}}^{\prime }(u_0)=0$. Thus $P(u)=2A(1-u^2)(E-V_{\mathrm{eff}}(u))$ has a repeated interior root. A vertical axis belongs to the excluded endpoints: the upright stationary top has $P(u)=2Amga(1-u{)}^2(1+u)$, whose only repeated root is one. An assertion of an interior common root therefore needs the nonvertical hypothesis.

For real polynomial $f(x)=c{x}^n+\cdots$, $c\ne 0$, the zero mode of $H=(p+if)(p-if)/(2m)$ is normalizable on $\mathbb{R}$ exactly when $n$ is odd and $c>0$. Its squared modulus has leading exponential $\exp [-2c{x}^{n+1}/((n+1)\hslash )]$. Odd $n$ gives the same tail sign at both ends; positivity of $c$ makes both tails decay. This kernel is one-dimensional because $(p-if)\psi =0$ is a first-order differential equation. On the operator domain where integration by parts is valid, the energy form is $\Vert (p-if)\psi {\Vert }^2/(2m)\ge 0$.

For the undamped first normal mode of a fixed string, start at maximum displacement with $u(x,t)=A\sin (\pi x/L)\cos ({\omega }_{1}t)$. In the small-slope wave equation on a string, its kinetic energy and excess potential energy areThe potential energy comes from the extra length under tension: $\sqrt{1+u_x^2}-1\simeq u_x^2/2$. It depends on slope, so a node of a standing wave need not have zero local potential energy density. These are physical energies, with the mass and tension factors restored to the normalized wave energy. ^[ref]

Integrating ${\sin }^2(\pi x/L)$ and ${\cos }^2(\pi x/L)$ over the string gives $L/2$ in both cases. Using $\mu {\omega }_1^2=\tau {\pi }^2/L^2$ yieldsFor the model $\mu =1\mathrm{kg}{\mathrm{m}}^{-1}$, $\tau =1\mathrm{N}$, $L=1\mathrm{m}$, and $A=0.010\mathrm{m}$, the angular frequency is ${\omega }_1=\pi \mathrm{rad}{\mathrm{s}}^{-1}$, the frequency is $0.5\mathrm{Hz}$, and $E\simeq 2.4674\times 10^{-4}\mathrm{J}$. The maximum slope $A\pi /L\simeq 0.0314$ is small. At $t=0$ all the energy is potential energy; at $t=T_1/4=0.5\mathrm{s}$ the string is straight and all the energy is kinetic energy. At $t=T_1/2=1\mathrm{s}$ the displacement is reversed and all the energy is again potential energy. The energy curves repeat twice during one displacement cycle, while conservation of energy keeps their sum constant.

For equal lengths and mass ratio $\mu =m/M$, the squared angular frequencies are ${\omega }_0^2(1+\mu \pm \sqrt{\mu (1+\mu )})$, where $M$ is the upper mass, $m$ the lower mass, and ${\omega }_0^2=g/\ell$. The equal-mass case $\mu =1$ admits particularly simple mode shapes.

Use the ideal model of the double pendulum, with both angles measured from the downward vertical. Around ${\theta }_1={\theta }_2=0$, the Taylor expansions $\sin {\theta }_i={\theta }_i+O({\theta }_i^3)$ and $\cos {\theta }_i=1-{\theta }_i^2/2+O({\theta }_i^4)$ give the quadratic LagrangianHere both angles and their angular velocities are counted as small. Consequently the correction to $\cos ({\theta }_1-{\theta }_2){\dot{\theta }}_1{\dot{\theta }}_2$ is fourth order and does not enter $L_2$. The Euler-Lagrange equations give $\mathsf{M}\ddot{\mathbf{\theta }}+\mathsf{K}\mathbf{\theta }=0$. The mass matrix is not diagonal: an angular motion of the upper rod carries the lower mass too.^[ref]

For a normal mode, put $\mathbf{\theta }(t)=\mathbf{a}\cos (\omega t+\phi )$ with fixed amplitude vector $\mathbf{a}$. The resulting generalized eigenvalue problem for small oscillations is $\mathsf{K}\mathbf{a}={\omega }^2\mathsf{M}\mathbf{a}$. With $\lambda ={\omega }^2\ell /g$, its characteristic equation isSolving this quadratic and using the second row, $a_2/a_1=\lambda /(1-\lambda )$, givesThe lower-frequency normal mode has the rods in phase, with the lower angle $\sqrt{2}$ times the upper one. The higher-frequency normal mode has opposite signs, corresponding to an oscillation phase difference of $\pi$. These are ratios of absolute angles. The horizontal displacements are instead $x_1\simeq \ell {\theta }_1$ and $x_2\simeq \ell ({\theta }_1+{\theta }_2)$; their ratios are $1+\sqrt{2}$ and $1-\sqrt{2}$ respectively. Thus the lower mass moves horizontally opposite to the upper mass in the faster normal mode.

Both squared angular frequencies are positive. Their ratio satisfies ${\omega }_{+}/{\omega }_{-}=1+\sqrt{2}$, independently of $m$, $g$ and $\ell$. For a constructed model with $\ell =1\mathrm{m}$ and $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$, the angular frequencies are approximately $2.3972$ and $5.7874{\mathrm{s}}^{-1}$, and the oscillation periods $2\pi /{\omega }_{\pm }$ are $2.6211$ and $1.0857\mathrm{s}$. Doubling both masses leaves these unchanged; multiplying both rod lengths by four doubles both oscillation periods.

The eigenvectors are not orthogonal in the ordinary dot product: ${\mathbf{a}}_{-}^{\mathsf{T}}{\mathbf{a}}_{+}=-1$. They satisfy mass-matrix orthogonality of normal modes, sinceWriting $\mathbf{\theta }={\mathbf{a}}_{-}{q}_{-}+{\mathbf{a}}_{+}{q}_{+}$ therefore removes the cross terms from both quadratic energies. The two normal coordinates satisfy ${\ddot{q}}_{\pm }+{\omega }_{\pm }^2{q}_{\pm }=0$. This is why the normal modes are useful: one coupled system becomes two independently evolving harmonic oscillators, and any small release can be decomposed into them. At larger angles, the discarded terms in the double pendulum equations of motion matter and this exact linear superposition no longer applies.

For a canonically quantized real scalar field, the quadratic Hamiltonian contains $\frac{1}{2}{\sum }_{\mathbf{p}}{E}_{\mathbf{p}}$ in a regulated box. Normal ordering subtracts this zero-point energy, leavingThe vacuum momentum is zero with an inversion-symmetric regulator. The canonical commutation relation then gives $[P^{\mu },a_{\mathbf{p}}^{†}]=p^{\mu }{a}_{\mathbf{p}}^{†}$, identifying the momentum and energy added by one particle. The un-subtracted energy differs by a constant and is not literally the same operator.

Spatial integration of the free Hamiltonian density cancels particle-antiparticle pair terms through $E_p^2-{\mathbf{p}}^2-m^2=0$. The raw Hamiltonian is $\int d{\Pi }_p{E}_p(a^{†}a+b{b}^{†})$. Normal ordering removes the field-independent vacuum energy and gives $H=\int d{\Pi }_p{E}_p(a^{†}a+b^{†}b)$. Both creation operators increase the energy by $E_p$, so a negative-frequency field mode represents a positive-energy antiparticle, not a negative-energy state.

Norman V. Klassen is a physicist whose work with Carl K. Ross connects water calorimetry, radiolysis and absorbed dose measurement. Their 1997 study compared calculated heat defects with experimental observations, including independently measured chemical products. This comparison tests the physical model through more than its agreement with a single temperature rise.^[ref]

The issue is that a water calorimeter measures a thermal response to ionizing radiation, whereas absorbed dose concerns deposited energy per unit mass. If $H=c_w\Delta T$ is the thermal energy per unit mass, the heat defect $k$ relates them byA useful sensitivity calculation follows directly. Holding $H$ fixed and differentiating with respect to $k$ givesFor an illustrative $k=0.02$, an absolute change $\delta k=0.001$ changes the inferred absorbed dose by approximately $0.001/0.98$, or $0.102\%$. Thus improving thermometer precision alone cannot remove the contribution of an uncertain heat defect to measurement uncertainty. The example is a local sensitivity estimate, rather than an uncertainty budget for their apparatus.^[ref]

In 2002, Klassen and Carl K. Ross reported that a literature reaction rate constant used in their earlier model was incorrect and supplied a revised model. Their correction also compared model predictions with calorimetric and chemical-product measurements. It illustrates why the inputs to a computer simulation are part of the scientific claim: agreement must be reassessed when those inputs are corrected.^[ref]

For a smooth gauge field on ${\mathbb{R}}^3$ with finite positive magnetic energy, dilate its connection one-form by $A_i^{(s)}(x)=s{A}_i(sx)$. Its gauge curvature scales as $F_{ij}^{(s)}(x)=s^2{F}_{ij}(sx)$, so the energy scales as $s^4{s}^{-3}E=sE$. A static stationary configuration must have zero derivative at $s=1$, forcing $E=0$ and hence $F=0$. This Derrick scaling argument assumes the dilation preserves the boundary conditions and that boundary terms vanish. It does not exclude four-dimensional Yang-Mills instantons, whose Euclidean action is scale invariant.

In two-body alpha decay of a parent at rest, momentum conservation makes daughter and alpha momenta equal and opposite. The nonrelativistic kinetic energies therefore satisfy $T_d/T_{\alpha }=m_{\alpha }/M_d$, givingAn excited daughter reduces the available Q value. To leading recoil order, an alpha line separation $\Delta T_{\alpha }$ corresponds to excitation $E_x=(1+m_{\alpha }/M_d)\Delta T_{\alpha }$. Thus ratios of excitation energies inferred from line separations do not need the common recoil factor. These formulas assume kinetic energies small compared with rest masses; an exact relativistic two-body decay calculation can replace them when necessary.

A nuclear chain reaction is a sequence in which particles released by one nuclear reaction induce further nuclear reactions. In a neutron-driven nuclear fission chain, emitted neutrons can cause new nuclear fission, be absorbed without causing nuclear fission, or leave the material.

In a simple generation model, let $N_g$ be the mean number of chain-carrying neutrons in generation $g$, and let $k_{\mathrm{eff}}$ be their effective reproduction ratio. Then $N_{g+1}=k_{\mathrm{eff}}{N}_g$. For $k_{\mathrm{eff}}<1$ the mean chain dies away; $k_{\mathrm{eff}}=1$ gives a steady mean population; and $k_{\mathrm{eff}}>1$ gives growth in this fixed-coefficient model. Thus a steady chain has reproduction ratio one, while a growing chain is also self-sustaining in the mean. This elementary generation model does not describe every timescale or spatial change.

Whether material is fissile material depends on reaction probabilities, neutron production and losses in a specified setting. A positive nuclear fission Q value establishes that the final products can release energy, but does not by itself establish the neutron balance.

For an ultrarelativistic electron scattered elastically by a static, spherical, spin-zero atomic nucleus, a plane-wave Born approximation retains a single electromagnetic interaction. The Fourier transform of its Coulomb potential is proportional to $ZeF(q)/q^2$. Relative to the same point charge, the scattering amplitude is therefore multiplied by $F(q)$ and the differential scattering cross-section by $|F(q){|}^2$:Here $E\gg m_e{c}^2$ is the incident total electron energy, $\theta$ the scattering angle, $d\Omega$ a solid angle element, and $\alpha$ the fine-structure constant. The cosine factor already accounts for the projectile's spin; a spinless projectile would give a different point-charge factor. The formula neglects nuclear recoil, electron rest-mass corrections, radiative corrections and Coulomb distortion in electron scattering. The latter can be substantial at large $Z\alpha$: electron waves then require treatment in the target's Coulomb potential, rather than undistorted plane waves.

Nonzero target spin can introduce magnetic and additional electric contributions, so division by this point-charge expression does not generally isolate one charge form factor for a proton or deuteron. Even for spin zero, finite energy resolution, recoil and Coulomb distortion in electron scattering can change a measured minimum. A zero of the uniform-sphere model is a clear illustration of amplitude cancellation, not a universal prediction of a zero experimental counting rate.

An elastic scattering event leaves the nuclear internal quantum state unchanged while transferring energy to its motion. Let a target of mass $M$ initially be at rest. Neglect the electron rest mass, but retain relativistic target recoil. Conservation of momentum and conservation of relativistic energy giveEliminating $q$ yields $M{c}^2(E-E^{\prime })=E{E}^{\prime }(1-\cos \theta )$ and henceThe static relation $q=2E\sin (\theta /2)/(\hslash c)$ follows only when $E/(M{c}^2)$ is negligible. For small recoil, $T_R\simeq {\hslash }^2{q}^2/(2M)$; this last approximation is nonrelativistic for the recoiling nucleus, not for the incident electron. The invariant spacelike transfer, expressed in energy-squared units, isso $Q/(\hslash c)$ and the spatial inverse length $q$ coincide only in the static limit. Recoil kinematics alone does not correct the static differential scattering cross-section: a finite-mass target also changes its kinematic prefactor and the appropriate current matrix elements.

In 1947, Cesare Lattes, Hugh Muirhead, Giuseppe Occhialini and Cecil Frank Powell reported charged particle tracks in which a stopping particle was followed by a secondary particle. Their paper treated the interpretation as provisional.^[ref] The eventual distinction between pions and muons made a striking use of photography: a tiny permanent record could distinguish particles by how they slowed and transformed.

The method already had a history. In their jointly authored 1937 report, Marietta Blau and Hertha Wambacher described photographic plates exposed for five months at Hafelekar, $2300\mathrm{m}$ above sea level.^[ref] Neither shared the 1950 Nobel Prize in Physics, awarded to Cecil Frank Powell.^[ref] This was a development of photographic particle detection, with earlier contributors, rather than its invention by one person.

In his own later account, Cesare Lattes recalls calibrating emulsion shrinkage and the particle range–energy relation with accelerator-produced particles. Giuseppe Occhialini took borax-treated and untreated plates to Pic du Midi for about a month. The untreated plates provided a control; the treated plates retained more events because latent image fading was reduced. Later, Cesare Lattes exposed plates at the higher Chacaltaya station in Bolivia and obtained more candidate events.^[ref] The Pic du Midi cosmic ray exposure was roughly $3\mathrm{km}$ above sea level.^[ref]

Nuclear engineering designs and studies systems that use nuclear reactions or manage materials with radioactivity. Its subjects include nuclear reactors, radiation detection, radiation shielding, nuclear fuel, the behaviour of structural materials under irradiation, and radioactive waste management.

A nuclear power plant shows why several areas of engineering must work together. Nuclear fission releases energy; a coolant transports the resulting heat; a heat engine converts part of that heat to mechanical work; an electrical generator produces electrical energy. In a pressurized water reactor, a heat exchanger transfers heat from the pressurized reactor coolant to a separate circuit that produces steam. Thus neutron physics, heat transfer, fluid mechanics and materials science constrain the same design.^[ref]

Nuclear evaporation is statistical particle emission from an excited compound nucleus after energy has been redistributed among many nuclear degrees of freedom. It is the nuclear analogue of preferentially losing particles energetic enough to escape a bound system, rather than a direct knockout by the initial projectile.

For a particle separation energy $S$ and relative kinetic energy $ϵ$, the daughter excitation is $E_f^{\ast }=E_i^{\ast }-S-ϵ$. The emission probability depends on the available daughter nuclear level density and the inverse reaction probability. Neutron emission avoids a charged-particle Coulomb barrier, while proton or heavier-particle emission competes under different thresholds and barriers. The emitted spectrum is therefore not determined by temperature alone.