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Two identical torsional loads of moment of inertia $I$ hang in series from massless fibres of lengths ${\ell }_1,{\ell }_2$ and common filament torsional stiffness $K_t$. Put $R={\ell }_2/{\ell }_1$ and ${\omega }_0^2=K_t/(I{\ell }_1)$. The elastic energy is $K_t{\varphi }_1^2/(2{\ell }_1)+K_t({\varphi }_2-{\varphi }_1{)}^2/(2{\ell }_2)$ and the kinetic energy is $I({\dot{\varphi }}_1^2+{\dot{\varphi }}_2^2)/2$. Their normal modes have the displayed squared angular frequencies. Writing $\lambda ={\omega }^2/{\omega }_0^2$, their amplitude ratio is $A_2/A_1=1/(1-R\lambda )$. The lower mode twists in phase and approaches a combined inertia $2I$ as $R\to 0$, giving ${\omega }_{-}^2\to {\omega }_0^2/2$. The upper mode approaches the original single-load oscillation as $R\to \infty$, giving ${\omega }_{+}^2\to {\omega }_0^2$.

A clock following a timelike circular Schwarzschild geodesic of radius $R_s$ advances at $\sqrt{1-3\mu /R_s}$ per unit Schwarzschild coordinate time. A static clock at radius $R_E>2\mu$ advances at $\sqrt{1-2\mu /R_E}$. Their ratio over the same coordinate-time interval gives the displayed result. For periodic signal comparisons, successive corresponding signals must also have the same propagation delay, as happens for repeated overhead passages in a fixed circular orbit observed from a static pole. To first order in $\mu /R$, the ratio is $1+\mu /R_E-3\mu /(2R_s)$.

A source on a timelike circular Schwarzschild geodesic at radius $r>3GM/c^2$ emits light to a static observer at infinity on the orbital symmetry axis. Such a ray has zero conserved axial angular momentum. The locally measured photon energy is the contraction of its four-momentum with the observer's four-velocity. Comparing the emitter with the observer at infinity therefore gives ${\nu }_{\mathrm{em}}=U^t{\nu }_{\infty }$, where $U^t=[1-3GM/(c^{2}r){]}^{-1/2}$, proving the formula.

Equivalently the gravitational redshift factor $\sqrt{1-2GM/(c^{2}r)}$ multiplies the transverse Doppler effect factor $1/\gamma$ measured by a local static observer. For $R_S=2GM/c^2\ll r$,The coefficient contains both gravitational and orbital time dilation. It assumes an observer at infinity on the symmetry axis; other viewing directions generally introduce a first-order longitudinal Doppler effect.

Atmospheric cosmic rays generate muons whose mean lifetime at rest is about $2.2\mu \mathrm{s}$. For a selected speed $v$ and production-to-detection distance $h$ in the terrestrial frame, special relativity predicts survival fraction $e^{-h/(\gamma v{\tau }_0)}$, where $\gamma$ is the Lorentz factor. Without time dilation, the exponent would instead be $-h/(v{\tau }_0)$. Comparing altitude-dependent counts at controlled energies, allowing for acceptance and energy loss, tests these alternatives. In the muon rest frame the terrestrial distance is shortened to $h/\gamma$ by length contraction, so the proper flight time and survival fraction agree with the terrestrial calculation. This frame comparison does not turn the same decay measurement into two independent tests.

Accelerator-produced muons circulating in a magnetic storage ring allow a direct comparison between their measured laboratory mean lifetime and $\gamma$ times the rest lifetime. The 1977 experiment at $\gamma =29.33$ measured positive- and negative-muon lifetimes near $64.4\mu \mathrm{s}$, consistent with time dilation. This is a controlled complement to atmospheric survival measurements, with different systematic effects.^[ref]

A two-state degree of freedom represented mathematically like spin one-half, without necessarily being physical spin. The sublattice pseudospin in graphene labels amplitudes on two sites per primitive cell.

A hopping integral is an off-diagonal Hamiltonian matrix element between localized orbitals on different lattice sites. It has units of energy and controls the dispersion in a tight-binding model. Its sign and phase depend on the orbital phase convention; observable band energies do not. The sign of a real nearest-neighbour $t$ in $E(k)=E_0+2t\cos (ka)$ determines the location of the minimum in that convention.

In the orthogonal tight-binding approximation, localized basis states satisfy $⟨i|j⟩={\delta }_{ij}$ while the Hamiltonian retains nonzero hopping integrals. These are independent matrix elements: orthogonal states need not have vanishing Hamiltonian coupling. One can construct localized orthonormal orbitals from overlapping atomic orbitals; truncating the resulting Hamiltonian to nearest neighbours is a further approximation. An overlap matrix must instead be retained for raw nonorthogonal orbitals.

For one localized orbital per site in a chain, take on-site matrix elements $H_{jj}=E_0$, nearest-neighbour matrix elements $H_{j,j\pm 1}=h$, overlaps $S_{jj}=1$ and $S_{j,j\pm 1}=s$, with other terms neglected and $h,s$ real. A Bloch state reduces the generalized eigenvalue problem $Hc=ESc$ toThe overlap metric must be positive for the retained model; $|s|<1/2$ is sufficient for all $k$. Dropping it yields the cosine band of the orthogonal tight-binding approximation.

In an orthonormal basis of localized atomic orbitals, a ring of three equivalent sites with real hopping $t$ has Hamiltonian $H=(E_0-t)I+tJ$, where $J$ is the three-by-three all-ones matrix. The normalized symmetric state $(|1⟩+|2⟩+|3⟩)/\sqrt{3}$ has energy $E_0+2t$. Its two-dimensional orthogonal complement has energy $E_0-t$. Thus the symmetric state is the unique spatial ground state for $t<0$, with an energy difference $3|t|$ to the excited eigenstates. Equivalently, the three discrete Bloch states have energies $E_0+2t\cos (2\pi l/3)$, $l=0,\pm 1$. Nonorthogonal atomic orbitals instead require a generalized eigenvalue problem.

For an identical-flavour pair in the symmetric spatial ground state of a baryon, the flavour factor is symmetric and the three-quark colour singlet is antisymmetric under their exchange. The Pauli exclusion principle therefore requires a symmetric pair spin factor. Two spin-one-half quarks then form a triplet state, $S_{qq}=1$, rather than a spin singlet. The conclusion applies to the identical pair in both spin-one-half and spin-three-half ground-state baryons.

Consider an original reduced model with a fixed number $N$ of active emitters. Let $u=N_2/N$ be the upper lasing-state fraction, $1-u=N_1/N$ the ground-state fraction, and $s=n/N$ the number of cavity photons per emitter. Assume the pump state empties rapidly, so its population can be neglected. Measure time by $x=t/\tau$, where $\tau$ is the upper state's spontaneous lifetime. Let $r=W\tau$ be the effective pump rate, $C>0$ the dimensionless stimulated-coupling coefficient, $K>0$ the photon-loss rate in these units, and $0\le \beta \le 1$ the fraction of spontaneous decays feeding this optical normal mode. The rate equations areEach term has a physical role. Pumping transfers emitters out of the ground state; spontaneous emission removes upper-state population. The net stimulated term includes emission from fraction $u$ and absorption by fraction $1-u$, giving their difference $2u-1$. A stimulated transition removes one stored excitation and creates one cavity photon, or does the reverse. The factor $\beta$ accounts for spontaneous emission into the selected mode without pretending that all spontaneous emission enters it.

For negligible cavity light, the upper-state equation is linear. A constant pump applied to $u(0)=0$ givesThus this model first permits population inversion at $r>1$. Positive optical gain is still insufficient for an oscillator: the gain must exceed the complete laser threshold, including photon loss. The small-signal photon growth coefficient changes sign whenIf $C\le K$, even complete inversion cannot exceed the assumed loss in this model. For chosen $C=100$ and $K=20$, the threshold population is $u_{\mathrm{th}}=0.6$ and the pump threshold is $r_{\mathrm{th}}=1.5$. At $r=2$, the unsaturated population first reaches $0.6$ at $x=(\log 10)/3\simeq 0.7675$. These numbers are stipulated examples, not ruby material parameters.

A ruby laser uses chromium ions in ruby as its amplifying medium. Maiman's report identifies a metastable state fed by optical excitation and red emission near $694.3\mathrm{nm}$.^[ref] The pump stores excitation in the medium; stimulated emission transfers it into the optical field. The simplified three-level laser rate equation explains why the stored population and the output pulse evolve together.

In a regular thermodynamic formulation, a state at absolute zero cannot be reached from positive temperature by finitely many cooling operations with finite control resources. The entropy-difference form of the third law of thermodynamics makes regular entropy curves approach a common limiting entropy. An isentropic process starting above that entropy cannot intersect the common zero-temperature endpoint.

For an illustrative pair of settings with $S-S_0=a_{1}T$ and $S-S_0=a_{2}T$, $0<a_1<a_2$, an isothermal entropy-reduction step followed by reversible adiabatic cooling gives $T_{n+1}=(a_1/a_2)T_n$. Every finite number of stages leaves $T_n>0$. The model illustrates the entropy geometry; a general finite-time impossibility statement additionally needs assumptions about available interactions and cooling rates.

At fixed pressure and material content, $S=-(\partial G/\partial T{)}_p$. If the equilibrium entropy tends to zero, the Gibbs free energy approaches its zero-temperature value with zero slope. When $C_p=a{T}^r+o(T^r)$ with $r>0$ and zero residual entropy, integrating $dS=C_{p}dT/T$ gives $S=(a/r)T^r+o(T^r)$ and $G=G(0,p)-a{T}^{r+1}/[r(r+1)]+o(T^{r+1})$. Nonzero residual entropy $S_0$ instead contributes $-S_{0}T$ to $G$; a horizontal limiting tangent requires the zero-entropy hypothesis.

For a closed system with fixed composition in an environment at temperature $T_R>0$ and pressure $p_R$, define $A=U+p_{R}V-T_{R}S$, where $U$ is internal energy, $V$ is volume and $S$ is entropy. The exergy is this availability relative to the equilibrium reference state, $A-A_{\mathrm{eq}}$. When the combined system and environment are isolated except for the specified exchanges, $\Delta S_{\mathrm{tot}}=-\Delta A/T_R$, so the Second law of thermodynamics makes $A$ decrease toward its constrained minimum. Away from equilibrium, the environmental $T_R,p_R$ must not be replaced by the system's instantaneous values.

The ideal Brayton cycle consists of adiabatic compression, constant-pressure heat addition, adiabatic expansion, and constant-pressure heat rejection. Its ideal working substance is an ideal gas with constant heat capacities, and the insulated stages are reversible thermodynamic processes.

For pressure ratio $r_p=p_h/p_l$ and heat capacity ratio $\gamma =C_p/C_V$, a reversible ideal-gas adiabat has $T\propto p^{(\gamma -1)/\gamma }$. Thus $T_2/T_1=T_3/T_4=k=r_p^{(\gamma -1)/\gamma }$, and the constant-pressure heat transfers have $Q_{\mathrm{out}}/Q_{\mathrm{in}}=(T_4-T_1)/(T_3-T_2)=1/k$. The thermal efficiency isThis formula requires constant heat capacities and internally reversible insulated stages. An arbitrary dissipative adiabatic process does not satisfy the same pressure-temperature relation.

Two equal sealed vessels of fixed volume, initially at $T_0,p_0$, exchange an ideal gas through a thin effusive opening. Hold one at $T_1$ and the other at $T_0$, with both interiors locally thermalized. Zero net particle transfer gives $p_h/\sqrt{T_1}=p_c/\sqrt{T_0}$. Writing $q=\sqrt{T_1/T_0}$, conservation of total particle number gives $p_h=2p_0{q}^2/(1+q)$ and $p_c=2p_{0}q/(1+q)$. Thus $\Delta p=2p_{0}q(q-1)/(1+q)$. The alternative $p_c=p_0$, $\Delta p=p_0(q-1)$ requires a cold reservoir whose pressure is externally held fixed; it is not the closed equal-volume problem.