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Let $q=\sigma /L$, with $L>0$, and let $Z=X^1+i{X}^2$. Direct differentiation gives $\ddot{Z}=Z^{\prime \prime }=-Z/L^2$, while the time coordinate also satisfies the wave equation. Using the Minkowski metric,Thus both Virasoro constraints hold, and the induced worldsheet metric is $g_{\mu \nu }={\cos }^{2}q\mathrm{diag}(-1,1)$. The configuration solves the Nambu–Goto action equations in its interior.

Place the fixed end at $\sigma =0$. The first free endpoint has $Z^{\prime }=e^{it/L}\cos q=0$ at $q=\pi /2$. For the usual single unfolded rotating Nambu–Goto string,This parameter interval need not equal the earlier interval $[0,\pi ]$, since that interval was a coordinate convention. The proper length is measured on a constant-$X^0$ slice; the local motion is perpendicular to the string, so there is no longitudinal Lorentz contraction.

All points have the same phase $t/L$, hence the segment rotates rigidly with angular velocity $\Omega =1/L$. At radius $r=L\sin q$, its speed is $v=r/L$. Therefore the free tip moves at the speed of light. This is consistent with its Neumann boundary condition: $X^{\prime }=0$ and the Virasoro constraints imply ${\dot{X}}^2=0$ at the free tip. The null endpoint is a boundary limit, not a nondegenerate interior point.

In this conformal gauge, $P_m=T{\dot{X}}_m$, so the energy density per $\sigma$ is $-P_0=T$. Equivalently, the energy density per proper length is $T/\sqrt{1-r^2/L^2}$. Thusand the rotational kinetic energy isThe angular momentum about the fixed point isTherefore the Regge trajectory isThe endpoint conditions alone also admit folded extensions with parameter length $(n+\frac{1}{2})\pi L$, $n\ge 0$. Their proper length, energy and angular momentum are $(2n+1)$ times the values above, and $\beta =1/[(2n+1)\pi T]$. The displayed answer uses the implicit unfolded-segment convention.

Differentiate the quadratic energy:For $x_1\ne 0$, put $r=x_2/x_1$. Immediate energy growth occurs precisely when ${\lambda }_2{r}^2+r+{\lambda }_1>0$. With both eigenvalues negative, such directions exist iff $1-4{\lambda }_1{\lambda }_2>0$. Writing $\delta =\sqrt{1-4{\lambda }_1{\lambda }_2}$, the growing initial conditions are the two opposite open wedgesIf $x_1=0$, energy decreases. No energy growth at any time, for any initial condition, is possible exactly whenIndeed, this makes the symmetric part of a matrix $H$ a negative semidefinite matrix, so $\dot{E}\le 0$ along every trajectory. If the product is smaller, the wedges already supply counterexamples at $t=0$. This is the instantaneous energy-growth criterion for a linear system.

At each fixed time, the maximum over nonzero initial conditions is the optimal energy amplification of a linear system:Write $a=e^{{\lambda }_{1}t}$, $d=e^{{\lambda }_{2}t}$ and $b=(a-d)/({\lambda }_1-{\lambda }_2)$. The symmetric matrix $A^{T}A$ has trace $\tau =a^2+b^2+d^2$ and determinant $a^2{d}^2$. Its largest eigenvalue isAn optimal initial condition is a right singular vector of $A(t)$ associated with its largest singular value.

If ${\lambda }_2>{\lambda }_1$, let $\alpha =1/({\lambda }_2-{\lambda }_1)$. Then $A(t)/e^{{\lambda }_{2}t}\to \left(\begin{array}{cc}0& 0\\ \alpha & 1\end{array}\right)$, so the precise long-time asymptotic statement isThe Cauchy-Schwarz inequality shows that the unit initial conditions achieving this leading factor areThis is an adjoint eigenvector, since $L^T(\alpha ,1{)}^T={\lambda }_2(\alpha ,1{)}^T$. It differs from the right eigenvector $v_2=(0,1{)}^T$: the initial condition maximizes the projection onto the slow mode, whereas the eventual state aligns with that right eigenvector. For ${\lambda }_2<0$, even the optimal energy ultimately decays to zero; the prefactor describes enhanced excitation of the slow mode.

For an infinitesimal test mass $\delta m$ on the surface of a homogeneous sphere, its Newtonian mechanical energy isThe spatially flat case corresponds to zero specific energy. Usingand dividing by $\delta m{R}^2/2$ givesSince $R$ is proportional to the scale factor $a$, this is the Newtonian derivation of the flat Friedmann equation

Now take $a(t_0)=1$ and $\rho ={\rho }_0{a}^{-4}$. PutThe contracting branch of the Friedmann equation isIntegrating $ada=-Cdt$ from $t_0$ givesThereforeon the contracting interval, reaching $a=0$ at $t=t_0+1/(2C)$.

The chemical potential is the change in internal energy when one particle is added at fixed entropy and volume:Equivalently, $\mu =(\partial F/\partial N{)}_{T,V}$ for the Helmholtz free energy.

Let a microstate $s$ have energy $E_s$ and particle number $N_s$. Maximizing the Gibbs entropy subject to normalization and fixed mean values of $E$ and $N$ is the maximum-entropy derivation of equilibrium ensembles. With Lagrange multipliers $\alpha$, $\beta$, and $-\beta \mu$, varyThe stationarity equation is $-\log p_s-1-\alpha -\beta (E_s-\mu N_s)=0$. Normalization therefore gives the grand canonical ensembleHere $\mathcal{Z}$ is the grand canonical partition function and $k_B$ is the Boltzmann constant.

For one fermionic quantum state of energy $\epsilon$, the Pauli exclusion principle permits occupation numbers only $n=0,1$. Its two grand-canonical weights are $1$ and $e^{-\beta (\epsilon -\mu )}$, so its mean occupation is the Fermi-Dirac distribution

For a free nonrelativistic particle, $\epsilon ={\hslash }^2{k}^2/(2m)$. In a region of area $A$, the number of wave-vector states in the annulus from $k$ to $k+dk$, including the two spin angular momentum states, isSince $kdk=(m/{\hslash }^2)d\epsilon$, the two-dimensional free-electron density of states is constant:

We now use units in which $k_B=1$, as in the question. At zero temperature, the Fermi-Dirac distribution is a step function, and henceAt positive temperature the same fixed particle number satisfies the exact relationThusThis is the low-temperature particle-number cancellation for constant density of states. Therefore, for $T\ll {\epsilon }_F$, the mean number of particles in the energy interval $[\epsilon ,\epsilon +d\epsilon ]$ is

Finally, compare the internal energy with its zero-temperature value. The thermally excited particles above ${\epsilon }_F$ and holes below ${\epsilon }_F$ have equal leading particle numbers, so their terms proportional to ${\epsilon }_F$ cancel. With $x=|\epsilon -{\epsilon }_F|/T$,where the integral follows from the fermion-to-boson thermal integral ratio. Differentiation gives the linear low-temperature heat capacity of a two-dimensional Fermi gas:The requested power law is therefore linear in $T$, with exponent one.

The coordinates $\varphi$ and $\psi$ are cyclic, so their canonical momenta are conserved:Because the Lagrangian has no explicit time dependence, the conserved energy isThese are the three integrals of motion requested.

The torque about the origin isSince $\dot{L}=\tau$, the angular momentum $L=mr\times \dot{r}$ is conserved. The force is the negative gradient of the inverse-square potentialso the conserved energy isDirectly, $\dot{E}=\dot{r}\cdot (m\ddot{r}+kmr/r^3)=0$.

The work done along the trajectory isBy Newton's second law, $F=m\ddot{x}$, soThis is the work-energy theorem: the work equals the change in kinetic energy.

A force is conservative when its work between two points is independent of the path, equivalently when it is the negative gradient of a potential energy. Here a potential with $V(0)=0$ isConservation of energy for $x>0$ givesThus

If $v_0>v_e$, the particle escapes with asymptotic speed $\sqrt{v_0^2-v_e^2}$; the graph decreases from $v_0$ to this horizontal asymptote. At the escape velocity $v_0=v_e$, it decreases as $v_e{e}^{-x/(2\lambda )}$ and approaches zero only at infinity. If $v_0<v_e$, it reaches the turning pointwhere $v=0$, then returns on the negative branch of the same curve and oscillates symmetrically about the origin.

For oscillation put $w=1-v_0^2/v_e^2$. One quarter of the period iswhere $u=e^{-x/\lambda }$. Therefore

Bosons have symmetric many-particle states, so any number may occupy the same one-particle state. At low temperature, an ideal Bose gas therefore places a macroscopic population in its lowest one-particle state; below the critical temperature this is Bose-Einstein condensation.

Fermions have antisymmetric many-particle states and obey the Pauli exclusion principle, so each one-particle state can hold at most one fermion of each available internal label. At zero temperature, a noninteracting Fermi gas fills distinct states in increasing order of energy up to the Fermi energy. Its many-particle ground state is thus the filled Fermi sea.

Let $x^{\mu }=(ct,\mathbf{x})$, let $\tau$ be proper time, and define the four-velocity and four-momentum byIf $F^{\mu \nu }$ is the electromagnetic field tensor, the covariant Lorentz force law isThe temporal component and three spatial components respectively giveHere $\mathcal{E}=\gamma m{c}^2$ is the relativistic energy and $\mathbf{p}=\gamma m\mathbf{v}$ is the relativistic momentum. In the nonrelativistic limit, $\gamma \to 1$, so the spatial equation becomesthe usual Lorentz-force law.

Photon absorption removes a photon and transfers its energy and momentum to matter. It contributes a negative photon-number collision source. It differs from Compton scattering, which redistributes a photon's energy and direction while preserving the count. The integrated source can vanish if photon emission and absorption balance.

For a free field, its positive-frequency part consists of the annihilation terms multiplying $e^{-i\omega t}$ with $\omega >0$; its negative-frequency part consists of the creation terms. The sign refers to positive energy, even though the exponential has a minus sign. Imposing the Lorenz gauge only through $(\partial \cdot A{)}^{(+)}|\mathrm{phys}⟩=0$ is the subsidiary condition in Gupta-Bleuler quantization, not an operator identity for the whole divergence.

A circular closed string contracts and re-expands with $X^0=Rt$ and $X^1+i{X}^2=R\cos t{e}^{i\sigma }$. Its proper length is $2\pi R|\cos t|$, while its energy remains $2\pi TR$. At collapse the induced worldsheet metric degenerates, but the Nambu-Goto phase-space action supplies a regular continuation.

For a resonator, the quality factor is $2\pi$ times the stored energy divided by the energy dissipated per cycle, using a specified steady-state stored-energy convention. A series RLC circuit at resonance has $Q=L{\omega }_0/R$, because its maximum magnetic energy is $L{I}_M^2/2$ and one cycle of Joule heating dissipates $\pi R{I}_M^2/{\omega }_0$.

A quantum time-reversal operator is an antiunitary operator reversing spatial momentum and angular momentum. For a time-reversal-invariant Hamiltonian operator, its conjugation of $i$ makes it intertwine forward and backward unitary time evolution without reversing the energy spectrum. A scalar momentum eigenstate may transform as $\hat{T}|\mathbf{p}⟩=e^{i\chi (\mathbf{p})}|-\mathbf{p}⟩$. With unit-modulus phases and a complete normalized basis, expansion of arbitrary states proves antiunitarity.

A quantum vortex has integer winding number of a condensate's complex argument. For atomic mass $m$, its superfluid velocity gives circulation $2\pi \hslash q/m$ for integer $q$. A straight vortex has a density-depleted core and, outside it, tangential velocity $q\hslash /(mr)$. Its long-range flow energy is proportional to $q^2\log (R/a)$, while its angular momentum has the sign of $q$.