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A bosonic coherent state is an annihilation-operator eigenstate,It has $⟨N⟩=\mathrm{Var}N=|\alpha {|}^2$, a selected complex amplitude, and Poisson-distributed particle number. A coherent-field condensate variational state extends this construction to a spatial field with $⟨\hat{\psi }(\mathbf{r})⟩=\Psi (\mathbf{r})$. Evaluating a normal-ordered interacting Bose Hamiltonian in that state gives a classical-field energy functional; its variational equation is the Gross–Pitaevskii equation. This connects a macroscopic phase to interference, quantized circulation and neutral Josephson effects. A fixed-number condensate instead has vanishing one-point field expectation; off-diagonal long-range order identifies the condensate without requiring exact number fluctuations.

The BCS ground state has the pair-coherent formIt is a product of Cooper-pair pseudospin coherent states: each paired momentum sector mixes its empty and doubly occupied configurations. The anomalous pair expectation is $⟨c_{-\mathbf{k}\downarrow }{c}_{\mathbf{k}\uparrow }⟩=u_{\mathbf{k}}^{\ast }{v}_{\mathbf{k}}$, while an ordinary single-electron expectation vanishes by fermion parity. Pair operators are not unrestricted canonical boson operators, so this state must not be identified with a naive oscillator coherent state of individual electrons. Number projection gives a fixed-even-number description with pair correlations. Phase differences between condensates enter the Josephson effect; in a charged condensate that phase difference must be gauge invariant.

Condensate depletion means particles outside a macroscopically occupied bosonic single-particle mode. Interactions can deplete that mode even at zero temperature: the Bogoliubov transformation mixes creation and annihilation operators, so its quasiparticle vacuum contains particles at nonzero momentum. A weak homogeneous three-dimensional gas with repulsive scattering length $a_s$ has leading quantum depletion of a Bose condensateThermal quasiparticles add another contribution. Depletion and normal-fluid density are distinct: a Galilean-invariant fluid can have quantum depletion while retaining its full superfluid mass response at zero temperature.

For the fermionic pairing problem, mean-field decoupling of the reduced BCS pairing Hamiltonian introduces the anomalous amplitude $\Delta$. Each paired Nambu block has diagonal entries ${\xi }_{\mathbf{k}},-{\xi }_{\mathbf{k}}$ and off-diagonal pairing entries of magnitude $\Delta$. A phase convention with negative off-diagonal entries gives eigenvalues $\pm E_{\mathbf{k}}$, where $E_{\mathbf{k}}=\sqrt{{\xi }_{\mathbf{k}}^2+{\Delta }^2}$. Its canonical Bogoliubov rotation yields the BCS coherence factorsThe partner transformation has the complementary sign required by fermionic anticommutation. At zero temperature all positive-energy BCS quasiparticle modes are empty. Expanding the supplied electron operator therefore givesSince $E_{\mathbf{k}}^2-{\xi }_{\mathbf{k}}^2={\Delta }^2$,The electron occupation is close to one far below the normal-state Fermi surface, equals one half at $\xi =0$, and decays toward zero above it. Pairing smooths the Fermi step over an energy range of order $\Delta$. The positive-energy quasiparticle occupation is instead zero everywhere at this temperature; the electron sea and the quasiparticle vacuum are different descriptions of the same state. At finite temperature, $n_{\mathbf{k}\sigma }=[1-(\xi /E)\tanh (E/(2k_{B}T))]/2$ in the same mean-field model.

For a three-dimensional parabolic normal band, the total density of states including both spins is $D(\mu )=3N/(2\mu )$. In the weak-gap limit, replace it by this constant across the narrow pairing region and use $\mu \simeq E_F$. The number of electrons above the normal-state Fermi surface isThe integral is $[\xi -\sqrt{{\xi }^2+{\Delta }^2}{]}_0^{\infty }/2=\Delta /2$. Therefore the BCS population above the normal-state Fermi surface isThe constant-density-of-states approximation and a small gap relative to the chemical potential are essential. This is the question's explicitly defined fermionic population measure; it is not the fraction of electrons outside a single bosonic condensate orbital.

For an order-of-magnitude mercury estimate, use $T_c\simeq 4.2\mathrm{K}$ and the weak-coupling scale ${\Delta }_0\simeq 1.76k_B{T}_c\simeq 0.64\mathrm{meV}$, with a representative metallic $\mu \simeq 7\mathrm{eV}$. These input approximations give $\Delta N/N\sim 7\times 10^{-5}$, of order $10^{-4}$. The use of a weak-coupling gap is an estimate, not an assertion of mercury's exact measured gap.

For a copper oxide, an illustrative gap $20\mathrm{meV}$ and effective chemical-potential scale $0.5\mathrm{eV}$ give about three per cent in this formula. Choosing $0.2$–$1\mathrm{eV}$ instead gives roughly $7.5$–$1.5$ per cent. These are explicit example inputs: the paper specifies no compound, doping or band parameters from which a unique number follows. A real cuprate's anisotropic gap, quasi-two-dimensional band and strong correlations require a band-resolved occupation integral, so the reliable contrast is the much larger gap-to-electronic-energy scale, rather than a universal cuprate percentage.

The initial momentum magnitude satisfies $p^2/(m^2{c}^2)={\gamma }_i^2-1=288$. Equality of longitudinal and transverse magnitudes therefore givesAn electric field along $z$ changes $p_z$, while the magnetic Lorentz force only rotates the transverse momentum. Its magnitude $p_{\perp }$ remains fixed. At the longitudinal turning point $p_z=0$, henceFor an electron initially moving toward positive $z$, the decelerating field points toward positive $z$ because its charge is $-e$. The electric work over $\Delta z=1\mathrm{m}$ givesThe radius is determined by transverse momentum rather than total energy:so the requested radius plot is horizontal. The cyclotron angular frequency changes as $e|B|/(\gamma m)$ and the helix pitch shrinks to zero. These statements neglect radiation, as stipulated.

For a vacuum electromagnetic wave, the four-wavevector is $K^{\mu }=(\omega /c,\mathbf{k})$ with $|\mathbf{k}|=\omega /c$ and $K^2=0$. Its phase can be writtenBecause $K$ and $X$ transform as Lorentz four-vectors, their contraction is a Lorentz scalar. This proves phase invariance, with the same phase convention in both frames.

For a hypothetical free emission $e\to e+\gamma$, four-momentum conservation gives $P=P^{\prime }+Q$. The initial and final electrons have $P^2=P^{\prime 2}=m_e^2{c}^2$, and $Q^2=0$. Squaring gives $P^{\prime }\cdot Q=0$. In the final electron's rest frame, however, $P^{\prime }\cdot Q=m_e{E}_{\gamma }>0$ for a nonzero future-directed photon. ThusAn external field, another body or a medium can exchange momentum and change this conclusion; those are different systems.

The centre-of-mass velocity follows from the total energy and momentum:A boost in this direction with ${\Gamma }_{\mathrm{CM}}=\sqrt{2}$ gives zero total spatial momentum and total energy $\sqrt{2}\hslash \omega$. The constituent photons remain moving at $c$ in that frame.

For the electric-dominated crossed-field orbit, choose a boost $V=c^{2}B/E$ along $x$. Since $|E|>c|B|$, it is subluminal. The Lorentz transformation of electromagnetic fields gives $B_z^{\prime }=0$ and $E_y^{\prime }=E/\Gamma$, where $\Gamma =(1-V^2/c^2{)}^{-1/2}$. The initially stationary electron has $U^{\prime }(0)=(\Gamma c,-\Gamma V,0,0)$, not zero velocity in this frame.

Let $q=-e$, $a=q{E}_y^{\prime }/m_e$ and $s=a\tau /c$. Neglect radiation reaction, as in the prescribed external-field model. Integrating the relativistic Lorentz force in this purely electric frame gives the exact trajectoryIndeed $U^{\prime 0}=\Gamma c\cosh s$, $U^{\prime x}=-\Gamma V$, and $U^{\prime y}=\Gamma c\sinh s$ satisfy both the force equation and $U^{\prime 2}=c^2$. The constant transverse momentum must be retained.

The inverse boost gives a useful exact laboratory parametrization as well:For $E,B>0$, the lab electron starts accelerating toward negative $y$ and bends toward positive $x$, the direction of $\mathbf{E}\times \mathbf{B}$. It never acquires a cyclotron orbit in this electric-dominated case. At late times, $v_x\to V$ and $v_y\to \mathrm{sgn}(q{E}^{\prime })c/\Gamma$, so its speed approaches $c$ along an oblique asymptote. Reversing field signs reverses the appropriate directions. The zero-$B$ limit is the ordinary electric-field hyperbolic motion.

With $c=1$, the total four-momentum is $P_{\mathrm{tot}}={\sum }_i{P}_i$. A sum of Lorentz four-vectors transforms as a four-vector, sois a Lorentz invariant. Its value is the squared invariant mass of the complete system, irrespective of whether its constituents interact. Isolation makes the total four-momentum constant in time.

For a photon gas, write ${\mathbf{p}}_i=W_i{\mathbf{n}}_i$ with $|{\mathbf{n}}_i|=1$. Its individual null norms do not make the total norm vanish:For an isotropic gas in its momentum rest frame, ${\sum }_i{\mathbf{p}}_i=0$ and $s^2=W_{\mathrm{tot}}^2$. Zero is possible only when all nonzero photons propagate in the same direction. Two equal-energy photons at right angles, for example, give $s^2=2W^2$, rather than zero. Without an angular distribution, the pairwise expression is the general answer; isotropy is an additional assumption.

The asserted necessity of equal masses is false. Equal unsigned outgoing angles imply equal longitudinal momenta when the transverse momenta cancel; they do not imply equal final speeds or energies for unequal masses.

An explicit counterexample uses $c=1$, masses $M_1=m$ and $M_2=2m$, initial projectile energy $E_1=2m$, and a stationary target. The initial total energy is $4m$ and momentum is $(\sqrt{3}m,0,0)$. Choose final four-momentaThey sum to the incoming four-momentum. Their squared norms are respectively $m^2$ and $4m^2$, so the collision is elastic. Both trajectories make the same acute angle with positive $x$:Thus no proof of the printed claim is possible.

The general centre-of-mass frame construction shows why. Let $p^{\ast }$ be the outgoing momentum magnitude there, ${\theta }^{\ast }$ the projectile scattering angle, and $E_1^{\ast },E_2^{\ast }$ the two energies. After a boost ${\beta }_{\mathrm{CM}}$ along $x$, the equal-angle condition for two forward nonzero trajectories isWith $E_1$ the initial projectile energy, its solution isprovided the right side lies between minus and plus one and both final longitudinal momenta are positive. Equal masses are one allowed case, giving ${\theta }^{\ast }=\pi /2$; they are not necessary. For the counterexample the right side is $3/16$. Even equal masses do not force equal angles for every scattering direction: the symmetric outgoing configuration is an additional choice.

For $P^{\mu }=(E/c,\mathbf{p})$, directly contract the supplied vector:Thus $\boxed{U_W\cdot P=0}$ whenever its stated normalization is defined, namely $m_0>0$ and $|\mathbf{s}|>0$. This is the Pauli-Lubanski orthogonality property.

The description as a unit vector, however, needs an additional assumption. From the given components and the mass shell,It equals minus one only for spin parallel to momentum, including the zero-momentum case. A concrete counterexample is $\mathbf{s}=(s,0,0)$ and $\mathbf{p}=(0,m_{0}c,0)$: the proposed normalized vector has square minus two. Orthogonality is still true. A genuine Pauli-Lubanski pseudovector constructed by boosting a fixed rest-frame spin has invariant norm $-m_0^2{c}^2|{\mathbf{s}}_{\mathrm{rest}}{|}^2$; the source's generic spatial expression and denominator do not supply that normalization for arbitrary $\mathbf{s}$.

Write $\Gamma =(1-v^2/c^2{)}^{-1/2}$ for the Lorentz factor of the inertial frame $S^{\prime }$, distinguishing it from the particles' changing Lorentz factors. Apply the Lorentz transformationto each world line in a. Both particles initially have velocity $-v$ in $S^{\prime }$. The switching events have coordinatesThus the Positron starts accelerating earlier in $S^{\prime }$. Its velocity first becomes less negative, vanishes when its laboratory velocity reaches $v$, and then becomes positive. The electron continues at $-v$ until $t^{\prime }=0$, then accelerates toward $-c$. The relativistic velocity-addition formula makes this precise:The electric field parallel to the Lorentz boost has unchanged magnitude, but its switching surface is $t^{\prime }+v{x}^{\prime }/c^2=0$, rather than $t^{\prime }=0$ everywhere.

In the laboratory, the total three-momentum is zero for all $t$: the two Lorentz forces cancel at equal laboratory times. In $S^{\prime }$, each active force is still $qE$, since this is one-dimensional motion with a parallel electric field. Define $t_p^{\prime }=-\Gamma vd/c^2$. For $v>0$ the total three-momentum, evaluated at equal $t^{\prime }$, isThis follows by integrating $d{p}_p^{\prime }/d{t}^{\prime }=F$ from its switching event and $d{p}_e^{\prime }/d{t}^{\prime }=-F$ from its later switching event. During the interval between them only the Positron feels the prescribed electric field, so momentum conservation for the particle pair fails. Once both forces act, the total three-momentum is constant again.

The equal-time momentum of an externally driven particle pair illustrates a distinction. The sum of particle four-momenta at a fixed pair of events is always a four-vector. However, a fresh sum at equal time in each inertial frame generally uses different events. Here transforming an equal-$t$ laboratory sum would give $-2\Gamma v\sqrt{m^2{c}^4+F^2{c}^2{t}^2}/c^2$, which varies with $t$; this differs from the equal-$t^{\prime }$ answer above. For example, with $c=m=F=d=1$ and $v=0.6$, the late-time equal-$t^{\prime }$ sum is $-0.75$, whereas transforming the equal-$t=0$ sum gives $-1.5$. This is a concrete failure of treating the pair's equal-time sums as components of a single frame-independent four-vector.

A closed system's total four-momentum, including the electromagnetic field and the apparatus supplying it, is conserved and independent of the choice of spacelike integration surface under the usual isolated-system boundary conditions. The particles alone exchange energy and momentum with an external system, so neither that independence nor universal particle-only momentum conservation is guaranteed.

With the reversed electric field, reflection about $x=d/2$ exchanges the electron and Positron. They therefore meet there with equal and opposite momenta. Each travels $d/2$, and the work-energy theorem gives its kinetic energy:Their combined kinetic energy is $1.000\mathrm{MeV}$, while each total relativistic energy is $\mathcal{E}=0.511+0.500=1.011\mathrm{MeV}$. The laboratory is the center-of-momentum frame of the colliding pair.

Local four-momentum conservation in the electron-positron annihilation excludes one real photon: the incoming total four-momentum is timelike, with zero spatial component and positive energy, while a single photon's four-momentum is null and has $|\mathbf{p}|=\mathcal{E}/c>0$. Two photons can balance their momenta by travelling in opposite directions, soThe equal photon energies follow from opposite equal momenta, not merely from sharing the incoming energy arbitrarily. Here $hc\simeq 1.23984\mathrm{MeV}\mathrm{pm}$. This is the free local annihilation idealization implicit in the question; significant momentum transfer to the external apparatus would change the kinematic system being considered.

Define ${\sigma }_{ij}={\mathbf{e}}_i\cdot \mathbf{t}({\mathbf{e}}_j)$: the second index labels the normal direction of a coordinate face, and the first labels the force component. These nine entries form the Cauchy stress tensor.

Take a small tetrahedron with oblique face of area $S$ and outward normal $\mathbf{n}$. First suppose $n_j>0$; its other three faces have normals $-{\mathbf{e}}_j$ and areas $S_j=n_{j}S$, by orthogonal projection of the oblique face. The total surface force isIf its linear size is $\ell$, its inertia is $O({\ell }^3)$ while its surface forces are $O({\ell }^2)$. Divide the momentum balance by $S$ and shrink the tetrahedron to the point. Smoothness of the stress and the odd-normal relation then give $\mathbf{t}(\mathbf{n})={\sum }_j{n}_j\mathbf{t}({\mathbf{e}}_j)$. Choosing coordinate faces with the appropriate reversed normals gives the same identity in the other octants. In components,The face-area relations used are exactly $S_j=|n_j|S$, with each coordinate face oriented opposite the corresponding component of the oblique normal.

For a material volume moving with the fluid, the Reynolds transport theorem gives, componentwise,For a scalar $q$ this is also ${\int }_{V(t)}(Dq/Dt+q\nabla \cdot \mathbf{u})dV$. Apply the component formula to momentum density $q=\rho u_i$. Constant density and incompressibility make the integrand $\rho D{u}_i/Dt$.

The traction $\mathbf{t}(\mathbf{n})$ is the force per unit area exerted by the material outside a surface on the material inside, with outward unit normal $\mathbf{n}$. Reversing the normal exchanges the two sides of the same internal surface. Thus $\mathbf{t}(-\mathbf{n})=-\mathbf{t}(\mathbf{n})$ expresses equal and opposite internal contact forces, the local form of Newton's third law.

With no body force, Newton's second law equates the rate of change of the material volume's total momentum to its surface force. Combining this balance with the transport theorem givesThis is a vector identity, since the component argument applies to each direction.

The only velocity component is axial, so incompressibility gives ${\partial }_{z}w=0$. Thus $w=w(x,y)$, and the steady convective acceleration is $w{\partial }_z(w\mathbf{k})=0$. The transverse momentum equations give $p_x=p_y=0$, so $p=p(z)$. The axial momentum equation is thereforeIts left side depends only on $x,y$, while its right side depends only on $z$. Equality throughout the pipe forces both to be one constant, say $G$. Consequently the axial pressure gradient is constant, and $p(z)=Gz+p_{\ast }$. This is absence of dependence on a coordinate, rather than probabilistic independence.

Take the tangential length and velocity scales as order one. In the boundary layer, continuity reads $u_x+({\delta }_2/{\delta }_1)V_Y=0$, so a nontrivial balance requires ${\delta }_2\sim {\delta }_1$. Tangential inertia is order one, and the normal viscous term has size ${\mathrm{Re}}^{-1}{\delta }_1^{-2}$. Balancing them givesup to constants absorbable in the scaled variables. Streamwise diffusion is then smaller by ${\mathrm{Re}}^{-1}$. The normal momentum equation has a pressure term of size ${\delta }_1^{-1}{p}_Y$ and inertial/viscous terms of size ${\delta }_1$, hence $p_Y=0$ at leading order. The remaining Prandtl boundary-layer equations areThe boundary conditions are no slip and no penetration at the plate, and matching to the outer tangential velocity:The outer inviscid tangential momentum equation supplies $p_x=-U_s{U}_s^{\prime }$; pressure is shared between the boundary layer and outer flow because it has no leading normal variation.

Linearize about rest, with the equilibrium forces already absorbed in the reduced pressure. The quadratic convective acceleration drops out. Since $\mathbf{k}\times (u\mathbf{i}+v\mathbf{j}+w\mathbf{k})=-v\mathbf{i}+u\mathbf{j}$, the component equations areThe scan prints $v_w$ in the continuity equation; the derivative required by incompressibility is $v_y$.

To eliminate pressure, put $\zeta =v_x-u_y$ and ${\nabla }_h^2={\partial }_x^2+{\partial }_y^2$. Taking the vertical component of the curl of momentum givesTaking horizontal divergence of momentum and substituting continuity givesDifferentiate this in $z$ and use $P_z/\rho =-w_t$:One more time derivative, followed by the vorticity relation, yields the Poincaré equation for inertial wavesThis derivation retains all three velocity components, including the transverse component coupled by rotation.

Write $\omega$ for the nonzero forcing frequency, to distinguish it from the vertical velocity $w$. Put $k_z=\pi /h$ and $Z(z)=\cos (\pi (2z+h)/(2h))=-\sin (k_{z}z)$. Then $Z(0)=Z(h)=0$ and $Z^{\prime \prime }=-k_z^{2}Z$, so it meets the rigid-wall conditions. Substituting $w=\mathrm{Re}(f(x)e^{-i\omega t}Z(z))$ into the scalar wave equation givesorThe radiation sign for an inertial wave must exclude any disturbance arriving from $x=+\infty$. This means an radiation condition for propagating waves, or bounded decay for an evanescent wave.

For $0<|\omega |<2|\Omega |$, put $q=k_z\sqrt{4{\Omega }^2/{\omega }^2-1}$. The solutions are oscillatory. The inertial-wave dispersion relation is ${\omega }^2=4{\Omega }^2{k}_z^2/(k_x^2+k_z^2)$; its horizontal group velocity is $c_{g,x}=-\omega k_x/(k_x^2+k_z^2)$. Positive energy transport into the channel therefore requires $k_x=-\mathrm{sgn}(\omega )q$. The solution isFor a positive frequency, the phase travels towards negative $x$ while the group velocity travels towards positive $x$. Choosing the outgoing sign from the phase velocity alone would give the wrong result.

For $|\omega |>2|\Omega |$, set $\kappa =k_z\sqrt{1-4{\Omega }^2/{\omega }^2}$. Excluding the exponentially growing solution givesThis disturbance decays away from the wavemaker and does not propagate horizontally. At the cutoff the ODE degenerates to $f^{\prime \prime }=0$; zero frequency is not a nonzero forced harmonic mode with this vertical shape.

There is a second printed inconsistency: the stated velocity $u\mathbf{i}+w\mathbf{k}$, if taken literally with no $y$ dependence and no transverse velocity, cannot support a nontrivial rotating inertial wave. Its transverse momentum equation would give $2\Omega u=0$, then continuity and the rigid walls would force $w=0$. The scalar solutions above are the intended inertial-wave solutions when two-dimensional means dependence on $x,z$ while retaining the transverse component $v$. For a mode, $-i\omega v+2\Omega u=0$ explicitly supplies the generally nonzero transverse velocity. Thus the scalar equation can be solved as requested, but the full vector field needs this correction to the printed two-component description.

Let the undisturbed state ahead have $({\rho }_0,p_0,u_0)=({\rho }_0,p_0,0)$, and the uniform compressed state behind have velocity $u_1=U$. The piston boundary fixes that downstream velocity. Mass and momentum jumps giveThus ${\rho }_1/{\rho }_0=V/(V-U)$. Equality of specific stagnation enthalpy in the shock frame isSubstitute $p_1=p_0+{\rho }_{0}UV$ and $1/{\rho }_1=(V-U)/({\rho }_{0}V)$, then multiply by $2(\gamma -1)V$. The resulting difference factors as $U[-2V^2+(\gamma +1)UV+2\gamma p_0/{\rho }_0]=0$. Since the piston moves into the gas with $U>0$, division givesThis determines the constant speed of the compression shock ahead of the piston.

Put $v=u-V$, the velocity in the shock frame. The mass jump condition gives a common mass flux $j=\rho v$. Subtracting $V$ times the mass balance from the momentum balance givesFor energy, transform the conserved energy to the shock frame by subtracting $V$ times momentum and adding $V^2/2$ times mass. The resulting energy flux isFor a genuine shock, $j\ne 0$; dividing its continuous energy flux by the same nonzero $j$ on both sides givesThe second term is the ideal-gas specific enthalpy. A contact discontinuity has zero relative mass flux and does not in general require equal specific enthalpies; the division here is appropriate for the shock specified in the question.

Differentiate with respect to the generalized velocities:The Lagrangian has no explicit $x$, so ${\dot{p}}_x=\partial L/\partial x=0$. This canonical momentum is also the system's total horizontal mechanical momentum. Horizontal translation symmetry conserves $p_x$ even though the support and bob separately accelerate.