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The real scalar field has canonical momentum $\pi =\dot{\varphi }$. Its Legendre transform givesThe first three terms constitute $H_0$, the free massive Hamiltonian operator. The remaining term is $H_I=\lambda \int {\varphi }^4{d}^{3}x/4!=-\int {\mathcal{L}}_I{d}^{3}x$. This sign is a consequence of having no time derivatives in the interaction: the canonical momentum is unchanged, while the interaction enters the Hamiltonian density with the opposite sign to the Lagrangian density.

In the interaction picture, operators evolve with $H_0$ and states evolve with $H_I(t)$ constructed from the free interaction-picture field. Their evolution operator satisfiesIntegrating once gives $U_I=I-i{\int }_{t_0}^t{H}_I(t_1)U_I(t_1,t_0)d{t}_1$. Repeated substitution orders the integrations as $t\ge t_1\ge \cdots \ge t_n\ge t_0$. Replacing the ordered simplex by the full integration cube introduces $1/n!$ and a time-ordered product, which proves the Dyson seriesTaking the asymptotic limits with the usual adiabatic or regulated scattering prescription yieldsThe field in this expression is the free field evolving in the interaction picture, not an independently prescribed classical field. For quantum fields the expansion is a regulated perturbation series; the time-ordering symbol cannot be dropped because the operators at different times need not commute.

Use relativistically normalized two-particle states $|p_1,p_2⟩=a^{†}(p_1)a^{†}(p_2)|0⟩$, with $⟨p|q⟩=(2\pi {)}^{3}2E_p{\delta }^3(\mathbf{p}-\mathbf{q})$. The connected first-order scattering contribution isHere the connected S-matrix uses a normalized vacuum and physical external one-particle states. At this order, normal ordering the vertex separates the four-external-leg contribution from vacuum and tadpole pieces; those disconnected or external-mass contributions are not the two-to-two scattering amplitude.

Two fields annihilate the incoming particles and two create the outgoing ones. There are $\left(\genfrac{}{}{0ex}{}{4}{2}\right)$ choices of the annihilating fields, two assignments of the incoming momenta and two assignments of the outgoing momenta. Hence the total factor is $\left(\genfrac{}{}{0ex}{}{4}{2}\right)2!2!=4!$, the external-leg factorial cancellation at a quartic scalar vertex. Each attachment contributes its plane-wave factor, soThe spacetime integral is the four-momentum conservation delta function, givingIn particular, the tree scattering amplitude is independent of angle.

For completeness, derive the relativistic two-body phase space in the centre-of-mass frame. Write $\sqrt{s}=2E_{\ast }$, $p_{\ast }^2=E_{\ast }^2-m^2$. The spatial delta function sets the two final momenta opposite, and the energy delta function is $\delta (\sqrt{s}-2\sqrt{p^2+m^2})$. Its radial Jacobian is $E_{\ast }/(2p_{\ast })$. ThusThe invariant flux factor is $\mathcal{F}=4p_{\ast }\sqrt{s}$ for equal initial masses. Using $d\sigma =|T{|}^{2}d{\Phi }_2/\mathcal{F}$, the labeled angular density is ${\lambda }^2/(64{\pi }^{2}s)$. The outgoing quanta belong to one real scalar field, so they are identical. A full-sphere integration counts each event twice and requires the identical final-state symmetry factor $1/2!$. ConsequentlyEquivalently one may integrate the labeled density over a hemisphere representing each event once. The final-state factor is separate from the vertex factorial: applying it at the vertex instead would give the wrong elastic scattering from a quartic scalar contact interaction cross-section.

An isotropic source's bolometric luminosity is distributed over a wavefront of physical area $4\pi a_0^2{r}^2$. Each photon loses an energy factor $(1+z{)}^{-1}$ through cosmological redshift, and successive photons arrive with a further rate factor $(1+z{)}^{-1}$ through cosmological time dilation. Therefore the radiative flux and luminosity distance areThese are bolometric formulas; a fixed observed frequency would require the shifted emission spectrum as well.

Normalize the scale factor by $a_0=1$. Summing the radiative flux over the source counts on an FLRW past light cone makes the transverse shell-area factors cancel:Thus ${\mathcal{F}}_0$ is power per effective detector area, summed over the whole sky; an all-sky detector of effective area $A_{\det }$ receives $A_{\det }{\mathcal{F}}_0$. The corresponding intensity per solid angle is ${\mathcal{F}}_0/(4\pi )$. Restoring the speed of light multiplies these flux integrals by $c$.

Spatially uniform $n(t)$ by itself does not specify its time evolution. For a persistent population comoving with the expansion, with neither creation nor destruction, the cosmological continuity equation for source number is $\dot{n}+3Hn=0$. Its conserved comoving number density is $n_0=a^{3}n$, so the cosmological bolometric background from conserved sources becomes $L{n}_0\int adt$. For the Einstein-de Sitter universe,The last equality uses $H_0=2/(3t_0)$. The early-time improper integral converges despite the diverging physical source density. If instead the proper number density were maintained at a constant value $n_0$ by continuous source creation, the same geometry would give $3L{n}_0{t}_0/11$. For a completely unspecified source history the general integral above is the determined answer; constancy of individual luminosity alone does not imply number conservation.

Write $B=b+ic$ with $b>0$. Normalization of the Gaussian wave packet gives $|A{|}^2\int e^{-2b{x}^2}dx=1$, hence $|A{|}^2=\sqrt{2b/\pi }$. The supplied Gaussian integral ratio then givesThe momentum operator obeys $p\Psi =2i\hslash Bx\Psi$, and direct second differentiation gives $p^2\Psi =(2{\hslash }^{2}B-4{\hslash }^2{B}^2{x}^2)\Psi$. Taking the expectation value and substituting the first result givessince $2b-B=\bar{B}$. These are the second moments of a complex Gaussian wave packet.

For the solution in part (b), the trigonometric identitygives, writing $D=\cosh (2t)+\cos (2\varphi )$,Thus normalizability requires $\sin (2\varphi )>0$, and the exact expectation values areSince $\cosh (2t)\sim e^{2t}/2$,Their difference is the constant $-\hslash \cot (2\varphi )$, so the mean energy $⟨H⟩=-\hslash \cot (2\varphi )/2$ stays constant despite exponential spreading in both observables.

Use the weighted inner product $⟨u,v{⟩}_r={\int }_a^{b}r(x)u(x)v(x)dx$. For real eigenfunctions, integration by parts with the zero endpoint values givesThus $({\lambda }_n-{\lambda }_m)⟨y_m,y_n{⟩}_r=0$. The regular separated Sturm-Liouville problem has simple eigenvalues: for a fixed eigenvalue, prescribing $y(a)=0$ leaves only the initial flux $p(a)y^{\prime }(a)$ free, so uniqueness of the associated first-order system makes any two eigenfunctions proportional. Distinct eigenfunctions are therefore orthogonal, and choosing their signs and scales so that $\int r{y}_n^2=1$ givesThe positive energy form $\int (p{y}^{\prime 2}+q{y}^2)$ realizes $r^{-1}\mathcal{L}$ as a positive self-adjoint operator in this weighted space, with compact inverse on the bounded interval. The spectral theorem for compact self-adjoint operators supplies a complete eigenbasis. This is the standard regular Sturm-Liouville eigenfunction expansion, also valid with continuous $p$ by working with the continuous flux $p{y}^{\prime }$.

For a source $F$, expand a zero-endpoint solution of $(\mathcal{L}-\mu r)y=F$ in that basis. Multiplication by $y_n$ and unweighted integration giveSince $\mu$ is not an eigenvalue, every coefficient is determined. The spectral Green function of a regular Sturm-Liouville problem is consequentlyWeighted completeness means $r(x){\sum }_n{y}_n(x)y_n(\xi )=\delta (x-\xi )$ as a distribution, so applying $\mathcal{L}-\mu r$ to this series produces the required delta source. The source integration is with $d\xi$, explaining the absence of an additional factor $r(\xi )$ in the stated kernel.

For the subsequent operator $L=d^2/d{x}^2$ the sign is explicitly reversed: $y_n=\sqrt{2/\pi }\sin (nx)$ and ${\lambda }_n=-n^2$. To cover every real non-eigenvalue $\mu$, define the homogeneous solutionIt satisfies $S_{\mu }(0)=0$, $S_{\mu }^{\prime }(0)=1$ and $S_{\mu }^{\prime \prime }=\mu S_{\mu }$. Matching the two endpoint solutions continuously at $x=\xi$ with derivative jump $G_x({\xi }^{+},\xi )-G_x({\xi }^{-},\xi )=1$ yieldsThe denominator is nonzero precisely off the Dirichlet eigenvalues $\mu =-n^2$. In particular $G_0=-x_{<}(\pi -x_{>})/\pi$. At $x=\xi =\pi /2$, its spectral expansion givesThis odd reciprocal-square sum from a Dirichlet Green function proves $\boxed{\sum _{j=0}^{\infty }(2j+1{)}^{-2}={\pi }^2/8}$.

The Time-dependent Schrödinger equation is $i\hslash {\partial }_t\Psi =H\Psi$. The normalized eigenstates of the Hermitian Hamiltonian $H_3$ with distinct eigenvalues are orthogonal, so they form an orthonormal basis of this two-state system. Substitute the expansion of $\Psi$ and compare the two coefficients:Hence $a_0(t)=a_0(0)e^{it/\hslash }$ and $a_1(t)=a_1(0)e^{-it/\hslash }$. For the normalized test state $\varphi =\alpha {\chi }_0+\beta {\chi }_1$, the Born rule givesThe complex conjugates are essential; relative phases can affect this probability even though the energy-state populations remain constant under $H_3$.

Let the mass per unit length be $m$ and the constant tension be $T$. Small transverse displacement obeys the wave equationWith characteristic coordinates $\xi =x+ct$ and $\eta =x-ct$, the equation becomes $y_{\xi \eta }=0$. Integrating once in each variable gives D'Alembert's formulaFor a smooth solution with vanishing endpoint energy flux, multiplication by $y_t$ and integration by parts giveThe fixed endpoint has $y_t(0,t)=0$. Decay of displacement alone does not guarantee either a pointwise derivative-flux limit or finiteness of the energy integral; the conservation statement is understood in the usual finite-energy class. The profile formula below proves it directly there without requiring such a pointwise limit at infinity.

The endpoint condition gives $g(-ct)=-f(ct)$, allowing the reflected wave to be expressed through one profile:For forward time one may define the negative-argument part of this profile using the original $g$; no restriction on the independent initial waves is lost. Differentiation gives $y_t=c[f^{\prime }(x+ct)-f^{\prime }(ct-x)]$ and $y_x=f^{\prime }(x+ct)+f^{\prime }(ct-x)$. HenceThe two substitutions partition the real line at $ct$, making conservation of energy explicit.

Let $M_{\ast }$ be the central mass, so ${\Omega }^2=G{M}_{\ast }/r^3$ and the specific orbital energy is $\epsilon =-G{M}_{\ast }/(2r)$. Define the inward accretion rate $\dot{M}>0$. The rate at which the gas acquires binding energy in moving from $b$ to $a$ isThe vertically integrated viscous heating per unit horizontal area, including the entire column, is $\bar{\nu }\Sigma (r{\Omega }^{\prime }{)}^2$. Since $r{\Omega }^{\prime }=-3\Omega /2$, the dissipated power in the annulus isHere the column integral already includes dissipation associated with both disk faces; adding another factor of two would double-count it. Performing the elementary integrals givesFor the whole disk, take $a=r_{\mathrm{in}}$ and $b\to \infty$. Both expressions giveIf $a\gg r_{\mathrm{in}}$, the bracket $1-\sqrt{r_{\mathrm{in}}/r}$ is uniformly close to one throughout the annulus. Comparing the positive integrals showswith the correction negative. Thus far from the inner edge the viscous dissipation is approximately three times the local binding-energy acquisition.

There is no violation of conservation of energy because the viscous torque in an accretion disk also transports mechanical energy. The outward viscous torque in an accretion disk isand its outward energy flux is $\Omega \mathcal{G}=G{M}_{\ast }\dot{M}[r^{-1}-\sqrt{r_{\mathrm{in}}}{r}^{-3/2}]$. The explicit expressions satisfyEnergy transported from the inner disk supplies the extra dissipation at larger radii. The torque power vanishes at the zero-torque inner boundary condition and tends to zero at infinity, leaving global balance. This is torque transport of energy in a steady accretion disk.

Write the gravitational potential with spherical symmetry as $\Phi (r)$ and let $\mathbf{n}$ be the unit vector normal to a circular orbit, oriented in the direction of its angular velocity. Put $\mathbf{\Omega }=\Omega \mathbf{n}$ and denote the specific angular momentum by $\mathbf{h}=h\mathbf{n}$. Circular force balance and the specific orbital energy giveDifferentiating along the family of circular orbits,A change in orbital orientation contributes $hd\mathbf{n}$ to $d\mathbf{h}$, but $\mathbf{n}\cdot d\mathbf{n}=0$. ConsequentlyThis circular-orbit energy gradient accounts for radial and inclination changes together; spherical symmetry makes the orbital energy independent of the plane.

For either variable-mass particle, $\mathbf{H}=M\mathbf{h}$ and $E=M\epsilon$. Thus $d\mathbf{H}=Md\mathbf{h}+\mathbf{h}dM$, soConservation of mass and conservation of angular momentum impose $d{M}_2=-d{M}_1$, $d{\mathbf{H}}_2=-d{\mathbf{H}}_1$. Adding the two particle contributions provesTo interpret the signs, label the inner orbit 1 and the outer orbit 2. For $f(r)=\epsilon -\Omega h$ the first identity gives $f^{\prime }=-h{\Omega }^{\prime }$. The assumed outward decrease of angular velocity therefore makes $f_1<f_2$, so the mass-exchange part lowers the energy when $d{M}_1>0$: mass is transferred inward. In a coplanar, co-rotating pair, ${\Omega }_1>{\Omega }_2$ and an outward angular momentum transfer has $d{\mathbf{H}}_1=-dH\mathbf{n}$ with $dH>0$. Its contribution is $-({\Omega }_1-{\Omega }_2)dH<0$.

The vector difference also identifies the inclination-reducing exchange. For a fixed small magnitude of admissible angular momentum exchange, its energy contribution is most negative whenby the Cauchy-Schwarz inequality. This gives $d{E}_{\mathrm{angular}}=-s|{\mathbf{\Omega }}_1-{\mathbf{\Omega }}_2{|}^2$. Let $q={\mathbf{n}}_1\cdot {\mathbf{n}}_2=\cos \iota$ and $H_i=|{\mathbf{H}}_i|$, taking fixed particle masses for this exchange. Projecting $d{\mathbf{H}}_i$ perpendicular to its own direction givesThereforeThe downhill exchange reduces the relative orbital inclination. Its inner angular momentum component is ${\mathbf{n}}_1\cdot d{\mathbf{H}}_1=-s({\Omega }_1-{\Omega }_{2}q)<0$; in the aligned limit the outer orbit receives that angular momentum directly. For strongly inclined orbits, part of the exchange instead cancels differently directed angular momenta as the planes align. Together these gradients explain the tendencies toward inward mass transfer, outward angular momentum transport and coplanarity.

The vertical hydrostatic equilibrium equation balances the gradient of the total pressure against gravity. In a thin disk with Keplerian rotation, expanding the central gravitational field about the midplane gives vertical acceleration $-{\Omega }^{2}z$, so $p_z=-\rho {\Omega }^{2}z$. Disk self-gravity is neglected in this approximation.

The flux-gradient equation is local thermal balance: the upward radiative flux increases with height because differential rotation dissipates energy. For a Newtonian effective dynamic viscosity $\mu$, the volumetric heating rate is $\mu (r{\Omega }^{\prime }{)}^2$, which becomes $9\mu {\Omega }^2/4$ for Keplerian rotation. The symbol $\mu$ here is dynamic viscosity, not kinematic viscosity; the latter is $\nu =\mu /\rho$.

The radiative diffusion equation applies to an optically thick medium in local thermodynamic equilibrium. Its diffusion coefficient is $c/(3\kappa \rho )$, and the equilibrium radiation energy density is $u_{\mathrm{rad}}=4\sigma T^4/c$. HenceThe minus sign means energy flows down the temperature gradient. Here $\sigma$ is the Stefan-Boltzmann constant, $\kappa$ the opacity per unit mass, and $c$ the speed of light. Constant Thomson scattering opacity is the adopted transport approximation; local thermodynamic equilibrium is an additional assumption behind the radiation equation of state.

Finally the equation of state adds ideal gas gas pressure $p_g=\rho kT/({\mu }_m{m}_H)$ and isotropic radiation pressure $p_r=u_{\mathrm{rad}}/3=4\sigma T^4/(3c)$. The Boltzmann constant is $k$, and ${\mu }_m{m}_H$ is the mean mass per gas particle. Both contributions supply the total pressure entering vertical hydrostatic equilibrium.

Choose a reference circular orbit at radius $r_0$, rotate the frame with $\mathbf{\Omega }=\Omega (r_0){\mathbf{e}}_z$, and introduce local Cartesian coordinates $x=r-r_0$, $y=r_0(\varphi -\Omega t)$ and vertical coordinate $z$. On scales small compared with $r_0$, neglect curvature of the coordinate directions but retain the leading differential rotation. The background velocity relative to the rotating frame isThus $A=3\Omega /4$ for a Keplerian disk. The shearing sheet retains Coriolis acceleration and the linear tidal acceleration. The radial shearing-sheet tidal potential is $-2\Omega A{x}^2$, whose force $4\Omega Ax{\mathbf{e}}_x$ balances the background Coriolis acceleration. If vertical tidal gravity is included, its equilibrium contribution is balanced by a background pressure $p_0(z)$ and disappears when that equilibrium is subtracted.

Write the total velocity as $\mathbf{u}={\mathbf{u}}_0+\mathbf{v}$. In the rotating-frame magnetohydrodynamic momentum equation, background advection is $({\mathbf{u}}_0\cdot \nabla )\mathbf{v}=-2Ax{\partial }_y\mathbf{v}$ and advection of the background by the perturbation is $(\mathbf{v}\cdot \nabla ){\mathbf{u}}_0=-2A{v}_x{\mathbf{e}}_y$. The uniform-viscosity diffusion of the linear background shear is zero. For a solenoidal magnetic field the Lorentz force per unit mass isAt constant density, absorb the magnetic-pressure gradient into $\psi =(p-p_0)/\rho +B^2/(2{\mu }_0\rho )$. Subtracting the background force balance then gives the stated momentum equation, with the remaining nonlinear advection $\mathbf{v}\cdot \nabla \mathbf{v}$, Coriolis acceleration $2\mathbf{\Omega }\times \mathbf{v}$, magnetic tension and viscous diffusion.

For constant finite electrical conductivity ${\sigma }_e$, put $\eta =({\mu }_0{\sigma }_e{)}^{-1}$. The resistive induction equation with uniform magnetic diffusivity is ${\partial }_t\mathbf{B}+\mathbf{u}\cdot \nabla \mathbf{B}=\mathbf{B}\cdot \nabla \mathbf{u}+\eta {\nabla }^2\mathbf{B}$. Substituting the background shear supplies the term $-2A{B}_x{\mathbf{e}}_y$ and the same background-advection operator as in the momentum equation. The two further conditions areThey express incompressibility and the absence of magnetic monopoles, respectively.

Now take real plane waves with common phase $\vartheta =\mathbf{k}(t)\cdot \mathbf{x}$ and real wavevector. The solenoidal conditions becomeEvery field depends on position only through $\vartheta$. Hence for either transverse vector field $\mathbf{P}$ and either wave field $\mathbf{Q}$, $(\mathbf{P}\cdot \nabla )\mathbf{Q}=(\mathbf{P}\cdot \mathbf{k}){\partial }_{\vartheta }\mathbf{Q}=0$. This proves that all the displayed quadratic advection and tension terms vanish pointwise, including products involving conjugate amplitudes. These waves are therefore exact solutions at finite amplitude, not merely a linearization.

Acting with $D_0={\partial }_t-2Ax{\partial }_y$ on the phase yieldsCanceling its position-dependent part requiresThe remaining amplitude equations areDifferentiating the velocity constraint determines the pressure amplitude:The resistive induction equation preserves its constraint because $\dot{\mathbf{k}}\cdot \tilde{\mathbf{B}}=2A{k}_y{\tilde{B}}_x$ cancels the corresponding shear term in $\mathbf{k}\cdot \dot{\tilde{\mathbf{B}}}$. The specified magnetic field has no spatially uniform part. Thus there is no background magnetic field producing an additional linear magnetic-tension coupling. The scalar $\psi$ is the modified pressure; the physical pressure can include a constant and a second spatial harmonic from $-B^2/(2{\mu }_0)$ while $\psi$ retains the stipulated plane-wave form. This is an exact zero-mean magnetic shearing wave.

For decay, take the real part of the scalar product of the velocity-amplitude equation with its complex conjugate. Pressure does no work because $\mathbf{k}\cdot \tilde{\mathbf{v}}=0$, and the real part of Coriolis work is zero. ThereforeThe resistive induction equation similarly givesusing $2|u_x{u}_y|\le |\mathbf{u}{|}^2$. Integrating these differential inequalities, defineThenA non-axisymmetric wave has $k_y\ne 0$. For nonzero differential shear and positive $\nu ,\eta$, the negative cubic terms dominate the linear upper bound on shear work. ConsequentlyThis cubic-exponent decay of a nonaxisymmetric shearing wave allows transient energy growth before the winding to short radial wavelengths makes diffusion dominant. The magnetic amplitudes can also be written explicitly aswhich confirms the damping independently. Positivity of the dissipative coefficients is necessary: with $k_z=0$, an inviscid vertical velocity amplitude can remain constant, and an ideal vertical magnetic amplitude can remain constant when $\eta =0$. The asserted decay applies to the viscous, finitely conducting model; taking an ideal limit before the long-time limit changes the conclusion.

Let $p$ be kinematic pressure, with the resting hydrostatic part subtracted, and let $\sigma$ be the buoyancy perturbation about background buoyancy $N^{2}z$. The nonrotating inviscid Boussinesq equations, with no buoyancy diffusion, areHere the third equation is mass conservation for incompressible flow, and the last expresses buoyancy conservation following a parcel. Linearization about rest removes the products of disturbances. For a plane internal gravity wave proportional to $e^{i(kx+mz-\omega t)}$, the resulting Linearized Boussinesq equations becomeWith $K=(k^2+m^2{)}^{1/2}$ and $k\ne 0$, the first and third equations give $\hat{p}=-m\omega \hat{w}/k^2$. Substitution in vertical momentum and buoyancy gives the dispersion relationSince the velocity is the time derivative of the displacement field, $\hat{\mathbf{u}}=-i\omega \hat{\mathbf{\xi }}$. Thus the complete buoyancy polarization of a plane internal gravity wave isBoth velocity and displacement are perpendicular to the wavevector; pressure and velocity are in quadrature with buoyancy and displacement.

For upward/rightward phase velocity, choose $k,m>0$ and $\omega =Nk/K$. Differentiating this dispersion relation gives the group velocity, while the vector phase velocity points along the wavevector:Their dot product is zero and ${\mathbf{c}}_p+{\mathbf{c}}_g=(N/K,0)$. Draw ${\mathbf{c}}_g$ from the tip of ${\mathbf{c}}_p$: the right-triangle geometry of internal-wave velocities has horizontal hypotenuse $N/K$. This hypotenuse is distinct from the horizontal phase velocity $\omega /k$.

At an instant with real complex amplitude $\hat{\sigma }=A$, write $\varphi =kx+mz-\omega t$. The spatial fields areThe constant-phase lines of an internal gravity wave slope down to the right, with $dz/dx=-k/m$. Buoyancy maxima lie a quarter wavelength from pressure maxima; the oscillating displacement and velocity run along the phase lines. The group velocity points down and right along them, despite upward/rightward motion of the wave phase.

Let $P=\overline{{\xi }_{x}u+{\zeta }_{x}w}$, using horizontal averaging over a wavelength. This is the displacement pseudomomentum of an internal gravity wave, whose sign convention is opposite to $k\overline{E}/\omega$. For complex coefficients, $\overline{ab}=\frac{1}{2}\mathrm{Re}(\hat{a}{\hat{b}}^{\ast })$. The above wave polarization consequently givesSince $c_{gz}=-\omega m/K^2$, these obey $\overline{{\zeta }_{x}p}=c_{gz}P$. With weak kinematic viscosity, each leading monochromatic velocity component satisfies ${\nabla }^{2}u=-K^{2}u$, ${\nabla }^{2}w=-K^{2}w$. The dissipative term is therefore $-\nu K^{2}P$, so the supplied averaged balance reduces toFor constant $k,m,N$ and a steady wave envelope, $c_{gz}{P}_z=-\nu K^{2}P$. As $P$ is proportional to squared wave amplitude, viscous attenuation of an internal gravity wave givesHere $c_{gz}<0$, so amplitude decreases below a source at $z_s$: the source is above the receiving level. The same conclusion follows directly because downward group velocity carries energy from the source downward. Upward phase velocity does not locate the source.