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To recover the printed dispersion relation, interpret the surface maintained flat as an imposed flat upper boundary for the perturbation, so its normal velocity is zero. This is the rigid-lid approximation. Work with perturbation velocity potentials ${\varphi }_{-}=A{e}^{ky}{e}^{ikx+\sigma t}$ below the interface and ${\varphi }_{+}=B\cosh [k(y-h)]e^{ikx+\sigma t}$ in the upper layer. They solve the Laplace equation, decay as $y\to -\infty$, and obey ${\partial }_y{\varphi }_{+}(h)=0$.

The linear kinematic boundary conditions at the interface give $(\sigma +ikU){\eta }_0=kA$ and $\sigma {\eta }_0=-kB\sinh (kh)$. Equality of pressures, using the linear Bernoulli equation, gives $(\sigma +ikU)A=\sigma B\cosh (kh)$. The equal fluid densities cancel the interface gravitational restoring term. Eliminating $A,B,{\eta }_0$ yieldsWith $q=\tanh (kh)$, the roots areFor every $k>0$ one root has positive real part. Thus the vortex sheet has a Kelvin-Helmholtz instability at every wavelength in this model. If $kh\gg 1$, $q\to 1$ and the growing rate is $Uk/2$. If $kh\ll 1$, $q\sim kh$ and it is $Uk\sqrt{kh}$.

There is a physical qualification to the wording. A dynamically deformable free surface at finite gravity is not exactly a flat lid. Its linear conditions combine to ${\sigma }^2{\varphi }_{+}(h)+g{\partial }_y{\varphi }_{+}(h)=0$. Writing the upper potential as $B\cosh (ky)+C\sinh (ky)$ and using the same interface conditions instead givesThe stated quadratic follows in the strong-gravity flat-surface limit, or with the flat condition imposed exactly. It is not the exact dispersion relation of a freely moving upper surface at arbitrary finite $g$. This makes explicit the boundary interpretation needed for the requested formula.

Use the Minkowski metric with signature $(-,+,+,+)$. The source point is the intersection of the worldline with the observation event's past light cone: $R^{\mu }{R}_{\mu }=0$ and $R^0>0$. In components this is $c(t-t_r)=|x-y(t_r)|$ with $t_r<t$. For a timelike worldline the retarded solution is unique in the usual complete-source setting.

Writing $R=|x-y(t_r)|$, $n=(x-y)/R$ and $u^{\mu }=\gamma (c,v)$ gives $R\cdot u=-\gamma cR(1-n\cdot v/c)$. Thus the Lienard-Wiechert potentials are $\varphi =q{\mu }_0{c}^2/[4\pi R(1-n\cdot v/c)]$ and $A=q{\mu }_{0}v/[4\pi R(1-n\cdot v/c)]$. At leading order in $1/r$ and first order in $v/c$, set $R\simeq r,n\simeq \hat{r}$ and expand the denominator:

Now use $t_{-}=t-r/c$ and let $a=\dot{v}(t_{-})$. Derivatives of $1/r$ and $\hat{r}$ give near-field terms of order $r^{-2}$; the radiation terms come from ${\partial }_t{t}_{-}=1$ and $\nabla t_{-}=-\hat{r}/c$. The electric field and magnetic field are thereforeThey are transverse to the observation direction. The Poynting vector isIntegrating over a sphere uses $\int {\sin }^2\theta d\Omega =8\pi /3$, yielding the nonrelativistic Larmor formula $P=q^2{\mu }_0|a{|}^2/(6\pi c)$. For simple harmonic motion, $|a{|}^2=A^2{\omega }^4{\cos }^2(\omega t_{-})$, whose period average is $A^2{\omega }^4/2$. Multiplication by the period $2\pi /\omega$ givesThis is the leading radiation energy in the nonrelativistic and far-field approximation.

The reciprocal lattice is ${\Lambda }^{\ast }=\{G:G\cdot \ell \in 2\pi \mathbb{Z}\text{ for every }\ell \in \Lambda \}$. With oriented primitive-cell volume $v_c=a_1\cdot (a_2\times a_3)$, its basis is $b_1=2\pi (a_2\times a_3)/v_c$ and the cyclic variants; thus $b_i\cdot a_j=2\pi {\delta }_{ij}$.

The total potential is $V(x)={\sum }_{\ell }U(x-\ell )$. Using the Fourier transform convention $\tilde{U}(q)=\int e^{-iq\cdot x}U(x)d^{3}x$, translation gives $\tilde{V}(q)=\tilde{U}(q)\Delta (q)$, withThe given Born approximation therefore supplies the amplitude $\boxed{-m\Delta (q)\tilde{U}(q)/(2\pi {\hslash }^2)}$. A finite geometric series gives, when the denominators are nonzero,At reciprocal-lattice vectors each phase is one, so $|\Delta |=N_1{N}_2{N}_3$, the number of atoms. For large finite crystals these narrow peaks represent coherent Bragg scattering, with peak intensity proportional to the squared number of atoms; their finite width reflects the finite crystal size.

For the specified face-centered cubic lattice, $v_c=a^3/4$ andThe reciprocal lattice have coordinates $(2\pi /a)(h,k,l)$ with $h,k,l$ all even or all odd. The two shortest nonzero length families are $G_1=2\pi \sqrt{3}/a$ for $(111)$ and $G_2=4\pi /a$ for $(200)$. Elastic scattering satisfies $|q|=2|k|\sin (\theta /2)$. The assumed incident magnitude exceeds $G_2/2$, so both families are kinematically accessible, andThis is the angular prediction for suitable crystal orientations, or for a powder containing those orientations. A fixed single-crystal orientation must additionally satisfy the vector Elastic Bragg scattering condition $2k\cdot G+|G{|}^2=0$; the magnitude bound alone cannot guarantee both peaks for every incident direction.

The interaction picture removes the evolution generated by $H_0$: $|{\psi }_I(t)⟩=e^{i{H}_{0}t/\hslash }|{\psi }_S(t)⟩$. Differentiating the Schrödinger equation gives $i\hslash {\partial }_t|{\psi }_I⟩=\lambda V_I(t)|{\psi }_I⟩$, with $V_I=e^{i{H}_{0}t/\hslash }V(t)e^{-i{H}_{0}t/\hslash }$. The integral equation, iterated once, givesDistinct-energy eigenstates are orthogonal. Their transition amplitude is therefore $-i\lambda {\hslash }^{-1}{\int }_0^t⟨\varphi |V(s)|\chi ⟩e^{i(E^{\prime }-E)s/\hslash }ds+O({\lambda }^2)$. Squaring its modulus givesThe first-order amplitude produces a second-order probability.

For the isotropic quantum harmonic oscillator, parity makes $\int x_i{x}_3|\chi {|}^2{d}^{3}x=0$ for $i=1,2$. Rotational symmetry and the given radial integral give $\int x_3^2|\chi {|}^2{d}^{3}x=\hslash /(2m\omega )$. Thus $⟨{\varphi }_i|{\hat{x}}_3|\chi ⟩={\delta }_{i3}\sqrt{\hslash /(2m\omega )}$, and the energy gap is $\hslash \omega$. The time integral is ${\int }_0^{\infty }{e}^{(-\mu +i\omega )s}ds=(\mu -i\omega {)}^{-1}$. ConsequentlyThe perturbation picks the excitation polarized along the third coordinate, rather than an arbitrary member of the degenerate first excited level.

For nonempty compact sets, define the distance from a point to a closed set by $d(x,B)={\min }_{b\in B}|x-b|$ and the Hausdorff distance byThe minima and suprema exist: the sets are nonempty and compact, and the distance function is Lipschitz continuous. The Hausdorff distance is finite, nonnegative and symmetric. If it is zero, every point of $A$ belongs to the closed set $B$ and conversely, so $A=B$; the converse is immediate.

For $a\in A$, choose $b\in B$ attaining $d(a,B)$ and then $c\in C$ attaining $d(b,C)$. The triangle inequality gives $d(a,C)\le d(a,B)+d_H(B,C)$. Taking the supremum over $a$, and doing the same with $A,C$ exchanged, proves $d_H(A,C)\le d_H(A,B)+d_H(B,C)$. Hence this is a metric.

The nested compact sets have the finite intersection property inside $K_1$, so $K=\bigcap K_n$ is nonempty. It is a closed subset of $K_1$ and therefore compact. Because $K\subseteq K_n$, only ${\sup }_{x\in K_n}d(x,K)$ contributes to the Hausdorff distance. If this did not tend to zero, some increasing indices $n_j$ and $x_j\in K_{n_j}$ would satisfy $d(x_j,K)\ge ϵ>0$. A convergent subsequence in $K_1$ has limit $x$. For each fixed $m$, its tail belongs to the closed set $K_m$, so $x\in K$. But then $d(x_j,K)\le |x_j-x|\to 0$, a contradiction. Thus $\boxed{d_H(K_n,K)\to 0}$.

A horseshoe for an interval map consists of two branches covering a common interval: there are an open interval $J$ and disjoint open subintervals $K_0,K_1\subseteq J$ with $f(K_0)=f(K_1)=J$. Equivalently one can work with two closed intervals with disjoint interiors whose images cover their union, and then restrict the branches. Glendinning chaos means that some positive iterate $f^m$ has such a horseshoe, rather than requiring one for $f$ itself.

For the three-cycle, use the intermediate value theorem on endpoint images. The interval covering relations are $I_a\to I_b$ and $I_b\to I_a,I_b$. Consequently both $f^2(I_a)$ and $f^2(I_b)$ cover $I_a\cup I_b$. This proves $f^2$ has a horseshoe. The directed covering graph of an interval map has exactly these forced arrows; a closed walk $a\to b\to b\to a$ has length three and a primitive itinerary. Pulling back the intervals successively and using the fixed-point property produces a point fixed by $f^3$ with this itinerary. It cannot be fixed by $f$, and a shared endpoint belongs to the already given three-cycle, so a genuine period-three orbit results.

For the increasing spatial order of the five-cycle, let $I_j=[x_{j-1},x_j]$, $1\le j\le 4$. The forced graph isThe primitive loops $I_4$, $I_3{I}_4$, $I_2{I}_3{I}_4$, and $I_1{I}_2{I}_3{I}_4$ force exact periods $\boxed{1,2,3,4}$. For periods below five the known five-cycle endpoints cannot be responsible, so the interval itineraries give distinct interior orbits. Both $g^2(I_3)$ and $g^2(I_4)$ cover $I_3\cup I_4$, so $g$ is guaranteed to be Glendinning-chaotic.

A horseshoe for $g$ itself is not guaranteed. As a counterexample put $x_j=j$ on $[0,4]$ and interpolate by $g(x)=x+1$ for $0\le x\le 3$, $g(x)=16-4x$ for $3\le x\le 4$. This has the required five-cycle. The unique fixed point is $16/5$, with $g(x)>x$ to its left and $g(x)<x$ to its right. Any nondegenerate closed interval whose image covers itself must contain this fixed point in its interior: an interval entirely on either side, or with the fixed point as an endpoint, fails to cover its extreme endpoint. Two intervals with disjoint interiors therefore cannot both cover themselves, as a horseshoe would require. Hence this example has no one-iterate horseshoe.

For the alternative spatial order, set $y_0=x_4,y_1=x_2,y_2=x_1,y_3=x_3,y_4=x_0$ and $J_j=[y_{j-1},y_j]$. Endpoint images giveThe self-loop at $J_2$, the two-cycle $J_1{J}_4$, and the four-cycle $J_2{J}_3{J}_1{J}_4$ force $\boxed{1,2,4}$; period three is not guaranteed. To see this last claim, take the connect-the-dots map at $y_j=j$ with values $(4,3,1,0,2)$. Its exact interval transitions are those displayed. Any length-three closed itinerary is the repeated self-loop in $J_2$; there the map is the affine decreasing function $5-2x$, whose third iterate has only its original fixed point. Endpoints belong to the five-cycle. Thus no exact three-cycle exists in this example.

Chaos is still guaranteed: from $J_2$ there are two different length-four closed walks, four self-loops and $J_2{J}_3{J}_1{J}_4{J}_2$. Pullbacks give disjoint branches of $g^4$ over a common interval in $J_2$, hence a horseshoe for an interval map for an iterate. A horseshoe for $g$ itself is again not guaranteed: the same connect-the-dots example is strictly decreasing on $[0,3]$ and increasing on $[3,4]$ with image $[0,2]$. Its unique fixed point is $5/3$, again with $g(x)>x$ to the left and $g(x)<x$ to the right. The same fixed-point argument rules out two self-covering intervals with disjoint interiors. These counterexamples distinguish the three separate questions about periods, chaos and a one-iterate horseshoe.

Take the logarithmic derivative of the proposed asymptotic expansion. Through order $z^{-1}$, $u^{\prime }/u=\lambda +\mu /z+O(z^{-2})$ and $u^{\prime \prime }/u={\lambda }^2+2\lambda \mu /z+O(z^{-2})$. Matching the constant and inverse-power terms givesThe second formula assumes a simple root, $2\lambda +f_0\ne 0$. At a repeated root the ansatz can fail: the next equation first requires $f_1\lambda +g_1=0$, and if that holds higher orders determine the remaining behaviour.

For the Bessel differential equation, $\lambda =\pm i$ and $\mu =-1/2$. Set $u=e^{\lambda z}{z}^{-1/2}v(z)$. Direct substitution leaves $v^{\prime \prime }+2\lambda v^{\prime }+(1/4-{\nu }^2)z^{-2}v=0$. With $v=1+a/z+O(z^{-2})$, matching $z^{-2}$ gives $a=(1/4-{\nu }^2)/(2\lambda )$. ThusTwo constant multiples of the Hankel functions normalize their leading coefficients to one and realize these solutions in their appropriate sectors.

For generic $\nu$, these are sectorial expansions, not unchanged expansions for a single branch over arbitrarily continued $\arg z$. The factors $z^{-1/2}$ require a branch, and crossing sector boundaries can introduce the other exponential by Stokes phenomenon and analytic continuation. More precisely, the standard normalized first Hankel solution has its expansion for $-\pi +\delta \le \arg z\le 2\pi -\delta$, and the second for $-2\pi +\delta \le \arg z\le \pi -\delta$, with any fixed positive $\delta$ inside these sectors. In particular both are valid in the common slit sector $-\pi <\arg z<\pi$, away from its boundaries. These sector statements are recorded in NIST's Hankel asymptotics. Exceptional half-integer orders have terminating elementary expressions, so a blanket assertion of nonzero Stokes mixing for every order would be too strong.