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In the Ffowcs Williams-Hawkings equation, the other source types are a surface acoustic monopole associated with acoustic thickness noise, and a volume acoustic quadrupole involving the Lighthill stress tensor. On an impermeable material surface, fluid and surface normal velocities agree. There is no through-surface mass-flux source; the remaining thickness source is $Q={\rho }_0{v}_n$. For a rigid body, the leading acoustic compact-source approximation to that source hasThus there is no leading net-volume acoustic monopole. To neglect thickness radiation beyond that leading cancellation, assume negligible volume displacement, as for ideal thin blades, or that its higher multipoles are small compared with the retained acoustic loading noise. Rigidity alone does not make a moving finite-volume body's local thickness source identically zero.

The volume acoustic quadrupole may be neglected for low Mach number motion when exterior turbulent or nonlinear stresses do not provide a competing strong source. We also assume small linear acoustics perturbations, a uniform reference sound speed, and negligible relevant viscous and entropy sources. These are source-strength approximations, particularly important if a loading contribution itself cancels by symmetry. Under them, the retained acoustic dipole is the force exerted by the object on the fluid, with the sign used in the previous solution.

Let $R=|x|$, $n=x/R$, and ${\tau }_0=t-R/c_0$. In the acoustic far field, $R$ is large compared with the object and $\omega R/c_0\gg 1$. For a source of size $\ell$ with $\omega \ell /c_0\ll 1$ and small surface Mach number, source-dependent delays and the Doppler factor can be neglected to leading order. The surface integral then contains just the total force $\mathcal{F}(\tau )={\int }_{S}FdS$. Differentiating its retarded time, rather than its $R^{-1}$ spreading factor, gives the radiating termThe sign follows from ${\partial }_{x_i}{\tau }_0=-n_i/c_0$. Differentiating $R^{-1}$ or the direction $n$ instead produces the lower-order near field. A constant total force does not radiate at this leading compact order.

The rotation introduces angular frequency $\Omega$ and, after combining the two blades, harmonics such as $2\Omega$. The acoustic compact-source approximation requires the propagation time $a/c_0$ to be small compared with ${\Omega }^{-1}$. ThusIt also bounds every blade element's Mach number by $\mu$. Consequently $|1-M_r|=1+O(\mu )$, and its leading value is one. The separate acoustic far field condition is $\Omega R/c_0\gg 1$.

Choose the positive rotation sense so that a first blade at phase $\alpha$ has radial and tangential unit vectorsIntegrating the given line force from $r=0$ to $a$ yields ${\mathcal{F}}_1=F{e}_{\varphi }(\Omega \tau )+D{e}_x$. Its axial component is constant, while ${\dot{\mathcal{F}}}_1=-F\Omega e_r$. Since $n=\cos \theta e_x+\sin \theta e_y$, the compact acoustic dipole sound from this blade isThe other blade has phase $\alpha +\pi$ and contributes the opposite rotating force. At a common compact retarded time, their total force is $2D{e}_x$, soThis cancellation calls for the next source-delay correction; it does not mean that the complete moving-source field vanishes.

Write $\alpha =\Omega {\tau }_0^{\ast }$ and $s=\sin \theta$, where ${\tau }_0^{\ast }=t-R/c_0$. Label the two arms by $\sigma =\pm 1$. Their positions are $y_{\sigma }=\sigma r{e}_r(\Omega {\tau }_{\sigma }^{\ast })$. To radiating far-field accuracy,The retarded time equation therefore gives, with ${\mu }_r=\Omega r/c_0$,For the first arm, $\sigma =1$, this is the requested phase expansion. The more explicit geometric error also makes the dimensional meaning of the printed $O(1/R)$ term clear.

Apply the Taylor theorem to the rotating line force. Since $d{e}_{\varphi }/d\alpha =-e_r$, its first two orders, expressed in a common reference frame, areAlso $v_{\sigma }=\sigma \Omega r{e}_{\varphi }(\Omega {\tau }_{\sigma }^{\ast })$, so the radial Mach number isThe absolute value causes no change of sign in this subsonic limit. Multiplying these two expansions before summing is essential: both the shifted force and the moving-surface retarded Jacobian contribute at the same order.

Denote the integrated numerator, including that Jacobian, by $\mathcal{B}$. Pairing the two blades cancels all terms odd in $\sigma$, including the first Doppler correction to the axial load. HenceThe factor $a/3$ comes from $a^{-2}{\int }_0^a{r}^{2}dr$. Only the time-dependent term radiates at order $R^{-1}$. Applying $-{\partial }_{x_i}$ to the integral now gives ${\rho }_{\mathrm{rad}}^{\prime }=(4\pi c_0^{3}R{)}^{-1}n\cdot {\partial }_t\mathcal{B}$, and thusThere are also nonradiating terms of order $R^{-2}$. Multiply by $c_0^2$ for the acoustic pressure. The stated coefficient uses the rotation convention fixed above; reversing the rotation reverses the corresponding signed load and phase convention.

This is a compact rotating two-blade loading source acting as an acoustic quadrupole. The compact total rotating force cancels, leaving the first spatial moment of the loading. Its two factors of the observer's projection into the rotor plane produce ${\sin }^2\theta$: there is no leading sound on the rotation axis and the density amplitude is maximal in the rotor plane. The configuration repeats after half a rotation, explaining frequency $2\Omega$. These statements concern amplitude; the corresponding acoustic intensity has a ${\sin }^4\theta$ factor.

If $F=0$, the displayed first-order contribution vanishes as well. For completeness, expanding the axial Jacobian to its next even order givesThe first remaining axial-load radiation is thenThus the constant axial total force does not produce the lower-order term, even though its moving spatial distribution can radiate at a higher order.

Use the Fourier transform pair $\hat{q}(k)=\int q(x)e^{ikx}dx$, $q(x)=(2\pi {)}^{-1}\int \hat{q}(k)e^{-ikx}dk$. The point force transforms to $F$. With the dispersion relation $D$ defined above, the sheet equation becomesThe point-force radiation from a fluid-loaded sheet is therefore represented exactly byThe causal contour and outgoing acoustic square-root branch are fixed first with $\mathrm{Im}\omega <0$, then continued to the desired real frequency. This prescription fixes how poles and the branch points are passed.

For the acoustic far field $x=r\cos \theta$, $y=r\sin \theta$, take $0<\theta <\pi$ bounded away from grazing and $k_{0}r\gg 1$. The method of steepest descent saddle point is $k_s=k_0\cos \theta$, with ${\gamma }_s=i{k}_0\sin \theta$. The supplied saddle point rule, including its $\sin \theta$ factor, gives, provided the contour deformation crosses no poles,A convenient simplification, free of division by $T$, isEquivalently, when $T\ne 0$,The expression printed in the PDF is missing sound-speed factors for general dimensional $c_0$. It agrees with this result if $c_0=1$ in fully normalized units; when $c_0$ is retained as an arbitrary sound speed, the numerator needs $c_0^{-3/2}$ and the structural term needs $T/c_0^2$ in the last form. These factors arise respectively from cylindrical spreading, the pressure-density relation, and $k_s=\omega \cos \theta /c_0$.

A direct countercheck is the transparent-sheet limit $m=T=0$. The sheet jump condition then gives ${\hat{p}}_{+}(k,0)=F/2$, so the saddle point rule requiresThe printed expression, interpreted continuously after multiplying out its structural factor, instead gives $F(2c_0{)}^{-1}\sqrt{\omega /(2\pi r)}\sin \theta$ times the same phase. It differs by a factor $c_0^{3/2}$; for example it is eight times too large when $c_0=4$. This limit also verifies the normalization of the corrected density field independently of the sheet's elastic-sheet tension.

To decide about poles, track the roots of $\Delta (k,\omega )$ on the chosen square-root sheet and deform the original causal contour to the steepest descent contour. A root contributes a residue exactly when it lies in the region swept out by that deformation; its sign is fixed by the contour orientation. Branch cuts must be retained throughout this comparison. Which roots are crossed can depend on observation angle, producing a change of the modal contribution when a pole meets the deformation boundary. A saddle point approaching a pole or a grazing endpoint requires an approximation uniform in that limit, rather than the isolated saddle point formula above.

The crossed poles are the free fluid-sheet modes of the preceding solution. Real subsonic roots represent evanescent acoustic surface waves carrying energy along the sheet, with normal decay; complex continuations represent leaky or radiating modes. Their residues must be added to the saddle point sound when the causal contour selects them. The specification “no poles contribute” is therefore a substantive condition on the contour, not permission to ignore zeros of the dispersion relation.

The area element in plane polar coordinates is $rdrd\theta$. Thus the axial moment of inertia isFor the removed material, the angular width is $2\pi /13$, andIts moment of inertia is therefore $I_{\mathrm{cut}}=\pi {\rho }_0{a}^5/160$, leavingThe torque equation $I\dot{\Omega }=\tau$ gives constant angular acceleration, so the time to reach the prescribed angular speed from rest is

To find the centre of mass after removing material, let ${\mathbf{e}}_1$ point along the bisector of the missing sector in the disc. The original disc has zero first mass moment about its centre. The missing sector has zero transverse first moment by reflection symmetry, while its component along ${\mathbf{e}}_1$ isbecause the radial integral is $11a^4/192$. The remaining first moment is $-Q_{\mathrm{cut}}{\mathbf{e}}_1$. Dividing by the given mass $k{\rho }_0{a}^3$ gives the body-fixed centre of massIt lies away from the missing sector. In the inertial frame, let $t_{\ast }$ be the moment the applied torque stops and let $\psi ={\psi }_{\ast }+\Omega (t-t_{\ast })$ give the orientation of that sector's bisector. Then, with $d=11a\sin (\pi /13)/(96k)$,The centre of mass has inward centripetal acceleration of magnitude ${\Omega }^{2}d$. The required real force is the constraint reaction exerted by the fixed rod and its support. Its horizontal resultant is $k{\rho }_0{a}^3\ddot{\mathbf{R}}$; a zero applied axial torque does not imply a zero resultant force.

Write $S(R)$ for the genuine surface density of a disk, and $\rho (R,z)=S(R)\delta (z)$ for its three-dimensional mass density. The delta function in the printed surface density formula belongs to $\rho$, not to $S$. Introduce a reference length $R_{\ast }$ to make logarithms dimensionless. The intended model is the infinite, self-gravitating, scale-free Mestel disk, with no extra source or imposed gravitational field.

The circular speed satisfies $v_c^2=R{\partial }_R\Phi (R,0)$. Thus the flat galaxy rotation curve givesA reflection-symmetric harmonic function with this midplane boundary value isVerify this continuation using the Poisson equation for Newtonian gravity. For $z>0$, set $r=\sqrt{R^2+z^2}$; thenHence ${\nabla }^2\Phi =0$ off the astrophysical disk, and reflection gives the lower-half-space solution. Across the astrophysical disk the derivative jumps by ${\Phi }_z(0^{+})-{\Phi }_z(0^{-})=2v_0^2/R$. The distributional Poisson equation for Newtonian gravity therefore givesAt the origin the enclosed disk mass tends to zero linearly with radius, so there is no additional central point mass. This verifies the Mestel disk potential-density pair. The astrophysical disk has infinite total mass and a logarithmic Newtonian gravitational potential, so it is not an isolated finite-mass model with Newtonian gravitational potential zero at infinity. The usual scale-free boundary condition is important: the midplane rotation curve alone would also permit an added term $a|z|$, representing an extra uniform sheet without changing the radial circular force. That contribution is excluded in the intended Mestel disk model.

Use a mass-weighted planar galactic distribution function, so $F(\mathbf{x},\mathbf{v})d^{2}x{d}^{2}v$ is the stellar mass in a small planar phase space element. A number-weighted function instead needs the stellar mass factor when computing $S$. Assume a steady collisionless stellar system and isotropy in the two in-plane velocity components. Put $u=|\mathbf{v}{|}^2/2$ and write $F=f(\mathbf{x},u)$. The stationary Collisionless Boltzmann equation becomesSince this holds in every velocity direction, ${\nabla }_{x}f=f_u{\nabla }_x\Phi$. In coordinates $(\mathbf{x},E)$ with $E=u+\Phi$, this is exactly ${\nabla }_{x}f{|}_E=0$. Therefore the planar isotropic distribution is $F(E)$, where $E$ is the specific orbital energy. Stationarity is essential; instantaneous isotropy alone would not imply this result.

Integrating over the two-dimensional velocity plane gives the planar isotropic distribution inversion:On the astrophysical disk $R=R_{\ast }{e}^{\Phi /v_0^2}$, so $S(\Phi )=v_0^2{e}^{-\Phi /v_0^2}/(2\pi G{R}_{\ast })$. Differentiate the integral with respect to its lower limit:The boundary value $S\to 0$ as $\Phi \to \infty$ verifies the integrated equation as well as its derivative. Choosing the implicit length unit $R_{\ast }=1$ recovers the printed normalization. Changing the additive energy zero changes this prefactor accordingly.

At a fixed radius, the normalized velocity density isIt is a product of centered Gaussian distributions. Differentiating the supplied Gaussian integral with respect to its coefficient gives the second moments, and odd moments vanish. Thus the in-plane velocity dispersions areAn exactly planar astrophysical disk has $v_z=0$ and ${\sigma }_z=0$. The nonzero circular speed is a property of the force field, not a statement that this hot stellar distribution has net rotation.

Reversing every retrograde star folds the azimuthal Gaussian to a half-normal distribution. Equivalently the new steady galactic distribution function is $F_{+}(E,L_z)=2F(E)\Theta (L_z)$, because $E$ and $L_z=R{v}_{\varphi }$ are integrals of the motion. Its density and even velocity moments are unchanged. Its streaming velocity isThis maximally prograde stellar distribution still has radial motion and a spread of azimuthal speeds; it does not place every star on a circular orbit. In particular $⟨v_{\varphi }^2{⟩}_{+}=v_0^2$ while ${\sigma }_{\varphi ,+}^2=v_0^2(1-2/\pi )$. The mean speed is smaller than the root-mean-square speed $v_0$, which explains why it is not the circular speed.

Use a local, homogeneous razor-thin disk approximation in a frame rotating with constant $\Omega$. The unperturbed planar velocity is zero in this frame, and the large-scale gravitational and centrifugal forces balance. Neglect viscosity, magnetic fields, thickness and background gradients across a wavelength. Take small planar disturbances, with wavelength short compared with the galaxy's background scale but long enough for a fluid description. The barotropic closure of a razor-thin disk is $P=K{\Sigma }^{\gamma }$, with $K$ fixed and a positive derivativeSolid-body rotation has no shear and has radial epicyclic frequency $\kappa =2|\Omega |$. By rotational symmetry of the local model choose a wavevector along $x$, and write perturbations proportional to $e^{i(kx-\omega t)}$.

Let ${\Sigma }_1,u,v,{\Phi }_1$ be the surface density, $x$-velocity, $y$-velocity and Newtonian gravitational potential amplitudes. The linearized continuity equation and Euler equations with the Coriolis force areThe perturbing Newtonian gravitational potential solves the Poisson equation for Newtonian gravityIts decaying solution is proportional to $e^{-|k||z|}$. The jump in its derivative is $-2|k|{\Phi }_1(0)=4\pi G{\Sigma }_1$, giving the razor-thin disk Poisson kernelEliminating $v$ from the two momentum equations, and using ${\Sigma }_1=k{\Sigma }_{0}u/\omega$, gives the density-wave branchThe complete linear system also has a zero-frequency balanced mode; it is not the growing density-wave branch, and division by $\omega$ in this elimination excludes it. The result is the uniformly rotating gas-sheet dispersion relation: pressure opposes compression at large wavenumber, self-gravity promotes compression, and rotation provides epicyclic support.

A mode is exponentially unstable when ${\omega }^2<0$. With $\Omega =0$ and $c>0$, this occurs atThe uniform $k=0$ perturbation is marginal rather than growing in this local calculation. With $c=0$, rotation fails to stabilize sufficiently short waves:These limits show why both pressure and rotation are needed for stability at all wavelengths.

For nonzero pressure and rotation, put $q=|k|$. Complete the square:Its minimum lies at $q_{\ast }=\pi G{\Sigma }_0/c^2$. Thus a growing mode exists precisely whenThe printed inequality has the opposite physical meaning: it is the condition for no exponentially growing density wave. Equality is marginal. In the usual gas Toomre stability criterion, $Q=\kappa c/(\pi G{\Sigma }_0)=2|\Omega |c/(\pi G{\Sigma }_0)$, so instability is $Q<1$ and stability is $Q\ge 1$. The unstable wavenumber band of a rotating gas sheet iswhen $Q<1$.

If ${\Sigma }_0$ and $\Omega \ne 0$ remain fixed while $c^2$ decreases slowly, first reach marginality atThe first wavelength to become unstable just below this threshold is the marginal fragmentation wavelength of a rotating sheet:Density maxima are separated by approximately this wavelength. The expected fragment size is of this order; an overdense half-wave has width about ${\ell }_{\mathrm{crit}}/2$. Linear theory fixes a preferred wavelength, not an exact nonlinear clump radius or shape. Further cooling shifts the fastest-growing wavelength to $2c^2/(G{\Sigma }_0)$. A rough fragment mass is consequently of order ${\Sigma }_0{\ell }_{\mathrm{crit}}^2$, with a geometrical factor depending on the nonlinear fragmentation pattern.

Use the paper's positive binding-energy convention $U>0$; the physical Newtonian gravitational potential energy is $-U$. Assume positive masses, an isolated system, and no collision during the time interval under consideration. Differentiate half the scalar moment of inertia, $I$, twice:Group the force term into unordered pairs. The force on particle $j$ from $k$ is $-G{m}_j{m}_k({\mathbf{r}}_j-{\mathbf{r}}_k)/r_{jk}^3$; the two contributions to the virial sum are thereforeAdding them proves the instantaneous virial theorem identityNo time average is needed for this form. A vanishing average of $\ddot{I}$ would require additional boundedness assumptions.

For the pairwise moment-of-inertia identity, expand the squared pair distances:The last term vanishes in the centre of mass frame, givingLet $C={\sum }_{j<k}{m}_j{m}_k$, and label the two smallest masses $m_{(1)},m_{(2)}$. If $r=\min r_{jk}$, every term in $U$ is at most $G{m}_j{m}_k/r$, so $U\le GC/r$. One pair attains $r$, and its mass product is at least $m_{(1)}{m}_{(2)}$, so $U\ge G{m}_{(1)}{m}_{(2)}/r$. HenceThese constants show the minimum-separation binding-energy bounds: inverse binding energy is comparable to the nearest-pair distance, independently of configuration.

Similarly, every separation is at most $R=\max r_{jk}$ and at least one pair attains $R$. The pairwise identity gives the maximum-separation inertia boundsBoth lower bounds rely on actual extremal separations, rather than an arbitrary numerical lower or upper estimate for all the pair distances.

Negative energy does not imply a positive lower bound on the minimum separation. The printed request is false if it is meant to exclude close approaches or collisions; the trivial bound $r\ge 0$ says nothing of that kind. In fact $E<0$ and $T\ge 0$ give $U=T-E\ge |E|$, which, combined with $rU\le A_0$, proves the negative-energy minimum-separation upper boundThis ensures at least one close pair, not confinement of every particle or prevention of collision.

An explicit negative-energy gravitational collision is a pair initially at rest with separation $a$. Its total energy is $E=-G{m}_1{m}_2/a<0$. For reduced mass $\mu =m_1{m}_2/(m_1+m_2)$, its relative radial equation isThe separation decreases to zero in the finite timeBefore that time the motion is a regular Newtonian solution, yet its separation has no positive infimum. Even at fixed negative energy, bound Kepler orbits with eccentricity approaching one have arbitrarily small pericentre distance. Thus a missing angular momentum or collision-exclusion hypothesis cannot be supplied by the energy sign alone.

For $E>0$, the instantaneous virial theorem gives $\ddot{I}=T+E\ge E$. Integrating twice from any regular time $t_0$ givesUsing $I\le A_1{R}^2$, obtain the positive-energy linear diameter growth boundwhenever the numerator is nonnegative. Consequently, for a solution existing for arbitrarily large future times,This is the precise at-least-linear expansion statement. It does not assert that $R$ is monotone at every instant, nor that all individual particles escape. The large-time conclusion presupposes continued existence of the trajectory; the inequality itself holds on every nonsingular time interval.

The dimensional linear friction coefficient is $\zeta$. Put $y=x-v_{T}t$ and $u(y)=F(y)/\zeta$. In overdamped particle dynamics the laboratory velocity is $\dot{x}=u(y)$, whereas the trap-frame velocity is $\dot{y}=u(y)-v_T$. The particle enters the optical tweezers at $y=X_R$ and leaves at $y=-X_L$ when the trap overtakes it. Write $w=X_R+X_L$ for this support width. The printed subtraction in the later definition of the width is a sign error: the distance between these endpoints is their sum. We use $w=2X_0$ in the circular calculation.

For a bounded continuous force, the strict conditionensures finite passage through a translating optical trap: the trap-frame coordinate decreases throughout the interaction. If $u(y)=v_T$ is encountered from the right, the deterministic trajectory cannot pass this stationary trap-frame point for a locally Lipschitz force; it approaches or stays at a locked state. At a smooth maximum the critical passage time diverges. More generally the passage criterion is $v_T-u>0$ on the traversed interval together with a finite integral below. Separating the trap-frame equation givesThese formulas account for both portions of the optical tweezers, including motion against the trap direction.

The forward-displacement result uses the ordinary localized potential energy interpretation of an optical trap: $F=-U^{\prime }$ and $U$ has the same value outside both ends, hence $\int Fdy=0$. Compact support of the force alone does not ensure that condition. Under the equal-endpoint condition,for a nonzero force and a passing trajectory. This is forward displacement from a translating localized potential. A zero force gives zero displacement. If the force is negative everywhere on its support, it instead gives $\Delta x<0$; thus an arbitrary compactly supported force does not satisfy the printed assertion.

There are two complementary explanations of the forward displacement from a translating localized potential. The forward push slows the relative passage and therefore acts longer, while the backward pull speeds up the relative passage and therefore acts for less time. An exact potential energy balance makes the same point: along the trajectoryThe moving optical trap supplies the positive work lost to linear drag, although the initial and final potential energy are equal.

For $v_T\gg \Vert u{\Vert }_{\infty }$, a uniformly convergent expansion of the passage integrals givesFor a localized potential energy well the first integral vanishes, soThe displacement decreases quadratically with the trap speed, rather than linearly. Without the equal-endpoint assumption the general expansion above remains valid.

For repeated kicks from a circular optical trap, let $L=2\pi R$ and interpret the force profile locally along the arc. This description requires a tangential constraint, a nonoverlapping force support $w<L$, and, if the straight profile is used geometrically, a small support compared with $R$. The condition $R\gg a$ alone does not specify that latter width hierarchy. During an encounter the particle advances by $\Delta x$ and the trap advances by $v_T\Delta t=w+\Delta x$. Between encounters the deterministic particle is stationary and the trap travels the remaining relative distance $L-w$. Thus the time between corresponding points of successive encounters isHere $f_T$ and $f_p$ are revolution frequencies, not angular velocities; the printed relation $f_T=v_T/(2\pi R)$ fixes this convention. The encounter frequency is $f_T-f_p$, so the same result follows from $L{f}_p=\Delta x(f_T-f_p)$.

The precise condition for the stated approximation is $|\Delta x|/L\ll 1$, in addition to finite passage and separated encounters. A sufficient high-speed regime is $v_T\gg \Vert F{\Vert }_{\infty }/\zeta$ with fixed $w<L$, for which the asymptotic displacement divided by $L$ tends to zero. ThenThis condition controls the encounter frequency correction; there is no need to discard the finite interaction time without accounting for the support width.

For the triangular optical-trap response, set $v_c=F/\zeta =L{f}_c$. In the passing regime $v_T>v_c$, the approach side has velocity $-v_c$ and the trailing side has velocity $+v_c$. ConsequentlyWith $\alpha =2X_0/L=X_0/(\pi R)$ and $\beta =v_T/v_c=f_T/f_c$, the repeated kicks from a circular optical trap formula becomesFor $0<\beta <1$ the usual ideal triangular-well dynamics locks at the cusp: the forces on its two sides direct the trap-frame particle towards the cusp. This is understood as the sticking limit of a rounded potential or of the overdamped differential inclusion at its discontinuous force. The particle follows the trap, giving $f_p/f_c=\beta$. At $\beta =1$ it can remain at a fixed point of the trailing segment and also has $f_p/f_c=1$. Thus the physical triangular optical-trap response is continuous at threshold, even though the isolated passing-kick displacement diverges there. For $\beta \gg 1$,and the more general kick approximation requires $\alpha /({\beta }^2-1)\ll 1$.