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For a nonrelativistic point charge, let $C_{ij}=⟨a_i{a}_j⟩$ be the time-averaged second moment of acceleration, without subtracting its mean. The outgoing radiation zone fields satisfy $\mathbf{B}=\mathbf{n}\times \mathbf{E}/c$ and $\mathbf{E}=q\mathbf{n}\times (\mathbf{n}\times \mathbf{a})/(4\pi {ϵ}_0{c}^{2}r)$. Their Poynting vector gives the displayed power per solid angle. Its integral is $⟨P⟩=q^2\mathrm{tr}C/(6\pi {ϵ}_0{c}^3)$, the averaged Larmor formula.

For periodic motion the mean acceleration vanishes and $C$ is also its covariance matrix. More generally the second moment, rather than the centered covariance matrix, is the relevant quantity. The angular formula is valid to leading nonrelativistic order and evaluates source motion at retarded time.

In the incompressible near field of a spherical bubble, $u_r=a^2\dot{a}/s^2$. Its volume-flux acoustic monopole is the displayed term at retarded time. The integrated surface pressure force vanishes by spherical symmetry, while the Lighthill stress tensor approximation ${\rho }_0{u}_i{u}_j$ has moment $S_{ij}=4\pi {\rho }_0{a}^3{\dot{a}}^2{\delta }_{ij}/3$. Its leading compact acoustic quadrupole contribution is ${\rho }_0(a^3{\dot{a}}^2{)}^{\prime \prime }/(3c_0^{4}r)$. For small fractional radius oscillation $\delta$ and compactness parameter $\alpha =\omega a_0/c_0$, its size relative to the leading monopole is $O(\delta {\alpha }^2)$.

Electromagnetic radiation is a propagating disturbance of the electric field and magnetic field that carries energy and momentum away from its source. It can be a sustained electromagnetic wave or a finite wave packet: periodic oscillation is not required. In vacuum the disturbance travels at the speed of light. A receiver responds to fields arriving now from an earlier source event, as expressed by retarded time. This is how electrical activity at one place can deliver energy and information elsewhere without a connecting wire.

Stored electromagnetic energy alone does not establish outgoing electromagnetic radiation. A stationary electric charge in electrostatics, or a steady localized electric current in magnetostatics, can store electromagnetic energy with zero Poynting vector. Static combinations can have circulating energy flux without net outward power. Near a changing source, reactive energy can return later. The near and far field distinction separates these contributions from outgoing radiation zone fields.^[ref]

A single traveling vacuum plane electromagnetic wave obeying the Maxwell equations has $\mathbf{B}=\hat{\mathbf{k}}\times \mathbf{E}/c$. Its Poynting vector gives the energy density and flux of a plane electromagnetic wave:Here $I$ is wave intensity; ${ϵ}_0$ and ${\mu }_0$ are vacuum permittivity and vacuum permeability. For a monochromatic wave with linear polarization and peak amplitude $E_0$, squared cosine averages to $1/2$. Pulses require the instantaneous Poynting vector; arbitrary incident-and-reflected superpositions need not obey this relation.^[ref]

The electromagnetic momentum density is $\mathbf{g}=\mathbf{S}/c^2$. Completely absorbing a unidirectional vacuum pulse of energy $U$ transfers momentum $U/c$; see radiation pressure on a perfectly absorbing surface. A calculated $1.0\mathrm{mJ}$ pulse transfers $3.34\times 10^{-12}\mathrm{kg}m/s$. Symmetric emission can lose energy without net recoil because opposite momenta cancel.^[ref]

The direction of emission matters as much as its total power. For an oscillating electric dipole moment along $z$, the angular distribution of electric dipole radiation is strongest across the axis and vanishes along it. Its origin is geometric: the electric dipole radiation field is the transverse projection of the second time derivative of the electric dipole moment, so its magnitude contains $\sin \theta$. Squaring this electric field to obtain wave intensity gives ${\sin }^2\theta$. The electric-dipole approximation for radiation requires a source much smaller than its wavelength and observation in the radiation zone. ^[ref]

For example, the power per solid angle at $\theta =30^{\circ }$ is one quarter of its equatorial value, while the electric field amplitude is one half. To normalize the angular distribution to total power, use spherical coordinates with $d\Omega =\sin \theta d\theta d\varphi$:The integral follows by substituting $x=\cos \theta$, giving ${\int }_{-1}^1(1-x^2)dx=4/3$. This is a probability-like normalization over direction, with units of inverse steradian; it does not make the dipole an isotropic radiator. The same angular integral underlies the Larmor formula for nonrelativistic accelerated electric charge.

Detection requires the incident electromagnetic field to do something observable. An antenna develops an electrical response; a thermopile converts absorbed energy into a thermal and then electrical signal; a photodetector can respond through photon absorption. Hertz spark transmission and detection experiment makes this conversion visible as a tiny spark. Detector orientation, sensitivity, and surrounding reflections all matter: failure to see a signal does not prove that no electromagnetic radiation reached the receiver.

A gravitational-wave pulse has curvature supported in a bounded interval of retarded time. Spacetime can be flat both before and after the pulse even though freely falling bodies retain a displacement or relative velocity produced while the pulse passed.

A Hertzian dipole is an ideal electrically short oscillating electric dipole used to model radiation from a localized current element. In vacuum its far-field electric dipole radiation has ${\sin }^2\theta$ angular power distribution and total power $P={\mu }_0|\ddot{\mathbf{p}}{|}^2/(6\pi c)$, evaluated at the retarded time. The electric-dipole approximation for radiation requires a source much smaller than its radiation wavelength.

The Liénard-Wiechert field is the electric field and magnetic field obtained from the retarded potentials of a moving point charge. In the nonrelativistic radiation zone, the electric radiation field is ${\mathbf{E}}_{\mathrm{rad}}=q\hat{\mathbf{n}}\times (\hat{\mathbf{n}}\times \mathbf{a})/(4\pi {ϵ}_0{c}^{2}R)$ at retarded time. Its outgoing Poynting vector integrates to the Larmor formula. Coulomb and induction fields must also be retained when a near-field calculation is required.

The potentials of a point charge $q$ arewith every source quantity evaluated at the retarded time.

For a source surface parameterized by $y(p,q,\tau )$, the retarded time equation is $t=\tau +|x-y(p,q,\tau )|/c_0$. Integrating a Dirac delta distribution enforcing that equation produces the reciprocal derivative $|1-M_r{|}^{-1}$ at each simple root. For subsonic motion there is a unique retarded root. Multiple roots require summation; a zero derivative is outside the simple-root rule. This factor is the moving-source counterpart of a delta change-of-variable Jacobian.

An outgoing radiation pulse carries a finite amount of electromagnetic energy through successively larger enclosing spheres. In vacuum, far from a localized source and without absorption or additional reflections, let $P_{\mathrm{emit}}(t)$ be the emitted power. The retarded time at radius $r$ is $s=t-r/c$. Define ${\overline{S}}_r$ as the angular average of the radial Poynting vector over the sphere, not a cycle average:Here $\overline{u}$ is the angularly averaged energy density in the leading radiation zone approximation. Multiplying by shell area removes geometric dilution: the energy per unit radial thickness translates outward without losing its shape. These angular averages do not assume that an actual electric dipole emits equally in all directions. Nor do they specify an exact spherically symmetric transverse solution of the Maxwell equations.

For an explicitly stipulated model, emitAfter emission ends, $t>\tau$, the pulse lies between $r=c(t-\tau )$ and $r=ct$, provided this shell is in the radiation zone. Its radial thickness remains $c\tau$. Substituting $s=t-r/c$ in the shell integral $\int 4\pi r^2\overline{u}dr$ gives $U=P_0\tau /2$ at every such time. The same substitution in the time integral of $4\pi r^2{\overline{S}}_r$ at any enclosing sphere gives that same total crossing energy. This is conservation of energy in two different views.

Along any fixed direction with nonzero outgoing power, compare the same part of the pulse at two radii, hence at different arrival times. The leading electric field amplitude falls as $1/r$, while its squared wave intensity falls as $1/r^2$. At $2r$ the field is one half and the energy flux one quarter; the signal arrives $r/c$ later. A small fixed area receiver therefore intercepts less energy at greater distance even though every enclosing sphere receives the same total pulse energy. The inverse-square law expresses spreading over area, rather than destruction of energy. Close to the source, the neglected near and far field terms prevent extrapolating these formulas down to $r=0$.

The moving point charge has distributional charge density and current densityLetThe given retarded fundamental solution identity immediately implies that the retarded potentialssolve $\square \varphi =-\rho /{ϵ}_0$ and $\square \mathbf{A}=-{\mu }_0\mathbf{J}$ in the sense of distributions. Integrating the spatial Dirac delta function first givesand the analogous integral for $\mathbf{A}$ has numerator ${\mu }_{0}q{\mathbf{v}}_0(t^{\prime })/(4\pi )$.

Assume the physical trajectory has subluminal speed and the retarded time exists. At its root $t_r$, with ${\mathbf{n}}_r=\mathbf{R}(t_r)/R(t_r)$,The positive factor $\kappa =1-{\mathbf{n}}_r\cdot {\mathbf{v}}_0/c$ also proves uniqueness of retarded time for a subluminal source. The Dirac delta function substitution formula gives $\delta (g(t^{\prime }))=\delta (t^{\prime }-t_r)/\kappa$, soThese are the Lienard-Wiechert potentials. The denominator includes the time-substitution factor, rather than just the retarded distance. Away from the trajectory the electric potential and magnetic vector potential are ordinary smooth functions; on the trajectory their wave equations have the indicated distributional sources.

For completeness, the charge continuity equation ${\partial }_{t^{\prime }}\rho +{\nabla }^{\prime }\cdot \mathbf{J}=0$ also verifies the Lorenz gauge. Using $\nabla K=-{\nabla }^{\prime }K$ and ${\partial }_{t}K=-{\partial }_{t^{\prime }}K$, integration by parts in the retarded integrals giveswith the retarded boundary prescription. Thus the construction also has the required Lorenz gauge condition.

Take the ideal electric dipole limit with a fixed smooth electric dipole moment $\mathbf{p}(t)$, so that ${\mathbf{r}}_0(t)=\mathbf{p}(t)/(2q)$; its required time derivatives are bounded on the time interval under consideration. Fix $r>0$ and put $\mathbf{n}=\mathbf{r}/r$, $t_0=t-r/c$. For $s=+1$ or $-1$, the source position is $s\mathbf{p}(t)/(2q)$. A Taylor expansion of the distance givesSubstitute this into the defining equation for retarded time:Boundedness first implies $t_r^s-t_0=O(q^{-1})$. The Taylor expansion $\mathbf{p}(t_r^s)=\mathbf{p}(t_0)+O(q^{-1})$ then changes the displayed correction only by $O(q^{-2})$. HenceThe nearer positive displacement reduces the propagation distance and consequently increases the retarded time. The Big O notation here is for $q\to \infty$ at a fixed observation event away from the origin. Keeping a fixed smooth $\mathbf{p}$ is essential: bounded amplitude alone, with arbitrarily increasing time frequency as $q$ varies, would not justify this Taylor expansion.

The leading electric dipole radiation fields areThe radiation electric field is transverse, $\mathbf{n}\cdot {\mathbf{E}}_{\mathrm{rad}}=0$. The vector triple product therefore gives the leading Poynting vectorIf $\theta$ is the angle between ${\mathbf{p}}^{\prime \prime }(t-r/c)$ and $\mathbf{n}$, thenUsing ${\mu }_0{ϵ}_0{c}^2=1$, the final result isThe leading Poynting flux is outward, vanishes along the instantaneous retarded ${\mathbf{p}}^{\prime \prime }$ axis, and is maximal in its perpendicular plane. If ${\mathbf{p}}^{\prime \prime }=0$, the radiation term is zero and no angle needs to be assigned. The large-$r$ expansion assumes bounded retarded electric dipole moment and time derivatives; it may also be read at fixed retarded time.

Use the conservative continuity equation and momentum balance, with ${\sigma }_{ij}$ the viscous stress:Differentiate the first equation in time and take the divergence of the second. Eliminating momentum givesSubtract $c_0^2\Delta \rho$, and define the Lighthill stress tensor byThe constant reference subtraction has no double divergence. The exact Lighthill acoustic analogy isThis is an exact rearrangement; linearization is not used to eliminate the nonlinear source tensor.

Convolving with the retarded acoustic Green function and transferring the source derivatives by integration by parts givesThe acoustic compact-source approximation requires source size $\ell$ to satisfy $\Omega \ell /c_0\ll 1$ for its characteristic frequency $\Omega$; the acoustic far field additionally requires $r\gg \ell$ and $\Omega r/c_0\gg 1$. Internal retardation is then negligible at leading order, and the radiating parts of the two observer derivatives act on the common retarded time. With $n_i=x_i/r$ and $S_{ij}=\int T_{ij}{d}^{3}y$, the far-field acoustic force and stress moments giveThe two spatial derivatives supply two factors of $1/c_0$, in addition to the $1/c_0^2$ in the Green function.

For a subsonic fluctuating source of velocity scale $V$, size $\ell$ and turnover time $\ell /V$, suppose its stress scale is $T_{ij}=O({\rho }_0{V}^2)$. Then $S_{ij}=O({\rho }_0{V}^2{\ell }^3)$ and ${\ddot{S}}_{ij}=O({\rho }_0{V}^4\ell )$. The compact acoustic quadrupole Mach-number scaling is thereforeThis fourth-power statement concerns the acoustic density amplitude under the specified eddy-time and stress scaling. With an independently imposed frequency, the general estimate instead retains $V^2{\Omega }^2{\ell }^3/c_0^{4}r$.

Convolution with the retarded acoustic Green function, followed by two integrations by parts in the source coordinate, yieldsFor the acoustic compact-source approximation, replace the distance and the source retarded time by $r=|x|$ and $\tau =t-r/c_0$ in the leading moment. Thus the integral is $S_{ij}(\tau )/(4\pi c_0^{2}r)$, where $S_{ij}=\int T_{ij}dV$. In the acoustic far field, each spatial derivative acts predominantly on retarded time: ${\partial }_{x_i}\tau =-n_i/c_0$, where $n_i=x_i/r$. Derivatives of $1/r$ and of $n_i$ give lower powers of $r$. The two retarded derivatives therefore giveThis is the compact acoustic quadrupole field, retaining its angular stress projection.

For the compact acoustic quadrupole Mach-number scaling, let the source length be $\ell$, typical fluctuation speed be $U$, and typical time be $\ell /U$. The low-Mach number stress is $T=O({\rho }_0{U}^2)$, so $S=O({\rho }_0{U}^2{\ell }^3)$ and $\ddot{S}=O({\rho }_0{U}^4\ell )$. ConsequentlyWith the geometric ratio separated, this is the stated fourth-power dependence. It assumes an advective source time scale and compactness $\omega \ell /c_0=O(m)\ll 1$. An externally driven source with an independent frequency does not acquire the same Mach power automatically.

Put $r=|\mathbf{x}|$, $\mathbf{n}=\mathbf{x}/r$, and take a compactly supported Lighthill stress tensor. The isotropic term in that tensor is $(p-c_0^2\rho ){\delta }_{ij}$: the Kronecker delta is implicit in the printed shorthand. Convolving the Lighthill acoustic analogy with the retarded acoustic Green function and integrating twice by parts in the source coordinates givesThe boundary terms vanish because of compact support. For source size $\ell$, the acoustic compact-source approximation requires $\omega \ell /c_0\ll 1$. At $r\gg \ell$, we may therefore replace the denominator by $r$ and the retarded argument by $t-r/c_0$ in the leading source integral. Define $S_{ij}(t)=\int T_{ij}(\mathbf{y},t)d^{3}y$. In the radiation region $\omega r/c_0\gg 1$, the leading two spatial derivatives act on the retarded time rather than on $1/r$ or $\mathbf{n}$:Thus the leading acoustic quadrupole density isFor the compact acoustic quadrupole Mach-number scaling, let $U$ be the source fluctuation speed, $m=U/c_0\ll 1$, and use the advective source time $\ell /U$. With $T_{ij}=O({\rho }_0{U}^2)$, one has $S_{ij}=O({\rho }_0{U}^2{\ell }^3)$ and ${\ddot{S}}_{ij}=O({\rho }_0{U}^4\ell )$. ConsequentlyThe fourth power refers to the radiation coefficient with geometric spreading separated. It assumes that the source frequency scales as $U/\ell$; an independently imposed frequency or an independently strong thermal source would change that Mach number estimate.

First retain the question's units $c=1$. Insert the retarded time Dirac delta:For a point charge, the spatial delta sets $x^{\prime }=x_0(t^{\prime })$. Put $R(t^{\prime })=|x-x_0(t^{\prime })|$, $\hat{R}=(x-x_0)/R$ and $v={\dot{x}}_0(t^{\prime })$. The retarded root $t_{\ast }$ satisfies $t-t_{\ast }=R(t_{\ast })$. The derivative of the delta's argument is $-1+\hat{R}\cdot v$, so for a subluminal trajectory its absolute value is $1-\hat{R}\cdot v$. Performing the time integral givesAll quantities in the denominator are evaluated at the retarded time. The retarded vector potential uses current $j=qv{\delta }^{(3)}(x^{\prime }-x_0)$ and the same Jacobian determinant, givingRestoring physical $c$, the root is $t-t_{\ast }=R_{\ast }/c$, both denominators become $R_{\ast }-v_{\ast }\cdot {\mathbf{R}}_{\ast }/c$, and ${\mu }_0=1/({ϵ}_0{c}^2)$. Thus $\mathbf{A}=v_{\ast }\varphi /c^2$. This is the Liénard–Wiechert potential; the Jacobian determinant, rather than merely substituting retarded position into a static Coulomb formula, accounts for the velocity factor.

The original spherical coordinates have $t\in \mathbb{R}$, $0\le r<\infty$, $0\le \theta \le \pi$ and $0\le \varphi <2\pi$, with $\varphi$ periodic. The angular coordinates degenerate at the polar axis and all angles identify at $r=0$; a regular spherical chart uses $r>0$, $0<\theta <\pi$ and suitable overlapping angular patches.

The retarded time $u=t-r$ and advanced coordinate $v=t+r$ both range over $\mathbb{R}$, with the joint restriction $u\le v$. Their inverse relations are $t=(u+v)/2$ and $r=(v-u)/2$, soSet $p=\arctan u$, $q=\arctan v$. ThenThe conformal factor is $\Omega =2\cos p\cos q>0$. Since $(\tan q-\tan p)\cos p\cos q=\sin (q-p)$, multiplication by ${\Omega }^2$ givesFinally $T=p+q$ and $R=q-p$ give $4dpdq=d{T}^2-d{R}^2$. HenceThis is a region of the Einstein static universe, with positive time and negative spatial signature. The exact joint coordinate domain isor equivalently $R\ge 0$ and $|T|+R<\pi$. The regular radial chart has $R>0$; $R=0$ is the ordinary centre. The angular ranges are unchanged. It is important not to treat the stated individual ranges of $T,R$ as independent. The inverse map and conformal factor provide a useful check:

The resulting Minkowski conformal compactification has a triangular radial Penrose diagram. The two sloping boundary edges areThese are future null infinity and past null infinity. The three vertices are future timelike infinity $i^{+}=(\pi ,0)$, past timelike infinity $i^{-}=(-\pi ,0)$ and spacelike infinity $i^0=(0,\pi )$. They are ideal boundary points, not events of physical Minkowski spacetime.

For a complete inertial timelike geodesic, $\mathbf{x}=\mathbf{b}+\mathbf{v}t$ with $|\mathbf{v}|<1$. As $t\to +\infty$, both $t-r$ and $t+r$ tend to $+\infty$, so $p,q\to \pi /2$ and the curve ends at $i^{+}$. As $t\to -\infty$, both tend to $-\infty$, giving $i^{-}$. These endpoints occur at infinite physical proper time.

For a complete null geodesic, write $\mathbf{x}=\mathbf{b}+\mathbf{n}t$, $|\mathbf{n}|=1$. To the future, $r=t+\mathbf{n}\cdot \mathbf{b}+O(1/t)$, so $u\to -\mathbf{n}\cdot \mathbf{b}$ is finite and $v\to +\infty$. Thus $q\to \pi /2$ with $p$ finite: the endpoint lies on ${\mathcal{I}}^{+}$. To the past, $v\to -\mathbf{n}\cdot \mathbf{b}$ remains finite while $u\to -\infty$, giving ${\mathcal{I}}^{-}$. Physical affine parameter is infinite at both ends. Along a radial null path, $dT=\pm dR$, so the diagram represents its null directions at $45$ degrees.

For a complete spacelike geodesic, take $t=t_0+a\lambda$, $\mathbf{x}=\mathbf{b}+\mathbf{w}\lambda$ with $|\mathbf{w}|>|a|$. At either end $r>|t|$ asymptotically, so $u\to -\infty$, $v\to +\infty$ and $(T,R)\to (0,\pi )$. Both ends therefore reach $i^0$, again at infinite physical affine parameter. These are the geodesic endpoints in Minkowski conformal compactification.

Only conformal preservation of null geodesic paths is guaranteed. The pictured timelike and spacelike curves are images of physical geodesics, not generally geodesics of the unphysical metric. For a null path, an unphysical affine parameter satisfies $d\tilde{\lambda }/d\lambda \propto {\Omega }^2$. Since $\Omega =O(1/|\lambda |)$ near a generic null-infinity endpoint, the unphysical parameter can converge while the physical parameter diverges. At $i^{\pm }$ and $i^0$, the usual smooth-boundary condition $d\Omega \ne 0$ degenerates; these vertices should not be treated as regular points of null infinity.

Complete Minkowski spacetime has no black-hole event horizon. Every physical event can send an outgoing null signal to ${\mathcal{I}}^{+}$, so the black-hole region $\mathcal{M}\setminus J^{-}({\mathcal{I}}^{+})$ is empty. There is likewise no particle or future event horizon for an eternal inertial observer: any finite event can eventually signal that observer. Null infinity is a conformal boundary, not a physical horizon. Uniformly accelerated observers can have Rindler horizons, but that is an observer-dependent restriction and is not a horizon of the full inertial Minkowski spacetime.

Let the observer have cylindrical coordinates $(r,\theta ,z)$ and total distance $R=\sqrt{r^2+z^2}$. At retarded emission time $\tau$, the source-observer distance isThe retarded time relation $t=\tau +|\mathbf{x}-\mathbf{y}(\tau )|/c_0$ consequently becomesSet $\Omega \tau =\theta -\pi /2+\Theta$. ThenThe observer-direction Mach number at emission is $M_r=\dot{\mathbf{y}}(\tau )\cdot \mathbf{n}/c_0=m\cos \Theta$. For a constant strength $q$ and $m<1$, the single retarded root has $1-M_r>0$. Substituting into the moving-source field gives the rotating acoustic point source resultIn particular, on the rotation axis $m=0$ and the result is $q/(4\pi c_0^{2}R)$.

For a genuinely varying source strength, replace $q$ by $q(\tau )$ in the last formula. If $m>1$, the retarded relation need not be one-to-one: each retarded root contributes with $|1-M_r|$ in the denominator. The general result is a sum of $q({\tau }_j)|{\dot{\Theta }}_j|/(4\pi c_0^2\Omega R)$. Dropping the absolute value and the root sum requires the subsonic assumption. At $1-M_r=0$ the ideal point-source expression has a caustic; finite source size and bandwidth regularize that idealization.

Let $\chi =H(F)$ use the Heaviside step function, and ${\delta }_s=|\nabla F|\delta (F)$, so that ${\mathbf{n}}_s=\nabla F/|\nabla F|$ points from the body into the fluid. The kinematic boundary condition gives $F_t+\mathbf{u}\cdot \nabla F=0$ on the surface. In the exterior, mass conservation and the Euler equations for an inviscid fluid areProducts with $\chi$ are understood distributionally, using ${\partial }_t\chi =F_t\delta (F)$ and ${\partial }_i\chi =F_i\delta (F)$. With ${\rho }^{\prime }=\rho -{\rho }_0$, the extended continuity equation isThe corresponding extended momentum equation isbecause its advective surface term is $\rho u_i(F_t+\mathbf{u}\cdot \nabla F)\delta (F)=0$. Differentiate continuity in time and substitute the divergence of momentum. Subtracting $c_0^2{\nabla }^2({\rho }^{\prime }\chi )$ gives the Ffowcs Williams-Hawkings equationwhere the inviscid Lighthill stress tensor isIf $p$ denotes total pressure, this tensor includes the constant ambient pressure $p_0$. Its contribution $p_0{\nabla }^2\chi$ cancels the corresponding surface term exactly. For radiation integrals it is convenient to subtract $p_0$ from both terms, making $T_{ij}=\rho u_i{u}_j+(p^{\prime }-c_0^2{\rho }^{\prime }){\delta }_{ij}$ with $p^{\prime }=p-p_0$. The displayed source identity is unchanged by this consistent pressure-gauge subtraction.

Assume no incoming sound and no free initial acoustic field. Put $R_y=|\mathbf{x}-\mathbf{y}|$, ${\tau }_y=t-R_y/c_0$, $Q={\rho }_0{u}_n{\delta }_s$, and $P_i=p^{\prime }{n}_{s,i}{\delta }_s$. Convolution with the retarded acoustic Green function gives the three source integrals, for an observer outside the body:These are the surface thickness, surface loading and volume quadrupole contributions. The retarded surface Dirac delta distributions must remain inside the integrals: integrating them over a moving retarded surface produces the moving-surface retarded Jacobian, and is not generally the same as simply substituting one retarded time into an instantaneous surface integral.

For an acoustic compact-source approximation, suppose the dominant sources have size $\ell$, source frequency scale $\omega$, and $\omega \ell /c_0\ll 1$. Also require $r=|\mathbf{x}|\gg \ell$ and the acoustic far field $\omega r/c_0\gg 1$. Then $R_y\simeq r-\mathbf{n}\cdot \mathbf{y}$, $1/R_y\simeq 1/r$, and the leading source integrals use the common time $\tau =t-r/c_0$. Slow source motion relative to $c_0$ is implicit in this leading retardation approximation. DefineThus $m$ is net displaced mass per unit time, $\mathbf{f}$ is the fluid pressure force on the body, and $S_{ij}$ is the integrated excess Lighthill stress tensor. The minus sign in $\mathbf{f}$ fixes the requested loading convention; the force on the fluid is opposite. On the radiating $1/r$ terms, ${\partial }_{x_i}\simeq -(n_i/c_0){\partial }_t$, givingSpatial derivatives of $1/r$ produce near-field terms of orders $r^{-2}$ and $r^{-3}$, omitted here. Neglecting source delays is a separate approximation from discarding those observer near-field terms. If a leading integrated source vanishes by symmetry, its first nonzero delay correction can be as important as another retained multipole.