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A finite spherical subvolume has radial-pressure virial surface term $4\pi R^3\rho (R){\sigma }_r^2$. The infinite singular isothermal sphere has divergent total mass and energies, so a finite isolated virial theorem cannot omit this term. Different orbital distributions with the same density can have different radial boundary pressures.

For the $r^{-2}$ singular isothermal sphere, the isotropic Spherical Jeans equation gives ${\sigma }_r^2={\sigma }_{\theta }^2={\sigma }_{\varphi }^2=v_0^2/2$, so $⟨v^2⟩=3v_0^2/2$. Randomly oriented circular orbits instead have ${\sigma }_r^2=0$, each transverse second moment $v_0^2/2$, and total mean square speed $v_0^2$.

For a singular isothermal sphere of dispersion $\sigma$ and gas fraction $f_g$, placing all gas inside radius $R$ onto a hole in time $R/\sigma$ gives ${\dot{M}}_{\mathrm{dyn}}\sim 2f_g{\sigma }^3/G$. This is an optimistic supply bound that omits angular momentum transport, star formation, and active-galactic-nucleus feedback. A numerical comparison with the Eddington accretion rate requires a gas fraction, reference radiative efficiency, opacity, and crossing-time convention.

For a singular isothermal sphere with $\rho =a_s^2/(2\pi G{r}^2)$, the initial mass inside the outward front $r_{\mathrm{ew}}=a_{s}t$ is $M_{\mathrm{ew}}=2a_s^{3}t/G$. This is not all immediately accreted by the central object. The inside-out collapse of a singular isothermal sphere instead has ${\dot{M}}_{\ast }=m_0{a}_s^3/G$, with $m_0$ fixed by its dynamical similarity solution; the remainder can stay in the infalling envelope. Equating front sweeping with central accretion loses that mass budget.

A test particle released from rest at $r_0$ in a fixed spherical mass density $\rho (r)={\rho }_0(r_0/r{)}^2$ obeys $\ddot{r}=-4\pi G{\rho }_0{r}_0^2/r$. Integrating once gives ${\dot{r}}^2=8\pi G{\rho }_0{r}_0^2\ln (r_0/r)$. The substitution $u=\sqrt{\ln (r_0/r)}$ then gives $t(r)=(8G{\rho }_0{)}^{-1/2}\mathrm{erf}(u)$, and the arrival time at the centre is $(8G{\rho }_0{)}^{-1/2}$. This is free fall in a frozen singular isothermal sphere, not the self-consistent pressureless collapse of the background. An initially resting material shell with the same initial enclosed mass instead arrives after $\sqrt{\pi /(32G{\rho }_0)}$, because its enclosed mass stays fixed while the cloud contracts.

An isothermal sphere is a spherically symmetric self-gravitating gas equilibrium with equation of state $p=K\rho$ and constant $K$. The hydrostatic equilibrium equations $p^{\prime }=-GM\rho /r^2$ and $M^{\prime }=4\pi r^2\rho$ yield the title equation. Its regular finite-central-density solution is distinct from the singular isothermal sphere. A finite outer radius requires confinement because the regular density does not vanish at finite radius.

For a low-mass singular isothermal sphere satellite on a circular orbit of radius $D$ in a singular isothermal sphere host, ${\Omega }^2=2{\sigma }_h^2/D^2$ and ${\Phi }_h^{\prime \prime }=-2{\sigma }_h^2/D^2$. The radial differential effective acceleration is $({\Omega }^2-{\Phi }_h^{\prime \prime })r$. Balancing it with satellite gravity $2{\sigma }_d^2/r$ gives the displayed Jacobi tidal radius and bound mass $m_t=\sqrt{2}D{\sigma }_d^3/(G{\sigma }_h)$. Equal mean density gives a different order-one approximation, $r_t=D{\sigma }_d/{\sigma }_h$; eccentric orbits require an orbital-history prescription.

Follow a fluid element with position $\mathbf{X}(t)$ satisfying $\dot{\mathbf{X}}=\mathbf{u}(\mathbf{X},t)$. The chain rule gives the material derivativeApply Newton's second law to a material element of volume $dV$ and mass $\rho dV$. The net pressure force is $-\nabla pdV$, and gravity gives $-\rho \nabla \varphi dV$. Dividing by the element mass, neglecting viscosity and using the chain rule componentwise gives the Euler equation

For a static spherical isothermal gas, $p=c_s^2\rho$ and $dp/dr=-GM(r)\rho /r^2$. Trial density $\rho =A/r^2$ gives enclosed mass $M(r)=4\pi Ar$, soThis is the singular isothermal sphere. Its central density diverges, although the enclosed mass tends to zero linearly and its total mass inside a finite radius is finite. It is not the finite-central-density regular isothermal solution.

Truncate at $r_0$ with boundary pressure $p_0$. Then $M=4\pi A{r}_0=2c_s^2{r}_0/G$ and $p_0=c_s^2\rho (r_0)$, givingThe external medium is being treated as a prescribed pressure boundary; its own gravitational field is excluded from the cloud self-energy calculation.

For a monatomic perfect gas, the equilibrium is isothermal but the internal-energy relation still uses its adiabatic index $\gamma =5/3$:The gravitational self-energy follows by assembling shells:Both integrals remain finite at the singular center. ConsequentlyThe pressure-confined singular isothermal gas sphere therefore satisfies the virial theorem including the work of the confining surface pressure. One can see its origin directly by integrating $\mathbf{r}\cdot \nabla p$: integration by parts gives $\oint p\mathbf{r}\cdot d\mathbf{S}-3\int pdV$, whose surface term is $4\pi r_0^3{p}_0$. An isolated vacuum-boundary version would omit that term and could not truncate this same positive-temperature isothermal profile at finite radius. The equality expresses equilibrium force and energy balance, rather than proving dynamical stability.

In a spherical self-gravitating equilibrium with positive mass density, $d\psi /dr=GM(r)/r^2>0$ away from the center. Thus $r$ can locally be replaced by $\psi$, so both $p(r)$ and $\rho (r)$ can be expressed as functions of $\psi$. Hydrostatic equilibrium further givesSince $p$ decreases outward wherever $\rho >0$, eliminating $\psi$ gives a barotropic equation of state along this equilibrium. This is a relation describing its structure; it does not require all thermodynamic perturbations of an arbitrary material to be barotropic.

For $p=K{\rho }^{1+1/n}$, integrate the polytropic enthalpy balance $dp/\rho =-d\psi$:For a finite-radius zero-pressure boundary with $K>0$ and $n>0$, $\rho (R)=0$ and therefore ${\psi }_T=\psi (R)$. With the gravitational potential zero at infinity, an isolated sphere has ${\psi }_T=-GM/R$. It is an integration constant tied to that boundary, not necessarily a zero of the gravitational potential.

Put $\theta =(\rho /{\rho }_c{)}^{1/n}$ and $\xi =\alpha r$, with ${\alpha }^2=4\pi G{\rho }_c^{1-1/n}/[(n+1)K]$. The preceding relation gives $\psi ={\psi }_T-(n+1)K{\rho }_c^{1/n}\theta$. Substitution into the spherical gravitational Poisson equation yields the Lane-Emden equationLet ${\xi }_1$ be the first zero of the regular Lane-Emden function and $Y_n=-{\xi }_1^2{\theta }^{\prime }({\xi }_1)>0$. Then the Lane-Emden mass formula and radius areFor a common fixed polytropic index with a finite first zero, their scalings areThe usual regular zero-pressure models have a finite first zero for $0<n<5$; the $n=5$ model has infinite radius, so a finite-radius formula cannot simply be applied to it.

For an ideal gas of fixed composition, $T_c\propto K{\rho }_c^{1/n}$. Consequently homologous polytropes at fixed central temperature have $K\propto {\rho }_c^{-1/n}$ and both $M,R\propto {\rho }_c^{-1/2}$, recovering $M\propto R$. The common central temperature, composition and dimensionless structure are the relevant assumptions. Keeping $K$ fixed instead gives the different polytropic mass-radius relation, and no finite index makes its generic fixed-$K$ exponent exactly one.

For adiabatic monatomic hydrogen, $\gamma =5/3$ and $n=3/2$. At a common fixed specific entropy, hence common $K$, the same relations giveHere $\bar{\rho }=3M/(4\pi R^3)$, and the omitted constant in the last relation depends on $K$. A population with different cloud entropies need not follow that mass sequence.

Efficient radiative cooling in molecular clouds prevents compression from forcing the temperature upward as steeply as in an adiabatic monatomic gas. Rotational transitions of molecules and radiation by dust provide low-temperature cooling channels; which dominates depends on composition, opacity and temperature. It is therefore possible to assemble dense massive cold clouds while disposing of compressional heat. This explains the direction of the idealized comparison, rather than proving a universal cloud-mass cutoff.

For a common isothermal sound speed $c_T$, a spherical isothermal equilibrium has $GM/R$ equal to $c_T^2$ times a dimensionless structural factor. Holding that factor fixed gives the homologous isothermal mass-radius relationFor example, the singular isothermal sphere has $M(R)=2c_T^{2}R/G$. A regular finite cloud requires a pressure-confined isothermal sphere or another outer boundary: an unconfined isothermal sphere does not end at a finite vacuum radius. Equal temperature alone, while allowing arbitrary truncation or density contrast, does not make the coefficient universal.

Hydrostatic equilibrium with self-gravity. A resting fluid obeys $\nabla p=-\rho \nabla \Psi$. Spherical symmetry and the Poisson equation for Newtonian gravity giveEliminate the Newtonian gravitational potential derivative to obtainEquivalently, the shell theorem gives ${\Psi }^{\prime }=GM(r)/r^2$ and $M^{\prime }(r)=4\pi r^2\rho$, which leads to the same hydrostatic equilibrium equation.

A uniform-density liquid. With constant $\rho$, $M(r)=4\pi \rho r^3/3$. There is no central point mass, so the interior gravitational acceleration magnitude is $4\pi G\rho r/3$. ThusUse $p(R)=0$ at a vacuum boundary, neglecting surface tension. The uniform-density planetary pressure profile isThe central value can also be writtenIt has units pascals and decreases smoothly from its maximum at the centre to zero at the surface.

An isothermal gas. Here $c$ is the constant isothermal sound speed, with $p=c^2\rho$. Try $p=A/r^2$, so $\rho =A/(c^2{r}^2)$. The two sides of the equilibrium equation becomeThus $A=c^4/(2\pi G)$, giving the singular isothermal sphere,This does not describe a finite gaseous sphere in vacuum. Both pressure and mass density diverge at the centre, and $p(R)>0$ for every finite outer radius $R$, so it cannot meet $p=0$ there. Extending it to infinity removes the finite-radius pressure jump but makes the total mass infinite. It is useful as an idealized profile over an intermediate radial range, or as a truncated model with nonzero external pressure. The vacuum boundary obstruction for an isothermal sphere also applies to the regular, nonsingular isothermal solution; removing the central singularity does not produce a finite vacuum surface.

Steady outflow. In the wind calculation, take the central mass $M$ to be fixed and neglect the wind's own self-gravity. Constant mass flow rate and the isothermal equation of state giveThe steady radial Euler equation isSubstituting the density derivative givesThis is the isothermal Parker wind equation. A smooth sonic point requires both sides to vanish when $u=c$, soMaintaining constant temperature during expansion generally requires heat input; it is an isothermal model, not the adiabatic blast-wave model of Question 5.

The escaping branch at large radius. Integrate the wind equation, using dimensionless logarithm arguments:Set $x=r/r_s$ and $w=u^2/c^2$. ThenFor the accelerating transonic branch, $w(1)=1$, which fixes $C_0=-3$. At $x\gg 1$ on its supersonic branch, $w\to \infty$ and $\ln w$ is smaller than $w$. ConsequentlyA first correction makes the ordering explicit:This is the large-radius asymptotics of an isothermal Parker wind. The leading expression also holds for other supersonic branches with finite $C_0$.

The source implicitly selects an escaping wind. Positive steady outward flow alone does not determine that branch: a subsonic isothermal breeze also satisfies the differential equation. Its $w\ll 1$ limit gives $w\sim e^{-C_0}{x}^{-4}$, hence $u\propto r^{-2}$ and $\rho \to {\rho }_{\infty }>0$. Such a breeze needs nonzero pressure at infinity. Selecting a regular outflow from a subsonic base into vanishing ambient pressure instead gives the accelerating transonic wind used above.

The asymptotic speed has the same units as $c$; the logarithm is dimensionless. Its unlimited growth is a formal feature of indefinite isothermal heating and a prescribed point-mass potential. Real heating, surrounding pressure, and eventually the neglected wind mass limit that extrapolation.

Set $\rho ={\rho }_c{e}^{-\Psi (\xi )}$ and $r=a\xi$. Then $d\ln \rho /dr=-a^{-1}{\Psi }^{\prime }$ and the equation from part (a) becomesWith $a^2=K/(4\pi G{\rho }_c)$, the coefficients cancel, giving the isothermal Lane-Emden equationThe regular central boundary conditions areThe first makes the central mass density equal to ${\rho }_c$. The second follows from smooth spherical symmetry and the vanishing central gravitational acceleration; a preferred radial slope at the origin would be singular. Substitution of a regular power series giveswhich supplies nonsingular starting data for numerical integration. The central conditions select the regular solution, not the singular isothermal sphere.

Interpret ${\sigma }_{\mathrm{DM}}$ as a constant one-dimensional, isotropic velocity dispersion. The effective pressure is $p=\rho {\sigma }_{\mathrm{DM}}^2$, corresponding to the stated gas-temperature analogy. For $\rho =A/r^2$, the spherical enclosed mass isThe two sides of the hydrostatic equilibrium equation are thenEquating them fixes the singular isothermal sphere:For collisionless dark matter this is an isotropic Jeans-equation construction, not a claim that the particles thermalize by gas collisions. The central singularity and infinite total mass show that the untruncated profile is an idealization.

For the singular isothermal sphere, integrate its projected surface mass density inside a circle of proper radius $\xi$:With $\xi =D_d\theta$ this gives the reduced deflection magnitudeindependent of $\theta$. Hence the singular isothermal sphere lens hasThe spherical three-dimensional mass $2{\sigma }^{2}r/G$ would give the wrong coefficient here; the thin gravitational lens equation needs the projected cylindrical mass.

Using the same ${\theta }_E=3\times 10^{-4}$ and $D_{ds}/D_s=0.5$, the one-dimensional velocity dispersion isThus $\sigma \simeq 2100\mathrm{km}{\mathrm{s}}^{-1}$ within the idealized SIS model.

A complete Einstein ring requires an axisymmetric gravitational lens and exact alignment of the source with the lens centre and observer: $\beta =0$. A finite aligned circular source gives a ring of finite width. The background source must lie behind the lens, so the lensing distance factor is positive.

With $\beta =0$ the thin gravitational lens equation givesThe Einstein radius here is angular, and the mass is projected inside the corresponding aperture. For an extended mass distribution the equation is implicit because the enclosed mass depends on ${\theta }_E$.

Using the supplied approximations, $D_{ds}/(D_s{D}_d)=1/(3\times 10^{25}\mathrm{m})$ and ${\theta }_E=3\times 10^{-4}$, the inferred mass isThusThis measures the projected aperture mass without requiring the singular isothermal sphere model used subsequently.

An integral of motion is a function of phase space that is constant along a trajectory. Put $\varphi =-\psi +\mathrm{constant}$, so the stellar acceleration is $\dot{\mathbf{v}}=\nabla \psi$. In a time-independent spherical Newtonian gravitational potential,The second equality uses the radial direction of $\nabla \psi$. Thus both the binding energy and the full specific angular momentum vector are integrals of motion.

The Collisionless Boltzmann equation isFor a stationary galactic distribution function, this says $df/dt=0$ on every characteristic, so $f$ is constant on each stellar orbit. Locally choose a complete set of orbit labels $I_j$, themselves integrals of motion; constancy along the characteristic means $f=F(I_1,\dots ,I_s)$. Conversely, for any such differentiable $F$, the chain rule gives $df/dt={\sum }_j{F}_{,j}{\dot{I}}_j=0$. This proves Jeans theorem in its orbit-invariance form. It does not claim that just two particular integrals classify every orbit in an arbitrary potential. For the specified spherical model, $F(E,L)$ is a rotationally invariant stationary choice.

For the velocity integration, write $v_r=v\cos \eta$, $v_{\perp }=v\sin \eta$, $L=rv\sin \eta$, and set $k=r/r_a$. At fixed position the velocity-dependent factor isUse ${\int }_0^{\infty }{v}^2{e}^{-A{v}^2/(2{\sigma }^2)}dv=\sqrt{\pi /2}{\sigma }^3{A}^{-3/2}$ and integrate over $\chi$. The mass density isSince ${\rho }_c=\rho (0)={\rho }_1{e}^{\psi (0)/{\sigma }^2}$,These integrals use the stated untruncated galactic distribution function over all velocities; an additional escape-energy cutoff would change the result.

In local orthogonal velocity components the same exponent is $-[v_r^2+(1+k^2)(v_{\theta }^2+v_{\varphi }^2)]/(2{\sigma }^2)$. Each component is a centred Gaussian distribution. The one-dimensional identity $\int v^2{e}^{-A{v}^2/(2{\sigma }^2)}dv/\int e^{-A{v}^2/(2{\sigma }^2)}dv={\sigma }^2/A$ yieldsIn particular the velocity-anisotropy parameter is ${\beta }_{\mathrm{an}}=1-({\sigma }_{\theta }^2+{\sigma }_{\varphi }^2)/(2{\sigma }_r^2)=r^2/(r_a^2+r^2)$. This is Gaussian angular-momentum suppression in a stellar distribution.

The radial laws also require self-consistency through the Poisson equation for Newtonian gravity:For $r\ll r_a$, outside a sufficiently small core, the model is almost isotropic and isothermal. Seek its scale-free envelope with $\rho =A/r^2$. Its distribution-density relation gives ${\psi }^{\prime }=-2{\sigma }^2/r$, and the Poisson equation for Newtonian gravity gives $A={\sigma }^2/(2\pi G)$. Thus the intended intermediate singular isothermal sphere behaviour iswhere a core scale is $r_c=\sigma /\sqrt{4\pi G{\rho }_c}$.

The stated $r^{-2}$ law cannot be the actual central limit of a regular finite-density model. For finite ${\rho }_c$ and $\psi (0)$, regularity and the Poisson equation for Newtonian gravity instead giveand $v_c^2=(4\pi G{\rho }_c/3)r^2+O(r^4)$. A broad intermediate $r^{-2}$ region exists only if $r_c\ll r_a$; the radius $r_a$ alone does not ensure it.

For the outer radially anisotropic isothermal halo asymptotics, put $s=\log (r/r_a)$, $y=[\psi (r)-\psi (0)]/{\sigma }^2$ and $\lambda =4\pi G{\rho }_c{r}_a^2/{\sigma }^2$. At $r\gg r_a$ the density relation and Poisson equation becomeThe slow outer balance is $y_s\sim -\lambda e^y$, so $d(e^{-y})/ds\sim \lambda$, giving $e^y\sim 1/(\lambda s)$. Its consistency check is $y\sim -\log (\lambda s)$, for which $y_{ss}/y_s\sim -1/s\to 0$. ThereforeThe reference radius $r_{\ast }$ changes only subleading logarithmic terms. The circular speed falls as $(\log r{)}^{-1/2}$, rather than remaining exactly flat.

Physically, the model has an almost isotropic central region and, if the scales separate, a flat-rotation isothermal envelope. Beyond $r_a$, suppression of large specific angular momentum leaves predominantly radial orbits: radial velocity dispersion remains $\sigma$, while transverse dispersion falls as $r^{-1}$. The outer density acquires a logarithmic suppression and the galaxy rotation curve declines slowly. The enclosed mass still grows approximately as $r/\log r$, so this untruncated model has infinite total mass.

Integrate the dwarf singular isothermal sphere only to its truncation radius:ConsequentlyThe dimensions are $(\text{length})(\text{speed}{)}^2/G$, as required for mass. At fixed host dispersion and separation the mass scales as ${\sigma }_d^3$, rather than the luminosity's fourth-power scaling used later. Under the rough equal-density tidal radius convention, the corresponding mass is $2D{\sigma }_d^3/(G{\sigma }_h)$, larger by $\sqrt{2}$.

An orbital prescription is needed to fix the coefficient in a tidal radius. Adopt a low-mass satellite on a circular orbit of separation $D$, with $r_t\ll D$. The host singular isothermal sphere hasIn the corotating frame, subtract the host force on the satellite centre and add the differential centrifugal force. Along the line joining the centres, the outward effective acceleration to first order in displacement $r$ is $({\Omega }^2-{\Phi }_h^{\prime \prime })r=4{\sigma }_h^{2}r/D^2$. Equate this to the dwarf's inward acceleration at its edge:This gives the Jacobi truncation of two singular isothermal spheresIt is the distance to the radial escape saddle, not an exact spherical surface: the actual tidal lobe is distorted. For an eccentric orbit, separation alone does not specify an instantaneous effective angular speed or the stripping history; a pericentric estimate is commonly used with further assumptions.

A simpler order-of-magnitude convention equates mean internal and host densities, $m_d/r_t^3=M_h(D)/D^3$, giving $r_t\sim D{\sigma }_d/{\sigma }_h$. Equivalently it retains the host gradient without the circular-orbit centrifugal contribution. Both reproduce the scaling $r_t\propto D{\sigma }_d/{\sigma }_h$, but their coefficients differ by $\sqrt{2}$. The familiar point-mass factor three is not appropriate for this extended host, whose enclosed mass grows linearly with radius. The circular-orbit Jacobi tidal radius will be used consistently below, with the rough convention's numerical values also given.

For a singular isothermal sphere, spherical integration of the mass density givesThe spherical shell theorem makes the radial attraction $GM(<R)/R^2$. Balancing it with circular centripetal acceleration yieldsThus the model has a flat galaxy rotation curve, resembling the outer galaxy rotation curves of many spiral galaxies. Central rotation curves usually rise, and not all real curves are flat throughout. A pressure-supported elliptical galaxy does not itself rotate at this test-particle circular speed.

The model has divergent central mass density, although finite enclosed mass at each finite radius, and an infinite total mass because $M(<R)\propto R$. Its potential is logarithmic, requiring an outer cutoff for a finite escape energy. A globally constant isotropic velocity dispersion, spherical symmetry, and one common profile for all gravitating components are further idealizations. Stellar light commonly has a quite different radial profile from the dark matter halo; real disks are flattened, and halos are neither exact isothermal spheres nor infinite. Tidal truncation also changes the distribution near the satellite boundary.

The host mass inside $D$ follows from the same singular isothermal sphere:Using the supplied host luminosity givesThis treats essentially all the stated optical light as contained in that aperture; a measured luminosity profile would refine that convention. Keeping consistent unrounded quantities,Thus the dwarf ratio is about twenty times the parent ratio; the rough equal-density tidal convention would give a factor thirty.

There is no logical contradiction in two objects sharing $L\propto {\sigma }^4$ but having different mass-to-light ratios. A virial theorem scaling gives $M\propto {\sigma }^{2}R$, so that $M/L\propto R/{\sigma }^2$. Equal ratios additionally require homologous relevant radii scaling as $R\propto {\sigma }^2$. Here the satellite's tidal radius scales as ${\sigma }_d$ in the fixed host field, not as ${\sigma }_d^2$. Equating the two ratios would require $r_t=D({\sigma }_d/{\sigma }_h{)}^2\simeq 56\mathrm{pc}$, far smaller than the derived tidal radius.

The heuristic derivation of the Faber-Jackson relation with common mass-to-light ratio and common surface brightness made auxiliary structural assumptions; the empirical relation alone cannot enforce them. Furthermore, the aperture dependence of mass-to-light ratio matters: a satellite's whole tidal mass and a host's $50\mathrm{kpc}$ halo mass are not homologous half-light-aperture masses. Dark matter need not follow stellar light, and a dwarf spheroidal galaxy is not automatically a scaled giant elliptical galaxy. The model calculates total gravitating mass: reading its density literally as stars alone would imply an implausibly large purely stellar ratio, exposing the limits of that interpretation.

A two-body annihilation requires two nearby particles, so the dark-matter annihilation radiation has emissivity proportional to the square of the mass density. With a spatially constant rate coefficient, writefor identical self-conjugate particles, where $E_a$ is the radiated energy per annihilation; the factor one-half counts unordered pairs. Particle and antiparticle populations instead require their respective number densities, without that identical-pair factor. For the singular isothermal sphere, the surface brightness becomesThe substitution $z=R\tan \theta$ gives $R^{-3}{\int }_{-\pi /2}^{\pi /2}{\cos }^2\theta d\theta =\pi /(2R^3)$. Hence annihilation gives the steeper projected profileFor the illustrative cored profile used above, $I_a=C_a{A}^2/[8(R^2+r_c^2{)}^{3/2}]$. These expressions assume a smooth dark matter halo, negligible absorption and constant $C_a$; unresolved density clumps or a velocity-dependent annihilation coefficient change the projected shape. The projection of a spherical power-law emissivity summarizes why three-dimensional powers $r^{-2}$ and $r^{-4}$ produce $R^{-1}$ and $R^{-3}$.