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For a positive diagonal quadratic potential $V=m{\sum }_i{\omega }_i^2{x}_i^2/2$, coordinate Hamiltonians commute and wavefunctions separate into one-dimensional oscillator factors. Equality of two squared frequencies gives rotation symmetry in their plane; equality of all three gives full rotation symmetry. Equalities create energy degeneracies and extra conserved angular momentum. Merely finding an accidental equality between particular level energies does not establish a global rotational symmetry.

Coma makes an off-axis optical point source appear asymmetric, often with a compact head and a flaring tail like a comet. Different parts of the optical pupil send its light rays to different transverse image positions. A sharp central star image therefore does not guarantee sharp images across an optical telescope's field of view. In an idealized, aligned system with rotational symmetry, coma vanishes on the optical axis; misalignment can introduce coma even at the nominal field centre.^[ref]

Coma differs from spherical aberration, whose dependence on optical pupil radius can blur an on-axis point without selecting a transverse direction. Astigmatism gives different optical foci in two perpendicular sections of an off-axis bundle; a through-focus sequence can change its elongated image's orientation. Diffraction persists when these optical aberrations vanish. A real asymmetric point spread function may mix several effects, so its appearance alone is not a unique diagnosis of coma.

The classical primary description belongs to the nineteenth-century Seidel aberration classification of five primary monochromatic image errors.^[ref] It remains useful in modern design: OpticStudio's Seidel Coefficients by ANSYS, Inc. documentation lists coma as $W_{131}$ and tabulates each surface's contribution as well as the system sum (Discussion). Cancellation of contributions can reduce coma without making every surface individually free of it. The same documentation limits this calculation to specified systems with axially symmetric surfaces; higher-order terms and broken rotational symmetry require more general ray tracing.

The twelve rotational symmetries of a tetrahedron act on its six edges. Their edge cycle patterns are six fixed edges for the identity, two three-cycles for each of eight rotational symmetries of order three, and two fixed edges plus two transpositions for each of three half-turns. Burnside's lemma therefore counts edge colourings with $q$ labelled colours by $(q^6+8q^2+3q^4)/12$.

A uniform electric current density outside a parallel cylindrical cavity can be constructed by the superposition principle: add a full cylinder carrying that density and a displaced cylinder carrying its negative. At an exterior point, each contribution is the external magnetic flux density of its own net electric current. The combined magnetic flux density generally lacks rotational symmetry even though Ampère's law fixes its line integral around the whole wire.

Each Killing vector field supplies a conserved quantity for freely moving test particles: it turns a geometric symmetry into a first integral of the geodesic equation. Let $\gamma (\lambda )$ be a geodesic with affine parameter $\lambda$, tangent vector $U^{\mu }=d{x}^{\mu }/d\lambda$, and ${\nabla }_{U}U=0$. Wherever a Killing vector field $K$ is defined along the geodesic,The statement applies to Riemannian manifolds and to timelike geodesics, null geodesics and spacelike geodesics in a Lorentzian manifold. It requires an affine parameter; for a massive test particle, proper time is one choice. See Tong, section 4.3.2, equation (4.32).

For a worked example in the Euclidean plane, take $\gamma (t)=(1+t,2-t)$ with affine parameter $t$ and velocity $U=(1,-1)$. The Euclidean Killing vector fields for translation and rotation giveThe rotational field along the geodesic is $K_{\mathrm{rot}}(\gamma (t))=(t-2,1+t)$: it changes from $(-2,1)$ at $t=0$ to $(-1,2)$ at $t=1$, yet its pairing with $U$ stays $-3$. With toy mass $m=1$ and $t$ interpreted as time, the first constant is the $x$ component of linear momentum, and the second is the signed $z$ component of angular momentum about the origin. Their numerical values use toy units. The geodesic is straight; the rotation flow orbits are circles.

This conservation law follows from metric compatibility and the Killing equation:The geodesic equation kills the acceleration term, and the symmetric product $U^{\mu }{U}^{\nu }$ selects the vanishing symmetric derivative. This tensor contraction is the whole proof outline; see Carroll, equation (5.43).

In a stationary spacetime, a chosen time-translation Killing vector field $K$ gives Killing energy. With metric signature $(-+++)$, $c=1$, and a massive test particle's four-momentum $p^{\mu }=m{U}^{\mu }$, the physical normalization isHere $\Phi$ generates the axial rotational symmetry when one exists. In coordinates with $K={\partial }_t$ and $\Phi ={\partial }_{\varphi }$, these are $E=-p_t$ and $L_z=p_{\varphi }$. Both are conserved quantities along the geodesic, even if the metric has a time-angle cross term. A future timelike $K$ makes $E$ positive for a future massive particle; the scale of $E$ depends on the normalization of $K$. Not every spacetime has a global timelike Killing vector field, and a stationary field can become spacelike in an ergoregion of a stationary spacetime. Conservation on the field's domain does not imply a universal global positive energy; see Tong, sections 4.3.1-4.3.2 on the timelike-domain qualification and conserved charges.

The Landau-de Gennes free energy is a rotational symmetry preserving Landau free energy of the nematic order parameter. One three-dimensional normalization through fourth order isThe cubic invariant generically produces a first-order phase transition. Its absence in two dimensions is an algebraic identity, not a guarantee that a mean-field approximation captures all fluctuations.

For point masses on concentric horizontal rings in parallel planes, set $I_1=m{a}^2$, $I_2=M{A}^2$, and $\delta =\theta -\phi$. A spring of zero natural length gives $V=\text{constant}-kaA\cos \delta$. The inertia-weighted angle $q=(I_1\theta +I_2\phi )/(I_1+I_2)$ separates the Lagrangian into a free common rotor with inertia $I_1+I_2$ and a relative rotor with inertia $I_r=I_1{I}_2/(I_1+I_2)$. Static equilibria have ${\delta }_0=0$ or $\pi$. Their normal modes are the common displacement $(1,1)$ with zero frequency and the relative displacement $(I_2,-I_1)$ with the displayed squared frequency. The aligned family has a restoring relative mode; the antipodal family has an exponentially growing one. The zero mode is required by rotational symmetry. Small nonzero total angular momentum produces common drift, so stability of the aligned family modulo rotations should be distinguished from Lyapunov stability of one fixed equilibrium point.

Parity violation means that a physical law fails to be invariant under the spatial reflection represented by the parity operator. A process and its fully inverted counterpart can then have different probabilities. For beta-minus decay with polarized parent atomic nuclei, a nonzero correlation between nuclear spin and emitted electron momentum is such a test. Rotational symmetry alone allows this correlation: a pseudoscalar is unchanged by proper rotations but changes sign under spatial reflection.

Work in the laboratory frame, with the fluid at rest at infinity. SetAt $r=a$, $f=1$ and $g=0$, so the no-slip boundary condition is $\mathbf{u}=\mathbf{U}$ on the translating sphere. As $r\to \infty$, $\mathbf{u}=O(r^{-1})\to 0$ and the gauge-fixed pressure tends to zero.

It is also possible to check the exterior Stokes equations directly. Differentiating the radial functions givesandThus the Newtonian fluid stress tensor $\sigma =-pI+\mu (\nabla \mathbf{u}+\nabla {\mathbf{u}}^T)$ satisfies $\nabla \cdot \sigma =0$ throughout $r>a$.

Let $\mathbf{n}$ point away from the common centre on each concentric sphere. Applying the divergence theorem to the intervening fluid shell, whose inner normal is $-\mathbf{n}$, yieldsThis is the force flux at infinity in Stokes flow. It is legitimate even though the stress tensor tends to zero pointwise, because the area of the large sphere grows as $R^2$.

For an explicit evaluation, put $A=3a/4$ and $\mathbf{n}=\mathbf{x}/R$. The leading terms are $\mathbf{u}=A[\mathbf{U}+(\mathbf{U}\cdot \mathbf{n})\mathbf{n}]/R$ and $p=2\mu A(\mathbf{U}\cdot \mathbf{n})/R^2$. Their velocity gradient isAfter symmetrizing and subtracting the pressure contribution,By rotational symmetry, $\int n_i{n}_{j}d\Omega =(4\pi /3){\delta }_{ij}$. HenceThis is Stokes's law, with the force on the sphere opposite its velocity. For a component calculation with $\mathbf{U}=U{\mathbf{e}}_z$, the same integral is $F_z=-(9\mu aU/2)\int {\cos }^2\theta d\Omega$.

Write ${ϵ}_k={\hslash }^2{k}^2/(2m)$, $E_k=\hslash \omega$, and take $\mu >0$. The Bogoliubov spectrum is gapless and positive for $k\ne 0$. Its small- and large-momentum limits areThe low-energy excitations are collective sound modes; high-energy excitations approach particles with a mean-field energy offset. With healing length $\xi =\hslash /\sqrt{2m\mu }$ and $q=|k|\xi$,The crossover occurs at $q$ of order one, not at an additional energy gap.

The critical velocity of a homogeneous condensate follows because the Landau criterion compares an excitation energy with the energy that flow can release by transferring its momentum. Creating a mode is energetically possible when $E_k-\hslash \mathbf{k}\cdot \mathbf{v}\le 0$. For this isotropic spectrum,The infimum is the long-wavelength speed of sound. This tests energetic stability against the stated excitations; it is not a calculation of a dissipation rate or a universal vortex-nucleation threshold.

For an external potential varying slowly on the healing scale, the Thomas–Fermi approximation for a condensate neglects density-gradient kinetic energy in the stationary Gross–Pitaevskii equation. In occupied regions it givesHere $\rho$ is the appropriate local number density; in the ring it is a linear density and $g$ is the one-dimensional coupling. This is the local-density sound speed of a condensate. The formula applies where $\mu >V$, not in an empty region where its square root would be imaginary.

Normalize the condensate order parameter to the atom number:up to an arbitrary constant phase. A wavefunction normalized to one instead has amplitude $1/\sqrt{2\pi R}$, with $N|\psi {|}^2$ then providing density. Acting with $L_z=-i\hslash {\partial }_{\theta }$ gives angular momentum $n\hslash$ per atom and $Nn\hslash$ in total for the uniform ring. The superfluid velocity isThe interaction chemical potential is ${\mu }_0=gN/(2\pi R)$; a flowing uniform state also has the constant kinetic contribution ${\hslash }^2{n}^2/(2m{R}^2)$ to its total chemical potential. Its speed of sound is set by the interaction term. Applying $|v|<\sqrt{{\mu }_0/m}$ givesFor a uniform ring condensate, this is the continuum phonon criterion used here. Quantization of the available excitation wavelengths in a very small ring can modify a finite-size threshold.

For the nonrotating equilibrium with a raised left half, let ${\rho }_L$ and ${\rho }_R$ be the bulk linear densities. Within the same Thomas–Fermi approximation for a condensate,When both halves are occupied, normalization fixesand thereforeThese expressions require $V_0<2{\mu }_0$. Beyond that threshold, extending them would give a negative left density and is unphysical. Instead the left bulk empties, all $N$ atoms occupy a segment of length $\pi R$, andAt equality the left bulk just reaches zero density. kinetic energy rounds the interface and produces tails beyond the Thomas–Fermi approximation for a condensate, so an exactly empty half is the idealized bulk result.

A nonconstant $V(\theta )$ breaks rotational symmetry:Consequently angular momentum is no longer a good quantum number, and an eigenfunction need not be $e^{in\theta }$ times a constant amplitude. Nevertheless, while the order parameter is nonzero around the whole ring, single-valuedness requiresThe integer is a phase winding of a condensate, not an assertion that the inhomogeneous state remains an angular-momentum eigenstate. A superfluid phase slip can change it if the amplitude vanishes locally.

Conservation of particles gives ${\partial }_t\rho +{\partial }_s(\rho v)=0$. For steady density the one-dimensional particle current $j=\rho v$ is constant. Since the two halves have equal lengths,The current bottleneck in an inhomogeneous condensate occurs because a lower-density region has faster flow at the same current; it also has a smaller speed of sound. This makes the raised half the limiting region.

For $V_0=2{\mu }_0/3$ the equilibrium approximation givesThe winding condition and equal lengths requireThus the left-hand sound-speed test isThe right-hand condition is only $n^2<3m{R}^2{\mu }_0/{\hslash }^2$, so it is less restrictive.

This local sound-speed estimate for ring-flow stability uses the step-potential equilibrium in a ring condensate densities together with the local sound-speed test, as suggested by the preceding approximations. A quantitatively self-consistent finite-current solution also includes $m{v}^2/2$ and the quantum pressure of a condensate in the stationary Gross–Pitaevskii equation, allowing flow to change the density. Smooth steps suppress reflection but do not alone remove those terms. The $3/8$ result is the stated local-density estimate; it should not be promoted to an exact stability theorem for every finite interacting ring.

Put the stagnation point at the origin and the local tangent plane at $z=0$. A regular Taylor expansion has no linear term because $\mathbf{u}(0)=0$. The impermeable tangent plane imposes ${\partial }_z\varphi =0$ at $z=0$ to leading order, excluding mixed quadratic terms $xz,yz$. For the illustrated axisymmetric approach, rotational symmetry in the tangent plane makes the $x^2$ and $y^2$ coefficients equal and removes $xy$. Finally the Laplace equation fixes the normal coefficient to minus twice the tangential one. Up to an irrelevant additive constant,For incoming flow from $z>0$, $W>0$. This is a local leading form under the stated symmetry, not the exact universal potential near every possible wall stagnation point; a generic tangential strain can have unequal principal rates.

A streamline solves $\dot{x}=2Wx$, $\dot{y}=2Wy$, $\dot{z}=-4Wz$, hence $x=A{e}^{2Wt}$, $y=B{e}^{2Wt}$, $z=C{e}^{-4Wt}$. Eliminating the parameter givesEach streamline lies in a meridional plane. With cylindrical radius $s=\sqrt{x^2+y^2}$, its shape is $z=C_1/s^2$. The symmetry axis and wall are limiting invariant streamlines. A particle exactly on the incoming axis approaches the stagnation point asymptotically.

Let $f(r)=1-2m/r$ and work in the exterior $r>2m$, with $G=c=1$. The affinely parametrized geodesic Lagrangian isIts polar-angle equation isThe initial data $\theta =\pi /2$, $\dot{\theta }=0$ therefore remain in that plane. Rotational symmetry allows the axes to be chosen so that any nonradial initial trajectory has those data: its conserved angular-momentum direction fixes an orbital plane through the centre. Radial trajectories can also be placed in an equatorial plane.

The coordinates $t,\varphi$ are cyclic in $L$, so their canonical momenta are constant:The corresponding Killing vector fields are ${\partial }_t$ and ${\partial }_{\varphi }$. With the displayed signature, $g({\partial }_t,\dot{x})=E$ and $g({\partial }_{\varphi },\dot{x})=-J$, explaining the sign in the second canonical momentum. Thus $E$ is conserved energy per unit rest mass for a massive particle parametrized by proper time, and $J$ is its axial angular momentum per unit rest mass. For a photon, affine rescaling rescales both constants; after the tangent is normalized to the physical four-momentum they represent its energy at infinity and axial angular momentum. Their ratio $b=J/E$ is the signed impact parameter.

In the equatorial plane, the photon satisfies the null geodesic constraintwhich givesFor $J\ne 0$, put $u=m/r$ and $\alpha =Em/J$. Thenand consequentlyThe $J=0$ radial case is not represented by $\varphi$ as an orbit parameter.

There is a circular photon orbit at $r=3m$, but it is unstable. The radial effective potential isA circular orbit requires $V^{\prime }(r)=0$ and $E^2=V(r)$, givingIts second derivative isIt is a maximum, rather than a minimum, so radial perturbations depart from the orbit. Equivalently the second-order orbit equation is $u^{\prime \prime }=3u^2-u$. Linearizing about $u_c=1/3$ gives $\delta u^{\prime \prime }=\delta u$, with modes $e^{\varphi }$ and $e^{-\varphi }$. The existence of a growing mode directly contradicts the printed stability claim. Exact initial data can remain circular, but that does not make the orbit stable.

The supplied implicit expression describes the critical-energy family with ${\alpha }^2=1/27$, as fixed by the circular orbit. DefineRationalizing once gives $F=(\sqrt{3}-s)/(2(\sqrt{3}+s))$ andDifferentiating $F(u)=A{e}^{\varphi }$ gives $F^{\prime }{u}^{\prime }=F$, soThis verifies the first-order equation for every allowed integration constant $A$ on its real branch. For $A=0$ the derivative argument is replaced by direct substitution: $u=1/3$ is constant and solves the critical equation. The displayed family does not solve the equation at an independently arbitrary value of $\alpha$.

For $A>0$, write $q=A{e}^{\varphi }$. On the physical exterior branch,As $\varphi \to -\infty$, $q\to 0^{+}$ andThus for $A=0$ the photon remains exactly on the circular orbit. For $A>0$ it approaches $r=3m$ from outside through indefinitely many revolutions in the negative-$\varphi$ limit. With increasing $\varphi$, this same branch peels away outward and reaches infinity when $q=1-\sqrt{3}/2$. Reversing the direction of motion gives an incoming critical ray asymptotically approaching the photon circle. These are separatrix trajectories, not evidence of stability.

The potential maximum and the outward-growing radial separation display the same instability in two complementary descriptions.

A motif can have fivefold symmetry. For example, repeat an identically oriented regular pentagonal molecular motif at every point of a square translation lattice. Each isolated motif has fivefold rotational symmetry, while the crystal structure as a whole does not: rotating about one motif fails to map the other motif centres onto lattice points. The symmetry of an atomic basis of a crystal need not be a symmetry of its Bravais lattice.

The material derivative follows surfactant carried by the moving interface. Tangential surface divergence changes the local surface area and hence dilutes or concentrates insoluble material. Normal motion changes area through the curvature: an expanding sphere has positive $u_n{\nabla }_s\cdot \mathbf{n}$ and its concentration decreases. The last term is surface diffusion, which smooths concentration gradients. There is no exchange with the bulk; this is conservation of insoluble surfactant on a moving interface.

The interfacial speed is of order $a|\mathbf{F}|$. Diffusion dominates redistribution when the surface Péclet number satisfiesThen $C^{\prime }/C_0=O({\mathrm{Pe}}_s)$, and advection of $C^{\prime }$ is a second-order effect. Write $S=\mathbf{n}\cdot \mathbf{Fn}$. Since ${\mathbf{u}}_s=\mathbf{Fx}-S\mathbf{x}$, the projected divergence of the first term is $\mathrm{tr}\mathbf{F}-S=-S$, whereas the second has divergence $2S$; its gradient contracted with $\mathbf{x}$ is zero. Hence ${\nabla }_s\cdot {\mathbf{u}}_s=-3S$.

The leading steady transport equation is $D_s{\Delta }_s{C}^{\prime }=C_0{\nabla }_s\cdot {\mathbf{u}}_s=-3C_{0}S$. Trace-freeness and the supplied surface Laplacian identity give ${\Delta }_{s}S=-6S/a^2$. Fixing the conserved total surfactant removes the uniform null mode and yieldsThis is quadrupolar surfactant distribution on a spherical interface; its spherical mean is zero because $⟨n_i{n}_j⟩={\delta }_{ij}/3$.

With $+$ outside, $-$ inside, and the normal directed out of the bubble, the interfacial stress balance with variable surface tension isNow ${\nabla }_{s}S=2I_s\mathbf{Fn}/a$, so $-{\nabla }_s\gamma =2A{\gamma }_1{I}_s\mathbf{Fn}/a$. Multiplying the curvature by the tension and retaining linear terms givesThe undeformed isotropic pressure jump is kept with its normal; the perturbation terms are evaluated on the reference sphere. Products of deformation and concentration perturbations are beyond the approximation.

The imposed tensor is symmetric and traceless, so $\mathbf{x}\cdot \mathbf{Ex}$ is a harmonic scalar of degree two. Its decaying partner is $\mathbf{E}:\nabla \nabla (1/r)$; the corresponding harmonic vector uses $\mathbf{E}\nabla (1/r)$. Linearity and rotational symmetry give precisely these tensorial modes, with the growing scalar chosen to reproduce the imposed strain. Thus in the unscaled convention one can takeThe coefficients multiply distinct decaying modes. No other angular harmonic is forced at this order.

No penetration in the steady state gives $P=3Q-1$. The given tangential surface velocity implies $\alpha =1+2Q$. The tangential stress balance, with $M=A{\gamma }_1/(\mu a)$, gives $1+P-8Q=M\alpha$, or $-5Q=M(1+2Q)$. ThereforeThe nonuniform normal stress is unaffected by the uniform inviscid interior pressure. Its balance isSubstitution and rearrangement giveAs $M\to \infty$, $\alpha \to 0$: the concentration-induced Marangoni stress suppresses tangential surface motion, making the exterior boundary behave like an immobile surface despite the inviscid interior. The deformation coefficient tends to $5\mu a/(2{\gamma }_0)$; the clean limit is $2\mu a/{\gamma }_0$. Both small surface Péclet number and small deformation, of order $\mu a|\mathbf{E}|/{\gamma }_0$, are required. This is Marangoni immobilization of a bubble in straining flow.

The normal form is invariant under $z\mapsto e^{i\phi }z$. This continuous rotational symmetry makes the axis invariant and reduces the nontrivial geometry to two amplitude variables. The axis connection from $E_{+}$ to $E_{-}$ persists without parameter tuning, while rotating the outer connection produces a family of connections from $E_{-}$ back to $E_{+}$. A two-dimensional autonomous amplitude flow cannot by itself produce the proposed saddle-focus return chaos: symmetry reduction hides the phase-dependent reinjection needed for that mechanism.

Generic higher-order terms need not retain this exact rotational symmetry. A normal form can preserve it to a finite truncation order while the full equations have small symmetry-breaking remainders. These unfold the special connection geometry into phase-dependent returns; nearby parameter curves can contain homoclinic orbits to a saddle-focus, and the Shilnikov bifurcation mechanism becomes available. Their exact placement depends on the omitted terms and is not fixed by the displayed truncation alone.

For example, at the perturbed positive-axis saddle, the real unstable eigenvalue is ${\lambda }_u=2a$ and the stable complex pair has real part ${\mu }_1-2a+a^2$. Near the calculated connection curve ${\mu }_1=-a^2/4+o(a^2)$, the contracting ratio isThis is the positive-saddle-value regime, explaining why the relevant saddle-focus return can be expanding.

In the Shilnikov return map, take a positive return coordinate $y$ and $0<\delta <1$, with nondegenerate $A\ne 0,B\ne 0$. At the connection parameter $\mu =0$, the fixed point condition isThe right side tends to zero, whereas the phase passes through infinitely many oscillations as $y\downarrow 0$. Hence there are infinitely many positive fixed points accumulating at zero, with logarithmically spaced amplitudes. For these points,whose magnitude tends to infinity along the accumulating fixed-point sequence. They represent long-period saddle periodic orbits of the full return, with period growing like ${\lambda }_u^{-1}\log (1/y)$, rather than infinitely many stable cycles.

Successive turning branches of this oscillatory return can stretch and fold across suitable intervals. The full saddle-focus Poincaré return map then admits invariant Smale horseshoes and chaotic itineraries near a generic homoclinic orbit at a Shilnikov bifurcation. Varying $\mu$ unfolds accumulating saddle-node and period-doubling events. Whether any resulting invariant set attracts trajectories depends on the other return direction and the omitted terms; the leading scalar map alone does not prove a chaotic attractor for every perturbation. Its domain also excludes negative iterates, which leave the chosen return branch. The degeneracies $A=0$ or $B=0$ would remove the oscillatory mechanism, so the conclusion uses generic, genuinely saddle-focus reinjection.

Take $V=|\mathbf{V}|$ to be the test body's instantaneous speed relative to the nonstreaming background, rather than an ensemble root mean square speed. Use a local isotropic galactic distribution function $f(v)$ normalized as a number distribution, and assume independent encounters and a large, approximately constant Coulomb logarithm in stellar dynamics.

For a background particle with relative incoming velocity $\mathbf{w}=\mathbf{V}-\mathbf{v}$, a gravitational scattering angle $\chi$ obeysIn the centre-of-momentum frame, the test body's share of the relative velocity is $m/(M+m)$. The azimuthal average cancels transverse changes, leaving a change parallel to the incoming relative velocity,The number of encounters per unit time in $d^{3}v$ and impact interval $db$ is $f(v)d^{3}vw2\pi bdb$. Integrating the longitudinal change from $b=0$ to $b_m$ givesAt leading logarithmic order replace the logarithm by $2\ln \Lambda$, evaluated at a representative speed. For isotropic $f$, the velocity-shell identity for dynamical friction isOne proof takes the divergence with respect to $\mathbf{V}$: the inverse-square vector field has divergence $4\pi {\delta }^{(3)}(\mathbf{V}-\mathbf{v})$, whose velocity integral is $4\pi f(V)$, and rotational symmetry plus the divergence theorem determines the radial field. Exterior velocity shells give zero field. Consequently the systematic Chandrasekhar dynamical friction is antiparallel to motion, withIn the massive-test-body approximation $M\gg m$, this becomesThe finite-test-mass dynamical friction coefficient shows the needed qualification: $M>m$ alone does not make $M+m=M$. The quoted expression also neglects diffusion and the velocity dependence of the logarithmic cutoff. For the lowered Maxwellian velocity distribution, set $f=0$ for $v>v_e$; continuing the printed subtraction beyond escape would give a negative distribution. If $V>v_e$, the integral stops at $v_e$.

Substituting $\eta =D_s\beta$ and $\xi =D_d\theta$ into the previous geometric relation verifiesFor a point mass gravitational lens, the given physical deflection yieldsAn aligned source has $\beta =0$, and therefore ${\theta }^2={\theta }_E^2$. Rotational symmetry turns these signed intersections into an Einstein ring, whose Einstein radius isFor an offset source, multiplying $\beta =\theta -{\theta }_E^2/\theta$ by $\theta$ gives a quadratic equation. Its two image positions areFor $\beta >0$, ${\theta }_{+}>0$ lies on the source side of the lens and ${\theta }_{-}<0$ lies on the opposite side.

Use the homogeneous equilibrium point condition $k({\rho }_0){\rho }_0=a_{0}f({\rho }_0)$. Evaluate $D_1,D_2$ at $(a_0,{\rho }_0)$ and abbreviateThe quantity $c$ is the chemical relaxation rate with the cell density held fixed. Derivatives of $D_1,D_2$ do not enter this linear stability analysis: they multiply spatial derivatives of the homogeneous background or products of perturbations. For a Fourier mode with time dependence $e^{st}$, put $z=|\mathbf{q}{|}^2$ and writeConsequentlyAt $z=0$, the eigenvalues are $0$ and $-c$. The neutral cell-density mode reflects mass conservation, not decay of every homogeneous perturbation.

Write $a=D_2>0$, $b=D_{\rho }>0$, $C=D_1{f}_0$, and interpret production physically as $f_0\ge 0$. If $c\ge 0$, the trace-determinant stability criterion shows that an unstable nonzero wavenumber exists precisely when $C>ac$. The unstable band isIf $c<0$, the chemical field is already unstable at zero wavenumber; arbitrarily small positive wavenumbers are unstable too. Since $C\ge 0$, this also implies $C>ac$. Thus, on the infinite plane, the undivided Keller--Segel aggregation threshold isFor $\kappa >0$, dividing by $D_2\kappa$ gives the requested formAt equality there is no strictly growing mode when $c\ge 0$; nonzero spatial modes decay. On a finite domain, a permitted nonzero wavenumber must actually lie in the unstable band. If one allows a signed production function, the displayed matrix and trace-determinant stability criterion remain valid, but the simplification using $C\ge 0$ must be revisited.

The printed hypotheses do not ensure $\kappa >0$. Positivity of the degradation rate $k$ alone is insufficient. For a concrete counterexample, take $k(\rho )=e^{-2\rho }$, $f(\rho )=e^{-2}$, $a_0={\rho }_0=1$, and $D_1=D_2=D_{\rho }=1$. The homogeneous equilibrium point condition holds, all rates and transport coefficients are positive, but $\kappa =c=-e^{-2}$. The printed left-hand side equals $-1$, although $\det M=z(z-2e^{-2})<0$ for $0<z<2e^{-2}$. These spatial modes grow. If $\kappa =0$, the printed expression is undefined. The undivided criterion and the matrix above resolve both cases.

To find the fastest-growing Keller--Segel mode, use the larger eigenvalueFor $C>0$ the square root is real. Put $p=a+b$ and $d=b-a$. Differentiation givesA positive maximizing wavenumber exists in either of two growing cases: $c\ge 0$, $C>ac$, or $c<0$, $C>-bc$. Equivalently, $C>\max (ac,-bc)$. In this range $C+dc>0$, andThus the stationary point is the unique maximum. The derivative condition yieldsSolving it, with the root that satisfies the unsquared derivative equation, givesFor $c=0$, the unequal-diffusivity expression simplifies to $z_{\ast }=C/[\sqrt{ab}(\sqrt{a}+\sqrt{b}{)}^2]$, also agreeing with the equal-diffusivity limit. The formula maximizes the full two-field growth rate; it makes no instantaneous-chemical approximation. Every direction of $\mathbf{q}$ with this length is equivalent by rotational symmetry.

If $c<0$ but $0\le C\le -bc$, the maximum instead occurs at $z=0$: the fastest mode is homogeneous, with $s_{\ast }=-c$ and infinite wavelength. For $C>0$, its initial derivative is $-b+C/(-c)\le 0$. If $C+dc\ge 0$ the derivative thereafter decreases, while if $C+dc<0$ it increases towards the still-negative large-$z$ limit; either way no positive-$z$ maximum is missed. At $C=0$, the two eigenvalues are simply $-az$ and $-c-bz$, giving the same conclusion. In a finite box, maximize $s_{+}$ over the allowed Fourier modes; excluding the homogeneous mode can change the selected length. In a stable parameter range there is no fastest-growing mode.