Set ${\mathbf{b}}_0=\mathbf{c}+h\mathbf{n}$, where $h=p-\mathbf{n}\cdot \mathbf{c}$. Then $\mathbf{n}\cdot {\mathbf{b}}_0=p$, so ${\mathbf{b}}_0$ lies in the plane. For any position $\mathbf{r}$ in that plane, $\mathbf{r}-{\mathbf{b}}_0$ is perpendicular to ${\mathbf{b}}_0-\mathbf{c}=h\mathbf{n}$. The Pythagorean theorem therefore givesOn the sphere the left side is $a^2$. Thus the intersection is the circle in the plane centred at ${\mathbf{b}}_0$, with squared radius $a^2-h^2$. Its centre must therefore be $\mathbf{b}={\mathbf{b}}_0$, andIn terms of the radii, $\lambda =\pm \sqrt{a^2-{\rho }^2}$, with the sign fixed by $\lambda =p-\mathbf{n}\cdot \mathbf{c}$. In the central-plane case $\rho =a$, both signs give zero. This is the center and radius of a sphere-plane intersection formula with its orientation made explicit.

The vector $\mathbf{r}-(\mathbf{r}\cdot \mathbf{u})\mathbf{u}$ is the perpendicular component of $\mathbf{r}$ relative to the axis $\mathbb{R}\mathbf{u}$, so its length is the distance from that axis. Hence the surface is an infinite circular cylinder of radius $\beta$ about the line through the origin parallel to $\mathbf{u}$. Here $\beta$ is the radius and $\mathbf{u}$ fixes the axis direction; reversing $\mathbf{u}$ leaves the same circular cylinder.

If ${\mathbf{v}}_1,{\mathbf{v}}_2$ are an orthonormal basis of the plane perpendicular to $\mathbf{u}$, a parameterization isThis verifies that all heights and all angular positions, but no interior points, belong to the surface.

For $\mathbf{r}\ne 0$, the dot product equation says that the angle $\theta$ between $\mathbf{r}$ and the oriented unit vector $\mathbf{u}$ satisfies $\cos \theta =\alpha$. Thus $\theta =\arccos \alpha$ is fixed. More explicitly, using the orthogonal projection from the introductory solution,Consequently the surface is one infinite nappe of a right circular cone, with apex at the origin, including the apex itself. The cone's axis points along $\mathrm{sgn}(\alpha )\mathbf{u}$, and its acute semi-angle is $\arccos |\alpha |$. Equivalently, $\alpha$ is the cosine of the angle from the oriented $\mathbf{u}$ axis. Squaring the original equation without retaining the sign of $s$ would incorrectly add the opposite nappe of a double circular cone. No base disk is included in this infinite lateral surface.

If $v=\pm u$, then $Au=u$. On the perpendicular plane, $u{u}^T$ vanishes and $[v{]}_{\times }$ is a signed right-angle rotation, whose square is minus the identity. Hence its complex eigenvalues are $\pm i$, and

If $u\cdot v=0$, use the positively oriented orthonormal basis $(u,v\times u,v)$. The cross product identities give $Au=u+v\times u$, $A(v\times u)=-u$ and $Av=0$. In this basis,Its characteristic polynomial is $\lambda ({\lambda }^2-\lambda +1)$, so

Gravitational lensing probes the cluster's projected mass through the deflection of background light. Strong gravitational lensing produces arcs and multiple images; weak gravitational lensing measures coherent distortions of many background galaxies. The lensing convergence is $\kappa =\Sigma /{\Sigma }_{\mathrm{crit}}$, withThus the image geometry and shear constrain the projected surface density $\Sigma$ without assuming the cluster is in dynamical or hydrostatic equilibrium. The reconstructed total mass again exceeds the visible stellar and gas contribution. Source-distance errors, line-of-sight structures and the mass-sheet degeneracy affect the reconstruction, but the independent agreement of lensing, galaxy motions and gas confinement is strong evidence for cluster dark matter.

The hot intracluster gas emits X-rays, which constrain its temperature and density profile. If its pressure is predominantly thermal and it is in hydrostatic equilibrium,Eliminating the pressure givesHere $\mu m_p$ is the mean mass per gas particle. A gas at the observed cluster temperatures would not remain confined without a deep gravitational potential; the inferred total mass exceeds the gas and stellar mass together. Nonthermal pressure, mergers, gas clumping and temperature-profile errors can bias this estimate, but do not turn X-ray luminosity itself into a direct measurement of total mass.

The three-gamma matrix $C{\gamma }^{mnp}$ is antisymmetric, and for three distinct indices it is nonzero. A pair of nonzero Grassmann components therefore produces a nonzero term exactly as in the scalar case. Consequently this three-form bilinear is not identically zero. Equivalently it is dual to the axial-vector Majorana spinor bilinear, which is allowed; it must not be confused with the vanishing vector in part (a).

Four-dimensional Hodge duality identifies ${\gamma }^{mn}{\gamma }_5$ with a nonzero convention-dependent multiple of ${ϵ}^{mnpq}{\gamma }_{pq}$. Every $C{\gamma }_{pq}$ is symmetric by the four-dimensional Majorana bilinear vanishing criterion; taking this linear combination preserves symmetry. HenceThe phase in the duality formula depends on the definition of ${\gamma }_5$ and orientation, but does not affect the vanishing result.

The matrix $C$ is antisymmetric, so the scalar Majorana spinor bilinear is not identically zero. For example, choose two independent odd Grassmann variables ${\lambda }_{\alpha },{\lambda }_{\beta }$ at a pair of indices with $C_{\alpha \beta }\ne 0$. Their contribution is $2C_{\alpha \beta }{\lambda }_{\alpha }{\lambda }_{\beta }$, which is nonzero as a Grassmann polynomial. This is why a Majorana mass term is possible.

A classical sequential probability ratio test addresses a fixed-start choice between two simple statistical hypotheses. It sums log-likelihood ratio increments until an upper boundary accepts the alternative or a lower boundary accepts the null; it then terminates that test. A CUSUM instead monitors indefinitely for a change at an unknown time. Its reset at zero discards sustained evidence from a previously satisfactory period, so a late change need not first undo all earlier negative scores. An upper boundary $h$ signals when the accumulated change evidence is sufficiently large.

A fast initial response CUSUM takes $0<X_0<h$, often $X_0=h/2$, in place of a zero start. Until its first reset or signal, it acts like a fixed-start sequential probability ratio test with a lower boundary at $-X_0$ and an upper boundary at $h-X_0$ for the subsequent cumulative increments. Upon reaching the lower boundary it resets and continues as an ordinary CUSUM. The head start gives faster detection if a change is already present when monitoring begins or restarts. It is useful after an intervention, a shutdown or a previous alarm when immediate renewed surveillance is desired. Its initial false-alarm behavior differs from the zero-start scheme, so the head start and signalling boundary must be calibrated together; the preceding numerical choice is a design convention, not a universal optimum.

Let $p_1$ be the alternative death probability. The odds ratio specifiesAssume $0<p_0<1$ and $k>0$. For observed batch count $y_t$, the common combinatorial factor in the two binomial likelihoods cancels, giving the CUSUM incrementThereforeFor $k>1$ large death counts add positive evidence of deterioration; for $k<1$ unusually small death counts add evidence for improvement. The score has nonpositive expectation under the null and nonnegative expectation under the specified alternative, as follows from the nonnegativity of Kullback-Leibler divergence. Independent, nonoverlapping batches avoid counting the same patient's outcome repeatedly.

The same pricing relation gives $x^{\top }{A}_0\ge 0$ whenever the terminal payment is nonnegative. Therefore negative initial cost also contradicts a strictly positive state-price density, even if the terminal payment is identically zero.

This second form is needed for directions invisible in the terminal payments. Let $L$ be the linear span of the essential values of $A_1$, equivalentlyIf $A_0\notin L$, let $v$ be its nonzero orthogonal projection on $L^{\perp }$ and set $x=-v$. Then $x^{\top }{A}_1=0$ almost surely while $x^{\top }{A}_0=-\Vert v{\Vert }^2<0$. Hence the second form exists in this case. Part (b) handles all remaining cases.

It is helpful to solve using the nonsingular total flux $J$ and volume flux $Q$ before recovering $M$. The relation $J=M/(1-Q^2/(R^{2}M))$ is quadratic:The plus sign selects the small-area pure-source branch. Its discriminant vanishes when $J=4Q^2/R^2$, givingThereforeThe specific upward plume and ambient momentum-flux contributions are equal, not opposite. The ambient's transported momentum has the opposite sign from its transported volume, so their product remains positive.

To show that the pure branch actually reaches this fold in finite height and that it is a genuine pole, introduceThen $s=q^2/m$ andThe pure-source condition is $j\sim (5/8{)}^{2/5}{q}^{4/5}$. Before the fold, $s<1/2$ and $m>2q^2$. Thus for every fixed positive $q_1$, the derivative $dj/dq$ is bounded above by a constant times $q^{-2}$ on $q\ge q_1$. If the branch continued for arbitrarily large $q$, $j$ would remain bounded, contradicting $j\ge 4q^2$. Hence it reaches $j=4q^2$ at a finite $q_o>0$. Its height is finite because $d\zeta /dq=(1-s)/(2\sqrt{m})$ is finite away from the source and behaves like a constant times $q^{-2/5}$ at the source, an integrable singularity.

For the numerator, write $m=j(1-s)$. ThenAt the fold $j=4q_o^2$, so $q_o^3\ge 5/256$. On the other hand the dimensionless numerator in the $m$ equation at $s=1/2$ isIt is strictly negative because $5/256>1/(64\sqrt{2})$. The denominator $1-2s$ approaches zero from above. Consequently $dM/dz\to -\infty$ at finite $z_o$, although the fluxes and velocities themselves remain finite. This proves the confined-plume momentum fold rather than merely noticing a possible zero denominator.

The dimensionless pure-source branch uniquely determines the remaining amplitudes. They can be specified without an elementary antiderivative by integrating the displayed scalar equation for $j(q)$ until $j=4q^2$, and settingA direct numerical integration gives $q_o\simeq 0.281610$, ${\zeta }_o\simeq 0.400049$, so the dimensional results for this specified entrainment convention areA convention using $\alpha w$ instead of $\alpha (w-v)$ would produce different numerical amplitudes; the relative-speed convention was stated in part (a).

Physically $z_o$ is a turnover or breakdown height for a slender steady plume surrounded by uniform counterflow. The opposing streams then occupy equal areas and have equal speeds; the single-valued small-area branch cannot continue smoothly. Above it one expects lateral spreading, stronger recirculation, mixing and unsteady buoyant structures rather than a continuation of the same slender top-hat column. A detailed flow or upper-layer evolution requires restoring pressure and stress dynamics and, eventually, the modified ambient stratification. The model does not predict an infinite physical velocity, nor does it require that all buoyant fluid forever remain below $z_o$.

Normalize all fluxes by $\pi$, rather than including that factor in only some of them. The specific plume volume flux, momentum flux and buoyancy flux areThe ambient transport is $-Q$ and its specific momentum flux isPut $s=b^2/R^2=Q^2/(R^{2}M)$, with $0<s<1$. The total specific momentum flux isThe first differential equation follows immediately from relative-speed fluid entrainment:For the second, differentiate $J=M/(1-s)$ and use $s^{\prime }=2s{Q}^{\prime }/Q-s{M}^{\prime }/M$. Since $J^{\prime }=b^2{g}^{\prime }=B_{s}Q/M$, this yieldsThis is the requested two-equation system, with $B=B_s$ and zero total volume flux as the additional conservation conditions. Near the pure source, $Q,M\to 0$ and $s\to 0$. Eliminating height in that limit givesThe zero integration constant selects a pure plume; a nonzero imposed source momentum would select a different branch and generally a different turning height. The leading radius is $b\sim 6\alpha z/5$, which recovers the unconfined pure-plume limit near the source.

Use the Boussinesq approximation and a top-hat plume model. Let $v(z)<0$ be the uniform ambient return velocity outside the plume. Negligible source volume flux and an impermeable cylindrical wall require zero net transport through every horizontal section:The entrainment speed must be specified relative to that moving ambient. Taking it to be $\alpha (w-v)$ givesThe surrounding unmodified fluid has zero buoyancy relative to the initial fluid. Neglecting thermal or compositional loss therefore givesLet $B_s$ denote the conserved specific source buoyancy flux: $B_s=B_o/\pi$ if $B_o$ is the actual integrated source flux, or $B_s=B_o$ if the source strength has already been normalized by $\pi$.

To obtain the usual integral-model singularity one must also state a pressure closure. Here take the whole-section mean pressure to have only its reference ambient hydrostatic pressure, and neglect wall force and extra vertical turbulent-stress fluxes in the integral budget. Internal momentum transfers between plume and ambient cancel in that whole-section balance. The remaining upward force is the plume buoyancy, givingThe two momentum fluxes add, even though $v$ is negative: vertical momentum transported downward also contributes $v^2$ to the flux through an upward-oriented horizontal section. These three flux balances, with the algebraic return-velocity constraint, govern $b,w,g^{\prime }$ in the confined plume with compensating return flow.

The pressure-neglecting closure is an integral approximation, not a claim that the return fluid has zero acceleration. If a mean pressure departure $p_{\ast }(z)$ is retained, the whole-section momentum balance instead contains $-(R^2/{\rho }_0)p_{\ast }^{\prime }(z)$, along with any wall-force or stress contribution. The entrainment hypothesis alone does not specify that function. The singularity calculated below belongs to the stated pressure-neglecting model; it is not forced by continuity alone or asserted to be an infinite velocity in the real fluid.

The filling box model requires the first outflow to spread across the enclosure and remain above unmodified fluid, while a narrow nearly vertical plume rises below it. Several mechanisms can defeat those assumptions. A tall narrow box makes the plume a substantial fraction of the width, so its compensating return velocity is no longer small and modifies both fluid entrainment and momentum flux. A fast ceiling current can descend the sidewalls and overturn the initially stable upper region. In both cases a horizontally uniform slowly descending filling-box first front is inappropriate. Strong plume displacement from the centre, imposed crossflow, large source volume, vents or substantial wall cooling likewise alter the volume and heat budgets.

A useful diagnostic is whether the ceiling outflow's inertial acceleration, of order $U_c^2/{\ell }_c$, overwhelms the restoring buoyancy of the nascent layer. If $U_c^2/(G_H{d}_c)$ is not small for its depth $d_c$, the outflow can penetrate or overturn the stratification rather than remain trapped above it. This is a diagnostic Froude number, not a universal threshold without specifying the wall-current geometry and mixing. Low Reynolds number invalidates a constant turbulent entrainment coefficient, and thermal diffusion or wall heat loss can erase the sharp first-front jump. Thus large aspect ratio, strong return circulation, overturning, ventilation or diffusion require a model beyond the standard filling-box approximation.

Let the closed box have ceiling height $H$ above the source. The printed width $2R$ alone does not specify this height, so it must appear as an additional geometrical parameter. Take $t=0$ here when the plume has first reached the ceiling and its thin outflow has spread across the box. The leading filling box model assumes a negligible source volume, insulated boundaries, fast horizontal spreading, slow evolution compared with the plume transit time, negligible diffusion, and $b\ll R$. Below the interface the ambient still has its original temperature, so the local plume flux remains $q(z)=2\alpha w_{\ast }z$ from part (c).

The ceiling outflow creates a warm upper region. Fluid entrainment draws the original lower fluid into the plume, making that upper region deepen. The filling-box first front is the lower material boundary of fluid already modified by plume discharge, separating it from the initially unmodified ambient; it need not bound a uniformly mixed upper layer. With negligible occupied plume area, volume conservation at its height $h(t)$ givesThereforeIf source switch-on is the time origin instead, replace $t$ by elapsed time after the initial transit and spreading stage; the quasi-steady model does not resolve that stage.

The buoyancy jump requires distinguishing ambient fluid at the front from the plume crossing it. Just after the first ceiling outflow, the ambient buoyancy immediately above the nascent front isOutside the narrow plume, ambient temperature is transported without diffusion. If $G_a$ is its buoyancy relative to the original fluid, its equation isThe first front follows precisely these material trajectories, so the temperature of the fluid immediately above it is conserved. The initially unmodified fluid below has $G_a=0$. Thus the ambient jump isBy contrast, the contemporaneous plume immediately below the front has $G_p(h,t)=w_{\ast }^2/h(t)$, which grows as the front descends. That plume value is not the ambient jump across the material first front. Equating the two would inadvertently impose a uniformly mixed upper layer and discard the stable density stratification of the standard filling box model. Diffusion or extensive interfacial mixing can smooth and alter the ideal jump.

Let ${\mathcal{B}}_0>0$ be the source buoyancy flux per unit length. The conversion from the heat-source strength depends on its units: if $F$ is a kinematic temperature-volume flux, ${\mathcal{B}}_0=g\beta F$; if it is a physical heat power per unit length, ${\mathcal{B}}_0=g\beta F/({\rho }_0{c}_p)$, where $c_p$ is specific heat capacity at constant pressure. This explicit convention avoids treating heat power as a buoyancy flux without conversion.

For uniform ambient $N^2=0$, so $\mathcal{B}={\mathcal{B}}_0$. The ideal pure line-source limit has negligible initial volume and momentum transport. The flux equations admit $w=w_{\ast }$ constant, $q=2\alpha w_{\ast }z$ and $b=\alpha z$. Substituting into the momentum equation gives $2\alpha w_{\ast }^2=2bG$, and the conserved buoyancy flux then fixes $2\alpha w_{\ast }^3={\mathcal{B}}_0$. ThusThis verifies all three differential equations, not just their dimensional exponents. The plume grows in width while maintaining constant upward velocity; its temperature anomaly decreases like $z^{-1}$. The zero-height singularity is the ideal point-source limit. A finite source or nonzero momentum input introduces a plume virtual origin or a forced near-source region; the heat strength alone cannot uniquely determine that finite-source region.

In the top-hat plume model, vertical velocity $w(z)$ and reduced gravity $G(z)$ are uniform over $-b<x<b$. The ambient has zero vertical velocity, negligible ambient turbulence and a prescribed $T_0(z)$. Thenand the equations becomeEquivalently, the ordinary differential equations for the three unknowns areThe last equation displays dilution by fluid entrainment as well as loss of relative buoyancy while rising through a stably stratified ambient. The model assumes a steady slender two-dimensional turbulent plume, large Reynolds number, the Boussinesq approximation, constant $\beta$ and $\alpha$, negligible heat loss, negligible imposed crossflow, and negligible pressure departure from ambient hydrostatic pressure. The factors of two count both entraining edges; $b$ is a half-width. A finite source also requires its volume, momentum and buoyancy fluxes as boundary data.