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The exponent of the point-mass weight is $\beta$, with $0<\beta <1$. At the origin the unreferenced relative potential has integrand $GB{b}^{\beta -1}$, whose integral diverges at its upper endpoint. In the origin-referenced potential the two terms cancel pointwise there. More generally, first form their difference at finite cutoffs and then pass to the limit. For any fixed nonzero $(R,z)$, the small-$b$ singular term is $-b^{\beta -1}$, which is integrable because $\beta >0$. At large $b$,so the weighted difference is integrable because $\beta <1$. Hence the referenced potential is well defined, and its value at the origin is zero. It is a potential difference for an infinite astrophysical disk, not a potential with zero at infinity. In fact rescaling all lengths shows ${\psi }_{II}(\lambda R,\lambda z)={\lambda }^{\beta }{\psi }_{II}(R,z)$; its finite values approach zero at the origin even though its force is singular there.

For $R>0$, differentiation under the convergent force integral in the plane givesUse $b=R\sqrt{t/(1-t)}$, for which $db=(R/2)t^{-1/2}(1-t{)}^{-3/2}dt$. The powers of $R$, $t$, and $1-t$ then givewhere the Euler beta function evaluates the constant asThe first gamma function argument is $(1+\beta )/2$: it follows from adding one to the exponent of $t$. Both Euler beta function arguments are positive in the stated range.

For a circular orbit in the plane, $V^2/R=-{\partial }_R{\psi }_{II}$. The prescribed circular speed therefore fixes $C=K^2$ andIndependently, the Kuzmin reflection method givesHere the power of $b$ is one larger than in the force integral, so its Euler beta function arguments differ. Substituting the normalization fixed by the circular speed yieldsThis scale-free Kuzmin superposition has nonnegative surface density, finite central enclosed mass, and infinite total mass. Its force and surface density approach the Mestel disk values as $\beta \downarrow 0$; that limiting disk requires a different potential reference at the origin.

Use gravitational acceleration $\mathbf{g}=\nabla \psi$. Through the plane with upward unit normal, the signed gravitational flux per unit area of the point mass isThus its downward flux magnitude is $Gmb/(R^2+b^2{)}^{3/2}$. Subtracting a constant from the gravitational potential does not change this flux.

The Kuzmin reflection method gives the even relative potential $\psi =Gm/\sqrt{R^2+(|z|+b{)}^2}$. It is harmonic above and below the plane because the corresponding point masses lie outside the respective half-spaces. At the plane the derivative jump isIntegrating the Poisson equation for Newtonian gravity ${\nabla }^2\psi =-4\pi G\Sigma (R)\delta (z)$ through a thin layer, with $\delta$ the Dirac delta, equates this jump to $-4\pi G\Sigma$. ThereforeThis is a Kuzmin disk. As a check, its total mass is $2\pi {\int }_0^{\infty }R\Sigma (R)dR=m$.

Replace the point mass by $dm=\mu (b)db$ and superpose the Kuzmin disks. Linearity of the Poisson equation for Newtonian gravity gives, whenever the force integral converges,Both choices of potential reference give this same surface density. If the unreferenced potential diverges, perform the reference subtraction in each component before taking the integral limit; the force and the surface density may still be finite.

Assume axisymmetry in a thin disk orbiting a fixed central mass $M$. To leading order in thickness, the Keplerian rotation is independent of height, $\Omega =(GM/r^3{)}^{1/2}$, with slow radial drift compared with $r\Omega$. Neglect radial pressure support in the orbital balance, changes in the central potential, mass loss through the disk surfaces and applied surface torques. The Cauchy stress tensor can represent viscous or effective turbulent/magnetic transport; only its vertically integrated azimuthal component enters the conservation of angular momentum equation.

Define the surface density and its mass-weighted radial velocity byand set $W_{r\varphi }=\int T_{r\varphi }dz$. Integrating mass conservation vertically givesThe azimuthal equation in cylindrical coordinates, multiplied by $r$ and combined with mass conservation, is the conservative equation for specific angular momentum $l=r^2\Omega$. The cylindrical stress tensor divergence contributes $r^{-1}{\partial }_r(r^2{T}_{r\varphi })+{\partial }_z(r{T}_{\varphi z})$. With vanishing surface torque, vertical integration therefore givesBecause $l(r)$ is fixed in time, subtracting $l$ times the mass equation yieldsDefine the effective viscosity defined by disk stress byThis sign convention treats $T_{r\varphi }$ as the shear stress appearing with $+\nabla \cdot \mathbf{T}$. For the Newtonian fluid stress tensor it reduces to the density-weighted viscosity of a disk, $\bar{\nu }={\Sigma }^{-1}\int \rho \nu dz$, rather than an unweighted height average.

Using $l=\sqrt{GMr}$ and $W_{r\varphi }=-(3/2)\bar{\nu }\Sigma \Omega$ in the conservation of angular momentum equation givesSubstitution into mass conservation proves the Keplerian viscous diffusion equation:The outward viscous torque in an accretion disk is $\mathcal{G}=-2\pi r^2{W}_{r\varphi }=3\pi \sqrt{GM}\sqrt{r}\bar{\nu }\Sigma$.

For $\bar{\nu }=Ar$ with $A>0$, make the heat-equation transform for linear-radius disk viscosityWriting $y=\sqrt{r}$, the diffusion equation becomes $g_t=(3A/4)g_{yy}$; the displayed rescaling of $y$ gives exactly $\boxed{g_t=g_{xx}}$, with the original time unchanged. The torque is $3\pi \sqrt{GM}g$, so the zero-torque inner boundary condition at the origin becomes $g(0,t)=0$ on the half-line $x>0$.

A positive similarity solution obeying this boundary condition isFor example, substitution of $g=t^{-1}F(x/\sqrt{t})$ reduces the heat equation to $F^{\prime \prime }+(\eta /2)F^{\prime }+F=0$, and $F=C\eta e^{-{\eta }^2/4}$ satisfies it and vanishes at zero. A replacement of $t$ by $t+t_0$ gives a finite-mass initial profile at time zero if desired.

To determine the global integrals, transform the mass and angular momentum:The Gaussian integrals giveThus the accreting similarity disk with viscosity proportional to radius loses mass as $t^{-1/2}$ while conserving angular momentum; its characteristic radius grows proportional to $At$. The accretion rate at the origin is $-{\dot{M}}_d=2\pi C\sqrt{3/A}{t}^{-3/2}$. Its radial velocity ${\bar{u}}_r=r/t-3A/2$ shows inward flow at small radius and outward spreading at large radius. The integrable inner density singularity is part of the idealized extension to $r=0$.

For the other profile, direct differentiation givesIt also satisfies the heat equation, but has $g(0,t)=t^{-1/2}\ne 0$ and $g_x(0,t)=0$. It therefore has a nonzero inner torque and zero mass flow at the origin, not the previous zero-torque inner boundary condition. HereAn arbitrary positive amplitude multiplies these quantities. This constant-mass disk with viscosity proportional to radius spreads outward under a central torque that supplies angular momentum, with no ongoing central mass supply. It has the reflecting Neumann boundary condition for the heat equation and can be regarded as the spreading of an initially central mass distribution driven by an applied torque. It is distinct from the accreting zero-torque solution; the ideal origin does not model a finite stellar surface or a relativistic inner disk.

Write $r=x-\bar{x}$, $A(t)={\int }_0^t\alpha (s)ds$, and use the usual reduced rigid-recoil representation $h=Z+r\theta -\alpha r^2$, with zero initial recoil and a smooth stroke returning $\alpha$ and $\dot{\alpha }$ to zero. For this representation,Define the fluid impulse and its first moment byThe force law from part (i) gives, by integration by parts,These are the endpoint momentum flux in elongated-body recoil. PutThe endpoint quantities are needed in addition to the integrals $M_j$ unless further end-shape assumptions are made. Lateral force and yaw torque balance areExpanding gives the requested simultaneous second-order equations, with the retained endpoint fluxes:Substituting $\ddot{Z}=\ddot{\gamma }-U\dot{\theta }$ and integrating once from the initially straight, unrecoiling state yieldsThese are first-order coupled equations for $\gamma ,\theta$ with known forcing.

The claimed mass-independent final angle requires a qualification. LetAfter the stroke, $\alpha =\dot{\alpha }=0$ and $A=A_T={\int }_0^T\alpha dt$. If the post-stroke dynamics are stable, the first-order system relaxes toLet $C=B_0(B_2-M_1)-B_1(B_1-M_0)$. Solving this two-by-two system givesThis is quadratic-bend turning with finite body mass. At the limiting state $\dot{\theta }=\dot{\gamma }=0$, so $\dot{Z}=-U{\theta }_{\infty }$. The forward velocity is $-U$ in the chosen $x$ direction, and its lateral component has precisely the body's slope: the fish glides parallel to itself. The angle is generally reached after a relaxation transient, not necessarily immediately at $t=T$.

The advertised result is obtained in the thin-body approximation that neglects $M_b$ relative to fluid added mass, or in the special geometry $B_1=M_0$:For example, take negligible nose added mass and nonzero tail added mass $m_T$, so $B_j=m_T(L-\bar{x}{)}^j$. In the zero-body-mass approximation, the post-stroke equations for $(\gamma ,\theta -2U{A}_T)$ are homogeneous. The inertia matrix is positive definite by Cauchy-Schwarz inequality, andThe two eigenvalues of ${\mathsf{M}}^{-1}\mathsf{K}$ therefore have positive real parts: their sum and product are positive, whether they are real or conjugate. For $U>0$ the transient decays, proving parallel gliding and the displayed turn in this approximation.

With finite $M_b$, the angle is not a consequence of the printed force law for arbitrary $m(x)$. A concrete stable counterexample uses $L=1$, $\bar{x}=1/2$, and $m(x)=m_{\ast }{x}^2$. Then $M_0=m_{\ast }/3$, $M_1=m_{\ast }/12$, $B_0=m_{\ast }$, $B_1=m_{\ast }/2$, $B_2=m_{\ast }/4$, and $C=m_{\ast }^2/12$. Hencewhich differs from $2U{A}_T$ for positive body mass. This example has positive determinant and positive matrix trace of the relaxation matrix. Another immediate check is that if both endpoint added masses vanish, integrating the force law conserves total lateral momentum. A finite-mass fish cannot then end with $w=0$ and a nonzero new transverse velocity. Thus endpoint and inertia assumptions cannot silently be omitted.

If $Z$ is literally the centre of mass position of a deforming body, use a mass-centered deformation rather than the raw quadratic. For body mass density ${\mu }_b(x)$, its mean is $-\alpha J_2/M_b$, with $J_2=\int {\mu }_b{r}^{2}dx$; in the slender approximation $J_2=I_b$. The reference translation above is then $Z_r=Z+\alpha J_2/M_b$. Likewise the active angular momentum contains $-\dot{\alpha }{J}_3$, where $J_3=\int {\mu }_b{r}^{3}dx$, which vanishes for symmetric mass distribution. These corrections modify derivative forcing during the stroke. Since the shape and its time derivative return to zero, they do not remove the finite-body-mass qualification of the final impulse relation.

The principal physical limitations are small slopes and small total turning angle, prescribed approximately constant forward speed, slowly varying slender geometry, and neglected viscosity, separation and detailed three-dimensional wake interactions. Fast large-amplitude C-starts need more than this linear theory. Neglecting body mass is most reasonable for a laterally thin body whose crossflow added mass greatly exceeds its displaced mass. Stable relaxation also requires an appropriate tail/end geometry; it is not guaranteed for every added-mass profile.

The Lighthill elongated-body theory is a reactive, nearly inviscid flow description of a long slender swimmer, appropriate when the longitudinal Reynolds number is large but its lateral displacement and slope are small. To leading order, crossflow past each section is the two-dimensional potential flow generated by lateral motion of that section. Longitudinal variation is slow on the cross-sectional scale, and axial speed $U$ is prescribed at this order. With $x$ increasing from nose to tail, fluid passes sections in increasing $x$, and the linearized relative lateral velocity isThe section's added mass per unit length $m(x)$ is defined by its crossflow kinetic energy $\frac{1}{2}m(x)w^2$. Equivalently, if $\Phi$ is the velocity potential for unit lateral section motion, then $m=\rho \int |{\nabla }_{\perp }\Phi {|}^{2}dA$ over the exterior crossflow plane. It has units of mass per length and depends on cross-sectional shape and density. For a circular section of radius $a$, $\Phi =-a^2\cos \phi /r$ gives $m=\rho \pi a^2$. It is not the body's mass per unit length.

The transverse fluid impulse attached to a section is $m(x)w$. A control segment has both changing impulse and an impulse flux $Umw$ through its ends. Taking their sum per unit length gives the reaction on the fluid:The force on the fish is $-F_z$. In particular, $m$ is inside the convective derivative; dropping its spatial derivative or the endpoint fluxes is not justified merely because the fish is slender.

The recoil correction in elongated-body theory adds unknown whole-body translation and yaw to a prescribed active bending shape:They are determined by total lateral linear momentum and yaw angular momentum balance, rather than by imposing that the deforming swimmer remains on its original axis. Body inertia and hydrodynamic added inertia have the same sign. If the translation variable is required to be the literal centre of mass, the active deformation must first have its mass-weighted mean removed: the reference displacement $Z_r$ is otherwise not the centre of mass displacement during the stroke. The corresponding deformation contribution to angular momentum must also be retained when it is significant.

The cell conservation equation has the form ${\bar{n}}_{\bar{t}}=-\bar{\nabla }\cdot \bar{\mathbf{J}}$, withThe first term is advection by the suspension's fluid velocity, the second is active swimming at fixed speed $V_c$ in direction $\mathbf{p}$, and the last is an isotropic diffusive flux due to random swimming. Here $D$ has units length squared per time, $\bar{n}$ is cell number per volume, and the divergence measures local loss by transport. No birth or death term is included.

For a spherical cell of radius $a_c$, its excess buoyancy force is $(4\pi a_c^3/3)\Delta \rho g$. The Stokes drag law gives passive settling speed $V_{\mathrm{set}}=2a_c^2\Delta \rho g/(9\mu )$. Neglecting individual settling requires $V_{\mathrm{set}}\ll V_c$, not exact neutral buoyancy. A dilute suspension can still generate appreciable collective buoyancy because many small excess masses contribute to the bulk density.

Use the half-rate convention for the gyrotactic reorientation time $B$, required by the coefficients printed in the question:The first torque tends to point the bottom-heavy cell upward; the second rotates a sphere with half the fluid vorticity. This is a quasistatic torque balance, assuming orientation relaxation is fast enough to follow the local flow. Keeping orientation dynamics instead would put $d\mathbf{p}/d\bar{t}$ on the left. The convention with aligning rate $1/B$ uses a reorientation parameter twice this $B$. Thus the factor of two is a definition, not a physical discrepancy. Close to the upward state the quasistatic gyrotactic balance with half-rate reorientation gives ${\mathbf{p}}_{\perp }=B\bar{\mathbf{\omega }}\times {\mathbf{e}}_z$.

For a dilute Newtonian suspension with the Boussinesq approximation, the other equations areThe background-water hydrostatic pressure is absorbed in $\bar{p}$. The cell volume is $\upsilon$, density is $\rho$, cell density is $\rho +\Delta \rho$, dynamic viscosity is $\mu =\rho \nu$, and $\upsilon \bar{n}\ll 1$ is the volume fraction. Extra cell mass produces the downward buoyancy term. Constant viscosity and no explicit swimming stress are part of this closure. Rigid impermeable plates impose no-slip boundary condition and zero normal total cell flux.

At rest, the stable cell direction is upward and the steady vertical flux is $V_c\bar{n}-D{\bar{n}}_{\bar{z}}$. It is constant in depth by conservation and zero at an impermeable plate, soThe pressure gradient balances its excess weight. The mean-concentration formula follows by averaging over $0\le \bar{z}\le h$ and is continuous at $\beta =0$.

Use $\mathbf{x}=\bar{\mathbf{x}}/h$, $t=D\bar{t}/h^2$, $\mathbf{u}=h\bar{\mathbf{u}}/D$, $n=\bar{n}/{\bar{N}}_0$, and $p=h^2\bar{p}/(\rho \nu D)$. ThenwhereThe dimensionless base concentration is $n_0=e^{\beta z}$.

Linearize with the horizontal Fourier mode $e^{\sigma t+ilx+imy}$ and $k^2=l^2+m^2$. Write the first-order orientation perturbation as $\mathbf{p}={\mathbf{e}}_z+ϵ{\mathbf{p}}_1+\cdots$. Unit length gives $p_{1z}=0$, while torque balance gives ${\mathbf{p}}_{1\perp }=G{\mathbf{\omega }}_1\times {\mathbf{e}}_z$. Incompressibility impliesLet $\mathcal{L}=d^2/d{z}^2-k^2$. The perturbation of fluid advection is $n_0^{\prime }W=\beta n_{0}W$, and swimming contributes $\beta N^{\prime }+\beta n_0\nabla \cdot {\mathbf{p}}_1$. Cell conservation therefore givesorThis explains the destabilizing concentration-advection term and the gyrotactic redistribution term with their signs.

To eliminate pressure from the fluid equations, take their divergence, giving $\mathcal{L}{p}_1=-(R/\beta )N^{\prime }$. Apply $\mathcal{L}$ to the vertical equation $S{c}^{-1}\sigma W=-p_1^{\prime }+\mathcal{L}W-(R/\beta )N$. The concentration second derivatives cancel, leavingNo-slip conditions give $W=0$ and $W^{\prime }=0$ at each plate; for nonzero $k$, the latter follows from the vanishing horizontal velocity and incompressibility. Zero perturbed normal cell flux givesThe normal orientation correction is zero at this order, and the fluid velocity is zero at the plate, so neither adds another term to this concentration condition.

Now use the stated distinguished scalings $k=\beta \alpha$, $\sigma ={\beta }^2{\sigma }_2$, $G={\beta }^{-1}{G}_{-1}$, $N=N_0+\beta N_1+{\beta }^2{N}_2+\cdots$, $W=\beta W_1+{\beta }^2{W}_2+\cdots$, and $R=R_0+\beta R_1+\cdots$. Here $\alpha$ is a scaled wavenumber, unrelated to Question 2's shear rate. At order one, the concentration equation and boundary conditions give $N_0^{\prime \prime }=0$, $N_0^{\prime }=0$ at both endpoints, so $N_0$ is constant. Absorb its nonzero value into the perturbation amplitude and chooseAt order $\beta$, the fluid equation is $W_1^{(4)}=-R_0{\alpha }^2$ with $W_1=W_1^{\prime }=0$ at both ends. Its unique solution isA homogeneous difference is cubic and the four clamped conditions make it zero.

The concentration equation at order $\beta$ givesso $N_1=z-G_{-1}{W}_1+C_1$, where $C_1$ fixes the amplitude normalization. At order ${\beta }^2$ it givesThe boundary conditions expand to $N_2^{\prime }=N_1$ at each endpoint. Integrating this equation from zero to one cancels $[N_2^{\prime }{]}_0^1$ against $[N_1{]}_0^1$. Also $[W_2^{\prime }{]}_0^1=0$ andThus the required solvability condition isSince ${\int }_0^1{z}^2(1-z{)}^{2}dz=1/30$, the clamped-quartic solvability in shallow bioconvection gives

For nonzero long horizontal waves, $R_0<720$ gives decay and $R_0>720$ gives growth; the leading neutral threshold is $R_0=720$. The leading instability is stationary and independent of $G_{-1}$ and $Sc$. The exactly uniform horizontal mode $\alpha =0$ has zero growth at this order because total cell number is conserved, so it does not determine the threshold. Upward swimming produces a dense upper layer; the unstable disturbance is its buoyancy-driven overturning, or bioconvection. Higher orders are needed for wavelength selection and finite-depth threshold corrections.

With fixed material parameters and mean concentration, $R\sim h^4{V}_c\upsilon \Delta \rho g{\bar{N}}_{00}/(\rho \nu D^2)$ for small $\beta$, so a sufficiently thin chamber is stable to this branch. Formally its threshold corresponds to $h_c\sim [720\rho \nu D^2/(V_c\upsilon \Delta \rho g{\bar{N}}_{00}){]}^{1/4}$ when the weak-stratification and distinguished-gyrotaxis assumptions apply. In particular, the assumed $G=O({\beta }^{-1})$ is a specified asymptotic analysis; at fixed $B,D,V_c$ and decreasing $h$, one instead has $G=O({\beta }^{-2})$. The leading result should not be asserted unchanged in that different ordering without redoing the expansion.

The mass statement needs a nonzero parallel spinor and an isolated asymptotically flat spacetime; the zero spinor solves the equation on every geometry and cannot imply anything about mass. Assume the usual complete regular spin initial data and dominant energy condition used by the positive energy theorem, with the spinor tending to a nonzero constant ${ϵ}_{\infty }$ at infinity. Let $V^a=\overline{ϵ}{\gamma }^aϵ$ be its future causal Dirac current.

The Nester two-form can be written, up to normalization, asIt is bilinear in $ϵ$ and $\nabla ϵ$. For a spacetime-parallel spinor it vanishes identically, and hence so does its asymptotic flux. The ADM boundary term of the Nester two-form identifies that flux, up to a fixed positive normalization, asThe positive energy theorem gives $E\ge |P|$. Since $V_{\infty }$ is nonzero future causal, a future timelike ADM momentum would have strictly positive contraction with it. Consequently the ADM momentum is zero or null:In a rest frame, when one exists, this immediately says $E=0$. The usual regular asymptotically flat rigidity conclusion also excludes a nontrivial null-momentum configuration and gives flat initial data. This is the spinorial boundary-charge proof, not an inference that every Lorentzian manifold with a parallel spinor is flat. Without the asymptotic and global hypotheses the local spinor equation alone does not define, let alone determine, an ADM mass.

For the modified connection, it is important to fix the Clifford algebra normalization. WriteThe compatible spinor curvature identity isIndeed, rescaling conventional gamma matrices by $\sqrt{s}$ rescales ${\gamma }_{ab}$ by $s$, while leaving the geometric spin connection unchanged. Since the connection is torsion-free and $\nabla \gamma =0$, the cross terms cancel in the modified commutator:Thus the Killing-spinor integrability with rescaled gamma matrices equation isA factor of two in the convention for antisymmetrization multiplies the whole zero equation and cannot change this relative coefficient.

The two printed coefficients are consistent with $s=2$, namely $\{{\gamma }_a,{\gamma }_b\}=4g_{ab}I$. In that convention the displayed integrability equation becomes $R_{abcd}{\gamma }^{cd}+16c^2{\gamma }_{ab}=0$ on each solution. With the customary convention $s=1$, the coefficient is instead $8c^2$, and the final Ricci coefficient below is $-12c^2$. The PDF does not state its Clifford normalization, so these alternatives must be distinguished rather than mixing them.

In four spacetime dimensions the complex Dirac spinor fibre has dimension four. Four independent solutions of $Dϵ=0$ span that fibre at every point: a solution vanishing at one point vanishes everywhere by parallel transport. Therefore the integrability matrix annihilates every spinor and is the zero matrix. The six bivector matrices ${\gamma }^{cd}$ are linearly independent, as is seen by taking traces against them; their trace pairing is a nondegenerate multiple of the metric on two-forms. Expressing ${\gamma }_{ab}=g_{ac}{g}_{bd}{\gamma }^{cd}$ therefore givesThis proves that maximal Killing spinors force constant negative curvature, not merely an Einstein Ricci tensor. Contracting in four dimensions givesFor $s=2$ this is the requested Einstein equation. Since $c>0$, its Ricci tensor automatically has rank four; under the maximal-spinor hypothesis the additional rank assumption is redundant. If the stated four-dimensional solution space is interpreted directly as the pointwise kernel of the algebraic integrability equation, the same spanning and trace argument applies.

Assume an asymptotically flat Reissner-Nordstrom spacetime, with $M>|Q|$, time normalized at infinity, and $G=\hslash =c_{\mathrm{light}}=k_B=1$. PutNear $r_{+}$, the Euclidean radial-time metric is $f^{\prime }(r_{+})(r-r_{+})d{\tau }^2+d{r}^2/[f^{\prime }(r_{+})(r-r_{+})]$. With $\rho =2\sqrt{(r-r_{+})/f^{\prime }(r_{+})}$, it becomesThe Euclidean black-hole regularity condition removes a conical defect only if $\tau$ has period $4\pi /f^{\prime }(r_{+})$. Therefore the Reissner-Nordstrom Hawking temperature isEquivalently this is the Reissner-Nordstrom horizon surface gravity divided by $2\pi$. The derivation assumes a nondegenerate horizon; the value zero at $M=q$ is obtained by the nonextremal limit, not by imposing conical regularity directly on the degenerate geometry.

For fixed nonzero $q$, write $y=r_{+}/q\ge 1$. The horizon equation givesThe mass is increasing with $y>1$, andThus the fixed-charge Reissner-Nordstrom temperature maximum occurs atThe curve starts at zero at $M=q$, rises to this maximum, then falls as $1/(8\pi M)$ for large mass. During evaporation from $M_0\gg q$, the mass moves from right to left: the temperature initially increases, reaches its maximum, and then decreases towards zero.