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Apply volume conservation, mass conservation and vertical momentum flux balance to a slice of the triangular-profile line plume. Ambient ingestion contributes $2e$ of volume and $2{\rho }_{0}e$ of mass per unit height and source length. The ambient is quiescent, so it supplies no leading vertical momentum. The integrated driving force is the stored buoyancy $\mathcal{B}$. Using the quantities from part (a), the non-Boussinesq triangular-profile line-plume balances areTogether with Batchelor entrainment, these are three equations for $W,\rho ,b$. They neglect viscous boundary stresses and streamwise pressure-force corrections within the integral-plume approximation.

Multiplying volume conservation by $g{\rho }_0$ and subtracting $g$ times mass conservation cancels the ambient sources exactly, givingThis is buoyancy conservation in the homogeneous ambient; it is a consequence of the first two balances, not an additional independent equation.

Now take the Boussinesq approximation in inertia and entrainment, retaining the small density deficit in buoyancy. Then $Q={\rho }_{0}bW$, $M=(2/3){\rho }_{0}b{W}^2$ and $F=(2/3)gbW({\rho }_0-\rho )$. ConsequentlyThe mass source is $2\alpha {\rho }_{0}W=3\alpha {\rho }_{0}M/Q$. Substitution into the mass conservation, momentum and buoyancy balances gives the unsteady Boussinesq triangular-profile line-plume balancesThese use density-weighted mass and momentum fluxes. Removing ${\rho }_0$ from some flux definitions while retaining it in the source would mix incompatible conventions.

For a smooth test function $\phi$, integration by parts gives the moment identityThe assumed decay removes all boundary terms. Apply it to $1$, $v_i$ and $|v{|}^2/2$. Writing $m_i=\int v_{i}f=M{u}_i$ gives the Ornstein-Uhlenbeck moment equationsAssume $M(0)\ne 0$ when using the normalized mean $u$; otherwise $u$ is undefined, while the unnormalized momentum equation still holds. Since $M$ is constant, $u_i^{\prime }=-u_i$. Their solutions areThus mass is conserved, the mean velocity tends to zero, and $E(t)\to d{M}_0/2$. This includes the mean velocity conclusion omitted from the converted TeX.

For positive smooth $f$ with the stated decay, integration by parts givesand, using unit mass,For zero values of $f$, these calculations can be made with the positive unit-mass approximation $(f+\epsilon \gamma )/(1+\epsilon )$ and then passed to the limit whenever the quantities are finite. Positive-time solutions also have the usual Gaussian smoothing.

Since $-\log \gamma =v^2/2+\frac{1}{2}\log (2\pi )$, mass conservation givesThe derivative of the first term is $\int {\partial }_t{f}_t\log f_t$, because $\int {\partial }_t{f}_t=0$. The two identities above and the energy equation in (b), with $d=M=1$, yieldNow expand the relative Fisher information:where $\int v{f}^{\prime }=-1$. Thus the entropy dissipation identity for Ornstein-Uhlenbeck flow is

For unit mass, the inverse-square force has potential energy $-k/r$, with zero potential at infinity. Conservation of angular momentum gives $r^2\dot{\theta }=h$. Differentiate the polar equation of the Kepler orbit:Consequently the total mechanical energy iswhere $k=h^2/\ell$. This also explains the sign distinction between an elliptic orbit, a parabolic Kepler orbit, and a hyperbolic Kepler orbit.

At the original periapsis, the parabolic Kepler orbit has $e=1$, $r=d=\ell /2$, and purely tangential speed $V$. ThusThe outward radial impulse has zero moment about the Sun, so it preserves angular momentum and hence $\ell$. Its additional velocity is perpendicular to the original velocity, so the new mechanical energy is $E^{\prime }={\alpha }^2{V}^2/2$. Applying the energy formula givesFor $\alpha >0$, the new orbit is a hyperbolic Kepler orbit; $\alpha =0$ leaves the parabola unchanged.

To fix the orientation of the sketch, put the Sun at the origin, the impulse point at $(d,0)$, and take the original motion there upwards. The new velocity is $(\alpha V,V)$. The eccentricity vector isIt points towards the new periapsis, which is rotated clockwise through $\beta =\arctan (2\alpha )$ from the original periapsis direction. Thus, measured using the original polar angle $\theta$, the new orbit isIn particular, the impulse point is already on the outgoing part of the new orbit and is not its new periapsis. The old parabola satisfies $x=d-y^2/(4d)$. Both orbits pass through $(d,0)$ and have the same focus, but the hyperbola's new periapsis axis is tilted.

Use Newton's second law, $F=ma$. Acceleration has physical dimension $L{T}^{-2}$, so multiplying by mass gives the dimension of forceThe bracket notation records a physical dimension, rather than the numerical value or the choice of units.

From Newton's law of universal gravitation, $F=G{m}_1{m}_2/r^2$, the gravitational constant has physical dimensionHere the two factors of mass and the inverse square of the separation determine the powers of $M$ and $L$.

Work in one fixed inertial frame, and take positive velocity in the rocket's direction. During $d\tau$, the rocket ejects the positive amount of mass $-m^{\prime }(\tau )d\tau$. Its exhaust has inertial velocity $v(\tau )-u$, by the Galilean transformation of velocities. Once expelled, that material retains its velocity because it experiences no external force. The system's momentum is therefore the rocket momentum plus the sum of all these exhaust momenta:The minus sign comes from the positive expelled mass being $-m^{\prime }(\tau )d\tau$.

Differentiate, applying the product rule to the rocket momentum and the fundamental theorem of calculus to the integral:The momentum conservation law thus gives the nonrelativistic rocket equation

The same bookkeeping for kinetic energy givesIts derivative iswhere the rocket equation cancels the first two terms. Thus kinetic energy increases during mass ejection, with $\boxed{K^{\prime }=-u^2{m}^{\prime }/2>0}$ when $u>0$ and $m^{\prime }<0$. This does not violate conservation of energy: the increase comes from stored internal energy.

For this question use the mostly-plus Dirac convention: the Minkowski metric is ${\eta }_{ab}=\mathrm{diag}(-1,1,1,1)$ andThis convention matches the printed plane-wave phase and final identity. It is related to the usual mostly-minus Dirac equation by reversing the metric and taking the negatives of the usual gamma matrices. In particular, the resulting Dirac action and its Dirac adjoint describe the same physical massive field. The Dirac gamma matrices are four complex $4\times 4$ matrices representing the spacetime Clifford algebra with quadratic form $-\eta$. The irreducible complex representation has dimension four. A convenient explicit choice is the negative of the standard Dirac representation of the gamma matrices:where ${\sigma }^i$ are the Pauli matrices. Their multiplication law verifies the displayed anticommutator.

In this representation the gamma matrix adjoint and transpose identities areandThe invariant way to express the latter pattern uses the charge-conjugation matrix:Individual transpose signs depend on the basis. More generally, a similarity transformation ${\gamma }^{\prime a}=M{\gamma }^a{M}^{-1}$ changes the Hermitizing matrix to $H^{\prime }=M^{-†}{\gamma }^0{M}^{-1}$ and the charge-conjugation matrix to $C^{\prime }=MC{M}^T$. Then ${\gamma }^{\prime a†}=H^{\prime }{\gamma }^{\prime a}{H}^{\prime -1}$ and $C^{\prime -1}{\gamma }^{\prime a}{C}^{\prime }=-{\gamma }^{\prime aT}$. Thus the simple formula with ${\gamma }^{\prime 0}$ itself presumes a compatible Hermitian basis, rather than an arbitrary nonunitary similarity transformation.

Applying $(i{\gamma }^a{\partial }_a-m)$ to the Dirac equation gives $(\square -m^2)\Psi =0$. Its mass shell is $p^2=-m^2$, so the frequencies are $p^0=\pm \sqrt{{\mathbf{p}}^2+m^2}$. The Dirac spinor transforms in the four-component Spinor representation of the Lorentz group. Under spatial rotations, the two upper and the two lower components each transform as a two-component spin-$1/2$ representation: the spin angular momentum matrices are $\mathrm{diag}({\sigma }^i,{\sigma }^i)/2$. At rest the positive-energy equation selects the upper two components, giving two independent spin polarizations, and the negative-frequency equation selects the lower two.

In the quantum theory a mode expansion isWith the mostly-plus Dirac convention, $e^{ip\cdot x}=e^{-iEt+i\mathbf{p}\cdot \mathbf{x}}$ is positive frequency. The negative-frequency coefficient obeys $(/p+m)v_s(p)=0$. The fermionic annihilation operators $b_s$ and $d_s$ satisfy the canonical anticommutation relations, with their respective fermionic creation operators. The $b_s^{†}$ excitations are particles; the $d_s^{†}$ excitations are antiparticles with the same positive energy, mass and spin-$1/2$ but opposite charge. After normal ordering, the Hamiltonian operator contains positive multiples of $b_s^{†}{b}_s+d_s^{†}{d}_s$. Reinterpreting the negative-frequency part as antiparticle creation supplies a spectrum bounded below rather than a physical tower of negative-energy particles.

For the printed $e^{ip\cdot x}$ wave, ${\partial }_a\Psi =i{p}_a\Psi$. Substitution givesThe spin label $s$ indexes the two states of a spin-$1/2$ particle, rather than varying the particle's total spin. For real on-shell $p$, Hermitian conjugation and ${\gamma }^{a†}={\gamma }^0{\gamma }^a{\gamma }^0$ giveThese are right and left null-vector equations for the same on-shell matrix.

To obtain the Gordon identity, take both external Dirac spinors to have the same real mass $m$. Their two equations implyDefine ${\gamma }^{ab}=[{\gamma }^a,{\gamma }^b]/2$. The Clifford algebra relation yieldsThereforeMultiplying the previous null-vector equation by $-1$ proves the required formula exactly:For $m\ne 0$ it can be solved for the vector-current matrix element, separating a momentum term from the antisymmetric Dirac spinor term. The identity before division also holds at $m=0$. The signs depend jointly on the metric, Clifford relation, Dirac mass term and plane-wave phase. In the mostly-minus convention of Questions 1 and 3, the printed phase instead gives $(/p+m)u=0$; the corresponding identity uses $(p_b-p_b^{\prime })$ in the antisymmetric term. Mixing that convention with the formula proved here would produce an apparent sign error.

For the finite-mass similarity solution above, the total disk mass isSubstitute $\xi =1/(\tau \sqrt{x})$, so $x={\tau }^{-2}{\xi }^{-2}$ and $x^{-3/2}|dx|=2\tau d\xi$. This givesThe requested linear increase of mass is therefore correct. It is supplied by the infinite-radius boundary, rather than by a source-free isolated finite disk. The inward mass accretion rate iswhich tends to zero at the origin and to $3\pi C{\bar{\nu }}_0{\Sigma }_0$ at infinity.

For Keplerian rotation, the specific angular momentum is $\ell =\sqrt{G{M}_{\star }r}$. The total disk angular momentum would beThis integral diverges at $\xi =0$, corresponding to infinite radius. There is no finite constant total angular momentum for the nonzero printed similarity solution. The seemingly time-independent last expression is an infinite quantity; it cannot prove a finite conservation law.

The problem remains visible under a cutoff. With $r\le r_{0}R$, write $J_R=J_{\ast }{E}_1(1/(\tau \sqrt{R}))$, where $J_{\ast }=4\pi C{\Sigma }_0\sqrt{G{M}_{\star }}{r}_0^{5/2}$ and $E_1$ is the exponential integral. ThenEven the finite change after subtracting a common logarithmic divergence is not zero. Direct integration of the disk diffusion equation gives the boundary identityfor this profile the outer expression has a nonzero limit, giving the same finite change rate. An inner zero torque alone does not impose zero outer angular-momentum flux.

More generally, the printed family $A+B{e}^{-\xi }$ can have finite angular momentum at both ends only if $A=0$ and $A+B=0$, hence it is trivial. Thus the request for a nonzero finite constant $J_d$ is incompatible with the stated viscosity exponent, similarity ansatz and infinite radial domain. A physical finite-domain or altered similarity problem would need different boundary data or assumptions; silently assigning a finite value to the divergent integral does not repair it.

Let $g=G(\xi )$ with $\xi ={\tau }^{-1}{x}^{-1/2}$. The derivatives of this similarity solution areThe first-derivative terms cancel in the dimensionless diffusion equation, leaving $G^{\prime \prime }+G^{\prime }=0$ for $\xi >0$. ThusAt the inner boundary $x\to 0$, $\xi \to \infty$. Interpreting the named variable as vanishing there sets $A=0$. Even if only the true viscous torque in an accretion disk $x^{1/2}g$ is required to vanish, a nonzero $A$ would give infinite total mass near $r=0$; finite mass again selects $A=0$. With nonnegative surface density, choose $B=C>0$; its value fixes the initial mass normalization. ThereforeThe surface density vanishes exponentially at the origin and has an $x^{-5/2}$ tail at infinity. Its peak occurs at $x_{\max }=1/(25{\tau }^2)$, with ${\Sigma }_{\max }/{\Sigma }_0=C(5\tau {)}^5{e}^{-5}$. The peak moves inward and grows as time increases.

The plot uses $C=1$ and shows the actual surface density, with logarithmic radius to resolve both peaks. The later profile is larger at every fixed radius. Its total mass is finite; its total angular momentum is not, as the next calculation establishes.

Write $\Sigma ={\Sigma }_0{x}^{-5/2}g$ because the prescribed kinematic viscosity is $\bar{\nu }={\bar{\nu }}_0{x}^{5/2}$. Substituting $r=r_{0}x$ and ${\partial }_t=(3{\bar{\nu }}_0/4r_0^2){\partial }_{\tau }$ into the viscous evolution of an accretion disk givesHence the dimensionless diffusion equation isFor the useful inverse-radius coordinate $y=x^{-1/2}$, this becomes simply $g_{\tau }=y{g}_{yy}$. The positive factor in $\tau =1+3{\bar{\nu }}_{0}t/(4r_0^2)$ makes increasing $\tau$ correspond to increasing physical time.

There is a distinction between the named torque variable and the physical viscous torque in an accretion disk. For Keplerian rotation, the outward torque magnitude isHere $M_{\star }$ denotes the central mass. The zero-torque inner boundary condition therefore requires $x^{1/2}g\to 0$ as $x\to 0$. Requiring $g\to 0$ is stronger. The nontrivial finite-mass profile used below satisfies both; this distinction matters when counting integration constants.

A central force has no component in the angular direction. The angular component of the acceleration in polar coordinates therefore gives $2\dot{r}\dot{\theta }+r\ddot{\theta }=0$, orFor unit mass, $h=r^2\dot{\theta }$ is the signed angular momentum perpendicular to the orbital plane, so its magnitude $|h|$ is conserved. Equivalently, the torque $\mathbf{r}\times \mathbf{F}$ vanishes, which also fixes that plane.

The given potential energy is $k/r$, whose radial force is $k/r^2$: positive $k$ is repulsive and negative $k$ attractive. The kinetic energy in polar coordinates is $({\dot{r}}^2+r^2{\dot{\theta }}^2)/2$, so conservation of energy givesFor $h\ne 0$ and $k>0$, the effective potential is positive, strictly decreasing from $+\infty$ to $0$, and has no stationary point. For $k=-\kappa <0$, it still tends to $+\infty$ at zero, crosses zero at $r=h^2/(2\kappa )$, and has its unique minimum atIt then approaches zero from below. The minimum corresponds to a stable circular orbit. If $h=0$, the centrifugal barrier disappears; the attractive effective potential is simply $-\kappa /r$ and has no minimum.

The rotating-frame derivative formula is $(d\mathbf{A}/dt{)}_S=(d\mathbf{A}/dt{)}_{S^{\prime }}+\mathbf{\omega }\times \mathbf{A}$. Applying it twice to the position vector, with constant angular velocity, givesSolving for ${\mathbf{a}}_{S^{\prime }}$ identifies the displayed terms: its left side is the acceleration relative to the rotating reference frame, the first right-hand term is the inertial acceleration, $-2\mathbf{\omega }\times {\mathbf{v}}_{S^{\prime }}$ is the Coriolis acceleration, and $-\mathbf{\omega }\times (\mathbf{\omega }\times \mathbf{r})$ is the outward centrifugal acceleration.

For a particle of mass $m$, replace ${\mathbf{a}}_S$ by the actual force divided by $m$ using Newton's second law. The remaining terms then appear as apparent forces in $S^{\prime }$. There is no Euler acceleration because $\mathbf{\omega }$ is constant and no translational acceleration term because the common origin is fixed in the inertial frame.

Write the total remaining rocket mass as $\mathcal{M}=M+m(t)$, with $\dot{m}=-\alpha$. Over an interval $dt$, an exhaust mass $d{m}_e=\alpha dt$ is expelled backwards with inertial velocity $v-u$ to first order. Applying momentum conservation to the rocket and this exhaust givesCancellation and division by $dt$ give the rocket equationWith $M>0$, $m(t)=m_0-\alpha t$, and $0\le t\le m_0/\alpha$, integration givesAt burnout,The logarithm depends on the mass ratio, not on the burn rate. This follows from momentum conservation of the complete rocket–exhaust system; applying a constant-mass equation directly to the rocket alone would omit the momentum carried away.

For $0\le k<1$, the first burn reduces the total mass from $M+m_0$ to $M+(1-k)m_0$. The rocket equation therefore givesDetachment exerts negligible impulse, so the surviving second stage keeps velocity $v_1$. Its initial and final masses are $(1-k)(M+m_0)$ and $(1-k)M$, respectively. A second application of the rocket equation givesFor the single-stage comparison from rest, $v_{\mathrm{single}}=u\log (1+m_0/M)$. HenceEquality holds at $k=0$; for positive fuel and exhaust speed it is strict when $0<k<1$. This two-stage rocket with proportional fuel and body masses gains speed by discarding inert body mass before the last burn.

As $k\to 1^{-}$ the expression tends to $2u\log (1+m_0/M)$, but the surviving second-stage mass also tends to zero. At exactly $k=1$ there is no second stage, so its final speed is undefined and that limiting formula cannot be assigned to it. The first stage alone reaches the single-stage burnout speed before detachment. If $m_0=0$, neither design gains speed.

Assume the particle masses $m_i$ are fixed and their total mass $M={\sum }_i{m}_i$ is positive. The centre of mass and the relative position vectors areTheir weighted sum is zero, so differentiation gives ${\sum }_i{m}_i{\dot{\mathbf{y}}}_i=0$. Expanding the kinetic energy with ${\dot{\mathbf{x}}}_i=\dot{\mathbf{R}}+{\dot{\mathbf{y}}}_i$ makes the mixed term vanish:This kinetic-energy decomposition about the centre of mass separates translation from internal motion without assuming any particular force law.

For the rigid body, its relative velocity is ${\dot{\mathbf{y}}}_i=\omega \mathbf{n}\times {\mathbf{y}}_i$. The cross product identity for a unit vector gives $|\mathbf{n}\times {\mathbf{y}}_i{|}^2=|{\mathbf{y}}_i{|}^2-(\mathbf{n}\cdot {\mathbf{y}}_i{)}^2$, the squared perpendicular distance to the rotation axis. Hence the moment of inertia of the particle is $I_i=m_i[|{\mathbf{y}}_i{|}^2-(\mathbf{n}\cdot {\mathbf{y}}_i{)}^2]$, and summing the rotational kinetic energy yields

Use a genus-zero Polyakov path integral, target signature $(-,+,\dots ,+)$ and all four momenta incoming. Define the Mandelstam variables by $s=-(p_1+p_2{)}^2$, $t=-(p_1+p_3{)}^2$, $u=-(p_1+p_4{)}^2$. The string mass-shell condition is $p_i^2=4/{\alpha }^{\prime }$ and momentum conservation is ${\sum }_i{p}_i=0$, soThe reduced amplitude omits the overall momentum-conservation Dirac delta distribution and any conventional overall scattering-matrix phase. Interpret each tachyon vertex operator as normal ordered; self-contractions must not be included.

The free-boson worldsheet propagator is $⟨X^a(z)X^b(w)⟩=-({\alpha }^{\prime }/2){\eta }^{ab}\log |z-w{|}^2$. Wick theorem then evaluates the normal-ordered exponential correlator as the Koba-Nielsen factorThe worldsheet zero mode supplies momentum conservation. Four vertices give $g_s^4$ and the sphere contributes $g_s^{-2}$ through the string genus expansion, leaving $g_s^2$.

The sphere's residual Möbius transformations fix three insertion points. In sphere gauge fixing for four string vertices, take $(z_1,z_2,z_3,z_4)=(z,0,1,\infty )$. The worldsheet ghost fields, equivalently the Faddeev-Popov determinant for this residual group, supply $|z_{23}{z}_{24}{z}_{34}{|}^2$. At finite $z_4$, this determinant grows as $|z_4{|}^4$, while the matter correlator decays as $|z_4{|}^{-4}$ by momentum conservation and the tachyon mass shell; their product has a finite limit. Equivalently the weight-$(1,1)$ matter operator at infinity is normalized with $|z_4{|}^4$, while the ghost pair uses the inverse factor. After stripping the fixed-position normalization, onlyremains. Here choose the standard complex-coordinate measure $d^{2}z=idz\wedge d\bar{z}=2dxdy$. With $dxdy$ instead, the reduced normalization must acquire a factor two to give the same requested amplitude. Absolute vertex/sphere normalization is a convention; this choice fixes it consistently with the displayed $2\pi$ prefactor.

Set $a=-1-{\alpha }^{\prime }s/4$, $b=-1-{\alpha }^{\prime }t/4$, $c=-1-{\alpha }^{\prime }u/4$. Then $a+b+c=1$ and the integrand is $|z{|}^{2a-2}|1-z{|}^{2b-2}$. To evaluate the complex beta integral first work where $\mathrm{Re}a,\mathrm{Re}b,\mathrm{Re}c>0$. This ensures convergence near $0$, $1$ and infinity. Set $v=1-a$ and $w=1-b$. Schwinger parameterization givesCompleting the square, the $dxdy$ Gaussian integral is $\pi (\lambda +\rho {)}^{-1}\exp [-\lambda \rho /(\lambda +\rho )]$. Change variables to $q=\lambda +\rho$ and $x=\lambda /q$; their Jacobian determinant is $q$. Integrating $q$ gives $\Gamma (v+w-1)[x(1-x){]}^{-(v+w-1)}$. The remaining integral is the ordinary beta function $B(1-w,1-v)=B(b,a)$. ThusRestoring $d^{2}z=2dxdy$ and substituting the Mandelstam variables proves the Virasoro–Shapiro amplitude:For physical scattering the original position integral generally fails to converge. The formula defines the amplitude by analytic continuation from the convergence domain, with the desired scattering boundary value at real poles; the convergent integral should not be claimed valid for every physical momentum.

The gamma function is a meromorphic function, with simple poles at nonpositive integers and no zeros, while its reciprocal $1/\Gamma$ is an entire function. Consequently it is a meromorphic function of the independent invariants and a crossing-symmetric scattering amplitude: permuting $s,t,u$ leaves it unchanged. Its generic channel poles areand the same tower in $t$ and $u$. They represent the exchanged bosonic string mass spectrum: the tachyon, massless states including the graviton, and an infinite sequence of massive closed-string levels. There are no threshold branch cuts at this tree order.