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Use a top-hat plume model per unit span. Define volume and kinematic momentum fluxes $Q=bw$ and $J=b{w}^2$, and set the pure-ice reduced gravity $g^{\prime }=g({\rho }_0-{\rho }_s)/{\rho }_0>0$. The suspension buoyancy is $g^{\prime }\varphi$. The supplied equations express vertical momentum conservation, conservation of relative enthalpy including latent heat, and conservation of excess salt, respectively. At the release height, prescribe compatible source volume/momentum, salt and enthalpy fluxes, or equivalent source width, velocity, composition and ice fraction; impose ambient entrained water at $(T_0,C_0)$ with no vertical momentum. A pure far-field similarity solution may instead be described using a virtual origin, but its singular near-origin values are not physical inlet conditions.

The missing balance is total mass, equivalently volume under the Boussinesq approximation. In a wall plume, only the exposed outer edge entrains, so the Batchelor entrainment hypothesis $v_e=ew$ givesIf two free edges are intended, replace $e$ by $2e$ everywhere; the numerical coefficient is a geometry convention. The glacier boundary itself has no normal flow. Local phase conversion does not add volume at retained Boussinesq order, and crystals are assumed to share the liquid velocity.

For a box of height $dz$, the upward momentum flux changes by ${\rho }_0[J(z+dz)-J(z)]$. The ambient hydrostatic pressure has been subtracted; the remaining force is the suspension buoyancy ${\rho }_{0}b{g}^{\prime }\varphi dz$. Entrained ambient water carries no vertical momentum, and drag is neglected, giving $J^{\prime }=b{g}^{\prime }\varphi$. Salt entering through the exposed edge is $C_0{Q}^{\prime }dz$, so $(\bar{C}Q{)}^{\prime }=C_0{Q}^{\prime }$ and thereforeThese control-volume derivations account for entrainment rather than treating $Q$ as constant.

Thermodynamic elimination. Let the conserved excess enthalpy and salt fluxes be $E$ and $S$:The ambient satisfies $T_0=-m{C}_0$, while equilibrium in the ice-bearing plume givesSubstitute this temperature in the enthalpy invariant. Exact rearrangement givesThus, at dilute-crystal order,The ice-bearing wall-plume thermodynamic invariant fixes the leading advected ice volume and buoyancy flux from the source deficits. A rising ice-dominated branch requires $I_0>0$. It is not determined from reduced gravity alone. The small-$\varphi$ expansion also presumes finite source fluxes and no cancellation making the discarded terms comparable to $-E-c_{p}mS$.

The three remaining leading equations are $Q^{\prime }=ew$, $(Qw{)}^{\prime }=\mathcal{F}/w$, and $\varphi =I_0/Q$. Seek a similarity form in $Z=z-z_v>0$ with $Q\propto Z^p$, $w\propto Z^q$. Continuity gives $p=q+1$, while momentum gives $p+2q=1$. Therefore $q=0,p=1$. The coefficients obey $e{w}^3=\mathcal{F}$, giving the self-similar ice-bearing wall plumeThe width grows linearly, the speed approaches a constant, and the crystals dilute inversely with height. This is a buoyancy-conserving wall line plume, with buoyancy supplied mainly by the ice fraction rather than a temperature anomaly. The dilute-crystal regime requires $Z\gg w^2/g^{\prime }$ and lies away from the formal virtual-origin singularity. If all source flux deficits vanish, $I_0=0$ and this nontrivial rising branch does not exist. The exact thermodynamic invariants above, rather than their truncated version, are needed near a concentrated source.

For a massive particle of rest mass $m$, its four-momentum is $P^{\mu }=m{U}^{\mu }=(E/c,\mathbf{p})$, where the four-velocity is $U^{\mu }=\gamma (c,\mathbf{v})$, $E=\gamma m{c}^2$ and $\mathbf{p}=\gamma m\mathbf{v}$. With metric signature $(-,+,+,+)$, $P\cdot P=-m^2{c}^2$. A photon has four-momentum $(E/c,(E/c)\hat{\mathbf{n}})$, a future-pointing null vector. Four-momentum conservation states that the total incoming and outgoing four-momenta agree for an isolated interaction; this includes both energy and momentum conservation in every inertial frame.

Take the incoming photon direction as $x$. Its energy is $\hslash \omega =\mu m{c}^2$, so the total initial four-momentum isThe invariant mass available after photon absorption is therefore $M=m\sqrt{1+2\mu }$. In the centre-of-momentum frame, the two equal daughters have opposite momenta. Their total energy is at least their combined rest energy $2\alpha m{c}^2$, with equality precisely when both daughters have zero momentum in that frame. Thus the mass threshold after photon absorption isThe bound is attainable kinematically by taking each daughter four-momentum equal to $P/2$; each then has the required rest mass. It immediately gives ${\alpha }_{\max }\to 1/2$ as $\mu \to 0$.

At this threshold both daughters move with the centre-of-momentum frame. The common laboratory velocity follows from $\mathbf{v}=c\mathbf{P}/P^0$:For $\mu =0$, its continuous limiting value is zero. At smaller allowed daughter rest masses, some of the centre-of-momentum energy appears as their relative kinetic energy, so their four-momenta need not be $P/2$.

First apply momentum conservation to the rocket and the infinitesimal amount of expelled gas. During $dt$, the exhaust mass is $\alpha dt$ and its inertial velocity, to first order, is $v-u$. With no dust, the momentum balance isIt gives the rocket equation $\boxed{m\dot{v}=\alpha u}$, while $\dot{m}=-\alpha$.

For a dust-collecting rocket, the incoming dust mass is $\beta vdt$ with zero initial momentum. The same balance has rocket mass $m+(\beta v-\alpha )dt$ after the interval, soConsequentlyThe second term in the acceleration equation accounts for the momentum required to bring collected dust up to rocket speed.

Set $\lambda =\sqrt{\alpha /(\beta u)}$ and $V=\lambda u$, so $\alpha u=\beta V^2$. As long as the engines operate and $0\le v<V$, the velocity increases and time can be eliminated:Integrating from $v=0$, $m=m_0$, givesThus the required mass–velocity relation is

If $\lambda >1$, this formula would give $m\to 0$ as $v\to V^{-}$. In this interval, $\dot{m}=\beta (v-{\lambda }^{2}u)<0$. A real rocket retains positive body mass even after its fuel is gone, and collected dust only increases that residual mass. It therefore cannot follow this branch all the way to zero mass: its fuel must run out while $v<\lambda u$. Indeed the hypothetical time integral $dt/dv=m/[\beta (V^2-v^2)]$ is integrable at $V$ when $\lambda >1$, so even the formal zero-mass endpoint would occur at finite time.

If $0<\lambda <1$, the same relation instead gives $m\to \infty$ as $v\to V^{-}$. The mass first decreases, reaches its minimum at $v={\lambda }^{2}u$, and then increases because dust arrives faster than fuel is expelled. The time integral now diverges at $V$: under indefinitely maintained thrust, $v$ approaches $\lambda u$ asymptotically while stored dust grows without bound. At that limiting speed, thrust $\alpha u$ balances the dust-loading term $\beta v^2$, and $\dot{m}\to \beta u\lambda (1-\lambda )>0$. This is the physical reason the total rocket mass no longer signals fuel exhaustion. A rocket with a finite initial fuel supply still runs out at finite time $t=M_{\mathrm{fuel}}/\alpha$ and before reaching $V$; increasing total mass does not create new fuel.

Take an oriented closed string in flat, critical bosonic string theory, with ${\alpha }^{\prime }=1/(2\pi T)$ and a mostly-plus target Minkowski metric. After continuation of the worldsheet to Euclidean signature, the Polyakov path integral sums over embeddings $X$ and worldsheet metrics, divided by worldsheet diffeomorphisms and Weyl transformations. In a flat target its kinetic action isTarget-time continuation or analytic continuation of external momenta defines the Lorentzian scattering amplitude; a naive real Euclidean Gaussian for timelike $X^0$ would not be convergent.

An external tachyon is represented by the tachyon vertex operator $V_p(z,\bar{z})=:e^{ip\cdot X}:$, with conformal weights $h=\bar{h}={\alpha }^{\prime }{p}^2/4$. The physical integrated vertex has $(h,\bar{h})=(1,1)$, so $p^2=4/{\alpha }^{\prime }$. Schematically its tree amplitude isThe Faddeev-Popov determinant from conformal gauge is represented by the worldsheet ghost fields. At tree level the worldsheet is a Riemann sphere, whose unpunctured complex structure has no moduli. Its residual conformal automorphisms are the Möbius transformations, $PSL(2,\mathbb{C})$. Fix three insertion points, accompanying their unintegrated vertices by the required $c\tilde{c}$ ghost factors. The remaining $N-3$ complex insertion positions are integrated over the sphere; equivalently one integrates all positions and divides by the residual conformal group.

The embedding fields are free, withTheir zero-mode integral gives momentum conservation, and their nonzero-mode Gaussian integral gives the Koba-Nielsen factorThis explains the sphere tachyon position integral without needing its evaluation. For four tachyons the resulting Virasoro–Shapiro amplitude displays the exchanged string spectrum directly.

Use all-incoming external momenta and introduce $u=-(p_1+p_4{)}^2/(8\pi T)$ alongside the two printed dimensionless Mandelstam variables. Since $p_a^2=8\pi T$ and ${\sum }_a{p}_a=0$, one obtains $s+t+u=-4$. The physical center-of-mass energy squared in the $s$ channel is $s_{\mathrm{phys}}=8\pi Ts=4s/{\alpha }^{\prime }$. The $t$ channel measures the analogous crossed momentum transfer, with sign determined by the mostly-plus convention. Rewriting the Gamma factors in a symmetric form givesThe Gamma function poles imply, at generic fixed values of the other invariant,The denominator Gamma factors can remove residues at special intersecting channel kinematics; the statement concerns a generic single-channel limit. An $s$-channel pole occurs when the intermediate momentum $p_1+p_2$ satisfies the mass-shell condition for a closed-string state:The pole at $s=-1$ exchanges the ground-state tachyon, the pole at $s=0$ exchanges massless states, and the positive integer poles exchange the infinite massive tower. The $t$-channel interpretation is the crossed version. Factorization means that each residue is a sum of products of couplings to intermediate physical states that couple to the chosen external particles. It need not expose every representation at that mass.

For example, the Gamma function recurrence and Gamma function residue at a nonpositive integer give the dimensionless Virasoro–Shapiro amplitude pole residueThe residue polynomial has degree $2n$, consistent with maximum spin $2n$ in the exchanged level. The amplitude also has the corresponding $u$-channel poles by crossing symmetry.

The massless fields can be treated as target backgrounds rather than separate asymptotic insertions. Write the target metric as $G_{mn}={\eta }_{mn}+h_{mn}$, introduce a Kalb–Ramond field $B_{mn}$, and a dilaton $\Phi$. In conventional Euclidean signs their string nonlinear sigma model action isHere ${ϵ}^{ab}$ is the antisymmetric tensor density. The three backgrounds correspond to the graviton, antisymmetric tensor and scalar states at closed-string level one. Expanding the vacuum functional in $h,B,\Phi$, then Fourier expanding the backgrounds, produces exactly their integrated string vertex operators. Its functional derivatives therefore generate the amplitudes with massless external strings. The connected vacuum functional organizes connected amplitudes; the spacetime effective action organizes the corresponding vertices after treating massless propagation consistently.

At momenta small compared with $1/\sqrt{{\alpha }^{\prime }}$, massive string propagators can be expanded in powers of momenta over their masses. The analytic part of the amplitudes consequently determines local higher-derivative interactions, ordered by powers of ${\alpha }^{\prime }$. Massless exchange poles are retained through propagation of the massless fields, rather than expanded into local contact terms. Up to field redefinitions, the leading massless-sector action in the string-frame metric isThe terms denoted $O({\alpha }^{\prime })$ contain additional derivatives, including curvature-squared terms in the bosonic theory. The expansion concerns the massless sector around the perturbative bosonic background; the tachyon instability remains and is not cured by omitting its field from this displayed action. Thus this is a formal perturbative effective description, not a claim of a stable bosonic vacuum.

Finally split the dilaton into a constant and its variation, $\Phi ={\Phi }_0+\varphi$. By the Gauss-Bonnet theorem, its dilaton Euler-characteristic weighting on a connected closed oriented surface follows fromDefine the string coupling by $g_s=e^{{\Phi }_0}$. A genus-$g$ path integral is weighted byThe sphere carries $g_s^{-2}$, the torus carries $g_s^0$, and each extra handle adds $g_s^2$. At higher genus one integrates over complex-structure moduli as well as insertion points, with the associated antighost insertions supplying the correct moduli measure. With canonically normalized external vertices an $N$-point genus-$g$ amplitude scales as $g_s^{2g-2+N}$.

Summing connected worldsheets of every genus of a surface yields the string-loop effective action expansionEach coefficient has its own low-energy ${\alpha }^{\prime }$ expansion. The two parameters have different roles: ${\alpha }^{\prime }$ resolves finite string size through higher derivatives, while $g_s^2$ counts additional string loops.

All dots in this solution denote differentiation with respect to conformal time $\eta$. This is required by comoving $k$, the streaming term $k{\Theta }_{\ell }$, and $\dot{\tau }=-a{\bar{n}}_e{\sigma }_T$. The PDF's reference to proper-time derivatives is inconsistent with these coefficients. Proper-time equations follow by dividing the conformal-time equations by $a$, replacing streaming $k$ by $k/a$ and the collision rate by $-{\bar{n}}_e{\sigma }_T$.

The monopole equation of the photon Boltzmann hierarchy is ${\dot{\Theta }}_0+k{\Theta }_1/3=\dot{\varphi }$; its collision term vanishes. Consequently the photon continuity equation isFor the dipole the hierarchy givesUsing the relations from the preceding part, the photon Euler equation becomesBecause $\dot{\tau }<0$, the collision term drives the photon velocity towards the baryon velocity, as Thomson scattering should.

For pressureless baryons, stress-energy conservation without energy exchange gives $\boxed{{\dot{\delta }}_b-k{v}_b-3\dot{\varphi }=0}$. At this order Thomson scattering changes momentum but not the local photon energy: energy transfer by the relative bulk motion is second order, and recoil/thermal energy exchange is neglected. Total momentum conservation fixes the baryon force to be the negative of the photon force. The momentum densities are proportional to their enthalpy densities, $4{\bar{\rho }}_{\gamma }/3$ and ${\bar{\rho }}_b$. ThusIncluding the pressureless expansion and gravitational terms yields the baryon Euler equation with Thomson dragThe expansion coefficient for photons cancels against the changing background pressure, while pressureless baryons retain the Hubble drag term. The opposite drag signs and the enthalpy ratio ensure that the combined photon-baryon fluid conserves momentum.

Use flavor hypercharge $Y=B+S$, where $B$ is baryon number and $S$ is strangeness. The isospin coordinate is $I_3$. The baryon octet has coordinatesHere the nucleons have $S=0$, the Sigma baryons and Lambda baryon have $S=-1$, and the Xi baryons have $S=-2$. The central Sigma baryon belongs to an isospin triplet while the central Lambda baryon is an isospin singlet; equal coordinates do not identify the states.

The pseudoscalar meson octet isThe pions have $S=0$, the upper kaons have $S=+1$, and their lower antiparticles have $S=-1$. The two central states are the neutral pion and the Eta octet state. This is the octet basis of flavor symmetry; the physical eta can also mix with the flavor-singlet state.

For the flavor SU(3) Cartan generators, choose the Hermitian physics convention and the inner product $⟨A,B⟩=\mathrm{tr}(AB)$. Then an orthonormal diagonal basis isStrictly, $i{h}_1,i{h}_2$ are elements of the anti-Hermitian SU(3) Lie algebra; $h_1,h_2$ are the corresponding Hermitian observables. In the quark basis $(u,d,s)$, the up and down quarks form an isospin doublet and the strange quark is a singlet. Their $I_3$ values are $(1/2,-1/2,0)$. Each quark has baryon number $1/3$, and their strangeness values are $(0,0,-1)$. It follows thatThese flavor hypercharge conventions differ from the electroweak hypercharge convention. If instead the inner product is $2\mathrm{tr}(AB)$, the orthonormal basis is $t_3=h_1/\sqrt{2}$, $t_8=h_2/\sqrt{2}$, and the same operators are $I_3=t_3$, $Y=2t_8/\sqrt{3}$.

In the ordinary quark model, the proton has valence content $uud$ and charge $+1$, while the neutron has content $udd$ and charge zero. Additivity of electric charge gives $2q_u+q_d=1$ and $q_u+2q_d=0$, hence $q_u=2/3$, $q_d=-1/3$. The Sigma baryon ${\Sigma }^{-}$ has content $dds$ and charge $-1$, giving $2q_d+q_s=-1$. Thus the quark triplet's electric charges, in units of the positive elementary charge, areThe Gell-Mann--Nishijima formula is consequentlyIt also reproduces every baryon octet charge from the first diagram. On antiquarks the additive quantum numbers reverse sign, and combining a quark with an antiquark reproduces the meson octet charges.

Because the down and strange quarks have identical electric charge, $Q$ commutes with the U-spin generatorsThey satisfy $[U_a,U_b]=i{ϵ}_{abc}{U}_c$; equivalently the anti-Hermitian matrices $i{U}_a$ span an $\mathfrak{su}(2)$ subalgebra. The entries on the $d,s$ block of $Q$ are equal, so $[Q,U_a]=0$ for all three U-spin generators. The electric charge is therefore constant within each irreducible U-spin multiplet. For example, U-spin relates ${\pi }^{+}$ and $K^{+}$, and relates $p$ and ${\Sigma }^{+}$, without changing their charge. It does not imply exact mass degeneracy: unequal down- and strange-quark masses break U-spin.

For pion-nucleon octet channels, assume the collision is governed by the strong interaction. The initial baryon number is one and strangeness is zero, so an outgoing meson-baryon pair must preserve $B=1$, $S=0$, and electric charge. Thus its total flavor hypercharge is $Y=1$. The allowed types areA kaon of $S=+1$ can accompany a Lambda baryon or Sigma baryon of $S=-1$. An antikaon cannot balance the nonpositive strangeness of an octet baryon. A Xi baryon would require a meson of $S=+2$, which the meson octet does not contain.

Resolving these types by electric charge gives all possible pairs:In the isospin-symmetric approximation, total isospin is conserved as well: the incoming $1\otimes \frac{1}{2}$ contains $I=\frac{1}{2},\frac{3}{2}$. The $\pi N$ and $K\Sigma$ channels contain both values, while ${\eta }_{8}N$ and $K\Lambda$ contain only $I=\frac{1}{2}$. The extreme-charge initial states are pure $I=\frac{3}{2}$, consistently excluding ${\eta }_{8}N$ and $K\Lambda$. Clebsch-Gordan coefficients relate amplitudes in different charge channels; the table establishes permission, not equal probabilities. Electromagnetism and unequal up- and down-quark masses introduce small violations of isospin symmetry.

Finally, energy and momentum conservation require $\sqrt{s}\ge m_M+m_B$ for a particular pair, where $s$ is the squared total four-momentum. Only channels above their own threshold can occur. Total angular momentum and parity symmetry in quantum field theory constrain the partial waves of the meson-baryon scattering: a pseudoscalar meson and a positive-parity spin-one-half baryon have pair parity $-(-1{)}^L$ and total angular momentum $J=L\pm \frac{1}{2}$ (only $J=\frac{1}{2}$ for $L=0$). Initial and final partial waves must have matching $J$ and parity. Sufficient energy can open a channel, but cannot remove the electric charge conservation, baryon number, or strangeness constraints.

Let $x_1,x_2,x_3$ be longitudinal displacements from equilibrium. The original diagram has masses $m,2m,m$, and no attachments to fixed walls. The Lagrangian isThe Euler-Lagrange equations giveThe forces sum to zero, consistent with momentum conservation for an isolated molecule.

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.

Now $m$ denotes the instantaneous rest mass of the rocket, including its fuel and stored internal energy, and $v$ is measured in a fixed inertial frame. Assume the exhaust is expelled backwards with speed $0<u<c$ in the rocket's instantaneous rest frame. By the relativistic velocity-addition formula, material emitted at time $\tau$ has inertial velocityIt coasts at this velocity after emission.

The crucial change is that the expelled material's rest mass is not simply $-dm$: some of the rocket's rest energy becomes exhaust motion. Instead, use conservation of relativistic energy. Put $q(t)=m(t)\gamma (t)$, so the rocket's energy is $c^{2}q(t)$. The exhaust emitted during $d\tau$ carries energy $-c^2{q}^{\prime }(\tau )d\tau$. For material moving at velocity $w$, the relativistic energy-momentum relation gives momentum $wE/c^2$, since $E={\gamma }_{\mathrm{ex}}\mu c^2$ and $p={\gamma }_{\mathrm{ex}}\mu w$. Hence the modified total momentum isMeanwhile the total relativistic energy is $c^2[q(t)-{\int }_0^t{q}^{\prime }(\tau )d\tau ]=c^{2}q(0)$. These expressions consistently account for both energy and momentum of the expelled material.

Differentiate the momentum formula. The momentum conservation law givesTo simplify it, useMultiplying by ${\gamma }^2(1-uv/c^2)$, the equation becomesThe two terms containing $uv/c^2$ cancel. Dividing by $\gamma >0$ proves the relativistic rocket equation

Define the open-string endpoint momentum flux along the open string by $J_m=e{T}^2{X}_m^{\prime }+u{P}_m$. Variation of the Nambu-Goto phase-space action givesThe two last equations are the Virasoro constraints. Integration of the momentum equation givesThus the necessary and sufficient condition for total momentum conservation is equality of the two endpoint fluxes, component by component. Vanishing of both fluxes is a sufficient local boundary condition.

The spatial boundary term in the action variation is $-\int dt[J_m\delta X^m{]}_0^{\pi }$. For free ends the endpoint variations are arbitrary, soIn a gauge with $u=0$ and nonzero $e$ this reduces to the usual Neumann boundary condition $X^{\prime m}=0$. The conclusion is conservation of total momentum, not that total momentum must vanish. The PDF contains the derivative in this conclusion; the TeX transcription drops it.

For endpoints confined to $x^1=0$, the Dirichlet boundary condition fixes $\delta X^1=0$. It therefore imposes no requirement that $J_1$ vanish. Preserving the fixed position requires ${\dot{X}}^1=e{P}^1+u{X}^{\prime 1}=0$ at each endpoint, but that is a different condition. For example, in $u=0$ gauge the endpoint momentum $P^1$ is zero while $e{T}^2{X}^{\prime 1}$ can be nonzero. Consequently ${\mathcal{P}}_1$ is generally not conserved: the normal endpoint forces transfer momentum to the hyperplane. A dynamical D-brane recoils and carries the missing momentum. If the hyperplane is idealized as fixed, its external support absorbs the momentum. The combined string-and-support system conserves momentum; freely varying directions tangent to the hyperplane retain their vanishing-flux condition.

Use the Fourier transform convention $\zeta (\mathbf{x})=\int d^{3}k{e}^{i\mathbf{k}\cdot \mathbf{x}}\zeta (\mathbf{k})/(2\pi {)}^3$. Statistical homogeneity and isotropy define the primordial bispectrum byThe Dirac delta function enforces momentum conservation; the three magnitudes describe a closed triangle. Set $\alpha =3f_{\mathrm{NL}}/5$. For the centered quadratic Gaussian transformation, its quadratic Fourier term isAt first order in $\alpha$, choose the quadratic field at one of the three external positions. For the first position, Wick contractions pair the two internal fields with the two remaining external fields in two ways. The self-pairing is precisely removed by the subtracted mean. The surviving contribution is $2\alpha P(k_2)P(k_3)$ times the momentum delta. Adding the other positions gives the tree-level local-type primordial non-Gaussianity resultThis expression is at leading order in $f_{\mathrm{NL}}$. The exact quadratic map also has a loop correction to the local primordial bispectrum, from three quadratic vertices:There is no quadratic-in-$\alpha$ bispectrum term, because it would involve five centered Gaussian fields. A regulator may be needed for this higher-order integral. The displayed leading formula uses the Gaussian power $P$, consistently neglecting its order-$f_{\mathrm{NL}}^2$ correction.

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.

Use momentum conservation and conservation of energy in the laboratory inertial frame. The photon has momentum magnitude $E/c$, while the outgoing Electron satisfies the relativistic energy-momentum relation. Its total energy and momentum obeySubstitute these into $E_e^2=m^2{c}^4+c^2|{\mathbf{p}}_e{|}^2$. Cancelling $m^2{c}^4$ and $(E_1-E_2{)}^2$ gives $m{c}^2(E_1-E_2)=E_1{E}_2(1-\cos \theta )$. Both photon energies are positive, so division yieldsThis is Compton scattering. The energies include the Electron's rest energy; using only its kinetic energy in the mass-shell equation would give the wrong result. The outgoing photon has unchanged energy at $\theta =0$ and smaller energy for a nonzero scattering angle.

Take Euclidean gamma matrices with $\{{\gamma }_{\mu },{\gamma }_{\nu }\}=2{\delta }_{\mu \nu }$, and independent Grassmann fields $\psi ,\bar{\psi }$. The minimally coupled Dirac action isIts mass term is the Dirac mass term, and the interaction is $-ig{\bar{\psi }}^a{\gamma }_{\mu }{A}_{\mu }^b(T_{\mathrm{ad}}^b{)}_{ac}{\psi }^c$. With Fourier convention $e^{ip\cdot x}$, the free kinetic operator is $i/p+m$. The Euclidean adjoint Dirac Feynman rules are thereforeThe vertex sign follows by expanding $e^{-S_{\psi }}$, not $e^{i{S}_{\psi }}$. Momentum conservation supplies the usual Dirac delta function at each vertex. There is only a one-gluon fermion vertex, with no two-gluon seagull because the fermionic covariant derivative is first order. A closed fermion loop has the usual fermionic sign $-1$. The identity $(i/p+m)(-i/p+m)=(p^2+m^2)I$ checks the Dirac propagator.