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At a fixed height, first determine the stored quantities per unit vertical height and per unit length along the line source. With $\Delta \rho =\rho -{\rho }_0$, the triangular-profile line plume has the useful profile integralThe geometric volume is $\mathcal{A}=2b$; the mass is $\mathcal{D}={\rho }_0{I}_0+\Delta \rho I_1=b(\rho +{\rho }_0)$. The upward momentum is $\mathcal{P}=W({\rho }_0{I}_1+\Delta \rho I_2)$, while the buoyancy force is $\mathcal{B}=-g\Delta \rho I_1$. ThusHere positive buoyancy corresponds to a rising light plume. If instead buoyancy is defined kinematically, divide $\mathcal{B}$ by ${\rho }_0$; the density-weighted convention here is the one matching the printed equations.

Multiply each local density by the vertical velocity to obtain the fluxes through a horizontal section, still per unit source length:In particular, the stored momentum equals the mass flux $Q$, and the buoyancy flux is $F=g({\rho }_{0}V-Q)$. The different triangular-profile factors in storage and flux must not be replaced by top-hat coefficients.

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.

Seek a nonzero-buoyancy, separable similarity solution of the unsteady Boussinesq triangular-profile line-plume balances, using powers of distance and time. Let $\tau =t+t_{\ast }>0$, allowing a time origin, and initially take distance $z$ from the virtual origin. Balancing the powers in the momentum and buoyancy equations and the entrainment balance gives the formOne way to recover the exponents is to let $Q\propto z^a{\tau }^p$ and $M\propto z^b{\tau }^q$. Matching the spatial powers in ${\partial }_t(Q^2/M)$, $Q_z$ and $M/Q$ gives $a=2,b=3$; matching the time powers gives $p=-1,q=-2$. The momentum equation then gives the powers of $F$. In the resulting solution, $Q^2/M$ is time-independent, so the storage term in the first equation vanishes exactly.

Direct substitution yields the coefficient equationsFor a rising buoyant solution $C>0$, the last equation gives $B=2A/3$. The first then gives $A=\alpha {\rho }_0$, and the second gives $C=B$. The separable decaying line-plume similarity is thereforeReconstructing the physical fields confirmsFor example, ${\partial }_t(FQ/M)=-2\alpha {\rho }_0{z}^2/{\tau }^3$ and $F_z=2\alpha {\rho }_0{z}^2/{\tau }^3$ cancel; the momentum balance similarly gives $-\alpha {\rho }_0{z}^2/{\tau }^2+2\alpha {\rho }_0{z}^2/{\tau }^2=FQ/M$.

The unshifted form uses $t_{\ast }=0$ and is defined for $t>0$, with a singular zero-time limit. To give a finite solution for all $t\ge 0$, choose $t_{\ast }>0$. The equations also permit replacing $z$ by $z+z_{\ast }>0$, a plume virtual origin; at a finite physical source distance this family has flux histories proportional to ${\tau }^{-1},{\tau }^{-2},{\tau }^{-3}$. The Boussinesq approximation requires $(z+z_{\ast })/(g{\tau }^2)\ll 1$ in the region modelled. This is a decaying similarity family; its time and virtual origins require source or initial data. The PDF supplies neither, so a unique forced-startup history cannot be selected from it. In particular, the singular unshifted field should not be presented as a regular solution at $t=0$.

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.

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

Use units $\hslash =c=1$, the Minkowski metric ${\eta }_{ab}=\mathrm{diag}(1,-1,-1,-1)$, and the Fourier transform convention $f(x)=\int d^{4}p{e}^{-ip\cdot x}\tilde{f}(p)/(2\pi {)}^4$. Here gamma matrices satisfy $\{{\gamma }^a,{\gamma }^b\}=2{\eta }^{ab}{I}_4$, and $\bar{\psi }={\psi }^{†}{\gamma }^0$ is the Dirac adjoint. A single Dirac field describes both Electrons and Positrons; these are the particle and antiparticle sectors of that field.

The free Maxwell Lagrangian and Dirac action, with a linear covariant gauge condition for the photon, giveBefore gauge fixing, the Maxwell Lagrangian has zero modes in field theory along $A_a\mapsto A_a+{\partial }_a\lambda$, so its quadratic operator cannot be inverted on all potentials. In the linear Lorenz gauge, the Faddeev-Popov determinant is $\det \square$. It is independent of $A$ and may be absorbed into the normalization; the corresponding Faddeev-Popov ghost fields have no interacting vertices in this Abelian linear gauge. A residual gauge symmetry is removed by the specified boundary conditions.

The free generating functional isThe Dirac field variables and their sources are independent Grassmann fields in this path integral. They anticommute; replacing them by ordinary commuting fields would give the wrong statistics and the wrong functional determinant. The bosonic Gaussian functional integral contributes an inverse square root of a determinant, and the Grassmann Gaussian integral contributes a determinant. If $K_A$ is the photon quadratic operator and $K_D=i{\gamma }^a{\partial }_a-m$, completing the square givesIn the last expression $D_F=i{K}_A^{-1}$ and $S_F=i{K}_D^{-1}$ are the photon propagator and Dirac propagator, with Feynman i-epsilon prescription. The products include the appropriate spacetime integrals and index contractions. Functional derivatives with respect to $J$, and consistently ordered left or right Grassmann derivatives with respect to the fermionic sources, generate the time ordering of the corresponding fields. In Feynman gauge, $\alpha =1$, the momentum-space two-point functions areThe Feynman i-epsilon prescription specifies vacuum boundary conditions rather than an arbitrary inverse of the differential operator.

The covariant photon propagator uses four potential components. Its operator formalism counterpart is Gupta-Bleuler quantization: impose $({\partial }_a{A}^a{)}^{(+)}|\mathrm{phys}⟩=0$ and take the Gupta-Bleuler null-state quotient. This leaves a positive physical state space with two transverse photon polarization vectors. The temporal and longitudinal oscillator components occur in intermediate covariant expressions; the Ward identity removes their dependence from physical amplitudes. The free Dirac field uses the canonical anticommutation relations, producing the same fermionic signs as its Grassmann fields in the path integral.

The operator-path-integral equivalence can be seen directly with a regulator. Divide time into small intervals and insert complete sets of field-coordinate states for bosons, and resolutions in fermionic coherent states for fermions. The bosonic matrix elements produce the phase-space factor $\exp i\sum [\pi \Delta \varphi -H\Delta t]$; integrating out the quadratic canonical momentum produces the bosonic action. The fermionic coherent state overlaps produce the first-order term $i{\psi }^{†}\dot{\psi }$ and the Berezin integral measure. Multiplying the short-time kernels recovers the path integral. Projecting the remote endpoints onto the Fock vacuum with an infinitesimal damping selects the same Feynman propagators as the operator formalism. Field insertions become time-ordered products under this construction. Conversely, their quadratic generating functional obeys the canonical free-field equations and has precisely the oscillator two-point functions, so its higher free correlators agree by the Wick theorem. This establishes the equivalence for the regulated free theory and order by order in the perturbation series.

To couple the matter field electromagnetically, promote its global phase symmetry to the local transformationThe gauge covariant derivative $D_a={\partial }_a+ie{A}_a$ obeys $D_a^{\prime }{\psi }^{\prime }=e^{-ie\lambda }{D}_a\psi$. Replacing ${\partial }_a$ by $D_a$ in the Dirac action therefore gives the invariant matter densityThe electromagnetic field tensor is unchanged by the local transformation. Thus the unfixed quantum electrodynamics action is gauge-invariant. The Dirac current $j^a=\bar{\psi }{\gamma }^a\psi$ is conserved by the matter equations, and the Electron and Positron excitations carry opposite charges. The added gauge fixing density selects a representative and is not itself invariant under arbitrary local transformations; it does not change gauge-invariant observables. At a free fermion vertex,For external on-shell Dirac spinors the corresponding current contraction vanishes. The quantum extension is the Ward identity, which makes physical amplitudes insensitive to adding a multiple of the photon momentum to its polarization vector. A gauge-compatible regularization preserves this vector-current identity.

With ${\mathcal{L}}_{\mathrm{int}}=-e\bar{\psi }{\gamma }^a\psi A_a$, expand $\exp (i\int {\mathcal{L}}_{\mathrm{int}})$ in powers of $e$. A term of order $n$ contains $n$ spacetime integrations and $1/n!$. Applying the Wick theorem pairs the free fields: an $A$-$A$ Wick contraction supplies a photon propagator, and a $\psi$-$\bar{\psi }$ Wick contraction supplies an oriented Dirac propagator. Each insertion supplies an interaction vertex. The permutations of contractions cancel the expansion factorials except for the Feynman-diagram symmetry factor. Interchanging Grassmann fields produces the fermionic sign, including a minus sign for every closed fermion loop. This is how Feynman diagrams arise from expectation values, rather than an extra dynamical assumption.

The normalized vacuum generating functional removes components with no external insertions. In the operator formalism, for an interacting-vacuum expectation value of an inserted product $\mathcal{O}$, the same cancellation appears asVacuum projection and the Feynman i-epsilon prescription are implicit. The linked-cluster theorem exponentiates all connected vacuum bubbles into the same factor in numerator and denominator, so it cancels. Normalize by $Z[0]$ to remove every vacuum component. This cancellation of vacuum bubbles still leaves products of disconnected diagrams that each contain external insertions. If only connected correlators are wanted, differentiate the connected generating functional $W=-i\log [Z[J,\bar{\eta },\eta ]/Z[0]]$.

The resulting momentum-space QED Feynman rules for the bare theory can be stated in Feynman gauge as follows:

The illustrated interaction vertex has incoming fermion momentum $p$, incoming photon momentum $k$, and outgoing fermion momentum $p^{\prime }=p+k$. Its solid-line arrows indicate fermion flow. The photon wavy line has no fermion-flow arrow. The propagators, the vertex $-ie{\gamma }^a$, momentum conservation, and the fermionic signs determine the perturbative amplitudes.

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.