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Continue with metric signature $+---$. The Pauli matrices obey $\{{\sigma }^i,{\sigma }^j\}=2{\delta }^{ij}{I}_2$. Block multiplication in the chiral gamma-matrix representation givesandHence $\{{\gamma }^{\mu },{\gamma }^{\nu }\}=2g^{\mu \nu }{I}_4$, the required Clifford algebra. Multiplying the Dirac equation by $i{\gamma }^{\nu }{\partial }_{\nu }+m$ yieldsThe antisymmetric part of ${\gamma }^{\nu }{\gamma }^{\mu }$ drops out because partial derivatives commute. Therefore every component of the Dirac spinor satisfies the Klein-Gordon equation. The converse is not true: four arbitrary scalar solutions need not obey the first-order Dirac equation.

The Lorentz group consists of real invertible matrices $\Lambda$ preserving the Minkowski metric: ${\Lambda }^{T}g\Lambda =g$. Writing $\Lambda =I+\omega +O({\omega }^2)$ gives ${\omega }_{\mu \nu }=-{\omega }_{\nu \mu }$, so there are six continuous parameters: three spatial rotation angles and three boost rapidities. The Proper orthochronous Lorentz group is the connected component with determinant $+1$ that preserves time orientation. Parity and time reversal are additional discrete operations and are not encoded by the six real parameters in an exponential near the identity.

For the real-generator convention used here, vector generators can be writtenTake ${\Omega }_{\rho \sigma }=-{\Omega }_{\sigma \rho }$ and use the same parameters in the spinor representation. The Lorentz-spinor generators from a Clifford algebra areTo verify the Lorentz algebra, first commute a generator with one gamma matrix:For example, this follows by moving ${\gamma }^{\tau }$ through each product using the Clifford anticommutator. Applying the commutator derivation rule to $\frac{1}{4}[{\gamma }^{\tau },{\gamma }^{\nu }]$ then givesNo extra factor of $i$ belongs in these generators with this commutator convention. Hermitian-generator conventions shift the factors of $i$ into the brackets and the exponential instead.

The same gamma commutator provesIndeed, to first order $S^{-1}{\gamma }^{\mu }S={\gamma }^{\mu }+\frac{1}{2}{\Omega }_{\rho \sigma }[{\gamma }^{\mu },M_{\mathrm{sp}}^{\rho \sigma }]$, exactly the infinitesimal vector action defined above. Exponentiating proves the finite relation. The spinor field transforms as ${\psi }^{\prime }(x^{\prime })=S(\Lambda )\psi (x)$ with $x^{\prime }=\Lambda x$; combining this relation with ${\partial }_{\mu }^{\prime }=({\Lambda }^{-1}{)}^{\nu }{}_{\mu }{\partial }_{\nu }$ preserves the Dirac equation. A $2\pi$ rotation gives $S=-I_4$, so the spinor matrices give the double-cover spin representation, rather than a single-valued ordinary representation of the Lorentz group itself.

For a boost in direction $i$, $M_{\mathrm{sp}}^{0i}=\frac{1}{2}{\gamma }^0{\gamma }^i$ is Hermitian, with eigenvalues $\pm \frac{1}{2}$. A real rapidity $\eta$ therefore produces eigenvalues $e^{\pm \eta /2}$ in $S$, whose moduli are not one. The finite-component spinor representation is not unitary for boosts, although the rotation matrices are unitary. The Lorentz group is noncompact; this does not forbid the unitary infinite-dimensional action on the physical Hilbert space of states.

Nevertheless, since $({\gamma }^{\mu }{)}^{†}={\gamma }^0{\gamma }^{\mu }{\gamma }^0$,This Dirac spinor pseudo-unitarity implies ${\bar{\psi }}^{\prime }(x^{\prime })=\bar{\psi }(x)S^{-1}$, and consequentlyThe Dirac adjoint is precisely the adjoint needed to form this Lorentz scalar; ordinary ${\psi }^{†}\psi$ alone is not a scalar.

Let ${\gamma }^5=i{\gamma }^0{\gamma }^1{\gamma }^2{\gamma }^3=\mathrm{diag}(-I_2,I_2)$. It anticommutes with every gamma matrix and commutes with every $M_{\mathrm{sp}}^{\rho \sigma }$, so $S^{-1}{\gamma }^{5}S={\gamma }^5$. Thus the axial current transforms as a four-vector under proper Lorentz transformations:Under parity it is an axial vector, acquiring the additional pseudovector sign. Contracting two axial currents cancels that sign and uses the invariant metric. Therefore $j_{5\mu }{j}_5^{\mu }$ is compatible with Lorentz invariance, including parity invariance of this contraction. Lorentz invariance of the interaction does not require the axial current to be conserved.

Keep only first order in the metric perturbation and use the Minkowski metric to raise and lower indices; set $G=c=1$. The Levi-Civita connection hasThe quadratic products of Christoffel symbols in the Riemann curvature tensor can be dropped at this order. Contracting the remaining derivatives giveswhere $\square ={\eta }^{\rho \sigma }{\partial }_{\rho }{\partial }_{\sigma }=-{\partial }_t^2+{\nabla }^2$ is the d'Alembert operator. In four dimensions the trace-reversed metric perturbation satisfies $\bar{h}=-h$ and $h_{\mu \nu }={\bar{h}}_{\mu \nu }-\frac{1}{2}{\eta }_{\mu \nu }\bar{h}$. Writing the Linearized Einstein equations in terms of it givesIn Lorenz gauge in linearized gravity, the first, second and fourth terms vanish. Therefore $G_{\mu \nu }^{(1)}=8\pi T_{\mu \nu }$ becomesThe divergence of this equation also gives ${\partial }^{\mu }{T}_{\mu \nu }=0$, the leading-order stress-energy conservation required by the contracted Bianchi identity.

Expand the Einstein tensor about the Minkowski metric, keeping a second-order metric perturbation as well:Here $G^{(2)}[h,h]$ includes all quadratic terms from the Einstein tensor, including the inverse-metric corrections in its contractions. In vacuum, the first-order Linearized Einstein equations give $G^{(1)}[h]=0$, while the next order gives $G^{(1)}[j]=-G^{(2)}[h,h]$. Define the quadratic gravitational effective stress-energy tensor byIt is symmetric because the Einstein tensor is symmetric, and is quadratic in $h$ and its derivatives. The linearized contracted Bianchi identity is the off-shell identity ${\partial }^{\mu }{G}_{\mu \nu }^{(1)}[j]=0$. The second-order vacuum equation therefore impliesOne can obtain the same conclusion directly by expanding the full contracted Bianchi identity to second order: every correction to the flat divergence multiplies $G^{(1)}[h]$ and vanishes on the first-order vacuum solution. Thus conservation holds for vacuum solutions of linearized gravity, rather than for an arbitrary off-shell $h$. Relative to the fixed Minkowski background, $t$ is a tensor under background changes of coordinates, but this does not make it invariant under the perturbative gauge freedom.

Use natural units and the Minkowski metric ${\eta }_{\mu \nu }=\mathrm{diag}(1,-1,-1,-1)$. For the real scalar field, choose ${\mathcal{L}}_0=\frac{1}{2}{\partial }_{\mu }\varphi {\partial }^{\mu }\varphi -\frac{1}{2}{m}^2{\varphi }^2$. The conjugate momentum is $\pi =\dot{\varphi }$. Under an active translation $\delta \varphi =\epsilon a^{\nu }{\partial }_{\nu }\varphi$, the density changes by $\delta {\mathcal{L}}_0=\epsilon {\partial }_{\mu }(a^{\mu }{\mathcal{L}}_0)$. The Noether theorem therefore gives the canonical stress-energy tensorIts divergence is $(\square \varphi +m^2\varphi ){\partial }^{\nu }\varphi$, which vanishes on the Klein-Gordon equation. With vanishing flux at spatial infinity, the charges are conserved. In particular,The minus sign is required because $P^i=\int T^{0i}$ and ${\partial }^i=-{\partial }_i$. These are the four-momentum of a free real scalar field.

For canonical quantization, impose $[\varphi (\mathbf{x}),\pi (\mathbf{y})]=i{\delta }^3(\mathbf{x}-\mathbf{y})$, with both equal-time field-field commutators zero. Inverting the mode expansion givesThe equal-time canonical commutation relation then yieldsFor example, the two mixed field-momentum terms in the first commutator have coefficients $\frac{1}{2}\sqrt{E_{\mathbf{p}}/E_{\mathbf{q}}}$ and $\frac{1}{2}\sqrt{E_{\mathbf{q}}/E_{\mathbf{p}}}$; the delta function sets their sum to one.

Insert the mode expansion into the quadratic energy. Spatial integration supplies $(2\pi {)}^3{\delta }^3(\mathbf{p}\pm \mathbf{q})$. In the $aa$ and $a^{†}{a}^{†}$ terms, the coefficient is proportional to $-E_{\mathbf{p}}^2+{\mathbf{p}}^2+m^2=0$. The remaining terms giveHere $E_0$ is the divergent zero-point energy, formally $\frac{1}{2}\int d^{3}p{E}_{\mathbf{p}}{\delta }^3(0)$. Thus the vacuum-free energy formula requires normal ordering, or equivalently subtraction of this constant. In a finite box with a cutoff this is the ordinary sum $\frac{1}{2}{\sum }_{\mathbf{p}}{E}_{\mathbf{p}}$, so the subtraction is explicit before passing to the continuum.

Likewise, use the Hermitian momentum expression $-\frac{1}{2}\int (\pi \nabla \varphi +\nabla \varphi \pi )d^{3}x$. Terms containing two annihilators or two creators vanish by antisymmetry under $\mathbf{p}\mapsto -\mathbf{p}$, leaving the symmetric number-operator expression. Its vacuum term is zero with an inversion-symmetric regulator. Therefore the normal-ordered free scalar four-momentum isThe creation operator commutators follow directly from $[a_{\mathbf{q}}^{†}{a}_{\mathbf{q}},a_{\mathbf{p}}^{†}]=a_{\mathbf{q}}^{†}(2\pi {)}^3{\delta }^3(\mathbf{q}-\mathbf{p})$:Starting from a vacuum annihilated by every $a_{\mathbf{p}}$, each creation adds a particle of mass $m$, energy $E_{\mathbf{p}}$ and momentum $\mathbf{p}$. The real scalar field has one spin-zero species: its antiparticle is the same species. Products of creation operators commute, so multiparticle states are invariant under exchange of their labels. In a normalized discrete mode, $(a^{†}{)}^r|0⟩/\sqrt{r!}$ exists for every $r\ge 0$; there is no exclusion restriction. These are bosonic statistics from commuting creation operators and prove Bose–Einstein statistics. For completeness, the single-mode thermal sum ${\sum }_{r\ge 0}{e}^{-\beta Er}$ gives $⟨r⟩=(e^{\beta E}-1{)}^{-1}$, the Bose-Einstein distribution at zero chemical potential.

Fix the Minkowski metric convention $\eta =\mathrm{diag}(1,-1,-1,-1)$ and the Levi-Civita symbol ${ϵ}_{123}=1$. The antisymmetry of the Lorentz algebra generators gives $M_{i0}=-K_i$ and the inverse relation $M_{ij}={ϵ}_{ijk}{J}_k$. Inserting one temporal index in each generator immediately yieldsTwo Lorentz boosts therefore generate a rotation through their commutator; the minus sign distinguishes this algebra from the rotation algebra in four-dimensional Euclidean space.

For a spatial generator and a boost, the same Lorentz algebra givesContracting with ${ϵ}_{iab}/2$ givesThus the three Lorentz boosts transform as a spatial vector under rotations.

Finally, the all-spatial bracket becomesUse $M_{ab}={ϵ}_{abr}{J}_r$ in the double contraction with ${ϵ}_{iab}{ϵ}_{jcd}/4$. The epsilon contraction identity reduces it to $[J_i,J_j]=i{ϵ}_{ijk}{J}_k$. One can check the sign directly: $J_1=M_{23}$ and $J_2=M_{31}$ give $[J_1,J_2]=i{M}_{12}=i{J}_3$; cyclic permutations give the other nonzero brackets. The requested coefficients areThese are rotation and boost commutators with a fixed metric signature. The PDF does not explicitly specify the signature. Keeping its generator convention but choosing $\eta =\mathrm{diag}(-1,1,1,1)$ reverses every displayed algebra coefficient: the pairs become $(1,0),(0,-1),(-1,0)$. More generally, if $s={\eta }_{00}=\pm 1$, the three nonzero coefficients are $A_1=-s$, $B_2=s$ and $A_3=s$. Specifying the Minkowski metric is therefore necessary to make the numerical signs unambiguous.

Assume $\lambda >0$ and $v>0$. The vacuum manifold is the sphere $|\varphi |=v$. Choose ${\varphi }_0=v{e}_3$. For the given generators, $t^3{\varphi }_0=0$ whereas $t^1{\varphi }_0$ and $t^2{\varphi }_0$ are nonzero and independent. The Adjoint representation of SU(2) rotates this sphere, and the full connected stabilizer subgroup of $v{e}_3$ is generated by $t^3$. Hence the classical symmetry pattern isThe central element of SU(2) acts trivially on the triplet and belongs to this stabilizer; it does not add a separate broken direction. The gauge-field-free representative has zero gauge field strength and constant scalar magnitude. Locally near this nonzero classical vacuum, unitary gauge aligns the triplet along the third axis.

First make the gauge normalization explicit. Write the gauge kinetic term as $-\mathcal{N}{F}_{\mu \nu }^a{F}^{a\mu \nu }/4$, where $\mathrm{Tr}(t^a{t}^b)=\mathcal{N}{\delta }^{ab}$ if $F=F^a{t}^a$ is the matrix used inside the printed trace. Canonical normalization of a gauge kinetic term givesFor a conventionally normalized component trace, $\mathcal{N}=1$ and $g_c=g$. If the trace is the ordinary matrix trace in the displayed three-dimensional representation, $\mathrm{Tr}(t^a{t}^b)=2{\delta }^{ab}$ and $\mathcal{N}=2$. The PDF does not specify which trace convention is intended; both are covered by this formula.

In unitary gauge, the gauge covariant derivative isExpanding this kinetic term and the scalar potential gives the complete physical-field LagrangianHere a square of a vector means its contraction with the Minkowski metric, in signature $(+---)$. The labels 1 and 2 represent massive vectors; 3 represents the surviving Abelian vector.

To display their charge and all interactions more clearly, put $a_{\mu }={\mathcal{A}}_{\mu }^3$, $W_{\mu }^{\pm }=({\mathcal{A}}_{\mu }^1\mp i{\mathcal{A}}_{\mu }^2)/\sqrt{2}$, and defineWith the sign of gauge field strength printed in the paper, ${\mathcal{F}}^3=f+i{g}_{c}C$ and $G^{\pm }=({\mathcal{F}}^1\mp i{\mathcal{F}}^2)/\sqrt{2}$. Thus the same Lagrangian becomesThe physical masses areIn particular, literal adjoint matrix trace gives $m_W=gv/\sqrt{2}$; the standard canonical component convention gives $m_W=gv$. These are descriptions with differently normalized couplings, not different physical spectra.

The complex vector pair carries opposite charges under the unbroken U(1) gauge symmetry. The gauge field strength terms contain $a{W}^{+}{W}^{-}$ cubic interactions, $aa{W}^{+}{W}^{-}$ quartic interactions, and four-vector interactions involving the charged fields. There is no pure Abelian cubic or quartic self-interaction. The Higgs mode $\eta$ has cubic and quartic scalar interactions and couples through $2g_c^{2}v\eta W^{+}{W}^{-}+g_c^2{\eta }^2{W}^{+}{W}^{-}$. It is neutral and has no tree-level $\eta aa$ interaction. This is the physical charged-vector Lagrangian for an adjoint SU2 Higgs model.

If the angular fields ${\pi }_1$ and ${\pi }_2$ were retained, their vanishing potential masses would identify the two Goldstone bosons of the ungauged triplet. In the gauge theory they mix with the broken-direction gauge fields and can be removed by gauge fixing; the Higgs mechanism uses them as the longitudinal polarizations of the two massive vectors. They are not additional physical massless scalars. The physical degrees of freedom are conserved: $3+3\times 2=9$ before rearrangement, and $1+2\times 3+2=9$ afterwards.

This is not the Standard Model electroweak interaction. It has three original gauge bosons, leaving two massive charged vectors and one massless neutral vector, with no massive neutral Z boson. The Standard Model instead has $SU(2{)}_L\times U(1{)}_Y$, a complex Higgs doublet, and three massive vectors plus the photon. Its charge is $Q=T_3+Y$, not just the surviving $T_3$. Adding fermions cannot supply the missing gauge generator or turn the surviving neutral vector into both a photon and a Z boson. In particular, the usual right-handed fermions in the Standard Model are SU(2) singlets; they would have zero charge if only $T_3$ were available, instead of the charges produced by hypercharge.

Use natural units and the Minkowski metric $(+---)$, and work at tree level with on-shell final particles. Write $M=m_H$ and, for a final particle of mass $m$, ${\beta }_m=\sqrt{1-4m^2/M^2}$. In the Higgs boson rest frame the Lorentz-invariant phase-space measure reduces toTo obtain this, integrate the momentum delta function to set ${\mathbf{k}}_2=-{\mathbf{k}}_1$; the energy delta function is $\delta (M-2\sqrt{k^2+m^2})$, whose radial Jacobian is $E/(2k)$. Hence for an angle-independent final-state spin sum $S=\sum |\mathcal{M}{|}^2$,There is no initial-state spin average for a scalar, and neither charged-particle pair here requires an identical-particle factor of $1/2$.

For Higgs decay to two W bosons, the Feynman vertex gives $\mathcal{M}=(2m_W^2/v){\epsilon }_1^{\ast }\cdot {\epsilon }_2^{\ast }$, up to an overall phase. Contracting the two massive vector polarization sums givesThe constant 2 follows from $4-1-1$ in the contraction, with the last term coming from the two momentum projectors. Putting $x_W=m_W^2/M^2$, the full massive result isThe on-shell two-body width is zero below this threshold; decays through off-shell W bosons into more particles are different channels.

For Higgs decay to a fermion pair, put $y_b=m_b/v$. The amplitude is $\mathcal{M}=-y_b\bar{u}(k_1)v(k_2)$, up to an overall phase and a color Kronecker delta. The fermion spin sums and gamma-matrix trace giveThe color multiplicity in a decay width is $N_c=3$: only a quark and antiquark with matching colors contribute, so the factor is three rather than nine. ThereforeAgain the on-shell two-body width is zero below threshold. This is a partonic tree-level answer with every mass retained; hadronization is outside the specified calculation.

The W boson channel dominates for large $M$ within the tree-level comparison. The widths scale as $M^3/v^2$ and $m_b^{2}M/v^2$, respectively, andThe enhancement comes from longitudinal polarization of a massive vector boson: its polarization vector grows as momentum divided by $m_W$. Thus the longitudinal pair survives the apparently small $m_W^2$ factor in the interaction. At masses so large that the scalar sector is strongly coupled, the tree-level extrapolation itself needs corrections.

Actions and equations. Use a mostly-plus Minkowski metric and define the induced worldsheet metric $h_{ab}={\partial }_{a}X\cdot {\partial }_{b}X$. For a nondegenerate timelike worldsheet, the Nambu–Goto action isVarying the independent metric in the Polyakov action sets its worldsheet stress tensor to zero:In two dimensions this says $h_{ab}={\Omega }^2{\gamma }_{ab}$ for a positive local factor. The factor drops out of $\sqrt{-\gamma }{\gamma }^{ab}$, and substitution gives the Nambu–Goto action. Conversely, any nondegenerate induced metric solves the auxiliary-metric equation up to a Weyl transformation. This classical equivalence of Polyakov and Nambu–Goto actions is a statement about classical embeddings; quantum equivalence additionally requires treatment of the metric measure and anomaly.

Variation of $X^m$ in the metric action, followed by elimination of $\gamma$, gives the Nambu–Goto equations of motionA closed string has periodic $X$ and no spatial endpoint variation.

For a background metric $G$, Kalb–Ramond field $b$, and dilaton $\Phi$, one consistent Lorentzian convention isThe orientation fixes the two-form sign; here the Lorentzian curvature convention is chosen so Wick rotation gives the positive Euclidean dilaton term $S_{E,\Phi }=\int \sqrt{\gamma }\Phi R^{(2)}/(4\pi )$. This avoids hiding the convention in the topology argument. For constant vacuum value ${\Phi }_0$, the Gauss-Bonnet theorem gives $S_{E,\Phi }={\Phi }_0\chi$. A connected closed oriented surface of genus $g$ has Euler characteristic $\chi =2-2g$, soThe connected vacuum amplitude has a string genus expansion ${\sum }_{g\ge 0}{g}_s^{2g-2}{\mathcal{A}}_g$; disconnected vacuum diagrams exponentiate the connected sum. Each additional handle supplies a factor $g_s^2$. This is dilaton Euler-characteristic weighting.

The rotating circle. For the specified embedding, the squared spatial speed and tangent length are both $1/2$, and their spatial inner product is zero. Including $X^0=t$ therefore givesThe metric is constant. Its equation reduces to $\ddot{X}-X^{\prime \prime }=0$, satisfied by the left- and right-moving trigonometric components and the linear time component. On a constant-time slice with $0\le \sigma <2\pi$, the proper length of a string isIt is time-independent.

Let $R(\theta )$ be an ordinary two-dimensional rotation matrix. Rotate the first coordinate pair by $R(t)$ and the second by $R(-t)$. In these time-dependent Cartesian coordinates both pairs become $(\cos \sigma ,\sin \sigma )/2$. A further fixed orthogonal change of basis givesThus every spatial slice is a planar circle of radius $1/\sqrt{2}$; its plane rotates in the ambient four-space. This is a rigid circular string in four spatial dimensions, not a pulsating circle. The rotating axes establish its spatial shape, not a transformation to an inertial spacetime frame.

The momentum density obtained from the Nambu–Goto action is $P_m=-T\sqrt{-h}{h}^{ta}{\partial }_a{X}_m$. Here $P_0=-T$, hence the conserved target-space energy isThe material velocity is transverse to the tangent and has magnitude $1/\sqrt{2}$. Its Lorentz factor is $\sqrt{2}$, so the excess over $TL$ is kinetic energy. The spatial circle being stationary in rotating axes does not eliminate this energy.

Constraints and endpoints. Varying the multipliers in the Nambu-Goto phase-space action imposesThese are first-class constraints generating normal and tangential worldsheet diffeomorphisms. They remove the two longitudinal embedding degrees of freedom; they are not extra physical force laws. Hamilton's equations areFor an open string, integration by parts leaves the spatial boundary variationAllowed variational open-string boundary conditions must make this vanish for every permitted endpoint displacement, and be preserved by the evolution. The bracketed expression is the open-string endpoint momentum flux. For free displacements, set that flux to zero. In a boundary-preserving gauge with $u=0$ and finite nonzero $e$, this is the Neumann boundary condition $X^{\prime m}=0$ at each end. This shows that free-end string boundary conditions are consistent.

At such an end, $C=0$ gives $P^2=0$, and $\dot{X}=eP$ gives ${\dot{X}}^2=0$. For a nontrivial endpoint trajectory with nonzero time velocity, its spatial speed is therefore one. This null motion of a free string endpoint follows from the boundary condition together with the constraints, not merely from the bulk wave equation. The induced metric may degenerate right at a free end; the result is understood as the endpoint limit or in the Polyakov description.

For a fixed spatial $p$-plane, split coordinates into $X^A$ along its $(p+1)$-dimensional worldvolume, including time, and $X^i$ transverse to it. Set $X^{\prime A}=0$ and $X^i=y^i$ at the ends. Tangential variations are free and their momentum flux vanishes; normal variations vanish, so their flux need not vanish. Evolution preserves the fixed normal values with $P^i=0$ at the boundary in the same gauge. These mixed Neumann boundary conditions and Dirichlet boundary conditions consistently restrict the endpoint worldlines to the plane.

The massless open-string excitations then separate into a gauge vector along the worldvolume and transverse scalar fields. A scalar displacement changes the corresponding $y^i$; its interpretation is a fluctuation of the plane's embedding. These worldvolume fields from open-string massless states support the interpretation as a dynamical planar D-brane. In the bosonic theory the still lower ground state is a tachyon, indicating an unstable brane; the interpretation does not require pretending that this tachyon is a stable massless field. Appropriate superstring sectors can remove that instability.

Use the Minkowski metric with signature $(+,-)$, so the Sine-Gordon equation is ${\varphi }_{tt}-{\varphi }_{xx}+\sin \varphi =0$. The light-cone coordinates here satisfyIn particular, there is no extra factor of four with this coordinate normalization. The Bäcklund transformation requires $a\ne 0$, because one of its equations contains $a^{-1}$.

Put $F=(\psi +\varphi )/2$ and $G=(\psi -\varphi )/2$. In the parameter convention of this paper the Sine-Gordon Bäcklund transformation gives $G_{+}=a\sin F$ and $F_{-}=a^{-1}\sin G$. Differentiating, for a twice differentiable transformed field, givesSince $\psi =F+G$, the addition formula yields ${\psi }_{+-}=\sin (F+G)=\sin \psi$. Therefore the transformed field satisfies the same Sine-Gordon equation:Subtracting the two differentiated equations also gives ${\varphi }_{+-}=\sin \varphi$, showing the compatibility with the seed equation. This proves the unheaded request before the numbered parts, without assuming a particular soliton form.

Choose the Fourier transform conventionThen ${\partial }_a$ becomes $i{q}_a$. Applying this to the Lorenz constraint in Proca theory yieldsThus every physical Fourier transform mode is orthogonal to its four-momentum in the Minkowski metric.

Use the Minkowski metric $\eta =\mathrm{diag}(-1,1,1,1)$, $\hslash =c=1$, and $\square ={\partial }_a{\partial }^a=-{\partial }_t^2+{\nabla }^2$. The Proca field has nonzero mass $m$. Vary its action, using the fact that the gauge field strength is an antisymmetric second-rank tensor:The integration by parts discards a boundary term, with the variation fixed at the boundary. Thus the Euler-Lagrange field equation is the Proca equationThe minus sign on $m^2$ is required by the chosen Minkowski metric; after imposing the constraint below, the plane waves have positive energy $E^2=|\mathbf{q}{|}^2+m^2$.