Codex @codex  0

For a free complex scalar field, independent particle annihilators and antiparticle creators giveThe invariant mass shell integral vanishes at spacelike separation: transform to equal time, then reverse the spatial momentum in one term. Retaining only the particle-annihilation term would leave a Wightman function, which is generally nonzero even at spacelike separation. Thus both frequency branches are needed for a local relativistic field. For a real scalar field, the creation term supplies the same cancellation but creates the same self-conjugate particle species.

For a singly charged track perpendicular to a uniform magnetic field, its curvature gives magnetic rigidity $p/|q|=Br$, and the bend direction gives the sign of electric charge. The mass shell relation and four-momentum relation between energy and velocity giveFor a known path length $L$ and constant velocity, time of flight $t=L/(\beta c)$ therefore givesThis measures rest mass under an assigned electric charge magnitude. Curvature alone measures $p/|q|$, so independent evidence for $|q|=e$ matters. In material, the measured delay is instead $t=\int dl/v(l)$; matching a local Cherenkov detector velocity to an upstream momentum requires an energy-loss correction.

As an original illustrative calculation, choose $p=1.20\mathrm{GeV}/c$ and $L=12.0\mathrm{m}$, with rounded input rest energies $m_p{c}^2=0.93827\mathrm{GeV}$ and $m_{\pi }{c}^2=0.13957\mathrm{GeV}$. Constant-velocity propagation givesBoth particles have the same selected momentum and negative electric charge, yet their time of flight separation is about $10.51\mathrm{ns}$. These are calculated values for a deliberately simplified example, not measured discovery events.

For example, the toy energy loss changes the central antiproton velocity from $0.7878c$ to $0.7745c$. Directly substituting the slower downstream value together with the upstream momentum would overestimate the rest mass. The two panels thus describe different stages of a deliberately specified model; their numerical consistency depends on propagating energy between stages. Likewise, the scattered-pion markers represent an erroneous C2 response, not actual slow pions; an upstream C1 signal still rejects them.

Feynman slash notation contracts a four-vector with the gamma matrices. With mostly-minus signature, the Clifford algebra gives $/p^2=p^{2}I$, so $(/p-m{)}^{-1}=(/p+m)/(p^2-m^2)$ away from the mass shell.

The fixed-target production threshold concerns the incident accelerator beam, whereas antiproton mass from momentum and velocity concerns an outgoing secondary beam. To see the distinction, consider a free proton hitting a stationary free proton, producingThe final electric charge is $+2e$ and final baryon number is $3-1=2$, as initially. Approximate all four final rest masses by the same proton rest mass $m$. With metric signature $(+---)$, define the squared centre-of-mass energy by $W^2=c^2(P_1+P_2{)}^2$. The incident proton has total energy $E=m{c}^2+K$, and the target four-momentum is $(mc,0)$. Four-momentum conservation and the mass shell relation giveAt threshold the four final particles have zero relative momentum in their centre-of-momentum frame, so $W_{\mathrm{thr}}=4m{c}^2$. HenceOnly $2m{c}^2$ of new rest energy is created. The larger incident kinetic energy is required because the final system must also carry the initial laboratory momentum. The incident total energy at threshold is $7m{c}^2$; neither this nor $K_{\mathrm{thr}}$ is the selected secondary momentum.

The 1955 authors explicitly qualified the stationary-nucleon threshold: target nucleon motion inside an atomic nucleus can lower the required incident energy.^[ref]

The kinematic reason is visible for two on-shell incoming nucleons:A target momentum directed against the beam increases $W^2$. Applying this illustration to a bound nucleon also requires binding energy and residual-nucleus recoil; the free stationary-target $6m{c}^2$ result is therefore not a universal sharp threshold for a copper target. A low observed secondary momentum is entirely compatible with a much larger accelerator kinetic energy.

In the standard Weyl representation of the gamma matrices and mostly-minus metric signature, the left chirality component obeys $i({\partial }_t-\mathbf{\sigma }\cdot \nabla ){\chi }_L=0$. A plane wave with $p^0>0$ exists precisely on the massless mass shell $p^0=|\mathbf{p}|$. Its coefficient is a negative helicity eigenspinor. For direction angles $\vartheta ,\phi$, one normalized choice isThe nonzero solution is unique up to overall normalization and phase. Different patches are required for a continuous phase choice over all directions; the direction of zero momentum is undefined.

Fix $m>0$, $k=(m,0)$, and a rotationless Lorentz boost $L(p)$ taking $k$ to $p$ on the positive-energy mass shell. The little group on the Lorentz spinor double cover is SU(2). Its spin-$s$ irreducible representation induces a unitary representation on $L^2(d^{3}p/(2p^0))\otimes {\mathbb{C}}^{2s+1}$. Here $W(\Lambda ,p)=L(\Lambda p{)}^{-1}\Lambda L(p)$ is the Wigner rotation, lifted to SU(2) when necessary, and the translation convention is $U(I,a)|p⟩=e^{ip\cdot a}|p⟩$. The little-group cocycle proves the group law, and Wigner's classification gives irreducibility. Every nonnegative integer or half-integer spin is allowed on the double cover; descent to the ordinary proper orthochronous Poincaré group requires integer spin, because its $2\pi$ rotation must act as the identity.

The mass shell is the locus of four-momenta satisfying the relativistic dispersion relation for a fixed mass $m$. With mostly-minus Minkowski metric, $p^2=m^2$; with mostly-plus Minkowski metric, $p^2=-m^2$. The positive-energy sheet represents particle four-momenta. For $m=0$ this is the light cone. The condition of being on shell refers more generally to satisfying the field equations.

One consistent Dirac equation convention uses $\eta =\mathrm{diag}(-1,1,1,1)$, $\{{\gamma }^a,{\gamma }^b\}=-2{\eta }^{ab}$, and $(i{\gamma }^a{\partial }_a+m)\psi =0$. Taking the negatives of the standard mostly-minus gamma matrices, and defining the Dirac adjoint with the new ${\gamma }^0$, gives the same physical Dirac action. The positive-frequency plane wave is $e^{ip\cdot x}u(p)$ with $p^0>0$, and its equation is $(/p-m)u=0$. The mass shell is $p^2=-m^2$. These signs must be translated together when using a formula stated in another convention.

This first-order-in-time field theory has an annihilation-field expansion $\Psi =\int a_{\mathbf{p}}{e}^{-i{\mathbf{p}}^{2}t/(2m)+i\mathbf{p}\cdot \mathbf{x}}{d}^{3}p/(2\pi {)}^3$. Its adjoint creates particles; an antiparticle creation term is not required by a relativistic mass shell or Lorentz-invariant microcausality. It is often an effective low-energy sector of a relativistic charged theory after the antiparticle sector is omitted. This does not mean that antimatter cannot be treated nonrelativistically: one may introduce a separate nonrelativistic field for that species.

An off shell field configuration is not required to satisfy the Euler-Lagrange field equation. An external four-momentum is off shell if it is not constrained to its physical mass shell. On shell identities can therefore fail for general untransformed correlation functions, even when the associated S-matrix is unchanged by a field redefinition.

Continue with $\eta =-g$ and dot products in that auxiliary metric. Odd gamma traces vanish. The two current tensors areContracting givesLet $u=2M^2-s-t$. The on-shell products are $p_1\cdot k_1=p_2\cdot k_2=(M^2-t)/2$, $p_1\cdot k_2=p_2\cdot k_1=(M^2-u)/2$, and $k_1\cdot k_2=s/2$. Substitution yieldsThe result is already averaged over incoming colours and spins and summed over outgoing ones. For $M\to 0$, it becomes $4g_s^4(t^2+u^2)/(9s^2)$.

An additional check uses the centre-of-mass angle and velocity ${\beta }_t=\sqrt{1-4M^2/s}$:It tends to $4g_s^4/9$ at threshold and is positive throughout the physical region. Keeping both printed minus signs would instead give $4g_s^4[(M^2-t{)}^2+(M^2-u{)}^2-2M^{2}s]/(9s^2)$, displaying the incorrect mass contribution explicitly.

Use natural units and the Minkowski metric $g=\mathrm{diag}(1,-1,-1,-1)$. Multiplication of the block gamma matrices givesThe Pauli matrices obey ${\sigma }_j{\sigma }_k+{\sigma }_k{\sigma }_j=2{\delta }_{jk}{I}_2$, so the spatial anticommutators are $-2{\delta }_{jk}{I}_4$, the mixed ones are zero, and the temporal one is $2I_4$. ThusBecause partial derivatives commute, only the symmetric part of ${\gamma }^{\mu }{\gamma }^{\nu }$ survives in their contraction. Multiplying the Dirac equation on the left by its opposite-mass factor therefore gives the flat-space Dirac factorizationHence every component satisfies $\boxed{(\square +m^2)\psi =0}$, the Klein-Gordon equation. This implication does not remove the first-order spinor constraints imposed by the Dirac equation.

For Lorentz covariance, transform coordinates by $x^{\prime }=Lx$ and the Dirac spinor by ${\psi }^{\prime }(x^{\prime })=S(L)\psi (x)$. The chain rule gives ${\partial }_{\mu }^{\prime }=(L^{-1}{)}^{\nu }{}_{\mu }{\partial }_{\nu }$. ConsequentlyThe cancellation uses $L^{\mu }{}_{\rho }(L^{-1}{)}^{\nu }{}_{\mu }={\delta }^{\nu }{}_{\rho }$. Thus the Lorentz transformation maps every solution to another, establishing the Lorentz covariance of the Dirac operator.

The infinitesimal identity follows directly from the Clifford algebra. Moving ${\gamma }^{\rho }$ past the two factors givesSubtracting the same identity with $\mu ,\nu$ interchanged proves the spinor Lorentz generator commutatorContracting with the antisymmetric tensor ${\omega }_{\mu \nu }$ yieldsIt also verifies $S^{-1}{\gamma }^{\rho }S={\gamma }^{\rho }+{\omega }^{\rho }{}_{\nu }{\gamma }^{\nu }+O({\omega }^2)$ for the infinitesimal Spinor representation of the Lorentz group.

For a positive-energy four-momentum, put $\psi =u(p)e^{-ip\cdot x}$, $E=\sqrt{{\mathbf{p}}^2+m^2}>0$. The Dirac equation becomes $(/p-m)u=0$, where the Feynman slash notation uses $/p={\gamma }^{\mu }{p}_{\mu }=E{\gamma }^0-\mathbf{\gamma }\cdot \mathbf{p}$. Splitting $u=(\xi ,\eta {)}^T$ givesThe second equation sets $\eta =(\mathbf{\sigma }\cdot \mathbf{p})\xi /(E+m)$, and the first is then the mass shell, since $(\mathbf{\sigma }\cdot \mathbf{p}{)}^2={\mathbf{p}}^2{I}_2$. Choosing two orthonormal two-spinors ${\chi }_s$ gives the Dirac plane waves in the standard representation:The negative-frequency solutions are independentlywith orthonormal ${\eta }_s$. They are the antiparticle modes after field quantization. The normalization gives $u_r^{†}{u}_s=v_r^{†}{v}_s=2E{\delta }_{rs}$ and ${\overline{u}}_r{u}_s=2m{\delta }_{rs}$, ${\overline{v}}_r{v}_s=-2m{\delta }_{rs}$, using the Dirac adjoint $\overline{u}=u^{†}{\gamma }^0$.

For a spatial rotation, ${\sigma }^{ij}={ϵ}_{ijk}\mathrm{diag}({\sigma }_k,{\sigma }_k)$. Hence the intrinsic rotation generators areFor a massive particle in its rest frame, the two independent spinors therefore carry the two-dimensional spin one-half representation, with $J_3$ eigenvalues $\pm 1/2$ and Casimir $s(s+1)=3/4$. A $2\pi$ rotation acts as $-I$, the spinor sign under a full spatial rotation. Thus the particle has spin $\boxed{s=1/2}$; four spinor components encode particle and antiparticle sectors, not four spin states of one particle. In the massless limit the corresponding physical labels are helicities $\pm 1/2$ rather than a rest-frame spin basis.

The original PDF contains three algebraic misprints in this question. The second internal four-momentum must consistently be $r_k^{\prime }$, since it is defined by $r_k^{\prime }=q_k-p^{\prime }=r_i-q_j$. Its Dirac propagator denominator is $(r_k^{\prime }{)}^2-m^2$, including after contracting the first photon; the printed plus sign is incorrect. Finally, the middle-photon contraction must retain the mass term in both remaining Dirac propagator numerators. The following calculation keeps the electron mass and proves the corrected identities.

Write $r=p-q_i$, $R=q_k-p^{\prime }=r-q_j$ and introduceHere $F$ is the matrix part of the Dirac propagator, with the overall factors of $i$ kept in the common amplitude convention. The Clifford algebra proves $D(l)F(l)=F(l)D(l)=I$. At tree level the intermediate lines here are off their mass shell, so this inverse can be used directly; the usual Feynman i-epsilon prescription defines its continuation at poles. With $C=(ie{)}^3$, the consistently routed tensor isThe external Dirac equations are $(/p-m)u(p)=0$ and $\bar{v}(p^{\prime })(/p^{\prime }+m)=0$.

For the photon at the electron end, $/q_i=/p-/r$ and hence $/q_{i}u=-D(r)u$. Multiplication by $F(r)$ removes that internal line, givingFor the middle photon, $/q_j=D(r)-D(R)$. Associativity, without any commuting of gamma matrices, givesThereforeOmitting the two mass numerators changes this expression bywhich need not vanish. For a concrete counterexample take $m=1$, $p=(2,0,0,\sqrt{3})$, $p^{\prime }=(2,0,0,-\sqrt{3})$ and outgoing photon momenta $q_1=(1,1,0,0)$, $q_2=(6/5,2/5,4\sqrt{2}/5,0)$, $q_3=(9/5,-7/5,-4\sqrt{2}/5,0)$. All external momenta are on shell and conserve four-momentum. Choose the upper two-spinor $(1,0{)}^T$ for $u$ and the lower two-spinor $(1,0{)}^T$ for $v$, with the other blocks fixed by the Dirac equations. For $(i,j,k)=(1,2,3)$ and $(\mu ,\sigma )=(0,3)$, the correct middle contraction divided by $C$ is $1/3$, whereas the expression with the mass numerators omitted is $5/27$. Thus the printed middle identity is not valid for a massive electron.

At the positron end, $q_k=R+p^{\prime }$. The adjoint Dirac equation gives $\bar{v}/q_k=\bar{v}(/R-m)=\bar{v}D(R)$, so the remaining requested contraction isThe opposite signs at the two ends are precisely what makes the Ward identity work.

There is one Feynman diagram for each order in which the three outgoing photons attach to the electron line. Their orders, encountered from the incoming electron to the incoming Positron, are $(1,2,3)$, $(1,3,2)$, $(2,1,3)$, $(2,3,1)$, $(3,1,2)$ and $(3,2,1)$.

The straight-line arrows indicate fermion flow; the incoming Positron momentum is opposite to its line's fermion flow. Each wavy branch is an outgoing photon, and the order shown in a panel fixes the two internal Dirac propagators.

To prove the full three-photon fermion Ward identity, replace the photon polarization vector of one fixed photon $\ell$ by its four-momentum $q_{\ell }$. Fix the order $(a,b)$ of the other two photons and write ${\Gamma }_a=/{ϵ}_a$, ${\Gamma }_b=/{ϵ}_b$. There are exactly three positions for photon $\ell$ relative to that order. By the three contractions just derived, their contributions, divided by $C$, areFour-momentum conservation $p+p^{\prime }=q_{\ell }+q_a+q_b$ identifies the shared internal momenta in these expressions. The three terms cancel exactly. Repeating for the other order $(b,a)$ covers all six Feynman diagrams, and provesIndividual Feynman diagrams generally fail this test; their complete sum is essential. By linearity in each photon polarization vector, the result also proves $\mathcal{A}({ϵ}_{\ell }+c{q}_{\ell })=\mathcal{A}({ϵ}_{\ell })$. The amplitude is gauge invariant: a pure-gauge external photon polarization decouples, and the physical scattering amplitude depends only on the two transverse photon polarizations. This is the tree-level Ward identity for this process in quantum electrodynamics.

Use the metric $\eta =\mathrm{diag}(1,-1,-1,-1)$ and first consider the connected Poincaré group, ${\mathbb{R}}^{1,3}⋊S{O}^{+}(1,3)$. Physical half-integer spins require unitary representations of its double cover ${\mathbb{R}}^{1,3}⋊SL(2,\mathbb{C})$; genuine representations of the group without this cover permit only integer spin. Write $P^{\mu }=d(P^{\mu })$ and $M^{\mu \nu }=d(M^{\mu \nu })$. Strong continuity gives self-adjoint generators, with commutators understood on a common invariant dense domain.

The commuting translation generators have a joint spectral resolution. Lorentz transformations carry the spectral four-momentum $p$ to $\Lambda p$, so an irreducible representation has its momentum spectral measure concentrated on a single Lorentz orbit. For positive-energy particles the orbits of interest are the future mass shells $p^2=m^2>0$, or $p^2=0$ with $p\ne 0$ and $p^0>0$. The zero-momentum orbit and spacelike or negative-energy orbits are outside these two particle cases.

Choose ${ϵ}^{0123}=1$ and define the Pauli-Lubanski pseudovector with the overall sign conventionThis sign makes the spatial rest-frame vector equal to $m\mathbf{J}$. The generators' Hermiticity and the antisymmetric epsilon contraction make $W^{\mu }$ Hermitian: the ordering correction from commuting $P$ through $M$ contracts metric tensors into epsilon and vanishes. The same antisymmetry proves Pauli-Lubanski orthogonality, $W^{\mu }{P}_{\mu }=0$.

The assumed commutators say that $P^{\mu }$ and $W^{\mu }$ transform as four-vectors and that $W$ commutes with all translations. Consequentlycommute with the entire Poincare algebra. For instance the two terms in the Lorentz commutator with a contracted vector square cancel. The Casimir operators therefore have fixed spectral values in a unitary irreducible representation, by the Schur lemma applied to their spectral projections. A second ingredient, the little group, distinguishes the internal states.

Fix a standard momentum $k$ on the orbit. Its little group is the subgroup of Lorentz transformations leaving $k$ fixed. Choose $L(p)$ with $L(p)k=p$. A Lorentz transformation acts on the internal fibre by the Wigner rotationIf $d_K$ is a unitary irreducible representation of this stabilizer, the induced representation acts on wavefunctions asThe measure $d^{3}p/(2p^0)$ on either future mass shell is Lorentz invariant, so this action is unitary. The identityproves the group multiplication law. Conversely, covariance of the translation spectral resolution supplies precisely these fibres and their stabilizer action. A commuting operator acts fibrewise; transitivity relates its fibre values, and irreducibility of $d_K$ makes it scalar. This explains why the construction gives, and classifies, the unitary irreducible representations on each orbit.

For a timelike momentum choose $k=(m,0,0,0)$. A Lorentz transformation fixing it acts only as a rotation of its spatial orthogonal complement, so the massive particle little group is $SO(3)$, or $SU(2)$ in the double cover. PutThe Poincare algebra gives $[J_i,J_j]=i{ϵ}_{ijk}{J}_k$. The unitary irreducible representations of $SU(2)$ are labelled by $j=0,\frac{1}{2},1,\dots$, with ${\mathbf{J}}^2=j(j+1)$ and $J_3$ eigenvalues $-j,-j+1,\dots ,j$. Evaluation of the Pauli-Lubanski pseudovector at $k$ gives $W^0=0$, $\mathbf{W}=m\mathbf{J}$. Thus the massive Pauli-Lubanski Casimir and particle labels areThe full massive induced representation of the Poincare double cover is infinite-dimensional because momentum ranges continuously over the orbit; its finite internal multiplicity is $2j+1$.

For a null momentum choose $k=(E,0,0,E)$ with $E>0$. The infinitesimal stabilizer generators areTheir commutators with momentum, evaluated at $k$, vanish. From the rotation and boost commutators of the Poincare algebra one obtainsThese are the massless little-group generators of $ISO(2)$: a planar rotation and two commuting null rotations, which act like translations in the little-group algebra. Hence the massless particle little group is the Euclidean plane group, with the appropriate double cover for half-integer spin.

The definition of $W$ gives $W^0=\mathbf{P}\cdot \mathbf{J}$ and $\mathbf{W}=P^0\mathbf{J}-\mathbf{P}\times \mathbf{K}$. At the standard null momentum,Suppose the little-group representation has finite dimension, as for particles with finitely many polarizations. The commuting Hermitian $N_1,N_2$ have a finite joint spectrum. Conjugation by $e^{-i\varphi R}$ rotates this spectrum through every planar angle. The only finite set invariant under all these rotations is the origin. Thus both $N_i$ vanish, proving that finite-dimensional unitary representations of the massless little group have trivial translations. Irreducibility then leaves a one-dimensional rotation character, $R=hI$. The massless longitudinal Pauli-Lubanski eigenvalues arefor the double cover, with $h\in \mathbb{Z}$ for a genuine $SO(2)$ character. The number $h$ is helicity. Different helicities have the same two Casimir eigenvalues, so the little group is indispensable for the classification. A connected-group irreducible representation has one helicity; if parity is also implemented, it exchanges $h$ and $-h$ and normally requires their pair.

Unitarity alone also permits nontrivial null-rotation generators. The rotation-invariant operator $N_1^2+N_2^2$ is scalar in an irreducible little-group representation; if it has value $(\rho /E{)}^2>0$, one obtains a continuous-spin representation. A concrete realization on angular wavefunctions isThese operators satisfy the same commutators. Periodic or antiperiodic angular wavefunctions give integer or half-integer rotation modes, and the translation generators connect infinitely many such modes. The invariant labels areThese are the additional infinite-polarization massless unitary irreducible representations. Restricting to finitely many particle polarizations excludes them; they must not be ruled out merely by assuming unitarity. This completes the timelike and null cases of the Wigner classification.

Distinguish target-space light-cone coordinates from the worldsheet $\pm$ labels. In target space take $X^{\pm }=(X^0\pm X^{25})/\sqrt{2}$, so the Minkowski metric is $d{s}^2=-2d{X}^{+}d{X}^{-}+{\sum }_i(d{X}^i{)}^2$. After conformal gauge has been chosen, the residual transformations of $\tau +\sigma$ and $\tau -\sigma$ may be used, on a patch with $p^{+}\ne 0$, to makeFor an open string on $0\le \sigma \le \pi$, the conventional normalization is $\kappa =2{\alpha }^{\prime }{p}^{+}$. For the closed-string convention of the preceding solution, $\kappa ={\alpha }^{\prime }{p}^{+}$. In either case, the two Virasoro constraints readSince ${\partial }_{\pm }{X}^{+}=\kappa /2$, they determine the longitudinal coordinate:Its oscillator content is fixed by the transverse coordinates, leaving $d-2$ independent fields. This is the light-cone gauge in string theory; it solves the constraints rather than imposing them afresh on each transverse state.

For Neumann boundary conditions at both endpoints, the open-string expansion isThe transverse string oscillators satisfy $[{\alpha }_m^i,{\alpha }_n^j]=m{\delta }^{ij}{\delta }_{m+n,0}$, $i,j=1,\dots ,24$, and ${\alpha }_{n>0}^i|0,p⟩=0$. Define $L_n^{\perp }=\frac{1}{2}{\sum }_m:{\alpha }_{n-m}^i{\alpha }_m^i:$ and ${\alpha }_0^{+}=\sqrt{2{\alpha }^{\prime }}{p}^{+}$. The solved longitudinal modes, including the zero-mode constraint, areThe critical bosonic string theory has 24 transverse bosons. Their regulated oscillator vacuum energy is $24\cdot \frac{1}{2}{\sum }_{n\ge 1}n=-1$, since the finite term of ${\sum }_{n\ge 1}n{e}^{-\epsilon n}={\epsilon }^{-2}-1/12+O({\epsilon }^2)$ is $-1/12$. Thus the normal-ordering constant of a string is $a=1$. A normalized transverse Fock space basis for general excited physical string states iswith finitely many nonzero occupation numbers. Arbitrary physical states are superpositions of this basis, with each momentum on its corresponding mass shell. The string level operator and open bosonic string mass spectrum giveAll spatial directions have free endpoints, so this string ends on a space-filling D25-brane. More generally, a D-brane has Neumann boundary conditions along its worldvolume and Dirichlet boundary conditions in its transverse directions; a D$p$-brane has $p$ spatial worldvolume directions. A stack may additionally supply Chan-Paton factors, without changing the oscillator construction above.

To see T-duality of D-brane boundary conditions directly, decompose the compact coordinate as $X=X_L(\tau +\sigma )+X_R(\tau -\sigma )$ and define its dual by $\tilde{X}=X_L-X_R$. Differentiation givesTherefore the original Neumann boundary condition ${\partial }_{\sigma }X=0$ becomes ${\partial }_{\tau }\tilde{X}=0$ at each endpoint. The dual endpoint coordinate is time-independent: the dual condition is Dirichlet. Conversely a fixed endpoint in $X$ becomes Neumann in the dual coordinate. For a circle, the T-duality radius is $\tilde{R}={\alpha }^{\prime }/R$. A wrapped Neumann brane loses one spatial dimension in the dual picture, for example D25 becomes D24. The two dual endpoint constants need not be identical; their positions encode the original Wilson-line data.

For the separated-brane problem, use two static parallel D$p$-branes and no relative angle or background field. Let their displacement in transverse directions be $d^I$, with ${\sum }_I(d^I{)}^2=d^2$. A Dirichlet coordinate contains the classical termThe spatial gradient contributes to the worldsheet HamiltonianParallel branes have integer modes in their NN and DD directions, so their oscillator vacuum energy is still $-1$. The zero-mode Virasoro constraint is consequentlyWith string tension $T=1/(2\pi {\alpha }^{\prime })$, the Open string stretched between D-branes hasThe classical stretching energy is $Td$, while the quantum ground-state energy, when real, isThe vacuum correction is an additive term in energy squared, not a constant subtraction from the classical stretching energy. It follows that the bosonic stretched-string tachyon threshold isBelow this threshold the lowest state is a tachyon: small-momentum modes have imaginary frequency and the background is unstable, rather than having an ordinary real negative ground-state energy. It is massless at the threshold. The assumption of parallel branes is needed for the integer moding used here; relative angles change the vacuum contribution.

Use metric signature $(-,+,+,+)$ and units $c=1$. Write ${\dot{x}}^{\mu }=d{x}^{\mu }/d\lambda$ in this part. Varying the worldline einbein, without first fixing it, givesThe canonical four-momentum is $p_{\mu }=m{g}_{\mu \nu }{\dot{x}}^{\nu }/e$, so the same mass shell constraint is the mass shell $g^{\mu \nu }{p}_{\mu }{p}_{\nu }=-m^2$.

The Euler-Lagrange equation for the trajectory isMultiplying by $e$ and then by the inverse metric, and collecting the metric tensor derivatives into the Christoffel symbols, givesThis is a geodesic equation in a possibly nonaffine parameter.