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The approximate equality of proton and neutron parton distribution functions after interchanging up quarks and down quarks, including their antiquarks. It follows from the corresponding isospin symmetry when electromagnetic and light-quark mass-difference effects are neglected. Isospin-singlet sea distributions are unchanged.

Scattering that resolves the constituents of a hadron at large spacelike momentum transfer, with an inclusive sum over final hadron states.

In electron scattering from a proton, write the reaction as $e^{-}p\to e^{-}X$: the outgoing electron is measured, while $X$ denotes all allowed final hadron states. Large momentum transfer supplies a short-distance probe; a large final invariant mass separates the deep continuum from elastic scattering and individual particle resonances. The useful surprise is that an extended proton can have a response characteristic of much smaller constituents. Deep inelastic scattering turns measurements outside the proton into constraints on its internal parton distribution functions.

The SLAC-MIT deep inelastic electron scattering experiment connects this definition to an actual measurement. Its evidence came from distributions of outgoing electron energies and scattering angles, rather than a photograph of a quark. Bjorken scaling concerns deep-inelastic structure functions after known scattering factors have been accounted for; it does not say that the raw differential scattering cross-section is constant.

The 1969 SLAC-MIT deep inelastic electron scattering experiment found approximate Bjorken scaling: an estimated deep-inelastic structure function depended mainly on a ratio of kinematic variables. Its restricted angular coverage did not separately determine both deep-inelastic structure functions, and the extraction assumed transverse virtual photons. This was motivation for a microscopic theory, not already a measurement of the QCD coupling.^[ref]

The 1973 David Gross–Frank Wilczek calculation supplied a quantum field theory explanation of approximately free partons with calculable logarithmic corrections. Asymptotic freedom therefore permits approximate Bjorken scaling without requiring exact scale independence. The QCD coupling governs the corrections, while parton distribution functions and their evolution connect the short-distance calculation to a proton.^[ref]

In a later direct test, H1 collided Positrons and protons at HERA and selected deep inelastic scattering events. A liquid-argon particle calorimeter identified the outgoing Positron with an associated track; calorimeter deposits and charged tracks reconstructed particle jets in the Breit frame. Vertex, timing and longitudinal energy-balance cuts suppressed backgrounds; two simulations checked detector corrections. Fits of particle jet cross sections to next-to-leading-order QCD extracted ${\alpha }_s(E_T)$ in transverse-energy bins, consistent with decreasing QCD coupling. Hadronization, parton distribution functions and renormalization scale choices limited the inference. This measures rates and fits a coupling, rather than directly reading a force meter.^[ref]

A nuclear target with equal proton and neutron contributions, or data corrected to that average. Its per-nucleon PDFs are $q_N=(q_p+q_n)/2$. Charge symmetry gives $u_N=d_N=(u_p+d_p)/2$; using proton down-quark density alone is not an isoscalar average.

For a massless neutrino scattering on an unpolarized up antiquark through the Fermi interaction, $d\sigma /dy=G_F^{2}s(1-y{)}^2/\pi$. The corresponding down-quark channel has $d\sigma /dy=G_F^{2}s/\pi$. Their difference follows from the chiral weak-current trace contraction. The parton distribution functions weight these rates in a hadron.

The scale at which collinear contributions are assigned to parton distribution functions rather than the hard scattering coefficient. The separate factors depend on this scale, while a consistently calculated physical differential scattering cross-section does not.

In the collinear approximation, a parton carries a fraction $0<x<1$ of its parent hadron's longitudinal momentum. A parton distribution function $f_i(x,Q^2)$ gives its distribution at the specified factorization scale. For production of one on-shell particle with no transverse recoil, $x_{+}{x}_{-}s=M^2$ and $x_{\pm }=(M/\sqrt{s})e^{\pm y}$, where $y$ is its rapidity.

First fix the photon momentum convention. With $Q^2=-q^2>0$, $\nu =P\cdot q>0$ and $x=Q^2/(2\nu )>0$, an absorbed photon gives final momentum $p_X=P+q$. This is the physical momentum support of a deep-inelastic hadronic tensor. The original PDF instead prints ${\delta }^{(4)}(P-p_X-q)$ while keeping that positive-$x$ convention. Taken literally, it would requirein the Bjorken scaling limit, leaving no physical final states. The sign is therefore inconsistent with the requested nonzero deep-inelastic structure functions. Below, use the physically consistent absorption delta function ${\delta }^{(4)}(P+q-p_X)$. Alternatively all photon-momentum and scaling-variable conventions would have to be reversed together.

Work in the leading parton model: the target is unpolarized; scattering is incoherent from massless collinear quarks and antiquarks; intrinsic transverse momentum and target-mass effects are neglected. The parton distribution functions count constituents and include their colour multiplicity. At this order, neglect strong radiative corrections and the resulting logarithmic scale evolution.

Let a quark carry momentum $p=\xi P$, using a lightlike reference $P$ at leading order, and let $r=p+q$. The spin-averaged electromagnetic trace isIntegrating its one-particle final Lorentz-invariant phase space, with the $1/(4\pi )$ tensor normalization, supplies ${\delta }_{+}(r^2)/2$. Thus the spin-averaged electromagnetic parton tensor isOn physical support, $r^2=0$ gives $\xi =x$ and $p\cdot q=Q^2/2$. Define the transverse tensorsThe bracket in $w_f$ becomesThis follows by writing $p=\xi \tilde{P}-q/2$ on shell and expanding; it also verifies transversality to $q$.

The correct parton convolution normalization for a hadronic tensor isThe $1/\xi$ converts number weighting into covariant target-state normalization: a parton response per constituent is normalized by $2p^0$, whereas the hadron tensor uses $2P^0$, and $P^0/p^0=1/\xi$. Antiquark charge changes sign but its square and symmetric spin trace are the same.

Use the parton on-shell delta identity,Matching the two tensor coefficients now givesIn particular $F_2=2x{F}_1$, the Callan-Gross relation for spin-one-half partons. Scaling here means the naive leading approximation; full Quantum chromodynamics produces scale-dependent distributions and corrections.

Each quark contributes $+1$ and each antiquark contributes $-1$ to its flavour-vector charge. Sea pairs therefore cancel in the quark number sum rule:Here $N_f$ is the net flavour number, not the total number including sea pairs. This condition by itself fixes no antiquark distribution.

Under the separately imposed valence-only approximation, ${\bar{q}}_f=0$ and only up/down distributions remain. The proton has $(N_u,N_d)=(2,1)$ and the neutron has $(1,2)$. With $Q_u=2/3$, $Q_d=-1/3$, the valence-only proton and neutron structure-function moments areNo extra factor of three belongs here: colour is already included in the constituent distributions and the flavour counts. These are conditional valence-model moments, not unrestricted asymptotic statements in a theory whose evolution generates a sea.

Use an incoming spacelike momentum transfer $q$, so the physical absorption process has $p_X=P+q$, $Q^2=-q^2>0$, and $P\cdot q>0$. There is a sign inconsistency in the original PDF: its delta function has $P-p_X-q$, while its positive Bjorken x definition and target formulas use incoming $q$. With that literal delta function and $x>0$, the final-state invariant would bewhich becomes negative in the stated high-energy limit and cannot represent a physical hadronic final state. We therefore use ${\delta }^4(P+q-p_X)$, preserving the printed $x=Q^2/(2P\cdot q)$ and structure-function conventions. Equivalently, keeping an outgoing momentum variable requires reversing the signs in those definitions as well.

Assume an unpolarized target, incoherent scattering from effectively free spin-one-half partons, and the massless collinear parton approximation. In the Bjorken scaling limit, take $Q^2,P\cdot q\to \infty$ at fixed $0<x<1$, neglecting target and quark masses and intrinsic transverse momentum relative to $Q$. Write the incoming struck-quark momentum as $k=\xi P$ in the leading lightlike-reference approximation. Let $f_f(\xi )$ and ${\bar{f}}_f(\xi )$ be quark and antiquark parton distribution functions, counting all colours. For the naive scaling result also neglect radiative scaling violations; in full QCD these distributions depend logarithmically on the hard scale.

The outgoing struck quark has momentum $k^{\prime }=k+q$. Its on-shell condition givesAverage its initial spin and sum its final spin. The spin-averaged electromagnetic parton tensor has numeratorThe supplied three-gamma identity gives this directly: contract its middle index with $(k+q{)}_{\lambda }$ and trace with $/k$; the ${\gamma }_5$ trace vanishes. Thus no antisymmetric term survives the unpolarized spin average.

Define transverse tensorsOn the cut, $2k\cdot q=Q^2$, so $\tilde{k}=k+q/2=\xi \tilde{P}$. Expanding the two transverse terms givesThis also explicitly verifies electromagnetic current conservation, $q_{\mu }{S}^{\mu \nu }=q_{\nu }{S}^{\mu \nu }=0$.

The one-particle final-state Lorentz-invariant phase-space measure obeysIncluding the current's charge factor and the given $1/(4\pi )$ normalization, a single quark therefore contributesAn antiquark has the same unpolarized tensor and the same squared charge $Q_f^2$.

The parton convolution normalization for a hadronic tensor isThe $1/\xi$ is important: the distributions count parton numbers, whereas covariantly normalized one-particle responses carry the state normalization $2k^0$. The incoherent sum of normalized responses is $W/(2P^0)={\sum }_f\int d\xi (f_f+{\bar{f}}_f)w_f/(2k^0)$; in the collinear frame $k^0=\xi P^0$, giving the factor above. There is no extra colour multiplicity because it is already included in the distributions.

On the physical cut,Comparing the two transverse coefficients in the hadronic tensor gives, with $n_f=f_f+{\bar{f}}_f$,Therefore the requested deep-inelastic structure functions areThe notation $f_f$ here is the question's $f$ with its flavour index made explicit. The calculation proves the Callan-Gross relation, $F_2=2x{F}_1$, rather than merely citing it. Finite masses and radiative corrections modify this leading approximation; at leading QCD order the same formulas use scale-dependent parton distribution functions.

The quark number sum rule follows because a distribution is a number density. Integrating $f_f$ counts quarks and integrating ${\bar{f}}_f$ counts antiquarks; their difference is the net conserved flavour charge:Every sea quark-antiquark pair contributes zero to this difference. More formally the flavour-vector charge counts quarks minus antiquarks and has a fixed eigenvalue on the hadron state. Thus $N_f$ is the valence or net quark number, not the total number including the sea.

Use both final assumptions printed in the question: only up/down flavours and no antiquarks. The second does not follow just from restricting the flavour list. Then $\int f_u=N_u$, $\int f_d=N_d$, with squared charges $Q_u^2=4/9$ and $Q_d^2=1/9$. The proton has $(N_u,N_d)=(2,1)$ and the neutron $(1,2)$. Hence the valence-only proton and neutron structure-function moments areThe integral limits assume the scaling approximation is valid for the indicated weighted moments, for example by dominated convergence of the weighted integrands. The same net-quark sum rule alone would not fix these individual moments in the presence of antiquarks, since $F_2$ involves the sum of quark and antiquark distributions.

In the parton model, the incoming parton has $k=\xi P_H$ and the outgoing massless parton has $k^{\prime }=\xi P_H+q$. Neglecting $P_H^2$ and imposing its on-shell condition givesThus the Bjorken scaling variable measures the struck parton's longitudinal momentum fraction in this approximation. The inelasticity is unchanged when the hadron momentum is replaced by its parton's collinear momentum:Let $S_H=(p+P_H{)}^2\simeq 2p\cdot P_H$. The partonic invariant in part (b) is $s(\xi )=\xi S_H$, and $Q^2=-q^2=xy{S}_H$. Insert its two parton cross sections into the parton distribution function convolution:Since the event's measured $x$ fixes $\xi =x$, the integrand is the cross-section density in $x$. No additional Jacobian from $y_q$ is needed because $y_q=y$. ThereforeEquivalently $S_H=2E_{\nu }{E}_H(1-\cos {\vartheta }_{\nu H})$ in the chosen massless-hadron frame. The explicit factor $x$ comes from the partonic energy $s=x{S}_H$, not from redefining the parton distribution functions. This expression uses the stated two-flavor and low-energy approximations, with masses and generation mixing omitted.

On the physical positive-energy branch, put $\omega =q^0+\xi E_P$. The massless outgoing parton obeysSince $E_{k^{\prime }}=\omega >0$ on the support and $\nu >0$, the positive-energy on-shell delta function identity givesInsert this into the parton model sum, and writeThe parton distribution functions $q_f,{\bar{q}}_f$ are number densities in momentum fraction. The contributing tensor becomesComparing with part (b), under the same leptonic tensor contraction, givesThis is the Callan-Gross relation for massless partons of spin angular momentum $1/2$ at leading order. The longitudinal deep-inelastic structure function is $F_L=F_2-2x{F}_1=0$ in this approximation. Target-mass effects and radiative Quantum chromodynamics corrections can change the relation. The leading parton model also gives Bjorken scaling: at this level the deep-inelastic structure functions depend on $x$ alone.

Theory. Quantum chromodynamics is a local, non-Abelian gauge theory with gauge group SU(3) acting on color charge. Its matter fields are spin-half quarks of six flavours, each in three colors; its eight massless spin-one gluons are the gauge fields. Unlike the neutral photon, gluons themselves carry gauge charge. Consequently there are three-gluon and four-gluon vertices as well as the quark-gluon vertex. A compact Lagrangian density iswhere $D_{\mu }$ is the gauge covariant derivative and the field strength contains the nonlinear term $g_s{f}^{abc}{G}_{\mu }^b{G}_{\nu }^c$.

The running coupling is weak at high momentum scales. At one loop, for $n_f$ active flavours,The positive coefficient $11-2n_f/3$ produces asymptotic freedom: short-distance, large-$Q$ processes admit perturbation theory. The gluon self-interaction is crucial to its sign. At low scales the coupling grows and isolated colored particles are not observed; the physical spectrum consists of color-singlet hadrons, the phenomenon of confinement. A hard-produced quark or gluon radiates and undergoes hadronization, giving a collimated jet. The one-loop formula is not valid at $Q\sim {\Lambda }_{\mathrm{QCD}}$ and is not, by itself, a mathematical proof of confinement.

Experimental tests. Different measurements test different ingredients of this account:

The convincing evidence is the quantitative consistency across these probes, with a common color group and coupling evolution. A jet by itself is evidence for a hard parton; its rates and correlations supply the more specific tests of Quantum chromodynamics.^[ref]

Production and identification. At a hadron collider, initial partons carry fractions of the proton momenta; observable production rates combine parton-level amplitudes with parton distribution functions.

Why the three masses form a test. At tree level, electroweak symmetry breaking gives $m_W=gv/2$ and $m_Z=\sqrt{g^2+g^{\prime 2}}v/2$. The Higgs mechanism relates masses to the vacuum field, but it does not by itself predict the top Yukawa coupling or the Higgs self-coupling, so the three masses are not fixed independently of measured parameters.

The stronger test comes from radiative corrections. With accurately measured $\alpha$, $G_F$ and $m_Z$, the electroweak precision test with the W, top and Higgs masses usesThe top-bottom loop contributes approximately $\Delta \rho =3G_F{m}_t^2/(8\sqrt{2}{\pi }^2)$, while the Higgs contribution has a characteristic logarithmic mass dependence. Hence direct measurements of $m_t$ and $m_H$ turn the precision calculation into a prediction for $m_W$. Comparing that prediction with the direct W-mass measurement overconstrains the Standard Model; disagreement beyond experimental and theoretical uncertainties could indicate additional particles or interactions in the loops. The test concerns the relation among independently measured quantities, rather than merely the existence of the three particles.

Use the parton model, neglect antiquarks, binding and target mass corrections, quark masses, muon mass, QCD scaling violation and the W boson propagator correction. The struck quark is effectively free during the short interaction and carries parton momentum fraction $x$, giving $\hat{s}\simeq xs$. The charged current reaction is ${\bar{\nu }}_{\mu }{u}_i\to {\mu }^{+}{d}_j$. Its helicity amplitude has the same $(1-y)$ factor as in a. Define the effective parton distribution function by $q(x)={\sum }_{i,j}|V_{ij}{|}^2{u}_i(x)$, summing only kinematically accessible final quarks; for valence-only protons and accessible $d,s$ this is approximately $u(x)$. It is not the unweighted sum $u+d$.Here $y=p\cdot (k-k^{\prime })/(p\cdot k)$ is the inclusive deep inelastic scattering variable, and $x=Q^2/(2p\cdot (k-k^{\prime }))$ in the scaling limit. Integrating $0<y<1$ gives $1/3$, henceThe moment measures the momentum fraction in the active quarks. Approximate linearity is evidence for scattering from pointlike spin-one-half constituents with approximately scale-independent parton distribution functions, rather than coherent scattering from an extended proton. It is an intermediate-energy prediction: thresholds, experimental deep inelastic scattering cuts, QCD scaling violation and eventually the finite W boson propagator modify it.

The parton model treats a fast proton as an ensemble of approximately collinear, nearly free quarks and antiquarks. A hard deep inelastic scattering collision resolves one parton on a time scale short compared with its binding motion. The probabilities are parton distribution functions; the simplest approximation neglects intrinsic transverse momentum, quark masses and the proton mass in the hard collision. Quantum chromodynamics introduces logarithmic violations of Bjorken scaling, so the parton distribution functions depend on a resolution scale as well as a parton momentum fraction.

Write the incident electron and proton four-momenta as $k,P$, the scattered electron four-momentum as $k^{\prime }$, and set $q=k-k^{\prime }$. With the $(+---)$ Minkowski metric, the Bjorken scaling variable and the inelasticity areAn incident parton with four-momentum $\xi P$ emerges on shell, giving $(\xi P+q{)}^2\simeq 2\xi P\cdot q-Q^2=0$. Thus $\xi =x_B$: the observable Bjorken scaling variable measures the struck parton momentum fraction in this approximation.

For clarity, measure the scattering angle $\vartheta$ from the incoming electron direction, and denote the outgoing electron energy by $E_e^{\prime }$. Direct four-momentum products giveThe expression uses the known beam energies and two measured quantities. An electromagnetic calorimeter measures $E_e^{\prime }$; a tracking detector, the collision interaction vertex and the position of the electromagnetic shower determine $\vartheta$. Coverage close to the incoming electron direction matters at small $Q^2$. A hadronic calorimeter supplies an independent consistency check and rejects backgrounds. If the angle is instead measured from the incoming proton, its sine and cosine half-angle factors interchange.

At the parton level, the dominant valence reactions are $e^{-}u\to {\nu }_{e}d$ through W boson exchange and $e^{+}d\to {\bar{\nu }}_{e}u$ through W boson exchange. The first also acts on $c$, while the second acts on $s$ with the appropriate Cabibbo-Kobayashi-Maskawa matrix factors; crossing supplies the antiquark reactions. The lines in the figure show fermion number flow; the Positron and antineutrino arrows point opposite to their momenta.

For unpolarized beams, the leading charged current cross sections, suppressing Cabibbo-Kobayashi-Maskawa matrix factors, have the formHere each parton distribution function is evaluated at $(x,Q^2)$. The proton has two valence up quarks and one valence down quark; moreover, its valence contribution in the Positron reaction carries the chiral angular factor $(1-y{)}^2$. Both effects favour $e^{-}p$. Increasing $Q^2$ at fixed collider energy enforces $x\ge Q^2/s$, progressively suppressing the sea relative to the valence quarks; large $y$ further suppresses the Positron valence term. There is no universal factor of two valid at every $(x,y)$.

The escaping neutrino prevents direct reconstruction from a scattered charged lepton. Instead use the Jacquet-Blondel reconstruction. With the $+z$ axis along the proton beam, measure the complete hadronic final state and formThese relations follow from four-momentum conservation, with the unobserved neutrino transverse momentum opposite to the measured hadronic sum. A nearly hermetic hadronic calorimeter, supplemented by tracking detectors, must measure jets, their energies and their directions. Forward losses, initial-state radiation and small $y_h$ cause significant reconstruction uncertainties.

The photon propagator gives approximately $d{\sigma }_{\mathrm{NC}}/d{Q}^2\propto 1/(Q^2{)}^2$, whereas a massive W boson gives $d{\sigma }_{\mathrm{CC}}/d{Q}^2\propto 1/(Q^2+M_W^2{)}^2$. Up to different coupling constant values, parton distribution functions and angular factors,Consequently the low-$Q^2$ charged current suppression disappears as $Q^2$ approaches $M_W^2$. The source graph shows comparable cross sections at $Q^2$ of order $10^4{\mathrm{GeV}}^2$, so the relevant mass scale is of order $10^2\mathrm{GeV}$.

A more quantitative, still rough reading uses the $e^{-}p$ ratio $R_1\simeq 0.006$ near $q_1=300{\mathrm{GeV}}^2$ and $R_2\simeq 0.17$ near $q_2=2000{\mathrm{GeV}}^2$, where $q_i$ denotes $Q^2$, not $Q$. If $C$ varies slowly over this interval, set $\kappa =\sqrt{R_2/R_1}{q}_1/q_2$. ThenThe reading is compatible with roughly $70$–$100\mathrm{GeV}$; it is not a precision mass determination. In particular, Z boson exchange contributes to the neutral current at high $Q^2$, and the parton distribution functions do not cancel exactly. Equality of two inclusive cross sections alone does not define $M_W$.

Treat a nucleon as incoherent collinear spin-half quarks and antiquarks carrying fractions $x$ of its momentum. Neglect their masses and transverse motion, use a charged-current four-fermion interaction, and take $Q^2\ll m_W^2$. The subprocess energy is $\hat{s}\simeq xs$. Antineutrino scattering on a quark has the same selected-current helicity factor $(1-y{)}^2$ as in part a; crossing to an antiquark changes the allowed spin pairing and gives a flat distribution. Multiplying the elementary strength by the respective parton densities givesHere $q,\bar{q}$ are the effective weak flavour combinations with the isoscalar proton-neutron average and CKM weights absorbed into their definitions. With only light flavours and charge symmetry, the relevant up and down combinations coincide on an isoscalar target. Rescaling those density definitions requires rescaling the prefactor consistently.

The equation is a leading-order scaling result: it neglects logarithmic QCD evolution, heavy-quark thresholds, higher-twist and target-mass corrections, nuclear modifications, and weak-propagator suppression. The current produces ${\mu }^{+}$ for an incident muon antineutrino. At higher momentum transfer the weak propagator supplies a factor approximately $[m_W^2/(m_W^2+Q^2){]}^2$, and at low energy resonance and quasielastic scattering invalidate a pure DIS description.

For a nucleon at rest and beam energy well above its mass scales, $s\simeq 2m_{N}E$. If the leading parton distributions are approximately independent of beam energy, integration over $x,y$ therefore makes both total cross sections approximately proportional to $E$. The plotted plateaus are evidence for scattering on approximately pointlike constituents in this range, with the departures expected from QCD evolution and threshold effects.

Define the momentum-weighted contents $Q={\int }_0^{1}xq(x)dx$ and $\bar{Q}={\int }_0^{1}x\bar{q}(x)dx$. Since ${\int }_0^1(1-y{)}^{2}dy=1/3$, the two cross sections obeyRead the original high-energy plateaus as approximately $0.67$ and $0.33$ in the plotted $10^{-38}\mathrm{c}{\mathrm{m}}^2/GeV$ units. With $R={\sigma }_{\bar{\nu }}/{\sigma }_{\nu }\simeq 1/2$ and $r=\bar{Q}/Q$,Using the literal readings $0.33/0.67$ gives $r=0.1905$, so twenty percent is an appropriate graphical estimate, not a precision fit. A purely valence target would have $R=1/3$, while a fully symmetric sea would give $R=1$.

The neutrino-antineutrino DIS sea momentum ratio is a ratio of momentum fractions in the effective flavour combinations, not an unweighted count of antiquarks. The cross-section integral contains $xq(x)$: it cannot determine $\int \bar{q}/\int q$ without additional assumptions about their shapes. A distribution growing at small $x$ can have a large number integral while its momentum integral remains finite.

The valence quark distributions describe net quark content: $\int u_v(x)dx=2$ and $\int d_v(x)dx=1$ for a proton. Their momentum densities $x{u}_v,x{d}_v$ are small at very small $x$, peak at moderate momentum fractions, and vanish toward $x=1$, where one parton carrying almost all proton momentum is unlikely. The larger $u$ distribution reflects two valence up quarks and its comparatively harder shape; its pointwise value need not be exactly twice the down distribution.

The sea quark distribution is generated largely by gluon radiation and $g\to q\bar{q}$ splitting. It grows strongly toward small $x$ and falls at large $x$. The plotted sea curve is scaled by $0.05$, so it must be multiplied by twenty before numerical use. The total up-quark PDF isusing the valence-up distribution and the up-sea distribution. The summed sea curve alone does not fix $\bar{u}$ without a flavour assumption. Under the later stated four-flavour symmetry, $\bar{u}=\Sigma /8$.