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The qualification concerns gravitational energy. Local matter energy exists: an observer $u$ measures $T_{\mu \nu }{u}^{\mu }{u}^{\nu }$. What is absent is a generally covariant local gravitational stress-energy density with the required universal conservation properties. The Equivalence principle allows the connection to vanish at a point in freely falling coordinates; standard gravitational energy expressions depend on coordinates. Curvature does remain, but supplies no unique gravitational energy tensor of the required kind. A nonstationary geometry also lacks a preferred time-translation Killing vector field. For a trial timelike $t$, stress-energy conservation gives ${\nabla }_{\mu }(T^{\mu \nu }{t}_{\nu })=T^{\mu \nu }{\nabla }_{(\mu }{t}_{\nu )}$, not generally zero.

An internal charge instead has a current ${\nabla }_{\mu }{j}^{\mu }=0$ independently of spacetime time translations. Its hypersurface flux is conserved with suitable boundary conditions. For electric charge, the Gauss law expresses it as a surface flux. Observer-dependent charge density does not prevent a well-defined total charge.

For an asymptotically flat spacetime, take asymptotically Cartesian slice coordinates with ${\gamma }_{ij}={\delta }_{ij}+h_{ij}$ and appropriate falloff. In $c=1$ units, the ADM energy isIt is an asymptotic gravitational energy. The dominant energy condition requires $-T^{\mu }{}_{\nu }{u}^{\nu }$ to be future-directed nonspacelike or zero for every future timelike $u$, in particular $T_{\mu \nu }{u}^{\mu }{u}^{\nu }\ge 0$. Together with the constraint equations, completeness and appropriate asymptotic/boundary hypotheses, it yields $E_{\mathrm{ADM}}\ge |{\mathbf{P}}_{\mathrm{ADM}}|\ge 0$. The energy condition alone is not the complete positive-energy theorem.

Take the divergence of the Ampère-Maxwell equation, $\nabla \times \mathbf{B}={\mu }_0\mathbf{J}+{\mu }_0{ϵ}_0{\partial }_t\mathbf{E}$. Since the divergence of a curl is zero vanishes, and Gauss's law gives $\nabla \cdot \mathbf{E}=\rho /{ϵ}_0$, the result isThis is the charge continuity equation.

At an internal point of a homogeneous Ohmic conductor, Ohm's law gives $\mathbf{J}=\sigma \mathbf{E}$. Since $\sigma$ is constant, the charge continuity equation becomes ${\partial }_t\rho =-(\sigma /{ϵ}_0)\rho$. ThereforeThis is charge relaxation. Here $\rho$ is the total charge appearing in the vacuum form of Gauss's law; in a homogeneous dielectric description of free charge the corresponding constant permittivity replaces ${ϵ}_0$.

For a finite isolated body, charge travels to the surface. The bulk formula applies at internal points, not across the interface where electrical conductivity jumps and surface charge density accumulates. Integrating the charge continuity equation over the body, including its surface charge, gives constant total electric charge because no electric current crosses the external boundary. Bulk decay and conservation of electric charge are therefore compatible.

In classical charged particle in a uniform magnetic field motion, the Lorentz force and Newton's second law giveAssume $\mathbf{B}\ne 0$ and choose an orthonormal coordinate system with $\mathbf{B}=B{\mathbf{e}}_z$, $B=|\mathbf{B}|$. Write $\Omega =qB/m$, retaining the sign of the electric charge. The velocity equations areThus the parallel velocity is constant, and the perpendicular velocity rotates at constant angular frequency. Choosing the time origin and perpendicular axes conveniently givesIntegrating the velocity, one obtainsThe first two coordinates describe a circle of radius $m{v}_{\perp }/(|q|B)$, while the third advances uniformly. This is helical motion in a uniform magnetic field: the helix has axis parallel to $\mathbf{B}$ through the guiding centre. Its cyclotron frequency isand its signed angular velocity vector about that axis is $-q\mathbf{B}/m$. The minus sign follows from the ordering $\mathbf{v}\times \mathbf{B}$ in the Lorentz force. The helix degenerates to a circle when $v_{\parallel }=0$ and to a straight line when $v_{\perp }=0$. If $q=0$ or $B=0$, there is only straight-line motion. In a relativistic extension with fixed Lorentz factor $\gamma$, the frequency becomes $|q|B/(\gamma m)$; the classical result above uses $m\mathbf{v}$ as momentum.

Take $D_{\mu }={\partial }_{\mu }+i{K}_{\mu }$, with the Pauli matrices acting on the Higgs doublet. For electroweak hypercharge $Y=1/2$,This specifies every component of the gauge covariant derivative for the electroweak interaction. Expanding the gauge-covariant kinetic term makes its derivative, trilinear and quartic interactions explicit:where the Pauli matrix multiplication law givesThe antisymmetric Pauli contribution vanishes because $W_{\mu }^a{W}^{b\mu }$ is symmetric in $a,b$. Reversing the sign convention for $D_{\mu }$ reverses the linear gauge interactions consistently, without changing the masses.

A nonzero vacuum expectation value requires ${\mu }^2<0$. Minimizing the Higgs potential gives $v^2=-{\mu }^2/\lambda$. By an gauge transformation choosein unitary gauge. The electroweak doublet gauge-boson mass matrix follows by inserting the vacuum expectation value in the gauge-covariant kinetic term:Define the charged electroweak gauge bosons and the neutral rotation through the Weinberg angle bywith inverse $W_{\mu }^3=c_W{Z}_{\mu }+s_W{A}_{\mu }$ and $B_{\mu }=-s_W{Z}_{\mu }+c_W{A}_{\mu }$. ThenThe factors differ because $W^{+}$ and $W^{-}$ are conjugate fields whereas $Z$ is real. The massless photon corresponds to the unbroken Lie algebra generator $Q=T_3+Y$, which annihilates $⟨\varphi ⟩$. Thus three of the four real gauge bosons acquire mass, with $m_W=m_Z{c}_W$. The three would-be Goldstone bosons provide their longitudinal polarizations; the remaining scalar $h$ is the Higgs boson.

Introduce the left-handed lepton doublet $L=({\nu }_{eL},e_L{)}^T$, with $Y_L=-1/2$, and the right-handed singlet $e_R$, with $Y_R=-1$. Use the chiral projectors $P_L=(1-{\gamma }^5)/2$ and $P_R=(1+{\gamma }^5)/2$ in the course's convention. The minimal Standard Model has no right-handed neutrino. The gauge-invariant fermion terms areThe doublet contraction in the Yukawa interaction is a singlet, and its total hypercharge is $+1/2+1/2-1=0$. A bare term ${\bar{e}}_L{e}_R$ would fail electroweak gauge invariance. The gauge covariant derivatives above already give all the requested fermion-gauge couplings. In terms of mass eigenstates, $e=g{s}_W=g^{\prime }{c}_W$ and they becomeThus the weak charged current is chiral and the neutrino has zero electric charge. The gauge-invariant electron Yukawa mass follows from the Yukawa interaction after electroweak symmetry breaking:Rephase $e_R$ to make $y_e$ real and positive. It givesThe electron Dirac mass is therefore compatible with the original gauge symmetry through the Higgs mechanism. The neutrino remains massless in this minimal renormalizable lepton sector.

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

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

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

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

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

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

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

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