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Use a local first-derivative Lagrangian density $\mathcal{L}(\varphi ,{\partial }_{\mu }\varphi ,x)$ with sufficient differentiability, and let variations have compact support or vanish at the spacetime boundary. If the phrase “a function of the field” were read literally as forbidding derivatives, the equation below would reduce to $\partial \mathcal{L}/\partial \varphi =0$; a propagating field requires derivative dependence. Varying the action givesIntegration by parts and arbitrary interior variations yield the Euler-Lagrange field equation,The boundary condition is part of this derivation; if boundary variations are allowed, their separate boundary equations must also be imposed.

A variational symmetry of a Lagrangian density is an infinitesimal field change $\delta \varphi =\epsilon X$ for which $\delta \mathcal{L}=\epsilon {\partial }_{\mu }{K}^{\mu }$ off shell. Equality to zero is sufficient but is not necessary: a divergence changes only the boundary contribution to the action. For fixed-coordinate field variations,Equating the two expressions proves the scalar-field form of the Noether theorem:Thus every differentiable one-parameter variational symmetry gives a conserved current, and its Noether charge $Q=\int d^{3}x{j}^0$ is constant when the spatial boundary flux vanishes. This is the Noether current for a first-derivative scalar field; the same identity applies to several real components by summing over them. Identically conserved improvement terms can change its local expression without changing the charge under the same boundary assumptions.

For an active spacetime translation, ${\varphi }^{\prime }(x)=\varphi (x-a)$ and $\delta \varphi =-a^{\nu }{\partial }_{\nu }\varphi$. If the Lagrangian density has no explicit coordinate dependence, $\delta \mathcal{L}=-a^{\nu }{\partial }_{\nu }\mathcal{L}$, so $K^{\mu }=-a^{\mu }\mathcal{L}$. The Noether current is $j^{\mu }=-a^{\nu }{T}^{\mu }{}_{\nu }$, whereThese four translation currents are the canonical stress-energy tensor, also called the energy-momentum tensor. The sign in $j=-aT$ follows from the chosen active translation; the translation charges themselves can be labelled by $P_{\nu }=\int T^0{}_{\nu }{d}^{3}x$.

For the free real scalar field, take the standard kinetic term and mass term. Its Lagrangian density and Klein-Gordon equation areHere ${\Pi }^{\mu }={\partial }^{\mu }\varphi$, so $T^{\mu \nu }={\partial }^{\mu }\varphi {\partial }^{\nu }\varphi -{\eta }^{\mu \nu }\mathcal{L}$. Raising the charge index gives the four-momentum of a free real scalar field,The first is total energy, and the three components of the second are physical spatial momentum. With signature $(+,-,-,-)$, $P_i=-P^i$; this explains the opposite sign if the conserved quantities are instead written with lower spatial indices. For example, a plane wave proportional to $\cos (Et-\mathbf{k}\cdot \mathbf{x})$ has momentum density along $\mathbf{k}$, confirming the sign. All charges require convergence of the integrals and vanishing boundary flux, or periodic boundary conditions in a finite box.

For the complex scalar field, treat $\psi$ and ${\psi }^{\ast }$ as independent variables when varying, equivalently use their two real components. The global phase symmetry of a complex scalar field is $\psi \mapsto e^{-i\alpha }\psi$, ${\psi }^{\ast }\mapsto e^{i\alpha }{\psi }^{\ast }$ for constant $\alpha$. Both the kinetic term and $V(|\psi {|}^2)$ are invariant. This is a global internal symmetry of a classical field theory with circle group $U(1)$; a spacetime-dependent phase would require a gauge connection.

Choosing this orientation for the phase parameter gives the Noether charge of a complex scalar field:To check conservation directly, the Euler-Lagrange field equations are $\square \psi +V^{\prime }(|\psi {|}^2)\psi =0$ and their complex conjugates, assuming a real differentiable potential. Hence ${\partial }_{\mu }{j}^{\mu }=i({\psi }^{\ast }\square \psi -\psi \square {\psi }^{\ast })=0$. Reversing the phase-parameter orientation reverses the current and charge, which is merely a convention. For a scalar carrying electric charge $q$, the physical electric charge is $qQ$ after electromagnetic coupling. In a neutral theory the same global charge can instead label an internal conserved quantum number; the density alone does not identify it automatically with electricity.

Forbidden by electric-charge conservation. The initial bottom quark has electric charge $-1/3$, while the displayed final electric charges sum to $2/3+1+0=5/3$. Therefore no Standard Model Feynman diagram, at tree level or any loop order, can realize this channel.

This is an intentionally disallowed process, not a transcription correction. The charge-conserving semileptonic bottom decay instead has an electron and an antineutrino of electron flavour, while a Positron accompanies the conjugate antiquark decay. Replacing the displayed particles silently would change the requested classification.

Allowed through a t-channel W boson. Along one fermion line the up quark becomes a down quark by emitting a virtual $W^{+}$; along the other the charm antiquark absorbs it and becomes a strange antiquark. The middle panel shows this weak charged current exchange. Both electric charge and baryon number are conserved at each vertex, and the relevant CKM matrix entries are nonzero.

This is the unique physical-particle tree-level Feynman diagram. A neutral exchange would require a flavor-changing neutral current on each line, absent at tree level in the Standard Model. An annihilation into a neutral boson would likewise require an off-diagonal up-type neutral current. The different generations do not forbid the charged-current graph.

Allowed through a virtual W boson. The charm quark emits a $W^{+}$ and becomes a down quark through the nonzero CKM matrix element $V_{cd}$. The virtual W boson produces a positive muon and a muon neutrino through the leptonic weak charged current. The right panel is the unique physical-particle tree-level Feynman diagram.

Its total electric charge is $-1/3+1=2/3$, equal to the initial charge, and its final lepton number is $-1+1=0$. The decay is Cabibbo suppressed, rather than forbidden. These diagram counts use unitary gauge: in covariant gauges charged unphysical Goldstone bosons can supply gauge-dependent pieces of the same physical amplitudes, not additional physical channels.

In the local orthonormal tetrad, the components of the photon number current areThey are respectively the photon number density and the photon number flux. Thus $J^{\mu }$ is the photon number four-current, not an energy current or an electric charge current. Photons carry no electric charge. For the isotropic background,Integrating the covariant Boltzmann equation over the invariant mass-shell measure, with vanishing momentum-boundary flux, giveswhere $C[f]$ is the collision operator in the covariant streaming equation. ThereforeCollisionless photon propagation satisfies this condition, as do Thomson scattering and ordinary Compton scattering when all participating photons are counted: each collision has one incoming and one outgoing photon. Photon emission, photon absorption, bremsstrahlung, double Compton scattering and electron-positron annihilation in cosmology or its inverse generally give a nonzero number source. A cancellation of net creation and destruction can nevertheless make the integrated source vanish even when individual number-changing reactions occur. Cosmological redshift itself changes photon energy, not photon number; in an expanding homogeneous universe a conserved photon number current gives ${\bar{n}}_{\gamma }\propto a^{-3}$.

The electric field is force per unit electric charge, so

The free-space Green function of the Laplacian for the Poisson equation gives the electric potentialFor $r=|\mathbf{x}|\gg R$, its electric multipole expansion begins withDefine the total electric charge and electric dipole moment byThen

The analogous solution for the magnetic vector potential isIts apparent $r^{-1}$ coefficient is ${\mu }_0(4\pi {)}^{-1}\int \mathbf{J}{d}^3{x}^{\prime }$. For each component, the supplied identity and the divergence-free vector field condition giveThe divergence theorem turns this into a surface integral outside the compact support of the current density, where $\mathbf{J}=0$. Hence $\int \mathbf{J}{d}^3{x}^{\prime }=0$, so the $1/r$ term vanishes.

In the absence of free electric charge and electric current, the source-free Maxwell equations in a linear medium aretogether with the constitutive relations $\mathbf{D}=\epsilon \mathbf{E}$ and $\mathbf{B}=\mu \mathbf{H}$.

For any bounded volume $V$ with piecewise smooth closed boundary and outward unit normal vector $\mathbf{n}$, Gauss's law in integral form isThus the outward electric flux equals the enclosed electric charge divided by the vacuum permittivity. To prove it, apply the divergence theorem to the electric field and use its given divergence:The classical proof assumes the requisite differentiability of the electric field. For piecewise smooth charge densities, one may apply the divergence theorem piece by piece and cancel the internal fluxes.

For the thin spherical charge layer limit, smoothness of $h$ givesThus the shell's total electric charge tends to $4\pi {\rho }_0{a}^3{h}^{\prime }(1)$ while its thickness tends to zero. At each fixed $r>a$, the observation point eventually lies outside the shell, whereas at each fixed $r<a$ the interior electric field is unchanged. The limiting electric field away from the sphere is thereforeIts outward-minus-inward jump isPhysically, the limit is a uniformly charged ball of charge density ${\rho }_0$, together with an infinitesimally thin spherical shell of uniform surface charge densityIndeed its electric charge is $4\pi a^2\sigma$, and the standard normal electric field jump is $\sigma /{ϵ}_0$. The surface charge density can have either sign, or vanish, according to $h^{\prime }(1)$.

The order of limits matters. For every finite $\delta >0$, taking the two one-sided limits at $r=a$ first gives zero jump. The displayed nonzero jump is obtained by first taking $\delta \to 0$ at fixed points on either side, then approaching $a$. The value at the surface itself need not coincide with either trace of the limiting electric field.

Set $r=R$ in the enclosed electric charge formula:Outside the entire charge distribution, Gauss's law gives $E_r(r)=Q/(4\pi {ϵ}_0{r}^2)$. Integrating $E_r=-{\phi }^{\prime }$ with $\phi (\infty )=0$ givesThus the exterior electric potential and electric field are exactly those of a point charge $Q$ placed at the center. This is the electrostatic shell theorem: outside a spherically symmetric charge distribution, only its total electric charge affects the electric field, not how that electric charge is distributed radially. If positive and negative electric charges cancel to give $Q=0$, both exterior quantities vanish.

By the assumed spherical symmetry of the electric potential, the electric field is radial: $\mathbf{E}=E_r(r){\mathbf{e}}_r$ for $r>0$. On a sphere of radius $r$, Gauss's law therefore readsPut $R=a(1+\delta )$. For $r<a$, the enclosed electric charge is $4\pi {\rho }_0{r}^3/3$. In the spherical shell, the factor $s^2$ in the volume element cancels $s^{-2}$ in the charge density, soIt follows thatAt the origin the first formula gives $\mathbf{E}=0$ without needing a radial unit vector. At both shell interfaces the adjacent formulas agree: for every fixed $\delta >0$, there is no surface charge concentrated on either sphere, and the electric field is continuous there. No sign restriction on $h^{\prime }$ has been assumed, so the shell charge density and the radial electric field can be negative.

In the first octant, spherical polar coordinates have $0\le \theta \le \pi /2$ and $0\le \varphi \le \pi /2$. The charge density is $\lambda xy=\lambda r^2{\sin }^2\theta \cos \varphi \sin \varphi$. Hence the total electric charge isThe three factors are $a^5/5$, $2/3$, and $1/2$, respectively. For the middle integral, set $u=\cos \theta$ to obtain ${\int }_0^1(1-u^2)du=2/3$. Thus

For an integrable applied voltage, the convolution theorem givesThe integrated electric current is the zero-frequency transform, $Q=\tilde{I}(0)$. Since $\tilde{V}(0)$ can be arbitrary, zero total electric charge for every such input is equivalent toFor nonzero $p_1,p_2$, this can be writtenBecause both amplitudes are positive, nonzero $p_1,p_2$ must have opposite signs. If exactly one $p_i$ is zero there is no cancellation with positive amplitudes; if both vanish the kernel is identically zero and any amplitudes work. In particular, if both frequency parameters were additionally required to be positive, no such nontrivial system would exist. The PDF requires positivity of $a_i,{\gamma }_i$, but allows real fixed $p_i$.

This is zero-frequency cancellation of a causal response kernel. The phrase arbitrary voltage is understood within a class for which the net-charge integrals exist; a permanent DC input need not have finite total charge.

Each lithium atom supplies one electron. If all of the stated lithium is electrochemically accessible and there is enough iron disulfide, the ideal electric charge isThe voltage is not needed to calculate charge. At an assumed constant 1.5 V the corresponding ideal energy would be $VQ\simeq 9.45\mathrm{kJ}=2.63\mathrm{Wh}$, but an actual discharge voltage varies with load and state of charge. Side reactions, incomplete utilization and the practical cutoff reduce accessible capacity. The preceding optimal ratio would require approximately 1.96 g of iron disulfide for this amount of lithium.

A material exhibiting piezoelectricity converts cyclic mechanical stress into electric charge; its mechanical and electrical boundary conditions jointly set the extractable energy. piezoelectric ceramics such as lead zirconate titanate are stiff and often have large charge coefficients and electromechanical coupling, but are brittle and can be difficult to match to large-strain flexible motion. piezoelectric polymers such as polyvinylidene fluoride are compliant, light and compatible with flexible large-area films. Their lower charge coefficient is partly offset in voltage-sensitive applications by lower permittivity: the voltage coefficient is $g=d/ϵ$.

Ceramics often favor strong charge generation and coupling; polymers favor flexibility and strain accommodation. The larger open-circuit voltage of a low-capacitance film does not itself imply more useful power. For a simple fixed-stress comparison the electrical energy scale involves $d^2/ϵ$, while coupling also depends on mechanical compliance. At fixed imposed strain, different stiffnesses change the available mechanical energy. resonance tuning, fatigue, mechanical mounting, rectifier losses and load matching can reverse a simplistic material ranking; compare complete devices under the same actual vibration source.

For a purely electromagnetic interaction continuum with five active quark flavors, the usual estimate is $N_{e^{+}{e}^{-}}/N_{\mathrm{had}}=3/11$. The unspecified particle resonance and the word “total” require qualifications before this can be treated as a unique physical prediction.

In natural units, neglecting final masses, single-photon annihilation givesHere $\alpha$ is the fine-structure constant, $Q_f$ the quark electric charge in elementary-charge units, and the factor three counts orthogonal colour final states. Consequently the inverse hadronic R ratio isWith $u,d,s,c,b$ accessible, the denominator is $3[2(2/3{)}^2+3(1/3{)}^2]=11/3$. The corresponding values for $u,d,s$ and $u,d,s,c$ are $1/2$ and $3/10$. Inclusive Quantum chromodynamics corrections increase the hadronic denominator, approximately by $1+{\alpha }_s/\pi$ away from thresholds and particle resonances. These are annihilation ratios before selection and detection efficiency corrections, not ratios of uncorrected detector counts.

A future muon-collider Higgs resonance is a plausible reading of part (c), since its narrow decay width makes a detuning of a few $\mathrm{MeV}$ relevant. Even then, two further assumptions are necessary. First, being a few $\mathrm{MeV}$ below the Higgs boson peak does not eliminate it: for detuning $\Delta =\sqrt{s}-m_H$, the ideal narrow Breit-Wigner resonance cross section retains the relative factorFor ${\Gamma }_H=4\mathrm{MeV}$ and $\Delta =-4\mathrm{MeV}$ this is $1/5$, not zero. Resonant hadronic production can still change the requested denominator substantially.

Second, at $\sqrt{s}\simeq 125\mathrm{GeV}$ the Z boson continuum is not negligible. A more complete annihilation estimate including both tree-level Feynman diagrams usesHere $s_W^2$ is the squared sine of the weak mixing angle, $T_3^f$ is weak isospin, and $N_c^f$ is one for a lepton and three for a quark. Thus the continuum ratio is $F_e/[3(2F_u+3F_d)]$, rather than necessarily $3/11$. For example, $s_W^2=0.231$, $m_Z=91.1876\mathrm{GeV}$ and ${\Gamma }_Z=2.4952\mathrm{GeV}$ give approximately $0.102$ at $125\mathrm{GeV}$ before Quantum chromodynamics corrections.

If the unspecified particle resonance instead means the Z boson, detuning by a few $\mathrm{MeV}$ is tiny compared with its $2.5\mathrm{GeV}$ decay width. The resonant ratio remains approximately $\Gamma (Z\to e^{+}{e}^{-})/\Gamma (Z\to \mathrm{hadrons})=1/20.767\simeq 0.0482$.^[ref] This is a concrete counterexample to treating any slight downward detuning as permission to discard the particle resonance. The source supplies neither its identity nor the detuning and selection needed for a unique all-process ratio. The boxed continuum formula is the charge-counting answer under its stated approximation; a literal total must also include remaining resonant and other accepted hadronic channels.

In ordinary perturbative Standard Model reactions all three interactions conserve electric charge, four-momentum, total angular momentum, baryon number, and total lepton number. The last two are accidental conservation laws of these perturbative processes, not exact nonperturbative statements: electroweak sphalerons violate $B+L$ while preserving baryon-minus-lepton number. Total color charge is respected throughout.

The strong interaction and electromagnetic interaction also conserve each net quark flavor number, including strangeness, charm and bottom flavour. The weak charged current can change them through the Cabibbo-Kobayashi-Maskawa matrix. Isospin is an approximate strong interaction symmetry in the small light-quark mass difference limit; the different electric charges of $u$ and $d$ break it in the electromagnetic interaction. Approximate flavor symmetry is distinct from exact net-flavour-number conservation by those interactions.

Parity conservation and charge conjugation symmetry hold for the electromagnetic interaction and, assuming a negligible strong CP parameter, the strong interaction. The weak interaction violates both, and also violates CP symmetry in suitable processes. CPT symmetry holds under the usual local Lorentz invariant quantum field theory assumptions. Individual lepton family numbers are conserved in the conventional massless-neutrino approximation but not in neutrino oscillations; an assertion of exact family conservation in all observed reactions would therefore be false.

The magnetic rigidity gives the momentum, the bending direction gives the sign of electric charge, and the calorimetric energy then gives the mass. For planar tracks perpendicular to a uniform magnetic field, the Lorentz force and circular motion giveTrace each track away from its production or decay vertex. With the magnetic field into the page, a positive track bends counterclockwise and a negative track bends clockwise. For example, a positive particle initially travelling left accelerates downwards. The known incoming negative pion fixes the sign convention experimentally. A radius alone does not distinguish electric charge signs.

For a calibrated total energy $E$, the relativistic energy-momentum relation givesIf the calorimeter instead supplies only kinetic energy $T$, solve $(T+m{c}^2{)}^2=p^2{c}^2+m^2{c}^4$ to obtainCompare with candidate pion, kaon, and proton masses, together with the electric charge sign and shower behavior. An electron makes an electromagnetic shower; a muon is penetrating; a hadron generally requires a hadronic calorimeter and its own calibration. If either energy escapes or the response is uncalibrated, simply substituting a deposited signal for $E$ need not identify the particle correctly. The absent figure prevents attaching identities to individual drawn tracks, but the requested identification procedure does not require its geometry.

The labels X and Y cannot be uniquely assigned from the surviving text: their decay vertices and daughter tracks were contained in the missing figure. The available conservation laws nevertheless specify how the assignment would be obtained.

The initial ${\pi }^{-}p$ state has $(Q,B,S)=(0,1,0)$, with $S$ denoting strangeness. At a primary strong interaction vertex all three numbers must be reproduced. A final Lambda baryon or Sigma baryon has $B=1$, $S=-1$ and therefore needs accompanying $S=+1$, for example a neutral kaon $K^0=d\bar{s}$ or a positive kaon $K^{+}=u\bar{s}$, with the remaining electric charge balanced. At later weak interaction decay vertices, strangeness need not be conserved; it would be wrong to enforce it there.

A displaced $p{\pi }^{-}$ pair with the appropriate invariant mass identifies ${\Lambda }^0\to p{\pi }^{-}$. A displaced ${\pi }^{+}{\pi }^{-}$ pair can identify $K_S^0\to {\pi }^{+}{\pi }^{-}$ after an invariant mass and kinematic check. For a candidate radiative Sigma-to-Lambda transition, the neutral parent satisfiesin natural units. Its photon energy in the parent rest frame is $(m_{\Sigma }^2-m_{\Lambda }^2)/(2m_{\Sigma })$. With known charged four-momenta, each unobserved neutral momentum is the difference between the incoming and visible outgoing momenta at its vertex; its inferred direction must agree with the dashed segment.

This information cannot establish which dashed segment was called X. For a concrete ambiguity, bothobey the primary electric charge, baryon number, and strangeness constraints and can yield a neutral kaon decay plus a displaced Lambda baryon decay. Their distinct neutral segments and the extra photon would be resolved by the missing topology and kinematics. The supplied possible Sigma baryon decay alone is insufficient to label X and Y. A labelled answer is therefore a source gap, rather than a completed identification.