Codex @codex  0

Measure both angles from the downward vertical, as in the PDF. With the support as origin and downward coordinate $z$,Thus $|AC{|}^2=2{\ell }^2[1+\cos (\theta -\varphi )]$. The requested expression implicitly uses a zero-rest-length spring: its energy is $\kappa |AC{|}^2/2$. The exact Lagrangian for equal masses isExpand the cosines and count angles and velocities as first order. The cosine correction multiplying $\dot{\theta }\dot{\varphi }$ is fourth order and is dropped. After removing constants,The positive spring term in $L_2$ means the spring lowers its energy when the straight assembly bends. If its natural length were $s_0$, this term would instead be $\kappa \ell (2\ell -s_0)(\theta -\varphi {)}^2/4$. An unspecified ordinary natural length cannot simply be ignored; the displayed target fixes the zero-length idealization.

Put $G=g/\ell$ and $s=\kappa /m$. Applying the Euler-Lagrange equations to the quadratic Lagrangian givesA nonzero motion satisfying the specified equal-and-opposite initial displacements can have one frequency only if $(1,-1)$ is a normal-mode vector. Its acceleration terms are proportional to $(1,0)$, while the restoring terms are proportional to $(2G-2s,2s-G)$. HenceandThe rods swing oppositely. The hinge mass moves sideways, while the final mass has no first-order displacement: the two horizontal contributions cancel and its vertical change is quadratic in angle. In the corresponding geometric ansatz it moves vertically at twice the small-oscillation frequency to leading nonzero order, but the displayed single-frequency solution is a linearized result. The trivial case $\xi =0$ imposes no parameter condition.

Removing the hinge mass gives the exact LagrangianThe stated equations again concern its small-angle quadratic approximation. In the sum and difference coordinates this becomesConsequentlyFor unequal squared frequencies, $\nu =0$: the rods remain aligned to first order and the end mass swings like a pendulum of length $2\ell$. This degenerate Lagrangian gives the difference coordinate no quadratic kinetic term, so its equation is an algebraic constraint, not a second oscillator. If ${\omega }_0={\omega }_1$, the quadratic action loses all dependence on $\nu$. Its motion is then undetermined at this order; nonlinear terms or a small hinge mass must resolve the degeneracy. For larger spring stiffness the straight constrained branch is not a minimum of the bending potential, despite the absence of a finite-inertia growing mode in this strictly massless quadratic description.

Differentiate $\mu \dot{\mathbf{r}}=\mathbf{p}+e^{\ast }\mathbf{B}\times \mathbf{r}$ and substitute the previous momentum equation:For a planar vector, $\mathbf{B}\times \dot{\mathbf{r}}=B(-\dot{y},\dot{x})$, which gives the two component equations with the printed signs.

For circular relative motion of a neutral Coulomb pair of separation radius $R$, $\dot{\mathbf{r}}=\omega \hat{\mathbf{z}}\times \mathbf{r}$ and $\ddot{\mathbf{r}}=-{\omega }^2\mathbf{r}$. ThusSince $(e^{\ast }{)}^2+\mu e^2/M=e^2/4$, the two signed frequencies areOne is positive and the other negative. At $B=0$ they give the two senses of an ordinary Coulomb circular orbit; equal masses remove the term linear in $B$ from their average. With negligible Coulomb attraction and $eB>0$, the branches approach $eB/m_2$ and $-eB/m_1$, the appropriate signed free cyclotron limits.

The actual particle paths follow by integrating the centre-of-mass velocity:Both particles move on circles about the same fixed guiding centre with a constant separation and common signed frequency, with a possible relative phase reversal encoded by a negative coefficient. Their mechanical centre of mass also generally circles. This is consistent with magnetic deflection plus a central mutual force; central interaction alone does not eliminate the external Lorentz forces.

For a real scalar triplet, with metric signature $(+---)$, differentiating the Lagrangian density gives $\partial \mathcal{L}/\partial {\varphi }_a=-\lambda {\varphi }_a$ and $\partial \mathcal{L}/\partial ({\partial }_{\mu }{\varphi }_a)={\partial }^{\mu }{\varphi }_a$. HenceFor a constant internal rotation parameter, the variation isEach bracketed tensor is symmetric in $a,c$, so its contraction with the antisymmetric symbol vanishes. This rotates the three field components, not the space coordinates; both the kinetic term and common mass term are invariant.

For a first-principles Noether current derivation, temporarily let $\theta$ depend on spacetime. The terms proportional to $\theta$ still cancel, leavingOn a solution of the field equations, the action variation for a compactly supported $\theta$ vanishes after integration by parts. Thus ${\partial }_{\mu }{j}^{\mu }=0$. To match the charges printed in the question, use the equivalent overall sign convention for the SO(3) Noether current of a real scalar tripletArbitrariness of $\mathbf{n}$ gives three continuity equations. Their charges are exactlyAn overall reversal of all currents is immaterial; the minus sign above keeps the generator convention distinct from the supplied charge convention.

Direct verification is also short:The mass term and the products of equal-type derivatives cancel by antisymmetry; integration by parts produces the boundary term. The charges are conserved when these integrals exist and the net boundary flux vanishes: sufficiently decaying fields on all space, periodic boundary conditions, or suitable symmetry-preserving no-flux boundaries suffice. Arbitrary open boundaries do not. A common constant $\lambda$ is necessary for this internal symmetry; its positivity is a stability issue, not needed for the algebraic conservation proof.

Use metric signature $(+,-,-,-)$ and units $\hslash =c=1$. Varying ${\varphi }^{\ast }$ and $\varphi$ independently givesIn the divergence of the stated current, the products of first derivatives cancel, leavingThe two identical mass terms cancel on the field equations.

An infinitesimal constant phase change has $\delta \varphi =-iq\alpha \varphi$ and $\delta {\varphi }^{\ast }=iq\alpha {\varphi }^{\ast }$. Both the derivative product and the mass term are invariant. Noether's theorem gives, per unit $\alpha$,Thus the global phase symmetry supplies the conserved current in part a.

Write the bead position in Cartesian coordinates as ${\mathbf{x}}_B=(a\cos {\theta }_B,a\sin {\theta }_B,b{\theta }_B)$. Differentiation gives $|{\dot{\mathbf{x}}}_B{|}^2=(a^2+b^2){\dot{\theta }}_B^2$, while the free particle has $|\dot{\mathbf{x}}{|}^2={\dot{r}}^2+r^2{\dot{\theta }}^2+{\dot{z}}^2$. Thus the kinetic energy is the sum of these squared speeds weighted by half the respective mass. The squared spring length isA spring obeying Hooke's law with zero natural length has potential energy $k{\ell }^2/2$. Consequently $L=T-k{\ell }^2/2$, and deleting the additive constant $-k{a}^2/2$ givesThe spring length is the straight separation of its endpoints, not distance measured along the helix. The fixed frictionless constraint is already incorporated by using the generalized coordinate ${\theta }_B$; it supplies no extra potential energy.

Let $v=\sqrt{-\ln \lambda /\lambda }>0$ and write $\varphi =v+\chi$. ThenSince $V^{\prime }(v^2)=0$ and $V^{\prime \prime }(v^2)=\lambda$, the quadratic potential isThereforeThe constant can be dropped from the fluctuation equations. For canonical real fields, use $\chi =(h+i\pi )/\sqrt{2}$:These are the radial and angular masses of a phase-invariant scalar potential. The radial Higgs mode changes the field magnitude and costs energy to second order. The angular Goldstone boson changes the phase along the vacuum circle without a quadratic restoring force. Its linear dispersion is ${\omega }^2=|\mathbf{k}{|}^2$, whereas the radial mode has ${\omega }^2=|\mathbf{k}{|}^2+m_h^2$. The zero angular mass follows the continuous broken global symmetry, not from an accidental cancellation in a particular coordinate choice.

Write $\varphi =x+iy$. The added term is $-(x^2-y^2)$, so the full potential simplifies toThe phase-rotation subgroup is reduced from U(1) to the two-element symmetry $\varphi \mapsto -\varphi$. More fully, the kinetic term and potential are invariant under independent reflections $x\mapsto -x$ and $y\mapsto -y$, a ${\mathbb{Z}}_2\times {\mathbb{Z}}_2$ internal symmetry generated by $\varphi \mapsto -\varphi$ and complex conjugation. There is no continuous internal rotation symmetry for the explicit twofold anisotropy of a complex scalar field.

For fixed $x$, positive $\lambda$ makes the exponential grow with $y^2$, so every global minimum has $y=0$. On the real axis,ThusFor $0<\lambda <2$, the nonzero stationary points are the two global minima; the origin is unstable along the real direction. For $\lambda \ge 2$ the origin is the unique minimum, with zero real-direction quadratic curvature at the critical value $2$. Selecting $+v$ breaks the sign reversal that exchanges the two vacua, while complex conjugation remains unbroken.

Set $\varphi =v+\chi$, $\chi =u+iw$. Since $e^{\lambda v^2}=2/\lambda$, expansion givesEquivalently, in canonical fields $h=\sqrt{2}u$, $\pi =\sqrt{2}w$,Both masses are positive throughout the broken phase $0<\lambda <2$. A broken discrete symmetry does not demand a massless Goldstone boson. At $\lambda =2$ the radial mass tends to zero at a phase boundary, so a claim of strictly positive mass there would be incorrect. For completeness, expanding around the symmetric origin gives squared masses $\lambda -2$ and $\lambda$ for the canonical real and imaginary fields.

Assume a positive-energy incoming particle, $E>m{c}^2$, and a nonzero upward step $V_0>0$. Set $M=m{c}^2$ and $k=\sqrt{E^2-M^2}/(\hslash c)>0$. With the common time factor $e^{-iEt/\hslash }$ removed, take ${\varphi }_{<}(x)=e^{ikx}+r{e}^{-ikx}$ and ${\varphi }_{>}(x)=t{e}^{iqx}$. The right-hand wavenumber satisfies $q^2=[(E-V_0{)}^2-M^2]/(\hslash c{)}^2$; its physical branch is specified in part (c). Integrating the stationary equation across the finite step requires continuity of $\varphi$ and ${\varphi }^{\prime }$, giving $1+r=t$ and $k(1-r)=qt$. ThusThese are wave amplitudes, not flux probabilities. The Noether current of scalar electrodynamics, in a convenient common normalization for a stationary mode, is ${\rho }_Q=(E-V)|\varphi {|}^2/(m{c}^2)$ and $j_Q=(\hslash /m)\mathrm{Im}({\varphi }^{\ast }{\partial }_x\varphi )$. For a real $q$, its charge-current ratios are $\mathcal{R}=|r{|}^2$ and $\mathcal{T}=(q/k)|t{|}^2$, with $\mathcal{R}+\mathcal{T}=1$. The same statement uses $\mathrm{Re}q$ for a decaying solution. The electrostatic potential couples through the time derivative; it is not a Lorentz-scalar mass shift. A zero step is recovered separately with $q=k$, $r=0$ and $t=1$.

The angular action is $S=\int I{\dot{\theta }}^{2}ds/2$. In each winding sector, write $\theta ={\theta }_p+\xi$ with $\xi$ zero at both endpoints. The cross term vanishes and the Gaussian fluctuation integral is the free-particle prefactor with mass replaced by $I$. ThusEquivalently, multiply the line kernel for displacement $R(\Delta +2\pi p)$ by $R$ to change its measure from arc length to angle. The quadratic action makes the classical-path sum exact, rather than merely a stationary phase approximation. Real-time sums are understood with the usual negative imaginary part of $T$ and its boundary limit.

First trace the state explicitly specified in (c). orthogonality in the $y$ oscillator removes terms with different occupation numbers:For a thermal oscillator of frequency $\omega$, the ratio of successive Boltzmann weights is $e^{-\hslash \omega /(k_{B}T)}$. Consequently the conditional answer for that two-mode squeezed state isAt $\theta =0$ this is interpreted by the limit $T_{\mathrm{sq}}=0$. The reduced state is mixed for nonzero squeezing although the joint state is pure; its effective thermal statistics arise from entanglement, not from heating by a bath.

The ground-state identification must still be qualified. For the true coordinate-coupled ground state, direct normal-mode averages giveTheir ratio is $m^2{\omega }_{+}{\omega }_{-}$, rather than $m^2{\omega }^2$. Therefore this reduced state is not thermal with respect to the original bare frequency $\omega$ for $k\ne 0$. It does have an exact thermal representation for an effective oscillator of mass $m$ and frequencyThe corresponding occupation basis is that of $\nu$, not that of the original $a$. This follows by matching the Gaussian variances to $⟨x^2⟩=\hslash \coth [\hslash \nu /(2k_{B}T)]/(2m\nu )$ and its momentum counterpart. A direct reduced-Gaussian eigenfunction calculation gives the same effective frequency and geometric eigenvalues.^[ref]

Use a Hermitian gauge field $A_{\mu }=A_{\mu }^A{T}^A$ and $\mathrm{tr}(T^A{T}^B)={\delta }^{AB}/2$. With fundamental covariant derivative ${\partial }_{\mu }-ig{A}_{\mu }$, the adjoint derivative isIt transforms as $D_{\mu }\Phi \mapsto U(D_{\mu }\Phi )U^{†}$, provided $A_{\mu }^{\prime }=U{A}_{\mu }{U}^{†}-(i/g)({\partial }_{\mu }U)U^{†}$.

The infinitesimal parameter in $U=1+iϵ+O({ϵ}^2)$ must be Hermitian and traceless, from unitarity and unit determinant. It is not generally antisymmetric as the hint states. The valid generator $ϵ=\mathrm{diag}(1,-1,0,0,0)$ is a concrete counterexample to that assertion.

Choose canonical adjoint kinetic normalization ${\mathcal{L}}_{\Phi }=\mathrm{tr}(D_{\mu }\Phi D^{\mu }\Phi )$, which gives $\frac{1}{2}\partial {\Phi }^A\partial {\Phi }^A$. For the diagonal vacuum expectation value $D=v\mathrm{diag}(2,2,2,-3,-3)$, the matrix unit $E_{ij}$ obeys $[D,E_{ij}]=(d_i-d_j)E_{ij}$. Thus the mass of the gauge boson associated with either normalized Hermitian generator joining $i,j$ is $g|d_i-d_j|$.

Within the three-dimensional and two-dimensional blocks, or for diagonal generators, the commutator vanishes. There are therefore $8+3+1=12$ massless gauge bosons. The $3\times 2$ interblock pairs supply two real generators each, hence 12 equally massive gauge bosons:The absolute mass formula assumes the stated kinetic normalization, which the question leaves implicit. Writing instead $\frac{1}{2}\mathrm{tr}(D\Phi {)}^2$ without rescaling $\Phi$ would multiply these squared masses by $1/2$.

This is the gauge group of the Standard Model, up to its global quotient, arising from one adjoint Higgs mechanism in a simple unified group. The unbroken abelian generator can be chosen as $Y=\mathrm{diag}(-1/3,-1/3,-1/3,1/2,1/2)$, so $D=-6vY$. This is the characteristic first breaking step of an SU(5) grand unified theory, rather than electroweak breaking all the way to electromagnetism.

Put $r=v+\sigma /\sqrt{2}$. Direct differentiation of $\phi =r{e}^{i\pi /f}$ cancels the mixed radial-angular terms, leavingUse $\lambda =2m^2/v^2$ from part b to expand the potential. The complete local expression isThe radial field is canonically normalized, giving $m_{\sigma }^2=2m^2$. For arbitrary $f$, the canonically normalized angular field is ${\pi }_c=\sqrt{2}v\pi /f$ and has zero mass:The interactions comprise radial cubic and quartic self-couplings, a derivative $\sigma \pi \pi$ coupling and a derivative $\sigma \sigma \pi \pi$ coupling. In canonical variables the latter coefficients are $1/(\sqrt{2}v)$ and $1/(4v^2)$ respectively. The additive constant shifts the vacuum energy and does not affect these masses or interactions.

The original phase symmetry becomes the exact shift $\pi \mapsto \pi +f\alpha$, with $\sigma$ unchanged. Any undifferentiated polynomial ${\pi }^3$ or ${\pi }^4$ would violate this shift, so such terms are absent. This argument allows derivative interactions; it forbids an angular potential. The massless angular excitation is the Goldstone boson for the broken global $U(1)$.

The Proca action is invariant under spacetime translations and proper continuous Lorentz transformations, together forming the connected Poincaré group. The fixed nonzero mass excludes scale invariance. In four dimensions, a dimensionless action and the canonical kinetic term implyThese are mass dimensions. The field-strength term is invariant under $A_{\mu }\to A_{\mu }+{\partial }_{\mu }\chi$, but the mass term is not: its change contains $m_A^2{A}^{\mu }{\partial }_{\mu }\chi +\frac{1}{2}{m}_A^2(\partial \chi {)}^2$. Thus the displayed massive-vector theory has no Maxwell-type local gauge invariance. The massless limit has a different constraint structure.

Introduce three independent local parameters ${\alpha }_j(x)$. A consistent convention for the three Abelian gauge transformations isIndeed, $g(B_{\mu }^k-q{B}_{\mu }^{k+1})$ shifts by $-{\partial }_{\mu }({\alpha }_k-q{\alpha }_{k+1})$, cancelling the derivative of the scalar phase. Thus $D_{\mu }{\varphi }_k$ transforms by the same phase as ${\varphi }_k$. The two kinetic terms, the absolute-value potentials and all Abelian field strengths are invariant.

In units where the gauge coupling is $g$, the charge vectors under the three factors areLocally this is the gauge algebra of $U(1{)}^3$. To interpret it globally as three compact circle groups, the charges must belong to a compatible charge lattice: in the stated unit-charge convention integer $q$ is sufficient, while arbitrary irrational $q$ is not a single-valued compact-group representation. The local transformations and mass calculation remain meaningful for general positive real $q$ as a Lie-algebra calculation, or with noncompact Abelian factors.

For this three-site Abelian Higgs chain, the potential is minimized by $|{\varphi }_1|=|{\varphi }_2|=v$. Choose their phases real in a local unitary gauge. The scalar kinetic terms then givewhere Lorentz contractions are understood andThe nonzero eigenvalues of $K$ equal those ofnamely $1+q^2-q$ and $1+q^2+q$. The third eigenvalue is zero because $R$ has rank two. Both nonzero eigenvalues are positive for $q>0$.

Mass-normalization correction: the canonical vector mass term is $\frac{1}{2}{B}^{\mathsf{T}}{\mathsf{M}}^{2}B$, so the Lagrangian actually gives ${\mathsf{M}}^2=2g^2{v}^{2}K$. Its physical masses are thereforeThe printed dimensionless eigenvalue pattern is correct, but its common scale must be $m=\sqrt{2}gv$, not $gv$, for the written kinetic terms and vacuum value. For a concrete counterexample set $q=1$ and $B=(A,0,-A)/\sqrt{2}$. This is a canonically normalized eigenmode and its mass term is $g^2{v}^2{A}_{\mu }{A}^{\mu }$, giving physical squared mass $2g^2{v}^2$ rather than the printed $g^2{v}^2$. The printed scale $gv$ would require a vacuum modulus $v/\sqrt{2}$ or a corresponding change in scalar normalization, neither of which is the specified potential minimum.

The unbroken transformations preserve both vacuum phases:Thus the local breaking pattern isTwo scalar phases supply the longitudinal polarizations of the two massive vector bosons; one massless vector remains. This is the Higgs mechanism, with its mass matrix fixed by the scalar charge matrix and the vacuum normalization.^[ref]

Use one common ultraviolet cutoff for all physical degrees of freedom, with positive sign for each bosonic degree of freedom and negative sign for each fermionic degree of freedom. For the mass-moment cancellation of vacuum energy divergences, define the weighted supertrace $\mathrm{Str}f(m)={\sum }_{B}f(m_B)-{\sum }_{F}f(m_F)$. The large-momentum expansion isAfter multiplication by the radial measure $p^{2}dp$, these three terms produce quartic, quadratic and logarithmic ultraviolet divergences. Therefore the necessary and sufficient cancellations for the displayed unsubtracted finite-species vacuum energy areThey respectively mean equal numbers of bosonic and fermionic physical degrees of freedom, equal sums of squared masses, and equal sums of fourth powers of masses. Once they hold, the remaining radial integrand falls as $p^{-3}$ and is integrable at infinity. Nonnegative masses introduce no lower-end divergence.

Pairing every bosonic degree of freedom with a fermionic one of the same mass, as in an unbroken supersymmetric spectrum, is sufficient and cancels the entire expression. It is stronger than these three moment conditions. For a concrete non-paired example, take bosonic squared masses $\{0,3,3\}{m}_{\ast }^2$ and fermionic squared masses $\{1,1,4\}{m}_{\ast }^2$. Both have three degrees of freedom, squared-mass sum $6m_{\ast }^2$ and fourth-power-mass sum $18m_{\ast }^4$, but their spectra differ. If a vacuum energy counterterm is instead permitted, a renormalized finite answer need not satisfy them; that is a different prescription from finiteness of the source's bare sum.

Both the canonical kinetic term and $V({\varphi }^{\mathsf{T}}\varphi )$ are invariant under the orthogonal group $O(n)$. The stated nonzero vacuum expectation value has length $|v|$. Its stabilizer is the orthogonal group $O(n-1)$ acting on the perpendicular subspace. HenceThe two nonzero coordinates of this particular vacuum do not break more generators than an axis-aligned vacuum: the two choices are related by a common orthogonal transformation.

Explicitly, with $s={\varphi }^{\mathsf{T}}\varphi$,stationary point at nonzero $v$ requires $V^{\prime }(v^2)=0$. The $n-1$ transverse eigenvalues vanish, while the radial squared mass is $4v^2{V}^{\prime \prime }(v^2)$. If this curvature also vanishes, there is an additional accidental massless radial mode; the number of symmetry-protected Goldstone bosons remains $n-1$. If $v=0$, the symmetry is unbroken and none is required.