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For uniform circular motion, the transverse Lorentz force supplies the change in direction of three-momentum:The magnetic rigidity relation gives $pc=0.299792458B[\mathrm{T}]r[\mathrm{m}]\mathrm{GeV}=2.99792458\mathrm{GeV}$. Using $m_e{c}^2=0.511\mathrm{MeV}$,The coordinate centripetal acceleration is $a=u^2/r$. For transverse acceleration, the magnitude of the four-acceleration, or proper acceleration, is $a_0={\gamma }^{2}a$. The supplied Larmor formula, in its proper acceleration form, therefore givesThe emitted power here is also the laboratory total emitted power: in the instantaneous rest inertial frame the dipole radiation pattern has zero net radiated three-momentum, so both emitted energy and elapsed coordinate time gain the same Lorentz factor on transformation. This does not assert that a directional detector flux is invariant.

Numerically $\mathcal{T}\simeq 1.048\times 10^{-5}\mathrm{s}$, $P\simeq 2.18\times 10^{-10}\mathrm{W}$ per electron, andThe synchrotron's radiofrequency accelerating system must replace this lost energy to keep $r$ and magnetic rigidity fixed over many revolutions. The fraction lost per turn is about $4.77\times 10^{-6}$, so a fixed-orbit calculation over one turn is self-consistent.^[ref]

Use the electromagnetic field tensor convention stated above, and lower indices with the Minkowski metric $g=\mathrm{diag}(-1,1,1,1)$. Lowering a single time index changes its sign. Summing both orders of each antisymmetric index pair gives the Lorentz scalarFor $T^{00}$, $F_j{}^0=F^{j0}=-E_j/c$ and $g^{00}=-1$. Direct substitution in the printed electromagnetic stress-energy tensor yieldsFor a spatial index $i$, $g^{0i}=0$ and $F_j{}^i=F^{ji}=-{ϵ}_{ijk}{B}_k$. HenceThus the full top row isHere $u_{\mathrm{EM}}$ is the electromagnetic energy density and $\mathbf{S}$ the Poynting vector, using ${ϵ}_0{\mu }_0{c}^2=1$. By the symmetry $T^{0i}=T^{i0}$, these components describe both energy flux divided by $c$ and $c$ times the electromagnetic momentum density $\mathbf{g}=\mathbf{S}/c^2$. Keeping the factor $c$ is essential because the time coordinate is $ct$.

Let starred quantities denote the capacitor's rest inertial frame, $u_0={ϵ}_0{E}^2/2$, $\beta =v/c$, and $\gamma =(1-{\beta }^2{)}^{-1/2}$. Work in the uniform interplate region of an ideal parallel-plate capacitor, neglecting its fringing field of a capacitor. To transform from that rest inertial frame to a laboratory in which the capacitor moves along $+x$, use the positive-sign Lorentz transformationFor a purely electric field, the spatial block of the electromagnetic stress-energy tensor isThis is the negative of the Maxwell stress tensor: it has $-u_0$ along the electric field and $+u_0$ perpendicular to it. The stress contribution to energy transport in a moving capacitor is essential to the Lorentz transformation of the energy density and flux.

Plates parallel to $xy$. Choose ${\mathbf{E}}_{\ast }=E\hat{\mathbf{z}}$ and ${\mathbf{B}}_{\ast }=0$. The electromagnetic field tensor and electromagnetic stress-energy tensor areThe transformed nonzero independent electromagnetic field tensor components are $F^{03}=\gamma E/c$ and $F^{13}=\gamma \beta E/c$. Thus $\mathbf{E}=\gamma E\hat{\mathbf{z}}$ and $\mathbf{B}=-\gamma vE\hat{\mathbf{y}}/c^2$, consistent with $\mathbf{B}=\mathbf{v}\times \mathbf{E}/c^2$. Transforming the electromagnetic stress-energy tensor givesIn particular, the laboratory Poynting vector is $\mathbf{S}=2u_0{\gamma }^{2}v\hat{\mathbf{x}}$, and the electromagnetic energy density is $u=u_0{\gamma }^2(1+{\beta }^2)$. The ratio $S_x/u=2v/(1+{\beta }^2)$ is subluminal for $v<c$, but differs from the capacitor's velocity. The difference from a simple convective flux isThe extra term is $v{T}_{\ast }^{xx}$, the rest-frame stress multiplied by the velocity. This shows why treating stored field energy as an unstressed parcel moving rigidly with the plates misses part of the energy transport.

Plates parallel to $yz$. Now ${\mathbf{E}}_{\ast }=E\hat{\mathbf{x}}$. The rest tensors areThe electric field is parallel to the Lorentz boost, so $F^{01}=E/c$ remains unchanged and $\mathbf{B}=0$. The cancellation between $T_{\ast }^{00}$ and $T_{\ast }^{11}$ givesThere is no Poynting flux in the ideal interior in this orientation, although the region containing the electric field moves. The longitudinal negative stress contribution cancels the convective term: $S_x={\gamma }^{2}v(u_0+T_{\ast }^{xx})=0$. Field energy is removed at the receding boundary and supplied at the advancing boundary by work on the charged plates; it need not flow through the interior to accompany the moving region. At edges, omitted fringing fields provide additional local structure.

For a finite idealized interplate rest volume $V_0$, length contraction gives $V=V_0/\gamma$. The field-only integrated laboratory energies would consequently be $U_{xy}=\gamma (1+{\beta }^2)U_0$ and $U_{yz}=U_0/\gamma$, with $U_0=u_0{V}_0$. Their different transformation laws are not paradoxical: the field alone is not a closed system. Include the plates and the supports holding them apart, whose mechanical stresses balance the Maxwell stress tensor. The total closed-system four-momentum then has the ordinary Lorentz transformation law. This is the same distinction between a subsystem and a conserved whole that appears in b.

By four-momentum conservation, $p_{\pi }=p_1+p_2$. With the $(-,+,+,+)$ Minkowski metric, the neutral pion has $p_{\pi }^2=-M_{\pi }^2$, while each photon has $p_i^2=0$ and $|{\mathbf{p}}_i|=E_i$. Taking the Minkowski norm givesThe half-angle identity therefore yieldsThis is a measurement of the diphoton invariant mass using two photon energies and their opening angle, independent of the neutral pion's laboratory velocity.

In the center-of-momentum frame, four-momentum conservation gives two opposite photon momenta, each of energy $M_{\pi }/2$. Let $z=\cos {\theta }_{\ast }$, with the axis along the Lorentz boost from that frame to the laboratory. Reversing the axis specified in the source changes $z$ to $-z$ and has no effect on the following product. The Lorentz transformation of four-momentum givesWriting $s=\sin (A/2)$, b now impliesThe isotropy assumption is $dN=a|dz|$ on $-1\le z\le 1$. Each interior opening angle has two inverse branches, $z$ and $-z$, both of which must enter the change of variables for probability densities. The positive branch has derivativeConsequently the diphoton opening-angle distribution isIts support is $2\arcsin (1/\gamma )<A<\pi$ for $0<\beta <1$, and its integral is $2a$, the integral of the original angular probability density before normalization. Thus the normalized probability density is $(2a{)}^{-1}dN/dA$. Omitting one inverse branch would incorrectly halve the total count.

Metric-sign qualification. With this paper's $(-,+,+,+)$ metric signature, b givesThe minus sign in the printed Larmor formula would therefore make a positive radiative loss negative. A concrete counterexample is an instant of rest with nonzero acceleration: $F^{\alpha }=(0,m\mathbf{a})$, so the printed expression is $-q^2{a}^2/(6\pi {ϵ}_0{c}^3)$. That sign is appropriate for the opposite metric signature. With the convention already fixed in question 1, positive emitted power instead isAlternatively the printed negative quantity can be interpreted as the particle's signed radiative energy change, rather than the positive loss magnitude.

For fixed rest mass, resolve the acceleration parallel and perpendicular to the velocity, as in b. This givesWriting $C_q=q^2/(6\pi {ϵ}_0{c}^3)$, the Liénard radiated-power formula becomes $P_{\mathrm{rad}}=C_q({\gamma }^6{a}_{\parallel }^2+{\gamma }^4{a}_{\perp }^2)$. For uniform circular motion, the acceleration is perpendicular to the velocity; for straight-line motion it is parallel. Comparing the same particle at equal relativistic energy and equal laboratory acceleration magnitude $a$ therefore givesThis particular comparison makes the linear loss larger. The familiar larger loss from bending applies to a different comparison: at equal applied laboratory force magnitude $F$, $a_{\perp }=F/(m\gamma )$ and $a_{\parallel }=F/(m{\gamma }^3)$, giving $P_{\mathrm{circular}}=C_q{\gamma }^2{F}^2/m^2$ and $P_{\mathrm{linear}}=C_q{F}^2/m^2$. Stating which quantity is held fixed is essential.

Both the muon and electron have the same electric charge magnitude. At equal kinetic energy $K$ per particle, their Lorentz factors are $\gamma =1+K/(m{c}^2)$, so the heavier muon has the smaller Lorentz factor. At equal total relativistic energy the same ordering follows from $\gamma =E/(m{c}^2)$, provided that energy is large enough to create either particle.

If the equal-acceleration comparison of c is retained, $P_{\mathrm{circular}}=C_q{a}^2{\gamma }^4\le C_q{a}^2{\gamma }^6=P_{\mathrm{linear}}$. Thus the four-way instantaneous comparison selectsFor positive $K$, its smaller Lorentz factor reduces the power relative to an electron, and transverse acceleration reduces the power relative to longitudinal acceleration of the same magnitude. This conclusion follows from the constraint actually stated in c.

For the customary engineering comparison at equal applied force, the result changes. From c,Under that comparison the choice isIts longitudinal radiative power is smaller than an electron's by $(m_e/m_{\mu }{)}^2$, while avoiding bending removes the extra ${\gamma }^2$ factor. At a fixed bend radius $R$, the circular loss is $C_q{\gamma }^4{v}^4/R^2$, giving the familiar $m^{-4}$ suppression at fixed ultrarelativistic energy and $R$.

An energy budget alone does not specify a unique optimization. The force, acceleration, bend radius, acceleration time, and length are otherwise unrestricted. For example, in a linear accelerator a constant small force $F$ gives radiative power proportional to $F^2$, and reaching a specified final three-momentum takes time proportional to $F^{-1}$; the total radiated energy therefore tends to zero as $F\to 0$. One cannot infer a unique accelerator choice without an additional constraint. The standard practical preference is the second boxed answer, with its stated comparison. Muon decay introduces a separate lifetime constraint if slow acceleration is contemplated, but it was not part of the specified radiative-loss calculation.

The first energy term is the rate of change of internal plus kinetic energy per unit volume; the second transports that energy with the moving gas; the pressure term is the mechanical work flux across the boundary. There is no heat flux or viscous work in the stated inviscid adiabatic model.

Use mass conservation to replace a conservative derivative of $\rho e$ by $\rho De/Dt$. Dotting the momentum equation with $\mathbf{u}$ gives $\rho D(|\mathbf{u}{|}^2/2)/Dt=-\mathbf{u}\cdot \nabla p$. Subtraction from total energy leavesNow $p=\rho RT$ and $D\rho /Dt=-\rho \nabla \cdot \mathbf{u}$ implyThus $\boxed{DS/Dt=0}$: smooth adiabatic inviscid trajectories are isentropic. Shocks need an entropy inequality rather than this smooth equality.

Adding $\nabla \cdot (k\nabla T)$ models Fourier heat conduction, with heat flux $-k\nabla T$. The same subtraction gives $\rho c_{v}DT/Dt+p\nabla \cdot \mathbf{u}=\nabla \cdot (k\nabla T)$, henceFor a fixed container use impermeability $\mathbf{u}\cdot \mathbf{n}=0$, insulation $k\nabla T\cdot \mathbf{n}=0$, and mass conservation. ThenThe boundary term vanishes and $\nabla (1/T)=-\nabla T/T^2$ supplies the positive sign. For $T>0$, $k\ge 0$, total entropy cannot decrease: this is entropy production by Fourier heat conduction.

Rigid walls require zero normal velocity, so ${\varphi }_x=0$ at $x=0,L$ and ${\varphi }_y=0$ at $y=0,b$. Separation of variables gives spatial modes $\cos (m\pi x/L)\cos (n\pi y/b)$, for nonnegative integers $m,n$. The wave equation therefore givesThe $(0,0)$ spatially constant mode has zero frequency and is not an ordinary positive-frequency acoustic oscillation.

A spatially uniform boundary forcing excites the $n=0$ transverse mode. Put $\varphi =X(x)\cos \omega t$. Then $X^{\prime \prime }+(\omega /c_0{)}^{2}X=0$ and $X^{\prime }(L)=0$, hence $X=C\cos [\omega (L-x)/c_0]$. The condition $X^{\prime }(0)=a$ fixes $C$ and givesArbitrary free oscillations can be superposed. The denominator vanishes at longitudinal resonances. Near them the lossless steady harmonic response ceases to be a uniformly small-amplitude approximation; damping from viscosity, heat transfer or wall losses regularizes the response, and large amplitudes can require nonlinear equations of motion or a more accurate moving-boundary model. At exact resonance a switched-on ideal forcing produces secular growth rather than a bounded harmonic particular solution.

For the semi-infinite duct expand $F(y)={\sum }_{n\ge 0}{F}_n\cos (n\pi y/b)$. Each axial mode hasFor a propagating mode a particular solution is ${\varphi }_n=(F_n/k_n)\cos (n\pi y/b)\cos (k_{n}x-\omega t)$; differentiating at zero gives the prescribed $F_n\cos (n\pi y/b)\sin \omega t$. A nonzero higher transverse coefficient propagates when $|\omega |>n\pi c_0/b$.

The outgoing radiation condition selects $c_0^2{k}_n/\omega >0$, its positive axial group velocity. Thus $\boxed{\omega k_n>0}$. This sign is selected by the absence of waves incident from infinity, not by the differential equation alone. If $F$ has a nonzero mean, its $n=0$ component can propagate even below every higher-mode cutoff.

For $F=\cos (\pi y/b)$ and $0<|\omega |<\pi c_0/b$, put $\kappa =\sqrt{(\pi /b{)}^2-{\omega }^2/c_0^2}$. The decaying solution isIt is evanescent and carries no mean energy to infinity. These mode and sign conditions are summarized by the radiation condition for a waveguide mode.

Let the bed be $z=0$ and the free surface $z=h(x,t)$. In a long, shallow layer, vertical acceleration is smaller than gravity, so the vertical Euler equation gives hydrostatic pressure $p=p_{\mathrm{atm}}+\rho g(h-z)$. Incompressibility makes the vertical speed smaller than the horizontal speed by the aspect ratio; irrotationality then makes the leading horizontal velocity depth-independent. The horizontal Euler equation becomesIntegrating $u_x+w_z=0$ across the depth, using bed impermeability $w(0)=0$ and the surface condition $w(h)=h_t+u{h}_x$, yieldsCombining these gives the conservative momentum equation $(hu{)}_t+(h{u}^2+g{h}^2/2{)}_x=0$. Integral mass and momentum balance across a stationary thin jump therefore requireThese conservation laws still apply across a jump where the smooth hydrostatic/irrotational derivation does not resolve the internal turbulent structure.

Put $A=h_{-}$, $B=h_{+}$ and $q=A{u}_{-}=B{u}_{+}>0$, orienting the flow from minus to plus. For a nontrivial jump the momentum condition gives $q^2=gAB(A+B)/2$. Its change in mechanical-energy flux isA passive jump cannot create mechanical energy, so the change is nonpositive. Since $q>0$, $\boxed{h_{+}>h_{-}}$ for a genuine jump; equality describes no jump. The upstream flow is supercritical and the downstream flow subcritical, since $F{r}_{-}^2=B(A+B)/(2A^2)>1$ and $F{r}_{+}^2=A(A+B)/(2B^2)<1$.

The gas-dynamic analogue is a compressive normal shock: density and pressure increase and flow changes from supersonic to subsonic. Entropy increases; directed mechanical energy is irreversibly converted into internal energy. Total gas energy, including internal energy, is conserved across an adiabatic shock. This distinguishes that total-energy balance from the reduced shallow-water mechanical-energy loss.

Insert the harmonic variables into the forward map and reduce the squared trigonometric functions. This givesIndeed $p=m\dot{x}$ at this order. The second harmonic and mean displacement of a weakly cubic oscillator show asymmetry of the motion: for positive $\lambda$ the mean shifts towards negative $x$. There is no first-order frequency shift, although the waveform is no longer purely sinusoidal.

The small parameter is $|\lambda |A/(m{\omega }^2)\ll 1$. A purely cubic perturbation is unbounded in one direction, so these formulas describe small local oscillations around the stable origin, not globally confined motion at arbitrary energy. Amplitude $A$ labels the transformed orbit and must be inferred using the inverse map if physical initial position and momentum are supplied.

The energy function $E={\sum }_i{\dot{q}}_i\partial L/\partial {\dot{q}}_i-L$ isThe velocity-linear magnetic term cancels. Since $L$ has no explicit time dependence, $dE/dt=-{\partial }_{t}L=0$. Equivalently, magnetic force does no work, and electric work is exactly minus the change of $e\lambda z^2$. Energy here is defined in the supplied time-independent gauge and equals physical kinetic plus electrostatic potential energy.

Use metric signature $(+,-,-,-)$ and units with propagation speed one. The scalar Euler-Lagrange equation givesWith the stated spatial Fourier convention, $-{\nabla }^2$ becomes $k^2$. Each transformed mode obeysHere $F$ and $V$ are the initial transformed field and velocity. Reality requires $F(-\mathbf{k})=F(\mathbf{k}{)}^{\ast }$ and the same for $V$; the Fourier convention places no extra $(2\pi {)}^3$ in these evolution formulas.

Including the interaction gives the two coupled real Klein-Gordon fields,Adding and subtracting diagonalizes them:Both squared masses are positive for $0<g<m^2$. For arbitrary initial data $F_i={\tilde{\varphi }}_i(0)$ and $V_i={\dot{\tilde{\varphi }}}_i(0)$, defineThen the general solution is ${\tilde{\varphi }}_1=(U_{+}+U_{-})/2$, ${\tilde{\varphi }}_2=(U_{+}-U_{-})/2$, with the corresponding reality conditions on initial data.

For the specified sinusoidal initial displacement, put ${\Omega }_{\pm }=\sqrt{q^2+m^2\mp g}$. Both normal fields initially have amplitude $A\sin (\mathbf{q}\cdot \mathbf{x})$ and zero time derivative. ConsequentlyWriting $\bar{\Omega }=({\Omega }_{-}+{\Omega }_{+})/2$ and $\delta \Omega =({\Omega }_{-}-{\Omega }_{+})/2$ gives ${\varphi }_1=A\sin (\mathbf{q}\cdot \mathbf{x})\cos \bar{\Omega }t\cos \delta \Omega t$ and ${\varphi }_2=A\sin (\mathbf{q}\cdot \mathbf{x})\sin \bar{\Omega }t\sin \delta \Omega t$. This is field-amplitude beating of coupled Klein-Gordon modes. The conserved energy includes the interaction term $-g{\varphi }_1{\varphi }_2$; separate uncoupled-field energies need not be conserved, and an envelope zero is not by itself exact complete energy transfer.

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.

Let $\psi$ be the small hoop's absolute spin angle with the same angular sign convention. Equating the two contact-point velocities cancels the large hoop centre's translation. The remaining tangential equality isThis rolling constraint inside a moving circular track adds small-hoop rotational energy $M{a}^2(2\dot{\theta }-\dot{\varphi }{)}^2/2$. At quadratic order its cross term cancels the translational cross term in part a. The potential is unchanged, soStatic friction enforces rolling; ideal no-slip contact dissipates no energy.

Separating $G=W(q)-\beta t$ gives $[W^{\prime }(q){]}^2/(2m)+m{\omega }^2{q}^2/2=\beta$. On the positive-momentum branch chooseWith $Q=\beta$ and $A=\sqrt{2\beta /(m{\omega }^2)}$, differentiation with respect to $Q$ givesSince $P=\gamma$ is constant,The displayed integral is a local positive-$p$ branch. Continuing the trajectory through turning points switches momentum branches and extends the sine solution over the complete orbit. $\beta$ is the oscillator energy, and $\gamma$ fixes its time phase.

At mean-field level, positive gradient stiffness favours a spatially uniform order parameter. Writing $r=|\varphi |$, minimization of $f=a{r}^2+r^4/2$ giveswith arbitrary constant $\delta$. The minimized density is zero above the transition and $-a^2/2$ below it. The order parameter vanishes continuously as $(-a{)}^{1/2}$; the first derivative of the minimized free energy is continuous while the second changes. This is the continuous mean-field transition.

The broken symmetry is global U(1) symmetry, not merely the Ising sign symmetry. Selecting one ordered phase breaks this continuous symmetry and leaves a soft phase direction. These are minima of the supplied Landau free energy, not an exact finite-temperature long-range-order claim in one dimension: phase fluctuations invalidate that inference for a short-range one-dimensional system.^[ref]

In the same Minkowski metric and electromagnetic units, $T^{\mu \nu }=-F^{\mu }{}_{\lambda }{F}^{\nu \lambda }+\frac{1}{4}{g}^{\mu \nu }{F}_{\alpha \beta }{F}^{\alpha \beta }$. Differentiating givesTo simplify the last two terms, contract ${\partial }_{\mu }{F}_{\nu \lambda }+{\partial }_{\nu }{F}_{\lambda \mu }+{\partial }_{\lambda }{F}_{\mu \nu }=0$ with $F^{\mu \lambda }$. Its first and third terms are equal, and its middle term is $-\frac{1}{2}{\partial }_{\nu }(F_{\alpha \beta }{F}^{\alpha \beta })$. Raising $\nu$ therefore gives $F^{\mu }{}_{\lambda }{\partial }_{\mu }{F}^{\nu \lambda }=\frac{1}{4}{\partial }^{\nu }(F_{\alpha \beta }{F}^{\alpha \beta })$. The cancellation leavesThis is the electromagnetic stress-energy divergence identity in rationalized units. As a sign check, $T^{00}=(E^2+B^2)/2$ is positive and $T^{0i}=(\mathbf{E}\times \mathbf{B}{)}_i$ is the Poynting vector. The integrated four-momentum $P^{\nu }=\int T^{0\nu }{d}^{3}x$ is conserved if the boundary flux vanishes. With matter sources, the electromagnetic field alone instead obeys ${\partial }_{\mu }{T}^{\mu \nu }=-F^{\nu \lambda }{J}_{\lambda }$: energy and momentum transfer to matter restore conservation of the total stress-energy tensor.

For the isotropic zero-field classical Heisenberg model of a ferromagnet, the analytic vector Landau expansion uses the Cartesian components of the vector order parameter near $\mathbf{m}=0$. rotational symmetry makes its scalar terms functions of $\mathbf{m}\cdot \mathbf{m}$. Global spin inversion symmetry also requires $f(\mathbf{m})=f(-\mathbf{m})$. Through fourth order this givesA positive quartic coefficient stabilizes the unconstrained truncated free energy. The disordered phase has $b>0$ and its minimum at $m=0$; the aligned phase has $b<0$ and a minimum at nonzero $m$. At the continuous transition $b=0$. The attractive alignment interaction supplies an energetic tendency toward negative quadratic curvature; orientational entropy favors the disordered phase at high temperature, so the sign of $b$ is a free-energy question rather than a consequence of the pair energy alone.

The analyticity assumption matters: the variable printed as $m$ is a nonnegative magnitude, so reversing $\mathbf{m}$ does not send that scalar to $-m$. Terms proportional to $|\mathbf{m}|$ or $|\mathbf{m}{|}^3$ would be rotationally invariant but nonanalytic at the vector origin. Standard analytic Landau theory excludes them. Without that assumption, the stated pair symmetry alone would not exclude odd powers of the magnitude.^[ref]

The independent head-tail reversal symmetry means that the squared dot product interaction energy is unchanged when any individual ${\mathbf{s}}_i$ is replaced by $-{\mathbf{s}}_i$. All configurations ${\mathbf{s}}_i={\sigma }_i\mathbf{n}$, with ${\sigma }_i=\pm 1$ independently, attain the aligned ground-state pair energy. There is consequently no energetic preference for either end of a molecular axis. Averaging over these equally weighted signs gives $⟨{\mathbf{s}}_i⟩=0$ and $⟨\mathbf{m}⟩=0$ even in the nematically ordered phase. The surviving order is the unoriented axis, represented by a nematic director with $\mathbf{n}\sim -\mathbf{n}$. A second-moment tensor can detect this order while a polar vector cannot.^[ref]

The word “always” needs the statistical ensemble or apolar interpretation: it is not an identity for each finite configuration. For example, choosing all ${\sigma }_i=+1$ is also a ground-state representative and gives $\mathbf{m}=\mathbf{n}$. In an unbiased sign statistical ensemble the typical residual polar vector has size of order $N^{-1/2}$, because $⟨|\mathbf{m}{|}^2⟩=1/N$ for these independent ground-state signs. This does not make polar order an equilibrium nematic order parameter. The tensor below is unchanged by each sign flip and thus removes the arbitrary labeling of rod ends.