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For RNS string matter normalized by $[{\dot{X}}^{\mu }(\sigma ),X^{\nu }({\sigma }^{\prime })]=-i\pi {\eta }^{\mu \nu }\delta (\sigma -{\sigma }^{\prime })$ and $\{{\psi }_1^{\mu }(\sigma ),{\psi }_1^{\nu }({\sigma }^{\prime })\}=\pi {\eta }^{\mu \nu }\delta (\sigma -{\sigma }^{\prime })$, put $A={\partial }_{+}X$ and $J={\psi }_1\cdot A$. Then $[A^{\mu }(\sigma ),A^{\nu }({\sigma }^{\prime })]=(i\pi /2){\eta }^{\mu \nu }{\partial }_{\sigma }\delta (\sigma -{\sigma }^{\prime })$. The noncentral part of $\{J(\sigma ),J({\sigma }^{\prime })\}$ is $\pi \delta A^2+(i\pi /2){\partial }_{\sigma }\delta {\psi }_1(\sigma )\cdot {\psi }_1({\sigma }^{\prime })$. The distributional derivative identity $f(\sigma ,{\sigma }^{\prime }){\delta }^{\prime }=f(\sigma ,\sigma ){\delta }^{\prime }+({\partial }_{{\sigma }^{\prime }}f)(\sigma ,\sigma )\delta$, with a vanishing coincident Grassmann bilinear, converts this to $\pi \delta {\theta }_{++}$. Use the chiral equation of motion to identify ${\partial }_{\sigma }{\psi }_1={\partial }_{+}{\psi }_1$. Quantum normal ordering adds a central contact term proportional to ${\delta }^{\prime \prime }$, the local form of the N=1 super-Virasoro algebra central extension.

Use local east, north, and up unit vectors $({\mathbf{e}}_E,{\mathbf{e}}_N,{\mathbf{e}}_U)$, with ${\mathbf{e}}_E\times {\mathbf{e}}_N={\mathbf{e}}_U$. In a spherical-Earth approximation at geographical latitude $\phi$, the angular velocity has components $\mathbf{\Omega }=(0,\Omega \cos \phi ,\Omega \sin \phi )$. Expanding the cross product for relative velocity $(v_E,v_N,v_U)$ givesThese are Coriolis acceleration components, not the total acceleration. For horizontal motion, the horizontal part is $(f{v}_N,-f{v}_E)$, where the Coriolis parameter is $f=2\Omega \sin \phi$. Thus northward motion in the northern hemisphere has an eastward Coriolis acceleration, to the right of its velocity; in the southern hemisphere the horizontal turning is to the left. At the equator $f=0$, but the full Coriolis acceleration need not vanish: vertical or eastward motion still couples through $\Omega \cos \phi$.

For a numerical example, take the Earth rotation rate $\Omega =7.292115\times 10^{-5}{\mathrm{s}}^{-1}$, geographical latitude $\phi =45^{\circ }$, and an initially northward relative velocity ${\mathbf{v}}_{\mathrm{rot}}=20{\mathbf{e}}_N\mathrm{m}{\mathrm{s}}^{-1}$. Then

For a local, short-time estimate with no other horizontal force and zero initial eastward velocity, hold the initial northward velocity in the leading term: $\Delta x_E\simeq \frac{1}{2}fu{t}^2$. At $t=10\mathrm{s}$ this gives about $0.1031\mathrm{m}$ eastward. It assumes nearly fixed local axes and geographical latitude, negligible changes of height, and $|ft|\ll 1$. It is not a complete wind trajectory: pressure gradients, friction, and evolving velocity must enter an atmospheric equation of motion. The effective gravity in a rotating frame already combines ordinary gravity with the steady terrestrial centrifugal acceleration.

A local real scalar field with Lagrangian density $\frac{1}{2}[({\partial }_t\varphi {)}^2-{\kappa }^2({\nabla }^2\varphi {)}^2-m^2{\varphi }^2]$ has the equation of motion ${\partial }_t^2\varphi +{\kappa }^2{\nabla }^4\varphi +m^2\varphi =0$. Two spatial integrations by parts produce the positive biharmonic operator in this equation. Its Hamiltonian is a sum of positive squares and each nonzero-frequency spatial Fourier mode quantizes as an oscillator with the displayed dispersion relation. Higher spatial derivatives do not require higher time derivatives. The theory has a preferred time and does not possess Lorentz invariance.

Two Lagrangian densities differing by a total derivative yield the same action under suitable boundary conditions. For example ${\partial }_{\mu }({\varphi }^3{\partial }^{\mu }\varphi )=3{\varphi }^2(\partial \varphi {)}^2+{\varphi }^3\square \varphi$. Likewise $\varphi {\square }^2\varphi \simeq -\partial \varphi \cdot \partial \square \varphi \simeq (\square \varphi {)}^2\simeq ({\partial }_{\mu }{\partial }_{\nu }\varphi {)}^2$. These are algebraic relations modulo boundary terms, not uses of the equation of motion.

The classical isotropic harmonic oscillator has the same angular frequency $\Omega >0$ in every Cartesian direction. Its equation of motion is $\ddot{\mathbf{r}}=-{\Omega }^2\mathbf{r}$, with solution $\mathbf{r}=\mathbf{A}\cos \Omega t+\mathbf{B}\sin \Omega t$. The trajectory lies in the span of $\mathbf{A},\mathbf{B}$ and is an ellipse centred on the force centre, degenerating to a line for dependent vectors. For nonzero angular momentum, $r^2$ oscillates at $2\Omega$ unless the trajectory is exactly circular. This distinguishes radial and Cartesian frequencies.

Orbit determination infers a body's position and velocity from observations and an equation of motion. It is an inverse problem; a residual of an inverse problem compares an observed readout with its modeled prediction.

Vary the action, taking the field variation to vanish at the boundary. Integration by parts givesThus the Euler-Lagrange field equation isFor the field redefinition in part f, define the functional derivative $E(\varphi )=\delta S_0/\delta \varphi =-(\square \varphi +m^2\varphi +\lambda {\varphi }^3/6)$. The sign does not affect the equation of motion, but it fixes the sign of an added redundant operator.

Let $a=\lambda /6$. Within terms already suppressed by ${\Lambda }^{-2}$, the leading equation of motion gives $\square \varphi =-m^2\varphi -a{\varphi }^3$. Equality here means equivalence under a perturbative field redefinition, together with integration by parts, rather than a pointwise identity inside arbitrary off shell correlation functions. HenceFor the second relation, factor the difference as $(\square \varphi -m^2\varphi -a{\varphi }^3)(\square \varphi +m^2\varphi +a{\varphi }^3)$, which is explicitly a redundant operator. Ifthe reduced correction isThe first two coefficients shift the already present mass and quartic coupling constant. After these shifts,This is the equation-of-motion reduction of scalar operators through $O({\Lambda }^{-2})$; terms of $O({\Lambda }^{-4})$ are outside that truncation. The lower-dimensional shifts should not be silently dropped when $m\ne 0$.

Take the initially expanding branch ${\dot{r}}_i=H_i{r}_i$; the printed squared velocity alone also permits an initially contracting branch, which would not reach a future maximum radius. Multiply the equation of motion by $\dot{r}$ and integrate. The conserved outer-shell specific orbital energy isThe initial mass follows from the background critical density:ConsequentlyIf $E$ means the energy of a test shell of mass $m$, $E=mϵ$. If it means the total energy of the entire uniform, homologously moving perturbation, direct integration instead gives $K=3M{\dot{r}}^2/10$, $U=-3G{M}^2/(5r)$ andThe common convention of writing the first integral as $E$ is therefore an energy per unit shell mass, not the total energy of the sphere.

At spherical-collapse turnaround, $\dot{r}=0$, so $ϵ=-GM/r_m$. HenceAny positive initial density contrast binds this pressureless perturbation in the stated Einstein-de Sitter universe.

There are two different physical interpretations. The supplied logarithmic integral signals free fall in a fixed singular isothermal sphere, with a test particle moving through a frozen background. Write $C={\rho }_0{r}_0^2$. By the shell theorem, the enclosed mass is $M(r)=4\pi Cr$ and the radial equation of motion isThe second equality follows by multiplying by $\dot{r}$ and using the initial rest condition. The free fall trajectory is given implicitly byThus the intended fixed-profile answer isA collapsing material shell gives a different answer. If the entire cloud undergoes pressureless gravitational collapse from rest, the edge follows the fixed enclosed mass $M_0=4\pi {\rho }_0{r}_0^3$, not the instantaneous fixed-profile value $4\pi Cr$. With no shell crossing, $\ddot{r}=-G{M}_0/r^2$, soThis is a concrete counterexample to treating the fixed-profile result as the exact collapse time of the whole nebula. The initial mean density inside $r_0$ is $3{\rho }_0$, consistent with the usual free-fall time of a uniform sphere expressed using that mean density. The logarithmic integral is valid for the explicitly frozen background approximation.

Use fixed masses in an isolated Newtonian $N$-body system, with origin at its centre of mass. Define the $xx$ moment $I_{xx}={\sum }_i{m}_i{x}_i^2$ and $K_{xx}=\frac{1}{2}{\sum }_i{m}_i{v}_{x,i}^2$. Twice differentiating givesInsert the gravitational equation of motion and pair each ordered term $(i,k)$ with $(k,i)$:Thus one component of the tensor virial theorem isA stationary or time-averaged bounded system has $⟨{\ddot{I}}_{xx}⟩=0$, yielding $2⟨K_{xx}⟩+⟨W_{xx}⟩=0$. If ordered and random motions are separated, $2K_{xx}=2T_{xx}+{\Pi }_{xx}$, where $T_{xx}=\frac{1}{2}\int \rho u_x^{2}dV$ and ${\Pi }_{xx}=\int \rho {\sigma }_{xx}^{2}dV$. Summing the three diagonal equations recovers the scalar virial theorem, with the pair gravitational potential energy counted once.

Throughout this question use natural units with $c=1$. For zero potential and nonzero kinetic energy, the homogeneous canonical scalar field has $P=\rho ={\dot{\phi }}^2/2$, so $\boxed{w=1}$. Its equation of motion becomes $d(a^3\dot{\phi })/dt=0$, giving $\dot{\phi }\propto a^{-3}$ and thereforeThe same scaling follows from cosmological energy conservation with constant equation-of-state parameter $w=1$. If the field is exactly constant, both $P$ and $\rho$ vanish and their ratio is undefined; that is the separate empty solution.

Neglecting gravity, the equation of motion is $m\ddot{\mathbf{x}}=a\dot{\mathbf{x}}\times \mathbf{x}/r^3$, where $r=|\mathbf{x}|>0$. Dotting with the velocity gives $d(m|\dot{\mathbf{x}}{|}^2/2)/dt=0$, because a cross product is perpendicular to its factors. Thus the speed is constant. The vector triple product givesHence the generalized angular momentum in a radial monopole field isDotting with $\mathbf{x}$ gives $\mathbf{L}\cdot \mathbf{x}=-ar$, or $\hat{\mathbf{L}}\cdot \hat{\mathbf{x}}=-a/|\mathbf{L}|$. For $\mathbf{L}\ne 0$, the direction of $\mathbf{x}$ therefore makes the fixed angle $\boxed{\cos \beta =-a/|\mathbf{L}|}$ with the axis $\mathbf{L}$, describing a right circular conical surface with vertex at the North Pole. Orthogonality gives $|\mathbf{L}{|}^2=m^2|\mathbf{x}\times \dot{\mathbf{x}}{|}^2+a^2$, so this cosine is in $[-1,1]$. Radial motion gives a degenerate cone; when $a=0$ and the ordinary angular momentum is nonzero the cone is the usual orbital plane. If $\mathbf{L}=0$, the norm identity forces $a=0$ and $\mathbf{x}\times \dot{\mathbf{x}}=0$, giving a radial straight line.

Use target-space signature $(-,+,\dots ,+)$ and let a dot between target vectors denote contraction with the Minkowski metric. Varying the string embedding map in the Polyakov action, with periodic variations around the closed string, givesThus the embedding equation of motion isThe independent worldsheet metric variation uses $\delta \sqrt{-\gamma }=-\frac{1}{2}\sqrt{-\gamma }{\gamma }_{ab}\delta {\gamma }^{ab}$. It sets the worldsheet stress tensor to zero:In conformal gauge, ${\gamma }_{ab}=e^{2\omega }\mathrm{diag}(-1,1)$, the Weyl factor cancels. Define worldsheet derivatives ${\partial }_{\pm }=({\partial }_{\tau }\pm {\partial }_{\sigma })/2$. The equations and Virasoro constraints reduce toEquivalently, $\ddot{X}-X^{\prime \prime }=0$, $\dot{X}\cdot X^{\prime }=0$ and ${\dot{X}}^2+X^{\prime 2}=0$. The free wave equation permits independent periodic left- and right-moving fields.

Take $0\le \sigma <2\pi$ and the uncompactified closed-string mode expansionThe Virasoro generators are the Fourier coefficients of the two quadratic worldsheet stress tensors. Their quantum expressions areThe canonical commutation relations are $[{\alpha }_m^{\mu },{\alpha }_n^{\nu }]=m{\eta }^{\mu \nu }{\delta }_{m+n,0}$, with an identical tilded algebra and commuting chiral sectors. Normal ordering places annihilation modes to the right. It gives $L_0={\alpha }^{\prime }{p}^2/4+N$, where $N={\sum }_{n>0}{\alpha }_{-n}\cdot {\alpha }_n$, and similarly on the other side. In old covariant string quantization, physical string states obeymodulo physical null string states. Negative generators are adjoint constraints on bras, not further annihilation conditions on the same ket. In critical bosonic string theory, $d=26$ and the string intercept is $a=1$. These closed-string physical-state Virasoro conditions implyThus the massless first closed-string level has $N=\tilde{N}=1$, and its general state isTo derive the polarization tensor constraints, the oscillator algebra gives $[L_n,{\alpha }_m^{\mu }]=-m{\alpha }_{n+m}^{\mu }$. Applying $L_1$ to this state replaces ${\alpha }_{-1}^{\mu }$ by ${\alpha }_0^{\mu }$; the independent tilded condition replaces the other oscillator by ${\tilde{\alpha }}_0^{\nu }$. All higher positive generators annihilate the state automatically. ThereforeThere is no condition that the tensor trace vanish: each chiral side has only one oscillator, so neither $L_2$ nor ${\tilde{L}}_2$ imposes a trace condition. The null-state quotient of a string also identifiesFor example, $L_{-1}({\xi }_{\nu }{\tilde{\alpha }}_{-1}^{\nu }|0,p⟩)$ is a momentum-longitudinal state with zero inner product against every transverse physical polarization. This explains the string-state gauge redundancy, rather than merely counting covariant components. For an ordinary nonzero null momentum, choose a frame adapted to $p$. The constraints remove one null direction in each index, and these null-state identifications remove the remaining longitudinal directions. The representatives are then ${\zeta }_{ij}$, $i,j=1,\dots ,24$.

Decompose that transverse matrix into its symmetric trace-free part, antisymmetric part and trace. They describe the graviton, Kalb–Ramond field and dilaton, respectively. Their particle polarization counts areThese are representations of the rotational part $SO(24)$ of the massless little group. The dilaton corresponds to the transverse trace; a covariant scalar polarization can be written proportional to ${\eta }_{\mu \nu }-p_{\mu }{\ell }_{\nu }-{\ell }_{\mu }{p}_{\nu }$, with ${\ell }^2=0$ and $p\cdot \ell =1$. It is transverse in both indices and has trace 24. The nonzero-momentum assumption is appropriate for propagating particle states; zero momentum requires separate treatment of gauge and background zero modes.

In a general background, these fields replace the flat-space free action by the string nonlinear sigma model. A convenient Euclidean convention isHere ${ϵ}^{ab}$ is the orientation density; reversing its convention changes the displayed sign of the two-form coupling. The metric tensor is $G_{\mu \nu }={\eta }_{\mu \nu }+h_{\mu \nu }+\cdots$, $B$ is the antisymmetric Kalb–Ramond field, and $\Phi$ is the dilaton. Their worldsheet couplings reproduce the corresponding massless closed-string vertex operators when expanded about flat space. The Polyakov path integral integrates $e^{-S_E}$ over embedding fields and metrics modulo worldsheet diffeomorphisms and Weyl transformations; gauge fixing supplies ghost determinants and moduli integrations. Quantum consistency of a general background additionally requires vanishing sigma-model beta functions, not just insertion of arbitrary backgrounds into a free theory.

Finally, separate the constant dilaton mode, $\Phi ={\Phi }_0+\hat{\Phi }$. By the Gauss-Bonnet theorem, its action isFor a connected oriented surface of genus $g$, its Euler characteristic is $\chi =2-2g$. Defining the string coupling by $g_s=e^{{\Phi }_0}$ givesThus the string genus expansion is a topology sum with two extra powers of $g_s$ per handle. An amplitude with $n$ canonically normalized external closed-string vertices has the additional vertex normalization $g_s^n$, so its overall factor is $g_s^{2g-2+n}$. The curvature term, rather than a guessed coupling inserted at each internal interaction, supplies this dependence directly from the functional integral.

Multiplying the tangential equation of motion by $r$ givesTherefore $h=r^2\dot{\theta }$ is constant. It is the signed specific angular momentum; reverse the angular orientation if necessary to take $h>0$ for a nonradial orbit. Set $u=1/r$ and denote differentiation with respect to $\theta$ by a prime. ThenSubstitution into the radial equation of motion and division by $-h^2{u}^2$ give the Binet equationIts general solution is $u=GM/h^2+A\cos \theta +B\sin \theta$. Write $A=(GM/h^2)e\cos {\theta }_0$ and $B=(GM/h^2)e\sin {\theta }_0$, where $e\ge 0$. This produces the Kepler orbitHere $e$ is the orbital eccentricity, and ${\theta }_0$ points toward periapsis when $e>0$. Only portions with $u>0$ describe positive radial distances. The expression assumes $h\ne 0$; a zero-angular momentum trajectory is radial and must be treated separately.

Let $\rho =3m/(4\pi R_0^3)$ be the uniform mass density. In spherical coordinates $(r,\vartheta ,\varphi )$ with the chosen diameter as polar axis, the squared perpendicular distance to that axis is $r^2{\sin }^2\vartheta$. The moment of inertia is thereforeThis derives the moment of inertia of a uniform solid sphere about any central axis by spherical symmetry.

Put $a=R_1-R_0$. The center of mass follows the circle of radius $a$, with position ${\mathbf{r}}_C=a(\sin \theta ,-\cos \theta )$. Let ${\mathbf{e}}_r=(\sin \theta ,-\cos \theta )$ and ${\mathbf{e}}_{\theta }=(\cos \theta ,\sin \theta )$. Its velocity is $a\dot{\theta }{\mathbf{e}}_{\theta }$, and the contact point is displaced from the center of mass by $R_0{\mathbf{e}}_r$. With positive spin angular velocity ${\Omega }_s$ counterclockwise, its contact-point velocity isThe cylinder is fixed, so rolling without slipping requires this velocity to vanish. Consequently ${\Omega }_s=-a\dot{\theta }/R_0$. If instead the spin angle $\psi$ is positive clockwise, opposite to increasing $\theta$, then the printed positive-coefficient convention isThe spin and orbital rotation have opposite directions when both are measured about the same oriented axis; the angular speed is $a|\dot{\theta }|/R_0$. This specifies the sign convention in the printed expression.

The instantaneous contact point is at rest, so the normal reaction and static friction do no work. Conservation of energy equates the loss of gravitational potential energy from release to the total kinetic energy:On the descent from $0<\alpha <\pi /2$, $\theta$ decreases, soIntegrating $dt=-d\theta /|\dot{\theta }|$ from release to the bottom givesThe integral is finite: at the release point its denominator is asymptotic to $\sqrt{\sin \alpha (\alpha -\theta )}$. Contact is maintained, since the inward radial equation of motion gives the positive normal reaction $N=m(g\cos \theta +a{\dot{\theta }}^2)$ throughout this range.

For frictionless sliding, the normal reaction passes through the sphere's centre and produces no torque; Newtonian gravity also exerts no central torque. With initial zero spin, the sphere remains nonrotating. All the lost gravitational potential energy becomes translational kinetic energy, soA hollow spherical shell of the same mass and radius places its material farther from the centre, giving a larger moment of inertia. Under rolling without slipping, a given centre speed therefore requires more rotational kinetic energy; the same gravitational drop yields a smaller speed. Replacing the solid sphere by a hollow shell increases $T_R$ and leaves $T_S$ unchanged. The sliding case remains nonrotating and has the same center of mass path and gravitational drop. This is the general comparison for rolling inside a fixed circular cylinder.

Put $M=m_1+m_2$ and let $\mathbf{R}=(m_1{\mathbf{r}}_1+m_2{\mathbf{r}}_2)/M$ be the center of mass. Applying Newton's second law separately to the two point masses givesThus the center of mass has constant velocity, $\mathbf{R}(t)=\mathbf{R}(0)+t\dot{\mathbf{R}}(0)$. For the relative position $\mathbf{r}={\mathbf{r}}_1-{\mathbf{r}}_2$, subtraction givesHere $\mu$ is the reduced mass; it converts the relative motion into a one-body equation of motion.

For the given linear force, the relative position obeys simple harmonic motion, $\ddot{\mathbf{r}}+{\Omega }^2\mathbf{r}=0$, where ${\Omega }^2=k/\mu$. The initial rest condition givesFor $d>0$, collision first occurs when the relative position vanishes, namely when $\Omega t=\pi /2$. ThereforeThe separation $d$ cancels because the frequency of this simple harmonic motion is independent of its amplitude.

During ejection, the remaining mass is $m(t)=m_o+m_w-Qt$, and the water is exhausted at $t_c=m_w/Q$. Consider the momentum of the octopus and the small portion of water expelled during $dt$. Initially it is $mu$. At the end of this interval, the remaining body has mass $m-Qdt$ and velocity $u+du$, while the expelled mass $Qdt$ has laboratory velocity $u-V$, to first order. Hence the total final momentum isThe external impulse from quadratic drag is $-\alpha u^{2}dt$. Equating this to the change in total momentum, and dividing by $dt$, proves the equation of motionThe outgoing water's momentum flux is essential: applying $d(mu)/dt$ to the remaining body alone would omit it.

For dimensional analysis, the independent physical inputs are $m_o,m_w,Q,V,\alpha$, with dimensionsThe scales $m_o$, $m_o/Q$ and $V{m}_o/Q$ fix units of mass, time and length. In these units, the only remaining independent dimensionless variables areFor example, with $s=Qt/m_o$ and $w=u/V$, the equation of motion becomes $(1+\lambda -s)dw/ds=1-\mu w^2$, with $w(0)=0$ and ejection ending at $s=\lambda$. Thus the dimensionless terminal value depends only on $\lambda ,\mu$, proving $u_c=Vf(\lambda ,\mu )$.

For its explicit value, put $m_0=m_o+m_w$ and $U=\sqrt{QV/\alpha }$. The initially stationary solution stays below $U$, since the acceleration vanishes at $U$ and is positive below it. Separation of variables givesThe two integrals yieldThis is constant mass-loss propulsion with quadratic drag. At $t_c$, the remaining mass is $m_o$, givingThus $f(\lambda ,\mu )={\mu }^{-1/2}\tanh (\sqrt{\mu }\log (1+\lambda ))$, verifying the required dimensional form. The limit as drag tends to zero is $V\log (1+\lambda )$, the ordinary rocket equation, while the positive-drag answer is below the terminal velocity $U$.

A two-component Weyl spinor is a complex spinor transforming in one of the two inequivalent group representations $(1/2,0)$ and $(0,1/2)$ of the Lorentz spinor double cover $\mathrm{SL}(2,\mathbb{C})$. An undotted left-handed field ${\chi }_{\alpha }$ transforms by a matrix $M\in \mathrm{SL}(2,\mathbb{C})$; its complex conjugate ${\bar{\chi }}_{\dot{\alpha }}$ transforms by $M^{\ast }$. Dotted and undotted indices distinguish the two chirality representations. Raising a dotted index uses the invariant epsilon tensor and converts the complex conjugate transformation into the equivalent inverse-adjoint form. Rotations act by the same SU(2) generators on the two chiralities, while boosts have opposite generators, so the two complex representations cannot be identified by a complex-linear Lorentz invariant map.

The antisymmetric invariant tensors ${ϵ}_{\alpha \beta }$ and ${ϵ}_{\dot{\alpha }\dot{\beta }}$ raise and lower spinor indices; $M^TϵM=ϵ$ proves invariance of the contracted bilinear $\chi \psi ={\chi }^{\alpha }{\psi }_{\alpha }$. The components of a fermionic Weyl spinor are odd Grassmann variables, so this bilinear is symmetric under exchange of the two spinors: the antisymmetric epsilon sign cancels the Grassmann sign. In particular, choosing ${ϵ}_{12}=1$ and ${ϵ}^{12}=-1$ gives $\chi \chi =2{\chi }_1{\chi }_2$, generally nonzero. This Weyl spinor bilinear exchange identity permits a Majorana mass term for a single suitable Weyl spinor. With mostly-minus metric signature and ${\bar{\sigma }}^{\mu }=(I,-\mathbf{\sigma })$, the free massless kinetic term is $i{\chi }^{†}{\bar{\sigma }}^{\mu }{\partial }_{\mu }\chi$, and its field equation of motion is $\boxed{{\bar{\sigma }}^{\mu }{\partial }_{\mu }\chi =0}$.

Here L-superfield means a left-chiral superfield: ${\bar{D}}_{\dot{\alpha }}\Phi =0$, and its complex conjugate is an antichiral superfield, $D_{\alpha }{\Phi }^{\ast }=0$. Interpret the two barred derivatives as the contracted scalar ${\bar{D}}^2={\bar{D}}_{\dot{\alpha }}{\bar{D}}^{\dot{\alpha }}$. The key point is chiral projection by squared supercovariant derivatives. Since the two components of $\bar{D}$ anticommute, any product of three barred derivatives repeats a component and is zero. Thusfor every superfield $X$, even when $X$ is antichiral. In particular, ${\bar{D}}^2{\Phi }^{\ast }$ is chiral.

The right side is chiral too. Because $\Phi$ is even, the graded derivation rule gives ${\bar{D}}_{\dot{\alpha }}({\Phi }^2)=({\bar{D}}_{\dot{\alpha }}\Phi )\Phi +\Phi ({\bar{D}}_{\dot{\alpha }}\Phi )=0$. Constant $m$ and $\lambda$ do not contribute derivatives. Consequentlyso applying the chiral constraint to the equation imposes no additional incompatible condition. This is the holomorphic closure of chiral superfields; a generic nonholomorphic right side would not satisfy it.

Both sides have zero Grassmann parity and are Lorentz scalars: the two dotted derivative indices are contracted with the invariant epsilon tensor, and $\Phi$ is a scalar superfield. Moreover, each covariant derivative anticommutes with every supercharge, so its squared contraction commutes with the supercharges. Hence the equation transforms covariantly under supersymmetry. It is the chiral field equation of motion of the Wess–Zumino model with superpotential $\mathcal{W}(\Phi )=m{\Phi }^2/2+\lambda {\Phi }^3/3$, for which ${\mathcal{W}}^{\prime }=m\Phi +\lambda {\Phi }^2$, using the normalization in the PDF. With mass dimensions $[\Phi ]=1$, $[\bar{D}]=1/2$, $[m]=1$ and $[\lambda ]=0$, each side has dimension two. No component field equations of motion are needed for this consistency argument.