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The flux representation gives $B_R=-{\psi }_z/R$ and $B_z={\psi }_R/R$. Therefore $\nabla \cdot \mathbf{B}=0$ and $\mathbf{B}\cdot \nabla \psi =0$. With magnetohydrodynamic mass loading $k=k(\psi )$,Thus the steady continuity equation is satisfied.

The cross product in the ideal magnetohydrodynamic induction equation simplifies toHence steady induction requires $\mathbf{B}\cdot \nabla (v/R)=0$. On a regular connected poloidal flux surface, this integrates locally toThis is the field-line angular velocity of an axisymmetric wind. It need not equal the gas angular velocity when $B_{\varphi }\ne 0$, because $u_{\varphi }=R\omega +k{B}_{\varphi }/\rho$. The regular-surface qualification matters: if $\psi$ is constant and the field is purely toroidal, induction alone cannot turn an arbitrary $v(R,z)/R$ into a function of $\psi$.

Put $\mathbf{F}=(\nabla \times \mathbf{B})\times \mathbf{B}$ and $E=|\mathbf{u}{|}^2/2+\Phi$. Dot the pressure-free momentum equation with $\mathbf{u}$. Since $\mathbf{B}\cdot \mathbf{F}=0$, axisymmetry givesThe azimuthal momentum equation in cylindrical coordinates similarly givesSubtracting $\omega$ times the second relation from the first, and using $\mathbf{B}\cdot \nabla \omega =0$, proves that $E-\omega R{u}_{\varphi }$ is constant along a flowing regular poloidal flux surface. For nonzero mass loading its corotating energy invariant of an axisymmetric magnetic wind isThis is the invariant labeled $ϵ$ in the question. It is a corotating mechanical energy, distinct from the total magnetohydrodynamic Bernoulli invariant, which includes electromagnetic energy flux. In standard notation it equals total specific energy minus $\omega$ times the total specific angular-momentum invariant.

Near the footpoint, negligible $k{B}_{\varphi }/\rho$ gives $u_{\varphi }\simeq R\omega$, while the poloidal velocity is parallel to the poloidal magnetic field. The gas approximately corotates and slides along a rotating field line as a bead on a wire, with effective potentialA decreasing effective potential converts corotating potential energy into poloidal kinetic energy. This is the mechanism of magnetocentrifugal acceleration; local launching does not alone guarantee global escape.

For an attractive point mass, take $\Phi =-GM/(R^2+z^2{)}^{1/2}$ and ${\omega }^2=GM/R_0^3$. Parameterize the outward straight line by $R=R_0+s\sin \alpha$, $z=s\cos \alpha$. The first derivative of ${\Phi }_{\mathrm{eff}}$ vanishes at $s=0$ because the footpoint is in circular radial balance. The meridional Hessian there is ${\omega }^2\mathrm{diag}(-3,1)$, soThus the thirty-degree magnetocentrifugal launching criterion in its strict negative-curvature form isAbove this angle the equilibrium is unstable to an outward displacement; below it the near-footpoint potential initially rises. The source's proportionality constant must be negative to describe attraction.

The marginal straight-line launch at thirty degrees has zero quadratic curvature. For precisely the straight line and point-mass potential here, let $x=s/R_0$ and $a=\sin \alpha$. Direct expansion givesAt $\alpha =\pi /6$, this is $-7x^3/16+O(x^4)$, so the outward direction is downhill at cubic order. Consequently the strict inequality is the nondegenerate quadratic criterion; it should not be interpreted as excluding every one-sided nonlinear marginal launch at equality. An exactly resting bead remains an equilibrium until displaced.

The central force has no transverse component. Its angular equation is thereforeThus the specific angular momentum is the constantThe radial equation isSubstitute $\dot{\theta }=l/r^2$:Consequently the radial motion for $r>0$ is exactly a unit-mass one-dimensional motion in the effective potentialIts conserved energy is $\frac{1}{2}{\dot{r}}^2+V_{\mathrm{eff}}(r)$; the second term in the effective potential accounts for the angular kinetic energy.

Chirality and the component expansion. A chiral superfield is constrained byUse left Grassmann derivatives. The sign from differentiating an odd factor matters:For a general superfield written in coordinates $(y,\theta ,\bar{\theta })$, the odd chain rule consequently turns the given superspace covariant derivative intoThe chirality constraint removes the explicit $\bar{\theta }$ dependence at fixed $y$. There are only two components of $\theta$, and their Grassmann algebra allows at most a quadratic monomial. Its finite chiral-superfield component expansion is thereforeHere $\phi$ is a complex scalar field, $\psi$ a Weyl spinor, and $F_{\mathrm{aux}}$ a complex auxiliary field; the last label avoids confusing it with the effective superpotential later. The factor $\sqrt{2}$ is the standard canonical component normalization. In ordinary coordinates the same statement isThis translation exponential terminates because its shift is nilpotent; it displays the full component dependence without an unstated convention for the barred spinor square.

A superspace covariant derivative obeys the graded product rule. Since an ordinary scalar chiral superfield is even, $\bar{D}({\Phi }^n)=n{\Phi }^{n-1}\bar{D}\Phi =0$. Linear combinations prove the polynomial claim. More generally, a nonsingular holomorphic function of chiral fields is chiral; inserting conjugate fields generally spoils this holomorphic closure of chiral superfields.

The superspace action. For a real Kähler potential and a holomorphic superpotential, the global chiral-field action has Lagrangian densityFull superspace integration gives a D-term, and chiral superspace integration an F-term. The action is real, and its supersymmetry variations are spacetime total derivatives. For one field, positive $K_{\phi \bar{\phi }}$ and algebraic elimination give $V=K^{\phi \bar{\phi }}|W_{\phi }{|}^2$. The Wess–Zumino model uses the canonical choice $K={\Phi }^{†}\Phi$ at tree level.

Scalar potential and the vertex. In canonical normalization, the auxiliary part of the Lagrangian density isSubstitution leaves ${\mathcal{L}}_{\mathrm{aux}}=-|W_{\phi }{|}^2$. Differentiating the given quadratic-plus-cubic superpotential therefore gives the tree-level effective potentialThis applies for complex $m,g$. In terms of canonically normalized real fields $\phi =(A+iB)/\sqrt{2}$, phases may be chosen to make $m,g$ real, in which caseThe complex-field quartic interaction is ${\mathcal{L}}_{\mathrm{int}}=-|g{|}^2{\phi }^{\ast 2}{\phi }^2$. There are two identical external legs of each field type. Differentiating with respect to those four fields, or counting the Wick attachments to the vertex, produces the factor $2!2!$:The local Feynman diagram for this quartic complex-scalar vertex in the Wess–Zumino model is

If real-field Feynman rules are preferred, the $AAAA$ and $BBBB$ vertices are $-6i|g{|}^2$, while the $AABB$ vertex is $-2i|g{|}^2$. These are the same interaction in a different component basis. They should not be confused with a convention that absorbs $2!2!$ into the coefficient of the complex quartic term.

Spurion symmetries. Treat $m,g$ as chiral spurions. An ordinary $U(1)$ acts on $\Phi ,m,g$ with charges $(1,-2,-3)$ and leaves $\theta$ neutral. An R-symmetry gives $\theta$ charge one and $\Phi ,m,g$ charges $(1,0,-1)$. Thus

Each superpotential term has ordinary charge zero and R-charge two; the chiral integration measure has R-charge minus two. In particular $m$ is neutral under the specified R-symmetry. These are formal transformations of fields and parameters together, not two exact symmetries of a theory with arbitrary fixed nontransforming numerical couplings. This Wess–Zumino spurion charge assignment is useful because it constrains possible quantum terms.

The general holomorphic form. Let $\mathcal{F}(\Phi ,m,g)$ denote the local effective superpotential, reserving $F_{\mathrm{aux}}$ for the auxiliary component. The holomorphy argument for superpotential non-renormalization permits dependence on the chiral spurions, not on their conjugates. The dimensionless combination $z=g\Phi /m$ is neutral under both formal symmetries, whereas $m{\Phi }^2$ has the required dimension and charges. Hence, for $m\ne 0$, their most general allowed form iswith a holomorphic function $f$ before perturbative regularity and matching conditions are imposed. Equivalently, a monomial $m^a{g}^b{\Phi }^c$ must obeySolving gives $a=1-b$, $c=b+2$. Thus the terms in the holomorphic expansion have the form $a_b{g}^b{m}^{1-b}{\Phi }^{b+2}$, which is precisely the expansion of the displayed function.

What is and is not renormalized. Apply the non-renormalization theorem to a local Wilsonian effective action retaining the elementary field and a nonzero infrared cutoff. Perturbative coefficients are regular as $g\to 0$ and $m\to 0$: no massless infrared modes have been integrated all the way to zero momentum. Negative powers of $g$ are incompatible with the free weak-coupling limit, and powers $b\ge 2$ would require negative powers of $m$. Only the quadratic and cubic structures survive this regularity test. Their coefficients cannot acquire a loop correction here: a quadratic term with no $g$ is the free-theory mass term, while a cubic term only linear in $g$ is already the tree interaction. A loop renormalizing that cubic term requires additional interaction insertions. Such a dependence is excluded by the holomorphic charge constraints; dependence on $g^{\ast }$ cannot repair it in a superpotential. Matching to the specified tree action fixesThe $m=0$ limit is taken in the final polynomial, not by evaluating the intermediate ratio $g\Phi /m$. The holomorphic Wilsonian superpotential and its parameters $m,g$ receive no independent perturbative vertex renormalization. Symmetries alone would only give the arbitrary function $f$; regularity and the free/tree matching are necessary to reach the stronger conclusion.

The Kähler potential is not protected by that theorem. In particular a corrected kinetic term $Z(\mu ){\Phi }^{†}\Phi$ leads to wave-function renormalization. Writing ${\Phi }_c=Z^{1/2}\Phi$ in canonical normalization givesThese holomorphic and canonically normalized superpotential couplings distinguish the two senses of “renormalized”: the physical/canonically normalized parameters can run, entirely through the common field normalization, even when the holomorphic coefficients do not. The scalar effective potential can consequently receive quantum corrections through the Kähler potential. Nor does the local perturbative statement automatically apply to infrared-singular one-particle-irreducible actions or to integrating out whole massive fields. Non-renormalization protects the local holomorphic F-term, not the complete quantum action or every physically normalized mass and coupling.

In a stationary axisymmetric gravitational potential, conservation of angular momentum gives $L_z=R^2\dot{\varphi }$. Eliminating $\dot{\varphi }$ introduces the effective potentialFor an equatorial circular orbit, radial balance is ${\partial }_R{\Phi }_{\mathrm{eff}}(R_c,0)=0$, givingVertical balance also requires ${\Phi }_{,z}(R_c,0)=0$.

For epicyclic motion, assume a twice differentiable stationary gravitational potential symmetric under $z\mapsto -z$, small displacements $|x|,|z|\ll R_c$, and a stable circular orbit. Choose the epicyclic guiding center using the conserved $L_z$. Reflection symmetry makes ${\Phi }_{,Rz}(R_c,0)=0$, eliminating radial-vertical coupling at first order. Expanding the equations about $(R_c,0)$ givesThe derivatives are evaluated at the guiding centre, and stability requires ${\kappa }^2,{\nu }^2>0$. Without midplane symmetry, a mixed Hessian term can couple the two oscillations. The radial epicyclic frequency and vertical epicyclic frequency are the frequencies of these independent linear oscillations.

Set $\Omega =L_z/R_c^2$. Along the family of equatorial circular orbits, ${\Phi }_{,R}=R{\Omega }^2$. Differentiating this relation and adding the centrifugal contribution givesTo interpret the common frequency range, let $q=d\log v_c/d\log R$. Since $v_c=R\Omega$, one has ${\kappa }^2=2(1+q){\Omega }^2$. A Keplerian disk has $q=-1/2$ and $\kappa =\Omega$, a flat galaxy rotation curve has $q=0$ and $\kappa =\sqrt{2}\Omega$, and solid-body rotation has $q=1$ and $\kappa =2\Omega$. Typical galactic rotation curves lie between these slopes. Thus $\Omega \lesssim \kappa \lesssim 2\Omega$ is a useful galactic range, not a theorem for every possible axisymmetric potential. As a precise sufficient example, epicyclic frequency bounds for monotone spherical density follow from ${\kappa }^2={\Omega }^2+4\pi G\rho$ and $0\le \rho \le 3M/(4\pi R^3)$.

Use a Cartesian frame rotating with the epicyclic guiding center, with $x$ pointing radially outwards and $y=R_c(\varphi -\Omega t)$ in the direction of rotation. To first order, conservation of angular momentum givesThe radial harmonic oscillator solution and its azimuthal integral areA constant in $y$ merely changes the azimuthal origin of the guiding centre. The epicyclic ellipse obeys $x^2/X^2+y^2/Y^2=1$. In the usual frequency range it is elongated azimuthally, and the star travels clockwise when $x$ points right and $y$ up: its small motion relative to the prograde guiding centre is retrograde.

For the Oort constants, subtract the defining expressions to obtain $\Omega =A-B$ and insert $R{\Omega }^{\prime }=-2A$ into the radial epicyclic frequency formula:The solar-neighbourhood values give $\Omega =26.5\mathrm{km}{\mathrm{s}}^{-1}\mathrm{kp}{\mathrm{c}}^{-1}$, $\kappa =\sqrt{1272}\mathrm{km}{\mathrm{s}}^{-1}\mathrm{kp}{\mathrm{c}}^{-1}$, andThe solar epicyclic ellipse is therefore about $1.49$ times longer azimuthally than radially.

A complete radial oscillation takes $T_r=2\pi /\kappa$. The oscillatory part of $\dot{\varphi }$ has zero average over this interval, so the epicyclic azimuthal advance isThis describes the advance between consecutive radial turning points of the same type; it need not be a full $2\pi$ revolution. For the Sun,The azimuthal advance estimate uses the same linear epicyclic motion approximation as the axis ratio.

At a circular orbit, $V^{\prime }=0$. Differentiating the Schwarzschild effective potential at fixed $h$ givesA timelike circular orbit requires $r>3r_s/2$. Evaluating the second derivative after this substitution givesA stable nondegenerate minimum therefore requires $r>3r_s$. The boundary $r=3r_s$ is the marginal innermost stable circular orbit, not a strict stable minimum.

Eliminate the cyclic velocities using $\dot{\varphi }=(L_z-L_3\cos \theta )/(I_1{\sin }^2\theta )$. The conserved energy becomesDifferentiating the effective potential giveswhich is also the $\theta$ Euler-Lagrange equation. Hence $I_1\ddot{\theta }=-V_{\mathrm{eff}}^{\prime }(\theta )$. Deriving it directly from the Euler-Lagrange equation ensures it also holds at turning points, where division by $\dot{\theta }$ in an energy argument would be invalid.

In a circular orbit, $r=a$ is constant, so the radial acceleration vanishes. The radial equation requiresFor every $a>0$, this equation permits a real nonzero specific angular momentum, $h=\pm \sqrt{k{a}^{n+3}}$, and hence a circular orbit. This chooses $h$ for the desired radius; it does not claim that one prescribed value of $h$ permits every radius.

To test circular-orbit stability for a power-law central force, keep that chosen specific angular momentum fixed and write $r=a+\rho$, with $|\rho |\ll a$. The radial equation becomesExpanding to first order and using $h^2=k{a}^{n+3}$ givesFor $n>-3$, the linearized equation is the harmonic oscillator equation, withThe radial disturbance undergoes bounded small oscillations, rather than growing. Equivalently, the effective potential has a strict local minimum: its second derivative at $a$ is $(n+3)k{a}^{n-1}>0$. By the effective potential stability criterion, sufficiently small radial disturbances remain small, proving stability for $n>-3$. For comparison, $n<-3$ gives exponential growth. At $n=-3$, the balancing value $h^2=k$ makes the radial acceleration identically zero, so a small radial velocity produces drift rather than a restoring oscillation.

The normalization of the circular restricted three-body problem implies ${\mu }_1+{\mu }_2=1$, by Kepler's third law. The two bodies have barycentric positions $(-{\mu }_2,0)$ and $({\mu }_1,0)$. ConsequentlyDifferentiate the effective potential directly:Since $x={\mu }_1(x+{\mu }_2)+{\mu }_2(x-{\mu }_1)$ and $y=({\mu }_1+{\mu }_2)y$, these becomeOn $y=0$, centrifugal acceleration is the straight line $a_{\mathrm{cf}}=x$. Each gravitational term points towards its source, diverges at that source, and tends to zero far away. Let $f(x)=U_x(x,0)$; away from the sources,Thus the net acceleration is strictly increasing on each of the three intervals separated by the bodies. On the left interval $f$ runs from $-\infty$ to $+\infty$; on the middle interval it runs from $-\infty$ just right of $M_1$ to $+\infty$ just left of $M_2$; on the right interval it runs from $-\infty$ to $+\infty$. The intermediate value theorem and strict monotonicity give exactly three collinear Lagrange points, $L_3,L_1,L_2$ in increasing $x$ order. The sketch plots gravity and centrifugal acceleration separately; their intersections with $a_{\mathrm{grav}}=-x$ locate the equilibria.

Write $x_{L2}={\mu }_1+\delta$, so $r_{2,L2}=\delta$ and $r_{1,L2}=1+\delta$. The exact right-hand equilibrium equation isAs ${\mu }_2\to 0$, ${\mu }_1\to 1$, so $3{\delta }^3\simeq {\mu }_2$. Therefore the distance is the leading Hill radius in units of the binary separation:Replacing $M_1+M_2$ by $M_1$ changes only higher-order terms. In dimensional units the right-hand side is multiplied by the binary separation.

Use the displacement vector $\mathbf{q}=(X,Y,\dot{X},\dot{Y}{)}^T$, where $X=x-x_{L2}$ and $Y=y$. The absolute coordinate $x$ cannot be the first entry of a homogeneous linear system about $L_2$; it must be translated. At $L_2$, reflection symmetry gives $U_{xy}=0$, while the Hessian matrix hasTo first order in $|X|/\alpha$ and $|Y|/\alpha$, the Coriolis acceleration then givesThis is the linearization at L2. The expression for $B$ is exact at the true equilibrium; in the small-secondary limit, ${\mu }_2/{\delta }^3\simeq 3$ and ${\mu }_1/(1+\delta {)}^3\simeq 1$, so $B\simeq 4$.

For an eigenmode proportional to $e^{\lambda t}$, eliminate the velocity coordinates. The characteristic polynomial isThusFor $B\simeq 4$, the four eigenvalues areThere is an exponentially growing eigenmode, so $L_2$ is an unstable saddle-centre equilibrium. A generic displacement has an unstable component whose distance grows as $e^{\sqrt{1+2\sqrt{7}}t}$. The L2 escape e-folding time isThis is a local exponential timescale, not the time to reach a prescribed distance from an unspecified initial offset. Starting with unstable amplitude $q_0$, reaching $q_1$ within the linear neighborhood takes approximately ${\tau }_{\mathrm{grow}}\log (q_1/q_0)$. Initial data exactly on the centre-stable subspace need not grow forward at this linear order; the existence of the growing eigenmode already proves instability.

A central force has no component in the angular direction. The angular component of the acceleration in polar coordinates therefore gives $2\dot{r}\dot{\theta }+r\ddot{\theta }=0$, orFor unit mass, $h=r^2\dot{\theta }$ is the signed angular momentum perpendicular to the orbital plane, so its magnitude $|h|$ is conserved. Equivalently, the torque $\mathbf{r}\times \mathbf{F}$ vanishes, which also fixes that plane.

The given potential energy is $k/r$, whose radial force is $k/r^2$: positive $k$ is repulsive and negative $k$ attractive. The kinetic energy in polar coordinates is $({\dot{r}}^2+r^2{\dot{\theta }}^2)/2$, so conservation of energy givesFor $h\ne 0$ and $k>0$, the effective potential is positive, strictly decreasing from $+\infty$ to $0$, and has no stationary point. For $k=-\kappa <0$, it still tends to $+\infty$ at zero, crosses zero at $r=h^2/(2\kappa )$, and has its unique minimum atIt then approaches zero from below. The minimum corresponds to a stable circular orbit. If $h=0$, the centrifugal barrier disappears; the attractive effective potential is simply $-\kappa /r$ and has no minimum.

The principle of stationary action, also called Hamilton's principle, requires $\delta {\int }_{t_1}^{t_2}Ldt=0$ for fixed-endpoint path variations. Here the original PDF has ${\dot{r}}^2$ in the radial kinetic energy; the TeX has lost that dot. With $m>0$ and $r>0$, the Lagrangian isThe Euler-Lagrange equation for each coordinate givesThe two conserved quantities are angular momentum $h=m{r}^2\dot{\varphi }$ and energy $E=\frac{m}{2}{\dot{r}}^2+\frac{m}{2}{r}^2{\dot{\varphi }}^2+km{r}^2$. By Noether's theorem, they arise respectively from rotational invariance and time-translation invariance of the action. The coordinate $\varphi$ is a cyclic coordinate, and $L$ has no explicit time dependence.

The conjugate momentum for each coordinate is $p_r=m\dot{r}$, $p_{\varphi }=m{r}^2\dot{\varphi }$. The Legendre transform $H=p_r\dot{r}+p_{\varphi }\dot{\varphi }-L$ gives the HamiltonianHamilton's equations areSetting $p_{\varphi }=h$ yields $m\ddot{r}=-V_{\mathrm{eff}}^{\prime }(r)$ with the effective potential $V_{\mathrm{eff}}=h^2/(2m{r}^2)+km{r}^2$.

For $k>0$ and $h\ne 0$, this potential tends to infinity both as $r\downarrow 0$ and as $r\to \infty$, and has its unique stationary point atThe effective potential stability criterion proves a stable circular orbit in a quadratic central potential, with $|\dot{\varphi }|=\sqrt{2k}$. For a small radial displacement $\eta$, $\ddot{\eta }+8k\eta =0$ to first order, so radial perturbations oscillate rather than grow. The diagram shows the centrifugal and quadratic contributions and their sum in dimensionless units:

The printed paper only says that $k$ is constant. Its claimed positive-radius stable circular orbit needs the additional attractive-force assumption $k>0$ and nonzero angular momentum. For $k<0$ there is no such minimum; for $k=0,h\ne 0$ the effective potential is strictly decreasing. With $h=0,k>0$, the minimum is at the origin and is not a positive-radius circular orbit. These cases are not represented by the diagram.

Put $h(r)=1-\mu /r^2$. For the equatorial geodesic, the geodesic Lagrangian is $\mathcal{L}=\frac{1}{2}[-h{\dot{t}}^2+h^{-1}{\dot{r}}^2+r^2{\dot{\varphi }}^2]$, where dots denote derivatives with respect to the affine parameter. The cyclic coordinates $t,\varphi$ give constants $p_t=-h\dot{t}=-E$ and $p_{\varphi }=r^2\dot{\varphi }=L$. A third constant is the tangent norm $g_{ab}{\dot{x}}^a{\dot{x}}^b=-\kappa$, preserved by the geodesic equation and metric compatibility. For a timelike geodesic rescale the affine parameter to proper time so $\kappa =1$; for a null geodesic $\kappa =0$.

Substitute $\dot{t}=E/h$, $\dot{\varphi }=L/r^2$ into the norm identity and multiply by $h$. It gives the effective potential equationOutside the horizon $r=\sqrt{\mu }$, its derivative is $V^{\prime }=-(L^2-\kappa \mu )/r^3+2\mu L^2/r^5$. A nontrivial circular null geodesic requires $L\ne 0$, givingFor a circular timelike geodesic,At it $V^{\prime \prime }=-4\mu L^2/r^6<0$, and its radius exceeds $\sqrt{2\mu }$. If $L^2\le \mu$ no finite circular timelike radius exists. Thus all finite circular null and timelike orbits in this five-dimensional exterior are radially unstable. The energy of a circular orbit is fixed by $E^2=2V(r)$, not independently arbitrary.

For comparison, with four-dimensional Schwarzschild horizon radius $r_s$, the photon circle is at $3r_s/2$ and is unstable. Circular timelike orbits exist for $r>3r_s/2$; they are unstable below $3r_s$, marginal at $3r_s$ and stable above $3r_s$. The five-dimensional potential has no corresponding stable outer circular timelike branch.

For every interacting $n\ge 3$, the one-loop scalar effective action already contains counterexamples. On a constant background, with a positive auxiliary mass $\mu$ to control infrared divergences, its field-dependent part isThe term of order ${\lambda }^V$ has $E=(n-2)V$ external scalar fields and is a one-loop polygon one-particle-irreducible Feynman diagram. Choose $V>n/(n-2)$ and $2V>d$. Then $E>n$, its loop momentum integral is ultraviolet convergent, and its coefficient is nonzero. Removing the ultraviolet cutoff does not remove this finite higher-point scalar vertex. For $d>2$, generic nonexceptional external momenta give the same conclusion without retaining the auxiliary mass. In $d\le 2$, the massless theory still needs an infrared prescription; removing the ultraviolet cutoff does not remove that need or the induced vertices. Hence there are no interacting pairs $(n,d)$ under the literal wording. For $n=2$ the functional determinant is field independent; $n=1$ merely shifts a Gaussian integral when an infrared prescription exists.

There is a different conventional interpretation: absence of new independent divergent counterterms. The superficial degree of divergence of a connected diagram isThe standard perturbative counterterm closure criterion is $[\lambda ]\ge 0$, with the same pairs as in part ii, provided one includes all symmetry-allowed lower-degree potential terms, the kinetic term, and the vacuum energy. A pure monomial family need not itself have counterterm closure: a sextic interaction in three dimensions generates a quartic counterterm, and a quartic interaction generates a mass counterterm. This interpretation concerns the local divergent part or the continuum defining action, not the complete quantum effective action with its finite interaction vertices and derivative expansion.

For the massive four-dimensional quartic scalar field theory, write ${\lambda }_B$ for the bare quartic coupling, ${\lambda }_R$ for its renormalized value, and $m>0$ for the mass held fixed by a renormalization condition. Define the scalar bubble integralThe quartic one-particle-irreducible correlation function, defined as a derivative of the Euclidean quantum effective action, isThere are three bubble diagrams, one for each pairing of external momenta, and each has Feynman-diagram symmetry factor $2$. The minus sign follows equivalently from the quadratic term in the functional determinant $\frac{1}{2}\mathrm{Tr}\log [1+({\lambda }_B/2)(-{\partial }^2+m^2{)}^{-1}{\varphi }^2]$.

Radial integration at zero external momentum givesThe bare quartic vertex is not finite at fixed bare coupling. Its logarithmic ultraviolet divergence must be subtracted. Impose the momentum-subtraction scheme condition ${\Gamma }_R^{(4)}(0,0,0,0)={\lambda }_R$. To one loop this requiresFor the finite difference, a Feynman parameter and a shift of loop momentum giveThe boundary error due to shifting a sharp cutoff vanishes in this limit. Substitution yields the finite renormalized quartic scalar vertexThus finiteness requires holding the renormalized coupling fixed and allowing the bare coupling to depend on ${\Lambda }_0$. The unsubtracted assertion would be false.