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The stated relation follows for an isolated uniform sphere ending at $r_0$, with the gravitational potential zero at infinity. Uniform density in a core alone does not fix the escape velocity: exterior shells add a constant interior gravitational potential while leaving the interior force unchanged.

Under the isolated-sphere assumption, the enclosed mass and circular speed areThe boundary value is $\Phi (r_0)=-Gm(r_0)/r_0$. Integrating ${\Phi }^{\prime }=Gm(r)/r^2$ inwards givesThus the escape and circular speeds inside an isolated uniform sphere satisfyThe finite central limit is $v_e^2(0)=4\pi G{\rho }_0{r}_0^2$, although $V_c(0)=0$. Adding an exterior spherical shell of mass $m_s$ at radius $s>r_0$ raises $v_e^2$ by $2G{m}_s/s$ and is a direct counterexample to deducing the boxed relation from a uniform core alone.

Interpret the barred velocities as second moments of a steady nonstreaming collisionless stellar system, and write ${\sigma }_r^2=\overline{v_r^2}$. In an isotropic model the three component dispersions agree, so the Spherical Jeans equation becomesPut $A={\rho }_0{r}_0^k$, so $\rho =A{r}^{-k}$. If this density supplies all of the Newtonian gravity and there is no central point mass, finite enclosed mass requires $k<3$ andIntegrate to an outer radius $b$ at which the pressure is $P_b=\rho (b){\sigma }_r^2(b)$. The exact Jeans-moment solution within the power-law region isand, for $k\ne 1$,For $k<1$ and $r\ll b$, the integral is dominated by outer radii. ThereforeThe coefficient cannot be determined from the local density cusp alone. Extending the power law to infinity gives a divergent pressure integral when $k<1$, so an outer scale or boundary condition is essential. For the usual shallow cusp $0<k<1$, the central velocity dispersion tends to zero even though the density diverges.

For $1<k<3$, inner radii dominate as $r/b\to 0$; equivalently one can send $b\to \infty$ with vanishing boundary pressure to obtainThus the central velocity dispersion decreases for $1<k<2$, is constant at $k=2$, and diverges for $2<k<3$. At the omitted threshold $k=1$, ${\sigma }_r^2=(P_b/A)r+2\pi GAr\ln (b/r)$. For $k\ge 3$ the assumed central density law has infinite enclosed mass, so the same unmodified model cannot determine a finite velocity dispersion. These boundary and domain restrictions are the boundary dependence of an isotropic self-gravitating cusp; specifying density without them does not specify a physical finite galaxy.

Let $p(r)=\rho (r){\sigma }_r^2(r)$ and use the velocity-anisotropy parameter $\beta =1-{\sigma }_{\theta }^2/{\sigma }_r^2$, with ${\sigma }_{\theta }^2={\sigma }_{\varphi }^2$. Here the tangential variance is per component; a definition using the sum of both tangential variances instead has an extra factor of two. The original PDF has this correct per-component ratio, which the TeX corrupts.

At a point along the line of sight, $r^2=R^2+z^2$. The radial direction makes an angle with the sightline satisfying ${\sin }^2\chi =R^2/r^2$. Spherical symmetry eliminates mixed moments, and projection givesUsing both sides of the galaxy and $dz=rdr/\sqrt{r^2-R^2}$,For this formula use $I$ as the projected density of the same tracer represented by $\rho$. If $\rho$ is a mass density and $I$ an observed surface brightness, a constant mass-to-light ratio rescales $I$ by that factor; otherwise replace $\rho$ by the luminosity density throughout the tracer equation. The gravitating $M(r)$ remains the total mass, not necessarily the tracer's mass.

The Spherical Jeans equation is $p^{\prime }+2\beta p/r=-\rho GM/r^2$. Eliminate $\beta p=-r[p^{\prime }+\rho GM/r^2]/2$ from the projected integrand:Rearranging provesNo constancy of $\beta$ was needed. Convergence of the projected moments and the steady spherical assumptions are required. This is the integral form of the projected spherical Jeans residual.

Although $\beta$ has disappeared explicitly, the unknown radial pressure $p$ still carries the orbital information. Thus the identity does not remove the mass--anisotropy--density degeneracy: projected brightness and one projected second moment do not uniquely determine both $M(r)$ and the radial/tangential partition. A central rise in velocity dispersion can reflect a compact central mass, radial orbital bias, or a changing mass-to-light ratio; flattening and departures from equilibrium add further ambiguities.

For example, within a point-mass-dominated region with tracer $\rho \propto r^{-k}$ and constant $\beta$, direct solution gives ${\sigma }_r^2=G{M}_{\bullet }/[(k+1-2\beta )r]$. For $k>1$ the projected velocity dispersion of a Keplerian cusp isThe beta function factors come from setting $r=R\sec \chi$ in the integrals along the line of sight. Different allowed $\beta$ generally change the mass needed to fit the same normalization; $k=2$ is a special cancellation, not the general case. Resolving the central black-hole sphere of influence and constraining orbital structure with higher velocity moments or additional kinematic directions strengthens a compact-mass inference. Brightness and velocity dispersion profiles alone do not establish a unique central black-hole mass, still less an event horizon.

Set $\mu =GM>0$ and interpret $L$ in the displayed equation as specific angular momentum, $L=r^2\dot{\psi }$, rather than total angular momentum. The gravitational potential is per unit mass and $a$ has dimensions of length. Spherical symmetry fixes the orbital plane and conserves the vector angular momentum, while the autonomous specific orbital energy is also conserved.

For the orbit equation for combined inverse-square and inverse-cube attraction, defineA regular nonradial bounded oscillation requires $L\ne 0$ and $L^2-2\mu a>0$. Solving the linear Binet equation then givesFor a noncircular bound orbit $0<e<1$; its specific orbital energy is $E={\mu }^2(e^2-1)/[2(L^2-2\mu a)]$. The effective potential $[L^2-2\mu a]/(2r^2)-\mu /r$ has a centrifugal barrier and a minimum in this case. If $L^2-2\mu a\le 0$, the oscillator frequency is zero or imaginary and there is no regular finite-radius bound oscillation: the attractive centre lacks a centrifugal barrier. Unbound branches with $e\ge 1$ are not closed ellipselike trajectories.

Successive pericentres differ in angle by $2\pi /\nu$. The full phase space state returns after $p$ radial oscillations and $q$ revolutions precisely whenwhere $p,q$ are coprime positive integers; the second formula applies when $a\ne 0$. Thus $a>0$ requires $0<p/q<1$ and $L^2>2\mu a$, whereas $a<0$ requires $p/q>1$. An irrational $\nu$ advances the apsidal direction through a dense set of angles and gives a nonclosed rosette, with only the four regular spherical invariants needed to label its two-dimensional closure.

To see the additional resonant label explicitly, let $u_c=\mu /(L^2-2\mu a)$ and $w=u-u_c-i{u}^{\prime }/\nu$. Then $w=C{e}^{i\nu (\psi -{\psi }_0)}$, so on a family with fixed resonant $L$,Its argument fixes the remaining apsidal phase modulo the symmetry of the closed curve; its magnitude is already determined by specific orbital energy and $L$. Choosing reference axes in the fixed orbital plane makes this single-valued under $\psi \mapsto \psi +2\pi$. This is the fifth independent orbit label on the resonant family described by closed orbits with an inverse-cube perturbation.

For $a\ne 0$, $\nu$ varies with $L$. The orbital resonance condition holds only on selected angular momentum surfaces, so this construction is not a fifth global smooth integral on an open set of arbitrary nearby $L$. If the question asks for a maximally superintegrable Hamiltonian system on an open region of noncircular bound orbits, the required case is $a=0$: then $\nu =1$ for every $L$, and the conserved Laplace-Runge-Lenz vector $\mathbf{A}=\mathbf{v}\times \mathbf{L}-\mu \hat{\mathbf{r}}$ fixes the apsidal direction globally. The relations $\mathbf{A}\cdot \mathbf{L}=0$ and $A^2={\mu }^2+2E{L}^2$ leave five independent constants. For the usual orbit-wise interpretation, noncircular bound orbits with rational $\nu$ have five isolating labels; irrational $\nu$ has four. Circular orbits are closed for any allowed $\nu$ but have no independent apsidal phase, so the generic noncircular count cannot simply be applied to them.

For completeness, regular unbound branches are different. When $e\ge 1$ and $\nu >0$, positive radius restricts the radial phase to $|\nu (\psi -{\psi }_0)|<\arccos (-1/e)\le \pi$. The apsidal invariant on an unbound central-force branch isThe principal argument is continuous on this scattering branch because it never reaches an apocentre across the negative-real-axis cut. Thus the periapsis direction can be a fifth single-valued scattering label for any positive $\nu$, including irrational $\nu$; the rationality condition applies to the noncircular bound case. For plunging or collision branches with $L^2-2\mu a\le 0$, local trajectory labels can still be defined away from the singular centre, but the preceding periodic-orbit criterion is inapplicable and no smooth continuation through $r=0$ is asserted.

The first equation is vertical hydrostatic equilibrium: the upward pressure gradient balances the central mass's vertical gravitational force. For $|z|\ll r$, the point-mass acceleration is $-GMz/(r^2+z^2{)}^{3/2}\simeq -{\Omega }_K^{2}z$, where Keplerian rotation gives ${\Omega }_K^2=GM/r^3$. The disc's self-gravity and vertical inertia are neglected.

The second equation is local thermal balance. The divergence of the vertical radiative flux equals viscous dissipation. In the convention of this alpha disk, $\nu =\alpha p/(\rho \Omega )$, and the dominant Keplerian shear has $rd\Omega /dr=-3\Omega /2$. ThereforeThe radiative flux increases from zero at the symmetric midplane to a positive outgoing value at the upper surface. Radial heat transport, thermal advection and rapid storage of thermal energy are excluded from this local balance.

The third equation is radiative diffusion, equivalentlyIt applies to optically thick material, with $\kappa$ the opacity per unit mass and $\sigma$ the Stefan-Boltzmann constant. Here constant electron-scattering opacity represents Thomson scattering; the formula uses a diffusive, nearly thermal radiation field. Positive upward flux requires temperature to decrease upwards.

The final equation is the ideal gas pressure law, with mean molecular weight ${\mu }_m$ in proton-mass units. Writing $\mathcal{R}=k/({\mu }_m{m}_p)$ gives $p=\mathcal{R}\rho T$. The pressure is gas pressure because radiation pressure is assumed negligible. The four equations express vertical support, local heating balance, diffusive cooling and the gas equation of state.

For dilute nonrelativistic nucleons in thermal equilibrium, their Maxwell-Boltzmann distribution gives $n_j=g_j(m_{j}kT/2\pi {\hslash }^2{)}^{3/2}\exp [({\mu }_j-m_j{c}^2)/kT]$. The neutron and proton spin degeneracies are equal, and their mass prefactor ratio is nearly one. ThusWeak reaction equilibrium imposes ${\mu }_n+{\mu }_{{\nu }_e}={\mu }_p+{\mu }_e$. Charge neutrality in an electron-positron plasma, together with the small baryon asymmetry, makes ${\mu }_e/kT$ very small. Assuming negligible neutrino lepton asymmetry also makes ${\mu }_{{\nu }_e}/kT$ small. Hence the difference, rather than the absolute baryon chemical potentials, is negligible, yieldingA large neutrino chemical potential would invalidate this last approximation; it is an assumption about the lepton asymmetry, not a general rule that baryonic chemical potentials vanish.

The weak-conversion rate per expansion time falls as $T^3$. Below the stated threshold it becomes less than one, so neutron-proton interconversion cannot track the rapidly changing equilibrium ratio: cosmological weak freeze-out occurs. Free neutrons subsequently decay, reducing their abundance until nuclei can survive photodissociation. Once the deuterium bottleneck is passed, nuclear reactions rapidly assemble neutrons into the strongly bound helium-4 nucleus. The short available time, low density and missing stable mass-five and mass-eight nuclei impede extensive production of heavier nuclei, leaving nearly all surviving neutrons in helium.

If $r=n_n/n_p$ just before helium formation, each helium nucleus uses two neutrons and two protons, so $n_{\mathrm{He}}=n_n/2$. Neglecting binding-energy corrections to nuclear masses,The remaining protons mainly become hydrogen. Counting all four helium nucleons, rather than only the captured neutrons, is essential to this mass fraction.

A pi-system is a collection closed under finite intersections. The uniqueness theorem says that two probability measures agreeing on a generating pi-system agree on its generated sigma-field. For finite measures one also fixes the total mass; for a sigma-finite version one requires a covering sequence from the pi-system with finite common masses. Adding the whole space to a pi-system of events causes no change to its generated sigma-field and preserves the stated independence condition.

Fix $H\in \mathcal{H}$. On $\sigma (\mathcal{G})$ the finite measures $A\mapsto P(A\cap H)$ and $A\mapsto P(A)P(H)$ agree on $\mathcal{G}$ and on the whole space. Uniqueness makes them agree for every $A\in \sigma (\mathcal{G})$. Now fix such an $A$. On $\sigma (\mathcal{H})$ the measures $B\mapsto P(A\cap B)$ and $B\mapsto P(A)P(B)$ agree on $\mathcal{H}$ and the whole space; a second application of uniqueness provesThus the two generated sigma-fields are independent. The extension is done one argument at a time, not by assuming the required independence for arbitrary unions.

For the independent random variables, let $\mathcal{G}$ be the rectangles ${\bigcap }_{i=1}^m\{Y_i\in B_i\}$ and let $\mathcal{H}$ be the analogous rectangles in the $Z_j$. Both are pi-systems. Joint independence of the full list factors each combined rectangular probability into the product of the individual probabilities, hence into $P(G)P(H)$. Their generated sigma-fields are exactly $\sigma (Y_1,\dots ,Y_m)$ and $\sigma (Z_1,\dots ,Z_n)$. The proved result therefore gives independence of the two vector-generated sigma-fields.

Stationarity gives the conserved Killing energy per unit rest mass $E=(1+x^2)\dot{t}$, where dots here denote proper-time derivatives. It measures energy relative to the time-translation Killing field; a static local observer measures $E/\sqrt{1+x^2}$. Timelike normalization isEliminating $\dot{t}$ and setting $a^2=E^2-1$ givesThus, on each increasing/decreasing leg respectively, $d\tau =\pm dx/\sqrt{a^2-x^2}$ and $dt=\pm Edx/[(1+x^2)\sqrt{a^2-x^2}]$. The printed positive differentials describe the increasing leg, not both legs at once. Differentiation along a leg gives $\ddot{x}=-x$, extending smoothly through the turning points. Hence $x=a\sin (\tau -{\tau }_0)$ and the proper-time period is $2\pi$.

The requested velocity sketches are the branches $\dot{x}=\pm \sqrt{a^2-x^2}$ and $dx/dt=\pm (1+x^2)\sqrt{a^2-x^2}/E$. They share the turning points and opposite signs on the two legs; their speed ratio is $(1+x^2)/E$, smaller than one near zero and larger than one near the endpoints. The supplied integral is the coordinate time from $-a$ to $a$, equal to $\pi$. The return leg takes the same time, so the coordinate-time period is also $2\pi$.

For the perturbed metric the same Killing conservation gives $\dot{t}=E/(1+x^2)$, and normalization becomes $(1+f(x){)}^2{\dot{x}}^2=a^2-x^2$. Since $1+f>0$,Both weights without $f$ are even in $x$. Oddness of $f$ makes its integral over the symmetric excursion $[-a,a]$ vanish. Each increasing leg therefore still takes $\pi$ in either clock; the decreasing leg takes the same integrals with reversed orientation. Both periods remain exactly $2\pi$, although the proper-time waveform is no longer generally sinusoidal. This is the odd radial perturbations preserve timelike-geodesic periods mechanism. Smoothness of the metric and positivity of $1+f$ ensure regular turning and monotone coordinate time.

Shannon-Fano coding orders symbols by decreasing probability and recursively splits each consecutive block into two groups with probability totals as close as possible, appending bits zero and one. Huffman coding instead repeatedly merges the two least probable nodes, constructs a binary tree upwards, and then assigns the branch bits. Both yield prefix codes, so a concatenation can be decoded without separators; Huffman coding minimizes expected length among binary prefix codes.

For the specified probabilities, the first Shannon-Fano coding split is $\{{\mu }_1\}$ of mass $0.45$ versus the remaining mass $0.55$. Split the latter into $\{{\mu }_2\}$ of mass $0.25$ and $\{{\mu }_3,{\mu }_4,{\mu }_5\}$ of mass $0.30$, and then split off ${\mu }_3$. One resulting code isIts expected word length is $0.45+2(0.25)+3(0.20)+4(0.05+0.05)=\boxed{1.95\text{ bits}}$.

The Huffman coding merges are $0.05+0.05=0.10$, $0.10+0.20=0.30$, $0.25+0.30=0.55$, and $0.45+0.55=1$. Reading the resulting tree gives the same code lengths $1,2,3,4,4$, and it may give exactly the displayed code if the branch bits are chosen accordingly. Its expected length is also $1.95$ bits. The methods agree here; their agreement on this particular distribution is not a claim that the two procedures always coincide.

There are two degrees of freedom, conveniently the azimuthal angles ${\theta }_1,{\theta }_2$ on the two circles. The positions are $(r_1\cos {\theta }_1,r_1\sin {\theta }_1,0)$ and $(r_2\cos {\theta }_2,r_2\sin {\theta }_2,c)$. Their squared separation is $d^2=r_1^2+r_2^2+c^2-2r_1{r}_2\cos ({\theta }_1-{\theta }_2)$. Thus the Lagrangian isThe constant terms may be omitted from the equations of motion. In this notation ${\omega }^2$ is the stated spring coefficient, with the units needed for potential energy; it is not automatically an angular frequency for arbitrary masses.

Work in units $\hslash =1$ and with the particle mass absorbed into the stated kinetic term. Divide the interval into $N$ pieces of length $\Delta =T/N$. For a suitable self-adjoint Hamiltonian operator, the Trotter product formula gives the regulated limitInsert $1=\int d{q}_j|q_j⟩⟨q_j|$ between neighboring factors. Because $V(\hat{q})$ is diagonal in the position operator basis, the given free-particle propagator yieldsThe exponent is the time-sliced action. Its continuum notation isThe measure is defined by the displayed finite-dimensional limits and their normalization, not by a translation-invariant flat measure on an infinite-dimensional space. The oscillatory limit needs a damping/analytic-continuation prescription. This is the time-sliced configuration-space path integral derived from the operator kernel.

For a quadratic potential energy, write $q=q_c+\eta$, where the classical path satisfies ${\ddot{q}}_c+V^{\prime }(q_c)=0$ and the endpoint conditions, and $\eta (0)=\eta (T)=0$. Expansion is exact:Integration by parts makes the linear term equal to $-[{\int }_0^T({\ddot{q}}_c+V^{\prime }(q_c))\eta dt]$, since the endpoint term vanishes. It is zero by the Euler-Lagrange equation. Also $V^{\prime \prime }$ is constant, independent of the endpoints. Thus classical-path factorization of a quadratic path integral is exact:Only the classical action carries the endpoint dependence.

Let $D={\partial }_t^2+V^{\prime \prime }$ on homogeneous Dirichlet boundary conditions. Expand $\eta$ in its normalized eigenfunctions, with eigenvalues ${\lambda }_n$. The fluctuation integral becomes a product of regulated integrals $\int d{a}_n\exp (-i{\lambda }_n{a}_n^2/2)$. Each is proportional to ${\lambda }_n^{-1/2}$, with a prescribed complex phase. Relative to $D_0={\partial }_t^2$, the normalization is thereforeThe subscript denotes the Dirichlet operator domain, and the square-root branch is continued from the free kernel with its convergence prescription. Equivalently one can use determinants of $-D$ and $-D_0$. For $V^{\prime \prime }={\omega }^2$, the Dirichlet oscillator determinant ratio isThis also checks the free limit. At conjugate times, the fluctuation operator has a zero eigenvalue and the ordinary finite-prefactor expression fails; the kernel is obtained by distributional continuation. Away from these times the classical boundary problem is unique and the stated factorization applies directly.

For the infinite-time oscillator with an external source, use vacuum boundary conditions and the Feynman i-epsilon prescription. For a smooth source of compact support, defineThe imaginary term damps the real-field integral. Integration by parts gives $S_{ϵ}[q]=\frac{1}{2}q{A}_{ϵ}q+Jq$, with the pairings understood as time integrals. Completing the square givesTranslate the integration variable. The source-independent Gaussian functional integral is $N$, soOn Fourier modes $e^{i\omega t}$, $A_{ϵ}$ has eigenvalue ${\omega }^2-m^2+iϵ$, henceFor $m>0$, closing the contour above for $t>0$ and below for $t<0$ gives $G(t)=-i{e}^{-im|t|}/(2m)$. The physical time-ordered oscillator quantum field theory propagator is $iG$, not $G$ itself. The shifted stationary path is $q_c=-GJ$ and obeys ${\ddot{q}}_c+m^2{q}_c=J$ with the same vacuum prescription. A retarded inverse would define a different boundary problem. Dividing by the zero-source integral removes $N$ and gives the vacuum source functional of a quantum oscillator. At zero mass an additional infrared prescription is needed for the inverse on constant modes.

In units $\hslash =c=1$, the action is dimensionless and a Lagrangian density has mass dimension $d$. The kinetic term therefore gives $2+2[\varphi ]=d$. Requiring each interaction density $g_n{\varphi }^n$ to have the same dimension givesThis is engineering dimension; quantum scaling can acquire an anomalous dimension.

For a connected Feynman diagram with no derivative interactions, each independent loop integration contributes $d$ powers of momentum, and each internal scalar propagator contributes $-2$. Thus its superficial degree of divergence is $D=dL-2I$. Let $V={\sum }_n{V}_n$. A connected graph has $L=I-V+1$ independent cycles, and counting the half-edges incident on vertices gives $2I+E={\sum }_{n}n{V}_n$. Substitution proves the scalar graph power-counting identity:For a graph with $C$ connected components, the first $d$ is instead $dC$, since $L=I-V+C$. The printed expression therefore uses the usual connected-graph convention. This is the overall degree under simultaneous rescaling of all loop momenta. Subgraphs can still have ultraviolet divergences when the whole graph has $D<0$, so their counterterms must be subtracted before declaring an integral finite.

When all retained couplings have nonnegative dimension, only finitely many external-leg/derivative structures can have nonnegative degree. Positive-dimensional relevant couplings lower $D$ as vertices are added, giving superrenormalizable interactions. Zero-dimensional marginal couplings give power-counting renormalizable interactions. Negative-dimensional irrelevant couplings instead permit $D$ to grow with the number of vertices, generally requiring an unbounded list of counterterms. Such nonrenormalizable interactions can still define an effective field theory to any fixed order in a low-energy expansion.

In $d=4$, $d_n=4-n$: cubic interactions are superrenormalizable, quartic interactions are marginal and renormalizable, and powers $n\ge 5$ are nonrenormalizable. In $d=3$, $d_n=3-n/2$: cubic, quartic and quintic interactions are superrenormalizable, sixth-order interactions are marginal and renormalizable, and $n\ge 7$ are nonrenormalizable. This is a power-counting classification, not a guarantee that an odd highest-power potential is bounded below. All lower-dimensional operators allowed by the symmetries must be included when required by renormalization; for example, odd interactions can require a linear counterterm, whereas the quartic model's $\varphi \mapsto -\varphi$ symmetry excludes it.

Dimensional regularization defines convergent Euclidean loop integrals in a suitable domain of $\mathrm{Re}d$, then continues their parameter formulas meromorphically to complex $d$. The Gaussian identity supplies $(4\pi {)}^{-d/2}$ and Schwinger parameterization supplies Gamma functions such as $\Gamma (n-d/2)$. Their poles encode the ultraviolet singularities near physical dimension. Noninteger complex $d$ is an analytic regulator of amplitudes, not a literal space with a noninteger number of coordinates. Masses or nonexceptional external momenta distinguish ultraviolet poles from infrared poles; scaleless integrals vanish in this convention and can hide that distinction.

For the quartic model let $d=4-ϵ$, ${\varphi }_0=Z^{1/2}\varphi$, $m_0^2=Z_{m^2}{m}^2$ and ${\lambda }_0={\mu }^{ϵ}{Z}_{\lambda }\lambda$. In renormalized fields, the bare Lagrangian isSeparate the terms with each $Z$ equal to one from their differences. The differences are field, mass and quartic counterterms. After subtracting divergent subgraphs, the remaining ultraviolet pole of a proper graph is a local polynomial in its external momenta. In four-dimensional phi-fourth theory, power counting and reflection symmetry restrict these local terms to $(\partial \varphi {)}^2$, ${\varphi }^2$, ${\varphi }^4$ and a field-independent constant. Choosing the Laurent coefficients of the three renormalization factors recursively cancels these poles at every perturbative order. The constant vacuum term is removed by the zero-source normalization. This is an order-by-order finite renormalized expansion, not a claim of convergence of the perturbation series.

The source renormalization of a scalar generating functional follows directly from rewriting the source coupling in bare variables:A general source attached to a single elementary field needs this field renormalization; its connected correlations are the ordinary renormalized Green functions. Normalizing the path integral at $h=0$ cancels source-independent vacuum diagrams. Subject to a separate infrared/volume prescription, their source-dependent generating functional $F$ is consequently ultraviolet finite order by order. For a constant source, this statement is interpreted as a formal connected expansion or a vacuum density, not as an automatically finite total logarithm in infinite spacetime.

Changing integration variables from $\varphi$ to ${\varphi }_0$ introduces a source-independent constant Jacobian that cancels between the numerator and its normalization. The resulting normalized integral depends only on ${\lambda }_0,m_0^2,h_0$ and fixed physical boundary/volume data. ThusDefine all scale derivatives at these fixed bare quantities:In a mass-independent subtraction scheme, logarithmic differentiation of the bare relations givesThe factors contain regulator poles, but the combinations determining the physical renormalization group functions have finite limits. In particular $\hat{\beta }=-ϵ\lambda +\beta (\lambda )$ and ${\hat{\gamma }}_h=ϵ/2+{\gamma }_{\varphi }(\lambda )$. The chain rule now gives the requested equation:As a sign check, minimal subtraction gives $Z_{m^2}=1+\lambda /(16{\pi }^2ϵ)+O({\lambda }^2)$ and $Z_{\lambda }=1+3\lambda /(16{\pi }^2ϵ)+O({\lambda }^2)$, with $Z=1+O({\lambda }^2)$. Hence $\beta =3{\lambda }^2/(16{\pi }^2)+O({\lambda }^3)$, ${\gamma }_{m^2}=\lambda /(16{\pi }^2)+O({\lambda }^2)$, and ${\gamma }_{\varphi }$ starts at two loops. The positive signs agree with the tadpole subtraction and the three four-point channels.

The dimensional rewriting needs care. The chosen source normalization gives $[h]=3$, $[m^2]=2$, and $[\lambda ]=0$, so $m^2/{\mu }^2$ and $h/{\mu }^3$ are the natural dimensionless parameters. For the full source functional, spacetime arguments must also be rescaled: with $y=\mu x$, write $\varphi (x)={\mu }^{(d-2)/2}\psi (y)$ and $\tilde{h}(y)={\mu }^{-3}h(y/\mu )$. The dimensionless functional then depends on $\lambda ,m^2/{\mu }^2$, the rescaled source profile, and the rescaled integration domain.

For a literal uniform source in a fixed physical spacetime volume ${\mathcal{V}}_d$, the uniform-source vacuum functional and spacetime volume relation is insteadFor example, even at $\lambda =0$ and $m^2>0$, completing the square for the constant mode givesThus the printed three-argument rewriting of the total $F$ suppresses necessary volume/source-profile data. It is valid as a dimensional shorthand only with those data understood; a dimensionless density has the three arguments, while its scale equation includes its engineering dimension. At infinite volume, the total constant-source logarithm is generally extensive and infinite even after ultraviolet renormalization.

The massless renormalization-group characteristics with a source solve the requested flow directly, independently of this notational issue. Since the mass flow is proportional to $m^2$, its zero value remains zero. Starting at a reference scale ${\mu }_{\ast }$, solveThenwith any physical volume or source-shape data retained. Where $\hat{\beta }\ne 0$, the running quantities are determined byThese equations transport matching data at one scale to any other scale on the same flow trajectory. At a zero of the beta function use the differential equations themselves, rather than divide by zero.

For the uniform-source density, set $y=h/{\mu }^3$ and $m^2=0$. At $d=4$ the corrected dimensionless equation and its local characteristic solution areThe arbitrary function $\Psi$ is fixed by boundary/matching data. If the volume is retained as a dimensionless argument $v={\mu }^4{\mathcal{V}}_4$, the full dimensionless functional obeys the same transport operator with $4v{\partial }_v$ in place of the additive $4$ term; extensivity makes the two statements equivalent. Massless zero-momentum source expansions can also have infrared singularities, so this RG solution is formal until an infrared prescription or a suitable background expansion is chosen.

An ultraviolet fixed point is a zero of the beta functions approached by the dimensionless couplings as $\mu \to \infty$. Near a single-coupling fixed point ${\lambda }_{\ast }$, perturbations obey $\delta \lambda \propto {\mu }^{{\beta }^{\prime }({\lambda }_{\ast })}$, so ${\beta }^{\prime }<0$ gives attraction in that direction. In four-dimensional quartic scalar theory, the positive one-loop beta function makes positive small coupling grow toward the ultraviolet: $1/\lambda (\mu )=1/\lambda ({\mu }_{\ast })-[3/(16{\pi }^2)]\log (\mu /{\mu }_{\ast })$. This exhibits a perturbative Landau pole, not an interacting ultraviolet fixed point. A perturbative result does not exclude hypothetical strong-coupling behavior beyond its regime.

Use $\mathcal{R}$ for the stellar gas constant, reserving $R$ for the stellar radius. The specific internal energy of a monatomic gas obeying the ideal gas law is $e=P/[(\gamma -1)\rho ]=3P/(2\rho )$. At fixed enclosed mass, the first law of thermodynamics gives the heat available to the outgoing luminosity as $ϵ=-{\partial }_{t}e-P{\partial }_t(1/\rho )$ when there is no stellar nuclear fusion heating. ThereforeThe second coefficient includes both the mass density dependence of $e$ and pressure work. This is gravitational energy generation in a homologously contracting ideal-gas star.

Write the characteristic scales asThen $\rho ={\rho }_{0}b$, $P=P_{0}p$, $T=T_{0}p/b$, $m=Mq$ and $L_r=Ll$. Since $d/dr=R^{-1}d/dx$, direct substitution in the stellar hydrostatic equation and mass conservation givesFor radiative diffusion in a star, dividing the temperature equation by $T_0/R$ givesIn particular the fourth power here is of $\mathcal{R}$, not of the radius.