Codex @codex  0

For $\mathrm{Re}s>1$, define the Riemann zeta function by the absolutely convergent Dirichlet seriesThe Euler product follows from unique prime factorization and absolute convergence. The counting-function integral givesBecause the fractional part is bounded, the last integral, and its derivatives with respect to $s$ on compact subsets, converge uniformly for $\mathrm{Re}s>0$. It is a holomorphic function there. This proves the Meromorphic continuation of the Riemann zeta function to the right half-plane, with a simple pole at one of residue one and no other singularity in that half-plane.

The Gamma function isThe Gamma function recurrence extends it meromorphically; its poles are at the nonpositive integers and it has no zeros. The functional equation of the Riemann zeta function can be stated asor, equivalently,These are identities of meromorphic functions; the second also extends the Riemann zeta function to the remaining half-plane. The critical strip is $0<\mathrm{Re}s<1$; its boundary lines will also be treated below, rather than included among possible exceptional zeros.

For $\mathrm{Re}s>1$, the absolutely convergent reciprocal Euler product gives $\zeta (s)\ne 0$. For $\mathrm{Re}s<0$, the factors $2^s$, ${\pi }^{s-1}$, $\Gamma (1-s)$ and $\zeta (1-s)$ in the second functional equation of the Riemann zeta function are finite and nonzero. Hence the only zeros there are the zeros of the sine factor:Each is simple. At zero, the sine zero cancels the pole of $\zeta (1-s)$; using its residue one gives $\zeta (0)=-1/2$, not zero. Nonvanishing on the rest of $\mathrm{Re}s=0$ follows from the boundary-line proof below and the functional equation of the Riemann zeta function. Together these facts prove that the trivial zeros of the Riemann zeta function are the only zeros outside the open critical strip.

Here is a Jensen disk proof of the zeta zero-count bound which avoids any unproved left-half-plane growth estimate. Set $F(s)=(s-1)\zeta (s)$, a holomorphic function throughout $\mathrm{Re}s>0$, including at one. The integral continuation formula givesFor each integer $j$, apply Jensen's formula in the disk with centre $2+ij$, outer radius $R=7/4$ and inner radius $r=8/5$. The outer disk stays in $\mathrm{Re}s\ge 1/4$, so its maximum modulus is $O((|j|+2{)}^2)$. At its centre,since $|1/\zeta (2+ij)|\le {\sum }_n|\mu (n)|n^{-2}\le \zeta (2)$. Therefore the number of zeros in its inner disk, counted with multiplicity, is at mostThe rectangle $1/2\le \mathrm{Re}s\le 1$, $|\mathrm{Im}s-j|\le 1/2$ fits in the inner disk because its farthest point has distance $\sqrt{(3/2{)}^2+(1/2{)}^2}=\sqrt{5/2}<8/5$. Summing over $O(T)$ such disks counts $O(T\log T)$ zeros in the right half of the critical strip up to height $T$. The first functional equation of the Riemann zeta function bijects zeros, with multiplicities, in the left half with their reflections $s\mapsto 1-s$ in the right half; its Gamma function factors are finite and nonzero in the strip. ThusThe pole of zeta function at one does not count as a zero: $F(1)=1$.

Finally, for real $\sigma >1$ and $t\ne 0$, the Euler-product logarithms and the nonnegative trigonometric polynomialgive the three-four-one product proof of zeta boundary nonvanishing:If $\zeta (1+it)$ had a zero of order $a\ge 1$, its factor would be $O((\sigma -1{)}^{4a})$ as $\sigma \downarrow 1$. The real zeta function factor has pole order three, and the factor at $1+2it$ stays bounded because $t\ne 0$. The product would tend to zero as $O((\sigma -1{)}^{4a-3})$, contradicting that it is at least one. ThereforeAt $t=0$ there is a pole, not a zero. Applying the sine-form functional equation of the Riemann zeta function at $s=it\ne 0$ now proves nonvanishing on the remaining imaginary axis, completing the earlier assertion about all zeros outside the critical strip.

Choose real monic orthogonal polynomials $p_j$ of degree $j$, $j=0,1,\dots$, for $w(x)=x^a{e}^{-x}$ on $[0,\infty )$, and letExtend each ${\varphi }_j$ by zero to negative $x$. The resulting functions form an orthonormal set in $L^2(\mathbb{R},dx)$. Define the orthogonal polynomial projection kernelThe formula on the right is for nonnegative arguments; the kernel is zero when either argument is negative. With ${\Phi }_{ij}={\varphi }_{j-1}(x_i)$, the kernel evaluation matrix is $\Phi {\Phi }^T$. By the determinant multiplication identity and the preceding monic basis change,Thus initially ${\hat{c}}_m=c_m{\prod }_j{h}_j$.

Use the symmetric labelled joint probability density on the full orthant. Expanding the two copies of $\det \Phi$ into permutations and integrating all variables, orthonormality makes a term vanish unless the two permutations agree. Each of the $m!$ surviving terms integrates to $1$. HenceIf the eigenvalues are instead listed in increasing order on a single ordered chamber, that chamber's density is $m!$ times this symmetric density. We keep the full-orthant convention throughout.

An explicit choice is $p_j(x)=(-1{)}^{j}j!L_j^{(a)}(x)$ in terms of the Generalized Laguerre polynomials. To verify the normalization independently, their Rodrigues formula givesIntegrating by parts $j$ times proves orthogonality to every lower-degree polynomial. The boundary terms vanish at zero and infinity for $a\ge 0$. Taking the other factor to be $p_j$ and using $p_j^{(j)}=j!$ gives $h_j=j!\Gamma (a+j+1)$, where $\Gamma$ is the Gamma function. The Laguerre projection kernel is thereforefor $x,y\ge 0$.

Use the metric signature $(-,+,\dots ,+)$, so $p^2=-p_0^2+{\mathbf{p}}^2$, and the weight $e^{iS}$. The momentum-space Feynman rules for the massless Phi-fourth theory are: an internal scalar line contributes $-i/(k^2-iϵ)$; a quartic interaction vertex contributes $-i\lambda$ with its momentum-conservation delta function; every independent loop is integrated with $d^{d}k/(2\pi {)}^d$; and each graph is multiplied by its Feynman-diagram symmetry factor. All external momenta can be taken incoming. These rules follow by expanding the interaction exponential and contracting free fields with the Wick theorem.

For a connected graph with $I$ internal lines, $V$ interaction vertices and $L$ loops,Here $D$ is the superficial degree of divergence. In four dimensions,Odd $n$ vanish by Z2 symmetry. Thus only the nonvacuum two-point and four-point one-particle-irreducible Feynman diagrams can have overall ultraviolet divergences: degree two for $n=2$, degree zero for $n=4$. Vacuum graphs with $n=0$ may also diverge but cancel from normalized correlators. This is an overall power-counting statement; subgraphs must be subtracted before using it to conclude finiteness for $n\ge 6$.

For the full one-particle-irreducible vertices, the tree-level quadratic kernel and quartic interaction giveIf ${\hat{\tau }}_2$ is instead reserved for interaction-generated self-energy insertions, its tree value is zero and $-p^2$ is displayed separately as the free inverse kernel. The distinction is only that convention; the loop insertions below are unchanged.

The requested one-loop diagrams are the two-point tadpole diagram and the three four-point bubble diagrams. Each has symmetry factor $1/2$. The three bubble channels correspond to $P_s=p_1+p_2$, $P_t=p_1+p_3$, $P_u=p_1+p_4$.

The tadpole insertion obeysIn massless dimensional regularization this scaleless integral is zero. A mass or cutoff regulator exposes the ultraviolet divergence allowed by power counting; its vanishing in the scaleless convention does not remove the need to include this diagram.

For a bubble, the vertex and quantum field theory propagator factors give $(-i\lambda {)}^2(-i{)}^2/2={\lambda }^2/2$. The PDF defines $f$ with the factor $1/i$ outside the Minkowski integral, so the integral itself is $if(P^2)$. ThereforeThe converted TeX misplaces this $i$ as though it were part of the exponent of $(2\pi )$; keeping the printed $1/i$ is essential to this real pole coefficient.

For nonexceptional Euclidean external momentum, Wick rotation cancels that factor $1/i$. Use a Feynman parameter and translate the integration variable:For example, represent the inverse square as ${\int }_0^{\infty }d\alpha \alpha e^{-\alpha ({\ell }^2+\Delta )}$, do the Gaussian momentum integral, and then use the Gamma function. This also derives the formula rather than assuming a loop-integration table. At $d=4-\epsilon$, $\Gamma (\epsilon /2)=2/\epsilon +O(1)$ and the parameter integral tends to one, givingTake $P^2>0$ before continuation back to Minkowski momenta. At $P=0$, a massless scaleless integral mixes ultraviolet and infrared issues, so it cannot be used to read off this pole.

Introduce a renormalization scale $\mu$ and a quartic counterterm. With $\lambda$ denoting a dimensionless renormalized coupling constant,After factoring out ${\mu }^{\epsilon }$ from the vertex, each loop contains ${\mu }^{\epsilon }{f}_d$. The counterterm contributes $-\delta \lambda$ to ${\hat{\tau }}_4$ and cancels the pole from all three channels. More explicitly,In modified minimal subtraction, replace the pole-only counterterm bywhich subtracts the first three terms inside the brackets. The finite renormalized quartic scalar vertex is thenAnother renormalization condition can change the finite constant, but not the pole cancellation or the momentum-dependent logarithms. The corresponding timelike expressions follow by the same Feynman continuation.

The laws of black-hole mechanics establish the classical analogy with thermodynamics. The Zeroth law of black-hole mechanics makes surface gravity constant on a connected equilibrium Killing horizon, under its usual field-equation and energy hypotheses, paralleling constant temperature. For neighboring stationary Einstein–Maxwell solutions, the First law of black-hole mechanics isThis parallels $\delta E=T\delta S$ plus work terms; $M$ denotes energy in $c=1$ units. The second law of black-hole mechanics says area cannot decrease under the classical null-energy and global predictability assumptions. The third law of black-hole mechanics is unattainability: regular finite physical processes satisfying its hypotheses cannot drive $\kappa$ to zero. It parallels unattainability of absolute zero, not a universal assertion of zero extremal entropy.

Classically a hole absorbs without emitting, so the analogy alone does not identify a measured temperature. Quantum Hawking radiation supplies $T_H=\hslash \kappa /(2\pi )$, with $c=k_B=1$ and future-horizon normalization at infinity. Comparing the area terms in the first laws gives the Bekenstein-Hawking entropyup to a conventionally fixed additive constant. In ordinary units it is $k_B{c}^{3}A/(4G\hslash )$. Entropy scales with area, not interior volume.

For the massless scalar in quantum field theory in curved spacetime, global hyperbolicity ensures well-posed Cauchy evolution. The conserved Klein-Gordon inner product isIts current has zero divergence by the wave equation, making it slice-independent with appropriate boundary behavior. Positive frequency relative to the asymptotic past and future Minkowski times defines complete bases $u_j,v_i$, with $(u_i,u_j)={\delta }_{ij}$, $(u_i^{\ast },u_j^{\ast })=-{\delta }_{ij}$ and vanishing mixed products. Expand the field asContinuous labels replace sums by integrals; wave packets avoid artificial normalization infinities. Particle creation by a nonstationary spacetime occurs when evolution mixes frequency signs through a Bogoliubov transformation:The inner product gives $\alpha {\alpha }^{†}-\beta {\beta }^{†}=I$ and $\alpha {\beta }^{\mathsf{T}}=\beta {\alpha }^{\mathsf{T}}$. For the initial vacuum annihilated by all $a_j^{\mathrm{in}}$, $\boxed{⟨N_i^{\mathrm{out}}{⟩}_{\mathrm{in}}=\sum _j|{\beta }_{ij}{|}^2}$. Thus nonzero negative-frequency mixing makes the in-vacuum non-vacuum for final observers. The vacuum state in a stationary spacetime depends on its positive-frequency splitting; this is not inconsistent with deterministic mode evolution. A unitary implementation on one fixed infinite-mode Fock space further needs the Hilbert–Schmidt condition on $\beta$.

For collapse forming a Schwarzschild black hole, a late outgoing ray traced back to past null infinity obeys the Hawking exponential ray mapPulling $e^{-i\omega u}$ back gives $(U_H-U{)}^{i\omega /\kappa }$ for $U<U_H$. Its past Fourier transform has both frequency signs. For $y=\omega /\kappa$ and damping $ϵ>0$, the Gamma function evaluates the Fourier integrals asAs $ϵ\downarrow 0$, the two denominator arguments approach $\mp \pi /2$, so $|I_{+}|/|I_{-}|=e^{-\pi y}$. Thus the thermal ratio of Hawking Bogoliubov coefficients is $|\beta {|}^2/|\alpha {|}^2=e^{-2\pi \omega /\kappa }$. Together with the bosonic normalization difference, this yields $1/(e^{2\pi \omega /\kappa }-1)$. Potential scattering adds a greybody factor:The collapse state has outgoing flux at future infinity without an incoming thermal bath and is regular for infall. It is not the eternal-hole equilibrium state. Horizon-entering modes complete the future basis; tracing over their correlated partners gives the approximately thermal exterior state. The collapsing geometry is not literally Minkowskian everywhere in the far future: asymptotically flat null-infinity modes are the relevant application of the earlier in/out construction.

Emission reduces the isolated hole's mass. Schwarzschild temperature is proportional to $M^{-1}$, giving the negative heat capacity of a Schwarzschild black hole: losing energy makes it hotter. Dimensional estimates give luminosity proportional to $M^{-2}$ and evaporation time proportional to $M^3$, with species and greybody factors determining coefficients. Area may shrink because quantum stress violates the energy assumptions of Hawking's area theorem. The generalized second law instead concerns $S_{\mathrm{gen}}=A/(4G\hslash )+S_{\mathrm{outside}}$, incorporating radiation entropy and preventing ordinary thermodynamic violations by disposal of entropy into a hole.

Pair correlations also distinguish thermal reduced states from the complete pure state. The black hole information paradox asks whether complete evaporation preserves quantum information: an exactly thermal final exterior with no remaining partners would appear inconsistent with unitary pure-state evolution. The semiclassical calculation controls late-time flux, not the Planck-scale endpoint, and does not itself settle this question. Quantum emission gives the temperature–entropy identification physical meaning; generalized entropy replaces the classical area alone when radiation back-reacts.