The Riemann Hypothesis can be reformulated as an infinite sequence of explicit positivity tests. This is not a proof. It turns a statement about complex zeros into inequalities that can be calculated, falsified at finite order, and compared with operator or physical models. The completed zeta function is $$\xi(s)=\frac12s(s-1)\pi^{-s/2}\Gamma\left(\frac{s}{2}\right)\zeta(s),$$ with $$\Xi(t)=\xi\left(\frac12+it\right).$$ The function $\Xi$ is even and real on the real axis. A nontrivial zero $\rho=\beta+i\gamma$ corresponds to $$t_\rho=\gamma-i\left(\beta-\frac12\right).$$ Therefore $$t_\rho\in\mathbb R \quad\Longleftrightarrow\quad \beta=\frac12,$$ and hence $$\boxed{\mathrm{RH}\quad\Longleftrightarrow\quad\text{every zero of }\Xi\text{ is real}.}$$ Folding the Zeros Because $\Xi$ is even, write $$\Xi(t)=\sum_{n\ge0}c_nt^{2n}.$$ Now define $$G(z)=\frac{\Xi(i\sqrt z)}{\Xi(0)}.$$ Although this contains $\sqrt z$, the even expansion makes $G$ entire: $$G(z)=...
In the static patch of de Sitter spacetime , the cosmological horizon behaves thermodynamically in ways closely analogous to a physical interface. One can assign it entropy, temperature, and even an effective surface tension . Remarkably, the familiar Young–Laplace pressure relation from surface physics appears naturally at the horizon. In the static patch of de Sitter spacetime, the cosmological horizon admits a useful surface-thermodynamic description. It has an effective surface tension and pressure satisfying a Young–Laplace-like relation. Use $G=c=\hbar=k_B=1$. The metric is $$ds^2=-f(r),dt^2+\frac{dr^2}{f(r)}+r^2d\Omega^2, \qquad f(r)=1-\frac{r^2}{L^2}, \qquad L=\sqrt{\frac3\Lambda}.$$ The horizon is at $r=L$. 1. Horizon Data The area, entropy, and temperature are $$A=4\pi L^2, \qquad S=\frac A4=\pi L^2, \qquad T=\frac{1}{2\pi L}.$$ The Misner–Sharp energy is $M(r)=\frac r2(1-f)=\frac{r^3}{2L^2}$, so at the horizon $M(L)=L/2$. This equals both th...