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The Riemann Hypothesis as a Stability Principle

  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)=...
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The fabric of space-time: stiffer than steel or weaker than jello?


Generated image 

The smart money is on cosmic jello, not comic steel  

The notorious rubber sheet analogy of spacetime teaches one concept and once concept only: Mass-energy causes curvature of space-time.

 When a gravitating mass recedes from a region of space-time the curvature diminishes. The field equations of General Relativity don’t have an explicit term for this elastic property, but the framework as a whole does have that property. As very large mass-energies are required to generate gravitational waves (ripples in space-time), the elastic property of space-time is generally regarded as very stiff

One other interesting consideration here is that elasticity is an emergent phenomenon. There is a great deal of interest in the idea that gravity is similarly emergent.  

In 2018, Kirk T. McDonald quantified the classical stiffness of space-time via an effective Youngs Modulus experienced by a gravitational wave frequency of 100 Hz. $$ Y_{\mathrm{GW}} = \frac{\pi c^{2} f^{2}}{4G}$$ Classical answer: Youngs Modulus of space-time $\sim$ 20 orders of magnitude greater than steel.  

DALL.E2 depiction of classical space-time.

If we think about it, we can see this effective modulus depends on frequency: $Y_{\mathrm{GW}} \propto f^{2}$. The coefficient changes with the frequency of the wave whose energy we are representing. This is a wave-energy effective modulus. 

Adrian Melissinos (2018) came at this problem differently. Rather than identifying gravitational wave energy with elastic strain energy, Melissinos considered the tidal acceleration produced by a gravitational wave passing through a region containing energy density $\rho$. For a test region of transverse size $L$, the resulting effective modulus is $$Y = \frac{\rho}{c^{2}} \left( 2\pi f L \right)^{2}$$ The local tidal approximation used in the derivation requires $$L \ll \frac{c}{2\pi f}$$ This means $$Y \ll \rho$$ The important result is that the density scale limiting this effective modulus is frequency independent. The local coefficient still depends on the frequency and size of the test region; its density bound does not. So, the now the effective Youngs Modulus of space-time is simply limited by the energy density of the medium through which the gravity wave propagates. 

In our current Universe, which includes matter, the natural energy density ($Y_{cosmic}$) is the critical density.  

Cosmological answer: Youngs Modulus of space-time $\sim$ 14 orders of magnitude less than jello! 

 

Why the massive discrepancy between these two answers?  

The immediate answer is that they the two approaches define the effective modulus differently. 

But there is a hidden connection! If we introduce the reduced gravitational wavelength  $$L_f = \frac{c}{2 \pi f}$$ and then substitute for $f$ into the McDonald modulus we get  $$Y_{\mathrm{GW}} = \frac{c^{4}}{16\pi G L_{f}^{2}} \sim \frac{c^4}{GL^2}$$ This means the apparent frequency dependence from McDonald is really an inverse length squared dependence  $Y_{\mathrm{GW}} \propto L^{-2}$. 

Now, consider the critical density of the Universe. The Hubble length is $$L_H=\frac{c}{H}$$ Substitute into the critical density equation $$Y_{cosmic}=\rho_c \sim \frac{c^4}{G L_H^2}$$ This is exactly the same inverse length squared dependence as the McDonald modulus. 

So, the steel and jello answers are both gravitational energy density scales, evaluated at vastly different characteristic lengths. 

  • For a 100 Hz gravitational wave, $L_f \sim 4.8 \times 10^5$ m.
  • For the present universe, $L_H \sim 1.4 \times 10^26$ m.

Small scale spacetime is effectively very stiff. Cosmic scale spacetime is effectively very soft!    

In field theory, spacetime is the geometric manifold (the stage), while the vacuum is the state of lowest energy of the quantum fields residing on it (the actors)   

These should not be casually identified. Nevertheless, in GR, vacuum energy gravitates and sets a curvature scale. In a previous post we discussed de Sitter space-time, which is space-time with no matter, just vacuum, and its associated energy-density, the cosmological constant. 

 In dS space, the energy density is the cosmological constant, so we can define an effective vacuum modulus as $$Y_\Lambda = \frac{\Lambda c^4}{8 \pi G} = \frac{3c^4}{8 \pi G l_{\Lambda}^2} \sim \frac{c^4}{G l_{\Lambda}^2}$$ Once again even in vacuum, we have the inverse length squared dependence.

Effective gravitational stiffness $\sim$ curvature energy density $\sim \frac{c^4}{G l_{\Lambda}^2}$

From Young’s modulus to the spring constant of the vacuum 

Youngs Modulus can also be related to the spring constant $k$ (the measure of a spring's stiffness), of a quantum harmonic oscillator via Hooke's Law, which holds up to the so-called proportionality limit i.e. when extension is linearly proportional to the load applied. The relation is:   \begin{equation}
\notag
Y_\Lambda = \frac{k \ L}{A}\
\end{equation} So, with a "cosmic string" of the de Sitter characteristic length $L=l_{\Lambda}$ and a cross-sectional vector area $A=1/ \Lambda$, with a Youngs Modulus of vacuum: $Y_\Lambda = \Lambda c^4 / 8 \pi G$, the spring constant of vacuum $k_L$ is:

\begin{equation}
\notag
k_L = \frac{c^4}{8 \pi G \ L}
\end{equation} The term $c^4 / 8 \pi G$ is the inverse Einstein gravitational constant in force units. This quantity is not a conventional modulus, which has units of force per area, so, a length scale is needed to convert it into either: a modulus, by dividing by an area $L^2$ or a spring constant, by dividing by a length $L$.   

Spacetime does not have one universal Young’s modulus or spring constant. Its effective elastic response is geometrically scale dependent.

Einstein's equations as a stress curvature relation  

If we look at the Einstein Field Equations in 4D $$G_{\mu\nu} + \Lambda g_{\mu\nu} = \frac{8\pi G}{c^4}\, T_{\mu\nu}$$ and re-arrange to $$T_{\mu\nu} = \frac{c^4}{8 \pi G} (G_{\mu\nu} + \Lambda g_{\mu\nu})$$ There is an obvious structural resemblance to Hooke's Law, which in classical elasticity relates stress (force per unit area) to strain (geometric deformation) via an elastic modulus: Stress = Modulus $\times$ Strain. But this analogy requires care.  

1. Stress

The stress-energy tensor $T_{\mu\nu}$ has units of energy density ($J/m^3$), which dimensionally reduces to ($N/m^2$), i.e. same dimensions as mechanical stress or pressure.   

2. Geometric response

The Einstein tensor $G_{\mu\nu}$ and the cosmological constant term $\Lambda g_{\mu\nu}$  have units of $L^{-2}$. They represent curvature, not dimensionless strain. 

3. Gravitational coupling 

The inverse of the Einstein gravitational coupling $\kappa^{-1} = \frac{c^4}{8\pi G}$ has units of force. Multiplying this force scale by curvature gives stress (force/area).  This yields the scale dependent effective modulus $$Y(L) \sim \frac{c^4}{8\pi G L^2}$$

The field equations therefore do not say that ${c^4} /{8\pi G}$ is itself a Young’s modulus. They say something more interesting:
The Einstein coupling is a fundamental force scale, and curvature supplies the inverse area needed to turn that force into stress. 

The spring coefficient reappears in horizon entropy 

The spring interpretation leads to one final result. For a dS horizon of radius $L$ standard horizon thermodynamics give 

$$ A_H = 4\pi L^2\qquad S_H = \frac{k_B c^3}{4\hbar G}\ A_H \qquad T_H = \frac{\hbar c}{2\pi k_B L} $$ So  $$T_H \frac{\partial S_H}{\partial A_H} = \frac{c^4}{8\pi G L} = k_L$$ We can see that the spring coefficient is also an entropic surface tension scale. 

The same gravitational coupling controls curvature response and horizon entropy. This supports a thermodynamic interpretation of gravity as Ted Jacobson first pointed out.

Then, with the areal volume  $V=4 \pi L^3 /{3}$  the cosmic event horizon energy $$E_{\Lambda}=T_HS_H = Y_{\Lambda}V = \frac{c^4L}{2G}$$ At the Planck scale, the spring's characteristic length is $L=L_p$. Identifying its squared displacement with the black-hole area quantum $(\Delta x)^2=\Delta A=8\pi L_p^2$  gives $$E_{\rm elastic} =\frac12k_{L_p}\Delta A =\frac12\frac{c^4}{8\pi GL_p}\,8\pi L_p^2 =\frac{c^4L_p}{2G}$$ Using $L_p^2=\hbar G/c^3$, we find $$E_{\rm elastic}=\frac{\hbar c}{2L_p} =\frac12\hbar\omega_p \qquad \omega_p=\frac{c}{L_p}$$ We recover both the horizon-energy form at the Planck length and the zero-point-energy form of a quantum harmonic oscillator. 

After all, a free scalar quantum field is mathematically equivalent to an infinite collection of independent quantum harmonic oscillators.   

 


 

 
 

 

 

  

 

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