Our cosmic event horizon on a string

A Cosmic Stringy Adventure!

 Requiring the reduced Compton wavelength to fit within a cosmic event horizon results in a "minimum mass" (this is not a minimum particle mass) $m_s = \frac{\hbar}{c l_{\Lambda}}$.  

In this post we interpret $m_s$ as a horizon energy gap and model the associated gravitational scale as a lightlike energy flow along an effective classical string.  

McDormand with a cosmic light-like energy flow $m_{s}$ along a string of minimal radius $2L_p$, giving a centripetal force $F_{max}=c^4/4G$.  

Mass-Energy Flux Along the String

 Let's think about that string for a bit. In fact, a great number of physicists have spent their entire careers tied up unravelling string theory. For a classical string associated with Nambu-Goto action, the the string tension $T_G$ is a local force, or energy per unit length (dimensions $MLT^{-2}$):
\begin{equation}
\notag
T_G = \frac{1}{2\pi \alpha_G \prime}
\end{equation}$\alpha_G \prime$ is the Regge slope parameter, set here with dimensions of inverse force. This relation is pointed out in Gibbons (2002) The Maximum Tension Principle in General Relativity. At the high energy limit: $\alpha_G \prime=4Gc^{-4}.$ We also discussed the minimum acceleration $a$ in this post. \begin{equation}
\frac{1}{\alpha_G \prime} = m_{s} \ a =\frac {c^4}{4G}=F_{local}
\end{equation}This is Gibbons Maximum tension conjecture in string theory and GR. This conjecture refers to an invariant local limit on observable force. This means the string tension $T_G$ can be written as: \begin{equation}
T_G = \frac {c^4}{8\pi G}
\end{equation} We see that $T_G$ is the inverse of Einstein's gravitational constant.

If we model the gravitational tension as driving an effective 1-D null energy channel at the speed of light $c$, its mass-energy flow rate (mass flux) (dimensions $MT^{-1}$) i.e. tension/velocity or horizon action $I$ per unit area $A$ as: $$\dot m_G = \frac{T_G}{c} = \frac{c^3}{8\pi G} = \frac{dI}{dA}$$ Thus the "string flow" is more precisely a null horizon action current: gravitational tension fixes the universal flux scale, while Euclidean horizon periodicity converts the transported energy into action.  

Illustration of volume flow rate. The surface transport coefficient (aka shear viscosity)  $\eta_{H}$ can be calculated by multiplying the volume flow rate by the mass density of the space-time "fluid", $\rho_{\Lambda} = \Lambda c^2 / 8\pi G$.  The volume flow rate is calculated by multiplying the effective drift velocity, $v_d=c/2$, by the cross-sectional vector area, $A_t=1 / \Lambda$. Image credit: Wiki

The classical relation can be expressed as a correspondence between a target-space flux and a fundamental geometric scale. In particular, the stringy mass–energy flux $\dot m$ combined with the B-H quantum of area (scalar) $$\Delta A = 8\pi L_P^2$$ defines a quantity with dimensions of action (action and angular momentum have the same dimension, but angular momentum is quantised, and is an axial vector, while the action is continuous, and is a scalar): $$\dot m_{\text{G}} \Delta A = \Delta I = \hbar.$$ This is a quantum of action, not angular momentum.  

The effective cosmic string gives a geometric picture: gravitational tension fixes a universal action flux, while horizon quantisation converts one area step into one action quantum. Requiring one horizon mode per cell: $T_{\rm dS}\Delta S_{\rm cell}=\frac{\hbar c}{l}$ makes maximum force and area quantisation equivalent: $E=\frac{A}{\Delta A}\frac{\hbar c}{l}, \qquad \frac{dE}{d(2l)}=\frac{c^4}{4G} \quad\Longleftrightarrow\quad \boxed{\Delta A=8\pi L_P^2}.$ Indeed, $\Delta A=8\pi L_P^2 \quad\Longrightarrow\quad E=\frac{c^4l}{2G} \quad\Longrightarrow\quad \frac{dE}{d(2l)}=\frac{c^4}{4G}$  

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Horizon shear viscosity

The membrane paradigm assigns the horizon a shear viscosity (aka surface transport coefficient) $$\eta_{H} = \frac{c^{3}}{16\pi G} = \frac{1}{2}\dot m_G$$ So, the minimal horizon-dissipation action per proposed area step is $$\eta_{H}\ \Delta A = \frac{\hbar}{2}$$

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String Theory: Regge Slope at the UV Crossover 

Now, while $\alpha_s \prime$ is still the Regge slope parameter, we set the dimensions of $\alpha_s \prime$ as the inverse of energy squared. The total angular momentum $J=L+S$ of the string (here $S$ is the intrinsic angular momentum, not the action), so we can write:
\begin{equation}
\label{eq:78}
\frac{J}{\hbar}= \alpha_s \prime E_{s}^2
\end{equation}$E_{s}$ is the string energy aka the horizon frequency gap (dimensions $ML^{2}T^{-2}$).
\begin{equation}
E_{s}=\frac{1}{\sqrt {\alpha_s \prime}} = \frac{\hbar c} {l_s}
\end{equation} This string energy is the same energy we derived above. The string tension $T$ is still a force (dimensions $MLT^{-2}$) except it is now written as:
\begin{equation}
\notag
T= \frac{1}{2\pi \hbar c \alpha_s \prime}
\end{equation}

 

McDormand now swinging a cosmic light-like mass-energy $m_{dS}$, with a "cosmic string" of radius $l_{\Lambda}$, giving a classical string tension $T=m_{dS} \ c^2/l_{\Lambda}$.  

$T$ is the string tension with the de Sitter mass-energy! In another previous post, we introduced the idea that our accelerating Universe has a cosmic event horizon (CEH), and that the temperature of this horizon was the minimal possible temperature of the Universe, the de Sitter (dS) temperature $T_{dS}$. If we express the dS temperature as a rest-energy $E_{dS}$ via Boltzmann's constant $k_B$: 

\begin{equation}
\notag
E_{dS}=k_{B}T_{dS}
\end{equation}This means we can also define a de Sitter mass-energy $m_{dS}$ via $E_{dS}=m_{dS}c^2$.

With this, we see that the string length (dimension $L$) as :
\begin{equation}
\hbar c \sqrt {\alpha_s \prime} = 2L_p = l_{s}
\end{equation}  

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Hagedorn Meets Hawking

The Hagedorn temperature is where the canonical partition function of a perturbative string gas diverges; the density of states grows exponentially and the system undergoes a phase transition (in the same sense that the boiling point of water is a "maximum" temperature). As we see, it is identical to the Hawking-Bekenstein black-hole temperature $T_{BH}$ (e.g. for a BH of radius $l_{UV} = 2L_P$):

\begin{equation}
T_{H}=\frac{1}{k_{B}4\pi \sqrt{ \alpha \prime}} = \frac{\hbar \kappa}{2\pi c \ k_{B}}= T_{BH} \approx 5.64\times 10^{30}K
\end{equation}Here $\kappa = \frac{c^2}{2l_{UV}}$ the so-called black hole surface acceleration. 

The Unruh (de Sitter) temperature, with maximum acceleration $a_{UV} = c^2/(2L_P)$: $$T_{UV} = \frac{\hbar\,a_{UV}}{2\pi c\,k_B} = \frac{M_P c^2}{4\pi k_B} = 2\,T_H.$$ The pesky factor of two difference is geometric: $$\kappa_{dS}=\frac{c^2}{l} \qquad \kappa_{Schwarzchild}=\frac{c^2}{2l}$$ The same radius does not imply the same surface gravity for different horizon geometries.  

 There is also a useful thermodynamic interpretation. Equality of the Hagedorn and Hawking temperatures matches their entropy slopes $$\frac{\partial S_{\rm string}}{\partial E} = \frac{\partial S_{\rm BH}}{\partial E}$$ By contrast, equality of the leading string and black-hole entropies introduces the Schwarzschild Smarr factor $$
E_{\rm BH}=2T_{\rm BH}S_{\rm BH}$$ and therefore gives$$
T_H=2T_{\rm BH}$$ Thus the factor of two may distinguish temperature matching from entropy matching. 

The Hagedorn temperature marks a phase transition, not a maximum temperature. Near this scale, the canonical ensemble of perturbative strings fails, and a thermal-winding or long-string condensate (which is precisely what a horizon-spanning string is). So, perhaps the de Sitter horizon is a post-Hagedorn object. Its thermodynamics would then be described not by the canonical ensemble of free strings but by the  microcanonical ensemble of the holographic screen.

Some of you might now be thinking: is the Hagedorn temperature the temperature of the reheating temperature, aka the Hot Big Bang? 

                                                                                Image credit: Hyperphysics

Well, sorry to disappoint, but no, the Hagedorn temperature is actually too hot. You can see the guesstimated temperatures for the inflationary epoch from the excellent diagram above. While the energy density during inflation can be greater than the reheating temperature, the reheating temperature cannot be larger than the GUT symmetry breaking scale ($10^{16}$ GeV ) otherwise relics might appear during the breaking of the GUT gauge group to the standard model gauge groups.  

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