Our cosmic event horizon on a string

In this post, we introduced the idea that in the presence of a positive cosmological constant, there is a minimum (local) mass $m_{GR}$

 

McDormand swinging a cosmic light-like mass-energy $m_{GR}$, with a "cosmic string" of radius $l_{\Lambda}$, giving a centripetal force $F_{local}=m_{GR} \ c^2/l_{\Lambda}$. 

 Let's think about that string for a bit. In fact, a great number of physicists have spent their entire careers tied up unraveling string theory. For a classical string, which lives in D = 10 dimensions, associated with Nambu-Goto action, the the string tension $T_G$ is a local force, or energy per unit length (dimensions $MLT^{-2}$):

$$\begin{equation} \notag T_G = \frac{1}{2\pi \alpha \prime} \end{equation}$$ $\alpha \prime$ is the Regge slope parameter, set here with dimensions of inverse force. 

The wavelength of the stringy mass-energy standing wave (such that it does not interfere with itself), is the circumference of the circle, and we know it moves at light-speed. From the wave relationship, we can define a local string frequency $f_{s}= c/2\pi l_{\Lambda}$. The mass-energy is really the string mass $m_{s}$, conceptualized as a point-particle with a massless string. We want our string mass to be as light as possible, giving $m_{s}= m_{GR}$, which also means that the angular momentum $L$ of the point particle is the reduced Planck constant: $L=m_{GR} \ l_{\Lambda} \ c=\hslash$. Therefore, $m_s=m_{GR}=\hslash/l_{\Lambda}c$ (at the end of inflation).

At the high energy limit: $\alpha \prime=4Gc^{-4}$. We also discussed the minimum acceleration $a$ in this post.

$$\begin{equation} \notag \frac{1}{\alpha \prime} = m_{GR} \ a =\frac {c^4}{4G}=F_{local} \end{equation}$$ Maximum tension conjecture in string theory and GR. This conjecture refers to an invariant local limit on observable force.   

Swinging that mass around also means that in a unit of time, there is a mass traveling past any particular spot. With this idea, we consider a local stringy mass-energy flow rate (mass flux) $\dot{m}$ (dimensions $MT^{-1}$) i.e. tension/velocity or mass$\times$frequency as:  

$$\begin{equation} \notag \dot{m} = \frac {T_G}{c} =m_{s} \ f_{s} = \frac {c^3}{8\pi G} \end{equation}$$

$$\begin{equation} \notag T_G = \frac {c^4}{8\pi G} \end{equation}$$ The string tension $T_G$ is also the inverse of Einstein's gravitational constant.

Illustration of volume flow rate. Mass flow rate $\dot{m}$ 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 flow velocity of the mass elements, $v=c$, by the cross-sectional vector area, $A=1 / \Lambda$. Image credit: Wiki

Another way to write the angular momentum (vector) is then the stringy mass-energy flux (scalar) times the B-H quantum of area $A=8\pi L_{Planck}^2$ (vector)

$$\begin{equation} \notag \dot{m} \ A = L = \hslash \end{equation}$$ The action $S$ (scalar) is then then the string energy times the inverse of the future Hubble constant (time) $S=E_s/ \omega_B = \hslash$. This string energy is of course the same system ground state energy we discussed here so $E_0=E_s$.

However, it is known that Nambu-Goto does not quantise properly, so the Polyakov action is used. This action is half the Nambu-Goto action, which then gives the correct least action (scalar) of the vacuum $\hslash/2$. We can also get this via $E_0/ \omega_0 = \hslash/2$. This rather implies that the angular momentum of the string itself should also be $\hslash/2$. If the radius is actually the Planck Length, we get $L=m_{GR} \ l_{\Lambda}/2 \ c=\hslash/2$.  
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Also, more typically in string theory, while $\alpha \prime$ is still the Regge slope parameter, the dimensions of $\alpha \prime$ are set 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:

$$\begin{equation} \label{eq:78} \frac{J}{\hslash}= \alpha \prime E_{s}^2 \end{equation}$$ $E_{s}$ is the string energy (dimensions $ML^{2}T^{-2}$).
$$\begin{equation} \label{eq:79} E_{s}=\frac{1}{\sqrt {\alpha \prime}} = \frac{\hslash c} {l_{\Lambda}} \end{equation}$$ 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:
$$\begin{equation} \notag T= \frac{1}{2\pi \hslash c \alpha \prime} \end{equation}$$

McDormand now swinging a cosmic light-like mass-energy $m_{dS}$, with a "cosmic string" of radius $l_{\Lambda}$, giving a classical string tension $T=m_{dS} \ c^2/l_{\Lambda}$. Image generated with Dall.E2 by SR Anderson

$T$ 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 $L$) is equivalent to the de Sitter characteristic length as :

$$\begin{equation} \label{eq:81} \hslash c \sqrt {\alpha \prime} = l_{\Lambda} = l_{s} \end{equation}$$ To end our stringy adventure, we will also connect the string Hagedorn temperature $T_H$ to the Hawking-Bekenstein cosmic event horizon temperature. The Hagedorn temperature is a maximum (critical) temperature in the same sense that the boiling point of water is a maximum temperature: namely, it is the maximum temperature for a particular "phase" of matter, and going beyond that temperature requires a "phase transition". 

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

In fact, we can also connect the de Sitter temperature we mentioned earlier, $T_{dS}$ with the Hawking-Bekenstein black hole temperature $T_{BH}$ and therefore also the Hagedorn temperature.

$$\begin{equation} \notag T_{BH}= \frac{T_{dS}}{2} \end{equation}$$ The pesky factor of two difference with the de Sitter (Unruh) and Hawking-Bekenstein temperature is due to the location of the evaluation of the temperature, locally (Unruh), or remotely-at-infinity (Hawking).

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 energy scale ($10^{16}$ GeV ) otherwise relics might appear during the breaking of the GUT gauge group to the standard model gauge groups. That is, the ‘melting point’ of quark-gluon plasma is the GUT scale.