$\Lambda$CDM and Holographic Dark Energy Holographic dark energy is usually written $$\rho_{\rm DE} = 3M_p^2\frac{c^2}{L^2}$$ where $c$ is a dimensionless HDE parameter and $L$ is an infrared cutoff. In Li's 2004 model, $$L=L_e \equiv a(t)\int_t^\infty\frac{dt'}{a(t')}$$ is the future event horizon. Li assumed constant $c$, giving $$w_{\rm DE} = -\frac13 \left( 1+\frac{2\sqrt{\Omega_{\rm DE}}}{c} \right).$$ However, with $c=1$ and $\Omega_{{\rm DE},0}=0.6889$, $w_0\simeq-0.887$, rather than $w=-1$. Does not match observations! Also, constant-$c$ observational fits often give $c<1$, leading to phantom evolution, null-energy-condition violation and a turning point in $H(z)$. That conclusion assumes $c'=0$. What about if $c$ is not constant? Then: $$c<1, \qquad c'=c-\sqrt{\Omega_\Lambda}>0, \qquad w=-1.$$ Therefore $\dot H = -4\pi G(\rho_m+\frac43\rho_r) \le0$, so there is no finite-redshift turning point and no phantom phase. Thus ...
Yes!
Also, it is not the 'Planck Power' (despite what you might have read in Misner, Thorne and Wheeler, P.980). The existence of black hole horizons implies a maximum luminosity (power) limit in General Relativity. Not even gravitational waves can escape a black hole. Consider
an (almost) black hole made of light (this is called a Kugelblitz) sphere
of radius
\begin{equation}
\notag
R \geq \frac{2Gp}{c^3}
\end{equation} which is filled with photons with a total mass-energy of momentum $p$ times speed of light $c$ \begin{equation}
\notag
E=p \ c
\end{equation} that leave after a time
\notag
R \geq \frac{2Gp}{c^3}
\end{equation} which is filled with photons with a total mass-energy of momentum $p$ times speed of light $c$ \begin{equation}
\notag
E=p \ c
\end{equation} that leave after a time
\begin{equation}
\notag
t=R/c
\end{equation} with average power (luminosity)
\notag
t=R/c
\end{equation} with average power (luminosity)
\begin{equation}
\notag
P_{max} = \frac{E}{t}=\frac{p \ c^2}{R}=\frac{c^5}{2G} \approx 1.8\times10^{52} \ W
\end{equation} This is maximum power in GR, regardless of the nature of the system. You might be tempted to call this half a 'Planck Power' but there is no $\hslash$ in this expression, it is purely classical. This is why you won't see an equation for 'Planck Power' in wiki
\notag
P_{max} = \frac{E}{t}=\frac{p \ c^2}{R}=\frac{c^5}{2G} \approx 1.8\times10^{52} \ W
\end{equation} This is maximum power in GR, regardless of the nature of the system. You might be tempted to call this half a 'Planck Power' but there is no $\hslash$ in this expression, it is purely classical. This is why you won't see an equation for 'Planck Power' in wiki
Source: Cardoso (2018)
OK, so what if there is a maximum power limit in GR? Well, for one it means that even in classical GR, a lab-scale Kugelblitz cannot be formed. From this paper, to overcome Schwinger dissipation (which is quantum) to form one with a 1m radius would require a power of around $P_{req}=10^{84}\ W$. This is greater than our derived maximum power!
To get an energy density low enough to avoid triggering the vacuum breakdown (Schwinger effect) to build a Kugelblitz that doesn't blow itself up via quantum effects, you would need to construct a sphere of light larger than our Sun. Getting $P_{max}$ watts of power coherently focused across a region 2 million kilometres wide is...effectively impossible.

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