Why does $\Lambda>0$ violate SEC (Strong energy condition) in GR We stated during lectures (without any further explanation) that existence of positive $\Lambda$ (cosmological constant) violates strong energy condition (SEC). I am trying to figure out where this comes from, aka proof or derivation.
I am using following formulation of SEC:
$$
\rho+3p \geq 0
$$
My initial guess was, that this can be derived out of Landau-Raychaudhuri equation:
$$
\frac{\ddot{a}}{a}=-\frac{4 \pi G}{3}(\rho+3p)+\frac{\Lambda}{3},
$$
but this actually allows for expanding universe with $\Lambda>0$ and SEC valid. Any idea where this violation comes from? Is this statement even true?
 A: I believe it follows from two things:
(1) In a Universe with only a positive cosmological constant $\Lambda>0$, $\rho+3p \propto \Lambda-3\Lambda = -2\Lambda < 0$.
(2) In a Universe with matter+radiation+a cosmological constant, at late times the cosmological constant will dominate the energy density budget of the Universe, according to the Friedman equation.

More details
The stress-energy tensor for a cosmological constant is
\begin{equation}
T^{(\Lambda)}_{\mu\nu} = -\frac{\Lambda}{8\pi G} g_{\mu\nu}
\end{equation}
where $g_{\mu\nu}$ is the metric.
This follows from starting from Einstein's equations with a cosmological constant
\begin{equation}
R_{\mu\nu} - \frac{1}{2} R g_{\mu\nu} + \Lambda g_{\mu\nu} = 8\pi G T_{\mu\nu}
\end{equation}
and moving the $\Lambda$ term to the opposite side:
\begin{equation}
R_{\mu\nu} - \frac{1}{2} R g_{\mu\nu} = 8\pi G \left( T_{\mu\nu} - \frac{\Lambda}{8\pi G} g_{\mu\nu} \right) = 8\pi G \left( T_{\mu\nu} + T^{(\Lambda)}_{\mu\nu} \right)
\end{equation}
Under the symmetries of cosmology (homogeneity and isotropy), in FLRW coordinates, the metric (assuming zero spatial curvature) is
\begin{equation}
ds^2 = -dt^2 + a(t)^2 d\vec{x}^2
\end{equation}
where $a(t)$ is the scale factor, and the stress energy tensor of a perfect fluid can be parameterized as $T_{\mu\nu}=(\rho+p)u_\mu u_\nu + p g_{\mu\nu}$, or in terms of components (with $u^\mu$ identified with the time direction in FLRW coordinates)
\begin{equation}
T_{\mu \nu} = \begin{pmatrix}
\rho & & & \\
& a^2 p & & \\
& & a^2 p & \\
& & & a^2 p
\end{pmatrix}
=
-\frac{\Lambda}{8\pi G} \begin{pmatrix}
-1 & & & \\
& a^2 & & \\
& & a^2 & \\
& & & a^2 
\end{pmatrix}
\end{equation}
Therefore, we identify $\rho=-p=\frac{\Lambda}{8\pi G}$.
The trace of the stress energy tensor is
\begin{equation}
T = g^{\mu\nu}T_{\mu\nu} = - \rho + 3 p
\end{equation}
The strong-energy condition states that for every timelike vector $X^\mu$, the following condition is satisfied
\begin{equation}
\left(T_{\mu\nu} - \frac{1}{2} T g_{\mu\nu}\right) X^\mu X^\nu \geq 0
\end{equation}
We can simply consider the time direction to define our timelike vector. Then in our context, the strong energy condition becomes
\begin{equation}
T_{00} - \frac{1}{2} T g_{00} = \rho + \frac{1}{2} \left(-\rho + 3 p\right) = \frac{1}{2} \rho + \frac{3}{2} p
\end{equation}
For a cosmological constant with $p=-\rho$,
\begin{equation}
T_{00} - \frac{1}{2} T g_{00} = - \rho = -\frac{\Lambda}{8\pi G} < 0
\end{equation}
which violates the SEC.
A: The strong energy condition requires that matter gravitates toward matter, s. Significances of energy conditions  .
While this is true within gravitationally bound systems like galaxies and galaxy cluster, larger systems e.g. galaxy supercluster start to expand under the influence of the cosmological constant which acts as "repelling" gravity. The energy density due to the cosmological constant causes the distances between galaxy cluster within a supercluster to increase.
