The cosmological constant $\Lambda$ can be written as part of the T-Tensor. It can then be considered as vacuum energy ($\rho_{vac}$) and vacuum pressure($p_{vac}$). $\rho_{vac}$ and $p_{vac}$ are the entries of the T-tensor in its diagonal, $T_{00}=\rho_{vac}$, $T_{11}=T_{22}=T_{33}=p_{vac}$. Because of the different sign of the time dimension it leads to $\rho_{vac}= -p_{vac}$.
Using $\Lambda$ in the field equations, vacuum (the space itself) gets a positive energy and a negative pressure.
For particles, $\rho_{vac}$ and $p_{vac}$ are independent from each other. Aren't they?
Why does no one complain when the vacuum energy density $T_{00}$ and the vacuum pressure are (with the introduction of $\Lambda$) restricted to have the same value?
Couldn't the energy density of the vacuum be totally independent from the pressure of the vacuum? Aren't those two totally different characteristics: The one is just the energy of the vacuum, the energy of space itself, the other is how space expands? Shouldn't there be two cosmological constants? One for $p_{vac}$ and one for $\rho_{vac}$?
Are there any papers using two different cosmological constants?
Or is it clear from other sources than a definition that $p_{vac} = -\rho_{vac}$?
Best Answer
Answer with no math
By definition, a cosmological constant has an energy density that is equal to minus the pressure. You could consider other types of matter in the Universe with a different relationship between energy density and pressure, but they won't be a cosmological constant.
Note that a constant energy density that did not have the energy density equals to minus pressure would break Lorentz invariance (or, in other words, break special relativity).
Answer with math
Einstein's equations with a cosmological constant $\Lambda$ (setting $c=1$) are \begin{equation} G^\mu_{\ \ \nu} + \Lambda \delta^\mu_{\ \ \nu}= 8 \pi G_N T^\mu_{\ \ \nu} \end{equation} We can move the cosmological constant term to the right hand side as follows \begin{equation} G^\mu_{\ \ \nu} = 8 \pi G_N \left( T^\mu_{\ \ \nu} + [T_\Lambda]^\mu_{\ \ \nu} \right) \end{equation} where we have defined an effective stress-energy tensor \begin{equation} [T_\Lambda]^\mu_{\ \ \nu} \equiv - \frac{\Lambda}{8 \pi G_N} \delta^\mu_{\ \ \nu} \end{equation} The definition of energy density is $\rho = - T^0_{\ \ 0} = \frac{\Lambda}{8\pi G_N}$, and pressure is $p = T^1_{1} = T^2_2 = T^3_3 = -\frac{\Lambda}{8\pi G_N}$. From this you can see that by definition, a cosmological constant has $\rho = - p$.
We were forced to set the cosmological constant energy density proportional to $\delta^\mu_\nu$ by symmetry. There are no other constant two-index tensors around for us to use. Any other constant tensor would break Lorentz invariance. Of course, generic dynamical matter fields will have stress-energy tensors that do not obey the relationship $p=-\rho$.