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 $p_{vac} = -\rho_{vac}$?

  • $\begingroup$ "For particles, $\rho_{vac}$ and $p_{vac}$ are independent from each other. Aren't they?" See here. $\endgroup$
    – J.G.
    Dec 6, 2021 at 21:27
  • $\begingroup$ This is an empirical question, known as the equation of state of dark energy. People define $p_{vac}=w\rho_{vac}$. Observation indicates that $w$ is close to -1, but there is an ongoing effort to reduce the observational uncertainty. $\endgroup$
    – Thomas
    Feb 4, 2022 at 4:34

1 Answer 1


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 $p=-\rho$.

  • $\begingroup$ Thank you, joseph h for making my question perfect! I like it! Thank you, Andrew, for this great answer! I'm still wondering: Why would for example rho_vac < -p_vac break Lorentz invariance? $\endgroup$ Nov 20, 2021 at 8:35
  • 1
    $\begingroup$ Instead of $\delta^\mu_\nu = \diag(1, 1, 1, 1)$ appearing in Einstein's equations, you'd have to have something with different entries like $\diag(2, 1, 1, 1)$. This is not a tensor, so doesn't have the same form in different reference frames. Or in more physicsy terms, the fact that "2" is different from "1" breaks the symmetry between space and time needed for Lorentz invariance. Again it's ok for dynamical fields to give you a different relationship between pressure and energy density, but there's no other constant tensor you can use. $\endgroup$
    – Andrew
    Nov 20, 2021 at 13:09
  • $\begingroup$ Please, where does the $\delta^{\mu} _{\nu} $ appear in Einstein's equations? I do not know this notation. $\endgroup$ Nov 20, 2021 at 19:31
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    $\begingroup$ $\delta^\mu_{\ \ \nu}$ is the Kronecker delta symbol, which is equal to 1 when $\mu=\nu$ and is 0 otherwise. It is essentially the identity matrix. You can see where it appears in the first equation under "Answer with math" above. Usually that term is written with lower indices, $\Lambda g_{\mu\nu}$ where $g_{\mu\nu}$ is the metric, but it's equally valid to write Einstein's equations with one upstairs and one downstairs index by raising indices with the inverse metric $g^{\mu\nu}$. I chose to write it that way because the connection between $T^\mu_\nu$ and $\rho, p$ is more direct that way. $\endgroup$
    – Andrew
    Nov 20, 2021 at 20:10
  • $\begingroup$ This is a good answer, by reading it I understood things better and so I voted it up. However, in the answer it is stated, that the cosmological constant is per definition something which leads to $\rho = -p$. The question, additionally, aims to understand why this definition is chosen, why $\rho$ and $p$ couldn't be independent constants/characteristics of our universe? It would be great to receive a more detailed answer on the equality of $\rho$ and $-p$. $\endgroup$ Dec 6, 2021 at 10:13

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