Using $\Lambda$ in the field equations, one can derive that then $p_{\text{vac}} =-\rho_{\text{vac}} $. The energy density of the vacuum is the same (with opposite sign) as the pressure of the vacuum. If that was not the case, I've been told, it would break Lorentz invariance. Why?

  • $\begingroup$ Are you asking why the term $\Lambda g_{\mu\nu}$ appears in the Einstein equations rather than a more general tensor $\tilde\Lambda_{\mu\nu}$? $\endgroup$
    – J. Murray
    Nov 20, 2021 at 19:22
  • $\begingroup$ Where does this $\tilde{\Lambda}_{\mu\nu}$ come from? Looks interesting, especially as you call it "more general" - do you have a reference or a keyword? $\endgroup$ Nov 20, 2021 at 19:51
  • $\begingroup$ Ah, you misunderstand. You say that $p_{vac}=-\rho_{vac}$ arises from the field equations - more specifically, from moving $\Lambda g_{\mu\nu}$ to the right-hand side and interpreting it as part of the stress-energy tensor. This immediately implies $p_{vac}=-\rho_{vac}$, so asking what would happen if this weren't true is equivalent to asking what would happen if something other than $\Lambda g_{\mu\nu}$ appeared in the field equations. I was trying to ascertain what precisely you meant. $\endgroup$
    – J. Murray
    Nov 20, 2021 at 20:12

1 Answer 1


These concepts $\rho_{vac}$ and $p_{vac}$ sometimes refer to diagonal components of cosmological tensor; if we accept this, then equality $|\rho_{vac}| = |p_{vac}|$ is obvious.

The cosmological tensor has the form

$$ \Lambda g_{\mu\nu} $$ where $\Lambda$ is a number and $g$ is metric tensor.

Since $g_{\mu\nu}$ is tensor too, $\Lambda$ has to have the same value in every frame.

In flat spacetime the cosmological tensor is then $$ \Lambda \left( \begin{matrix} -1~~ & ~0 & ~~~0 & ~~~0~~ \\ ~0 & ~1 & ~~~0 & ~~~0~~ \\ ~0 & ~0 & ~~~1 & ~~~0~~ \\ ~0 & ~0 & ~~~0 & ~~~1~~ \\ \end{matrix}\right). $$ So it is obvious that all components have the same magnitude.

The harder part of this question is, why do we call these components energy and pressure of vacuum.

Probably because components of energy-momentum tensor at the same positions often have this interpretation; $T^{00}$ is energy density of matter+EM field, $T^{11}$ is pressure of matter+EM field. So similar names are used for components of the cosmological tensor, even though this tensor is not part of the energy- momentum tensor, only now the terms are labeled by "vacuum".

  • $\begingroup$ Ah, thank you very much! Flat spacetime. This was the point I was missing: That the whole universe has to be flat. $\endgroup$ Nov 20, 2021 at 19:45

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