2
$\begingroup$

This note by Alan Guth says that

The false vacuum, however, cannot rapidly lower its energy density, so the energy density remains constant and the total energy increases. Since energy is conserved, the extra energy must be supplied by the agent that pulled on the piston.

  1. Why is it that the false vacuum cannot lower its energy rapidly?

  2. Do we know of a substance which behaves in this unusual fashion? In short, why is the energy density doesn't dilute with expansion?

$\endgroup$
1
  • $\begingroup$ Minor comment to the post (v2): Please consider to mention explicitly author, title, etc. of link, so it is possible to reconstruct link in case of link rot. $\endgroup$
    – Qmechanic
    Commented Jun 16, 2017 at 18:14

1 Answer 1

1
+50
$\begingroup$

From continuity equation ($\varepsilon$ - energy density, $p$ - pressure, $H$ - Hubble parametre) \begin{equation} \dot{\varepsilon}=-3H(p+\varepsilon) \end{equation} you can see that for a substance with negative pressure $p=-\varepsilon$, above equation gives $\varepsilon=const$, with or without expansion. Such equation of state is exactly what we need from the inflaton field during the inflation. Since pressure and energy density are defined as $$ \varepsilon=\frac{1}{2}\dot{\phi}^2+V(\phi)~,~~p=\frac{1}{2}\dot{\phi}^2-V(\phi)~, $$ when potential energy dominates, $V(\phi)\gg\frac{1}{2}\dot{\phi}^2$, we have $p\approx-\varepsilon$.

For further details see "Physical foundations of Cosmology" (inspire-hep link) by Mukhanov, or Baumann's lecture notes (pdf link, on pages 19-20 derivation of the continuity equation is given).

$\endgroup$
2
  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – rob
    Commented Dec 10, 2018 at 4:36
  • $\begingroup$ Hey, that's fine, but there should be a notice sent. The entire conversation disappeared from my history. $\endgroup$
    – user32023
    Commented Dec 11, 2018 at 17:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.