0
$\begingroup$

In chapter 13 of D'Inverno's Introducing Einstein's Relativity, question 4 in the exercises section instructs the reader to show that flat space is not a solution of

$(13.5): G_{ab} - \Lambda g_{ab} = 8\pi T_{ab}$

Which is, of course, the regular field equation with the inclusion of a cosmological constant where $G_{ab} = R_{ab} - \frac{1}{2}Rg_{ab}$.

In my attempts at showing this, I try the following:

1) In flat space, the Minkowski metric $diag(-1,1,1,1)$ for an arbitrary co-moving frame may be chosen.

2) Following that, the Ricci tensor $R_{ab}$ may be re-expressed as:

$R = tr(R_{ab}) = g^{ab}R_{ab} = \eta^{ab}R_{ab}$

However, I keep arriving at a similar conclusion: $\eta_{ab}=kT_{ab}$, which gives the familiar solutions $\rho = wp$.

Is there something incorrect about my setup here?

D'Inverno states that "The full field equations with the cosmological term are Machian in the sense that they no longer admit flat space as a solution."

How should I interpret this? Shouldn't this mean that it should admit flat space solutions since it must also describe relatively equivalent scenarios?

I'm fairly new with tensor algebra/calculus; I have spent the summer working through D'Inverno's book and Hobson's General Relativity: An Introduction for Physicists (the latter I would recommend over the former).

$\endgroup$
  • 3
    $\begingroup$ I'm not sure I see the point of this. We known that for the Minkowski metric $G_{\alpha\beta}=0$ so it cannot be a solution of the full equation unless $T_{\alpha\beta}+\Lambda\eta _{\alpha\beta}=0$. Or are you being asked to prove $G_{\alpha\beta}=0$ when $g=\eta$? $\endgroup$ – John Rennie Aug 2 '16 at 7:01
  • $\begingroup$ Nope, just to show that flat space isn't a solution. D'Inverno is not a good text. That's why I keep getting that the SE tensor is proportional to the metric, which just gives me the well-known boundaries for the energy conditions. $\endgroup$ – hebetudinous Aug 2 '16 at 14:32
  • $\begingroup$ Are you given a specific form of $T_{ab}$? $\endgroup$ – Ryan Unger Aug 2 '16 at 22:26
  • $\begingroup$ No. Nothing else was given. $\endgroup$ – hebetudinous Aug 2 '16 at 23:08
  • $\begingroup$ While I can't read the author's mind, I'd guess he meant a vacuum solution. $\endgroup$ – Javier Aug 5 '16 at 18:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.