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The problem

Consider the following euclidean action $$ S_E = - \int_{\mathcal{M}} d^4x \sqrt{g} \left [\frac{R}{2 \kappa} +\mathcal{L}_M \right ] + S_{GHY},$$ where $S_{GHY} = -\int_{\partial \mathcal{M}}\frac{K-K_0}{\kappa}$ is a boundary term. Consider the weak gravity limit by expanding terms in powers of $\kappa = 8 \pi G$, \begin{align} R = R^{(1)} \kappa + \mathcal{O}(\kappa^2), \quad S_{GHY} = S_{GHY}^{(0)} +\mathcal{O}(\kappa). \end{align} Similarly, to first order, only the flat metric $g^{(0)}_{\mu \nu}= \eta_{\mu \nu}$ is considered in the matter part, $\mathcal{L}_M[\phi, \partial \phi, g^{(0)}_{\mu \nu}]$. The action then reduces to \begin{align} S_E &= \int_{\mathcal{M}} d^4x \sqrt{g^{(0)}} (-\mathcal{L}_M[\phi, \partial\phi, g^{(0)}_{\mu \nu}])+ \left(\int_{\mathcal{M}} d^4x \sqrt{g^{(0)}} \left [\frac{-R^{(1)}}{2} \right ] + S^{(0)}_{GHY} \right) + \mathcal{O}(\kappa). \end{align} The matter part yields the equations of motion for matter fields that would be obtained ignoring gravity from the start. However, the actual value of the action is different in the limit $\kappa \to 0$, it is corrected by the second term. In quantum tunneling, this would make a difference, as the euclidean action of an instanton enters in the tunneling rate, $\Gamma \sim e^{-S_E}$. This issue is not fixed by subtracting a background field configuration. Anybody knows what is wrong with this line of reasoning or why the second term is harmless?

Some relevant examples

False vacuum decay with gravity

The second term of this first order action is harmless if it vanishes when evaluated on a physical configuration. This is the case in Coleman and de Luccia's article about false vacuum decay with gravity [1]. The action of the bounce computed correctly reduces to the result without gravity as $\kappa \to 0$. It can be shown that, in this case, that $$ \int_{\mathcal{M}} d^4x \sqrt{g^{(0)}} \left [\frac{-R^{(1)}}{2} \right ] = S^{(0)}_{GHY} = 0.$$

Action of a planar domain wall

However, it doesn't seem to always work out this well. Consider the action of a planar domain wall in a closed universe as described in Caldwell, Chamblin and Gibbons' Pair creation of black holes by domain walls [2]. Suppose the wall is perpendicular to the $z$ direction (which is a proper distance), then you can define the energy density as $$ \sigma = \int dz \left( \frac{1}{2}(\partial_z \phi)^2 + U(\phi) \right). $$ The action without gravity is then $$\int d^4x \sqrt{g^{(0)}} (- \mathcal{L}_M) = \sigma \int d^3x \sqrt{\gamma^{(0)}} \equiv \sigma \mathcal{A}$$ where $\gamma_{ij}$ is the metric induced on hypersurfaces with normal in the $z$ direction. The full action is $$- \int_{\mathcal{M}} d^4x \sqrt{g} \left [\frac{R}{2 \kappa} +\mathcal{L}_M \right ] = -\frac{\sigma}{2} \mathcal{A}. $$ The trace of Einstein's equation $-\tfrac{R}{2\kappa} = T$ is used to get this result. This doesn't reduce to the correct action as you turn off gravity because it has the wrong sign. Maybe it's because you can't have a closed spacetime as you turn off gravity, and a boundary term would save the day. But this seems far fetched.

A final note: Actually, $R^{(1)}$ is just a total derivative (see the linearised Ricci scalar in MTW's gravity, section 18.1): $$R_{lin} = h_{\mu \nu}^{ \quad ,\mu ,\nu} - h^{,\mu}_{\;,\mu} $$ so one could hope that $\int_{\mathcal{M}} R^{(1)}$ would cancel with the boundary term $S^{(0)}_{GHY}$. I failed to prove this.

References

[1] Coleman, S. & De Luccia, F. Gravitational effects on and of vacuum decay. Physical Review D 21, 3305 (1980).

[2] Caldwell, R. R., Chamblin, H. A. & Gibbons, G. W. Pair creation of black holes by domain walls. Physical Review D 53, 7103 (1996).

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    $\begingroup$ Hello Éric, would you consider, if it is possible, to ask your question explicitly? Users may ( /will) misread such a large amount of text. Thank you. $\endgroup$ – user108787 Sep 23 '16 at 20:37
  • $\begingroup$ Hi CountTo10, thank you for your remark. I've shorten my post a bit, and I hope the question is more clear now. Best. $\endgroup$ – Éric Dupuis Sep 24 '16 at 22:17
  • $\begingroup$ Hi Éric, the very best of luck with your question, it's a credit to the website. All the best. $\endgroup$ – user108787 Sep 24 '16 at 22:29
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in my opinion there is a flaw in your reasoning. I'll assume for simplicity that $\hbar=c=1$.

You have to carefully keep care of the dimensionality of the quantities in these equation. We have a dimensionless action $S$. Since we are in four dimensions, $\kappa$ has dimensions of $[E]^{-2}$. Then $g_{\mu\nu}$ is dimensionless and $R$ has dimensions of $[E]^{2}$. The same reasoning applies for the GHY term.

With this in mind you can expand the metric around flat spacetime: $g_{\mu\nu}\simeq\eta_{\mu\nu}+h_{\mu\nu}$.

Then you can define $\tilde{h_{\mu\nu}}=\sqrt{\kappa}\,h_{\mu\nu}$ obtaining something like $S \sim \dfrac{1}{2}\int d^4x \left[ (\partial \tilde{h})^2+\kappa \, (\partial\tilde{h})^2\tilde{h}+\dots\right]$

Managing the gravitational coupling constant is a tough task: as you may know GR is a non renormalizable theory!

Hope that my comment may help you!

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