on "Gravitation foundations and frontiers" at page 263, Padmanabhan claims that

[...] the gravitational field has only two genuine degrees of freedom per event

and he shows that by considering two infinitesimal coordinate transformations: the first one forces $g_{00}=1$ and $g_{0\alpha}=0$ and the second keeps $g_{00}=1$ and $g_{0\alpha}=0$. Given the second coordinate transformation as $$t\to t'=t+f(\textbf{x}) \\ x^{\alpha}\to x'^{\alpha}=x^{\alpha}+\lambda^{\alpha}$$ where $x^{\alpha}$ for $\alpha=1,2,3$ define the spatial coordinates, the author reaches formulae $\dot{f}=0$ and $\partial_{\alpha}f=-g_{\alpha\beta}\dot{\lambda}^{\beta}$, which I am happy about apart from a minus sign, but this is not the problem for me. The problem comes right after those formulae, because the second equation can be integrated to give $$ \tag{1} \lambda^{\alpha}=p^{\alpha}(\textbf{x})-\partial_{\beta}f\int^tdt g^{\alpha\beta}$$ where $p^{\alpha}(\textbf{x})$ represent three arbitrary functions that correspond to the initial conditions. (Here again I have some doubt about the sign, but this does not bother me too much). The author, then, concludes by writing that, given (1), on any spatial hypersurface we have the freedom to choose the four functions $f(\textbf{x})$ and $p^{\alpha}(\textbf{x})$ in order to bring four out of the six components of $g^{\alpha\beta}$ to pre-assigned values.

Here is my question: isn't formula (1) a system of three equations with six unknown $g^{\alpha\beta}$? Even though I have four functions ($f(\textbf{x})$ and $p^{\alpha}(\textbf{x})$) that I can choose, because I can choose the coordinate transformation, don't I need a fourth equation [...] in order to bring four out of the six components of $g^{\alpha\beta}$ to pre-assigned values?

To me this sounds like a very general proof (no weak field limit has been taken) that the gravitational field has locally only two independent components, but I am not able to convince myself about it because I feel that one condition on $g^{\alpha\beta}$ is missing.

  • $\begingroup$ I don't understand why this would be described as "per event." Isn't it more like "per spacelike surface?" If I'm understanding what Padmanabhan has in mind here (maybe more context is needed?), the metric has zero degrees of freedom per event, because the metric can always be put into the canonical form diag(1,-1,-1,-1) at any given event. $\endgroup$ – Ben Crowell Jun 3 '18 at 13:51

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