In this paper (https://arxiv.org/abs/astro-ph/0603302) the authors derive the field equations for [itex]f(R)[/itex] gravity considering a spherically symmetric and static metric. Now the Ricci scalar only depends on [itex]r[/itex] so you could write [itex]f(R(r)) = g(r)[/itex] for some [itex]g[/itex]. However what it seems the authors have done is in the field equation [tex](adsbygoogle = window.adsbygoogle || []).push({});

f'(R) R_{\mu \nu} - \tfrac12 f(R) g_{\mu \nu} + g_{\mu \nu} \Box f'(R) - \nabla_\mu \nabla_\nu f'(R) = 0

[/tex]

is rewrite it as [tex]

g'(r) R_{\mu \nu} - \tfrac12 g(r) g_{\mu \nu} + g_{\mu \nu} \Box g'(r) - \nabla_\mu \nabla_\nu g'(r) = 0

[/tex]

rather than [tex]

\frac{g'(r)}{R'(r)}R_{\mu \nu} - \tfrac12 g(r) g_{\mu \nu} + g_{\mu \nu} \Box \left (\frac{g'(r)}{R'(r)} \right) - \nabla_\mu \nabla_\nu \left (\frac{g'(r)}{R'(r)} \right) = 0.

[/tex]

The reason I say this is that when implementing this into Mathematica I first attempted in a field equation function to use an arbitrary [itex]f[/itex] and the Ricci scalar you calculate from the metric as [itex]R[/itex]. This results in equations that were very different and far more complicated, involving fourth derivatives of [itex]B(r)[/itex] which you would expect as the Ricci scalar has second derivatives and the field equations have second derivatives.

On the other hand replacing [itex]R[/itex] with [itex]r[/itex] resulted in equations very similar to their ones (off by signs). The contracted equation was identical. Now I may have made a mistake in the Mathematica notebook (although I did check against a known solution), but it seems too coincidental. At best I think my method wouldn't change fundamentally.

I do not see how it is possible they could make a mistake like this, but I also don't see where I could be wrong so I would appreciate any help.

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# I [itex]f(R)[/itex] gravity field equation derivation mistake?

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