- #1
Dazed&Confused
- 190
- 3
In this paper (https://arxiv.org/abs/astro-ph/0603302) the authors derive the field equations for f(R) gravity considering a spherically symmetric and static metric. Now the Ricci scalar only depends on r so you could write f(R(r)) = g(r) for some g. However what it seems the authors have done is in the field equation <br />
f'(R) R_{\mu \nu} - \tfrac12 f(R) g_{\mu \nu} + g_{\mu \nu} \Box f'(R) - \nabla_\mu \nabla_\nu f'(R) = 0<br />
is rewrite it as <br /> g'(r) R_{\mu \nu} - \tfrac12 g(r) g_{\mu \nu} + g_{\mu \nu} \Box g'(r) - \nabla_\mu \nabla_\nu g'(r) = 0<br />
rather than <br /> <br /> \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.<br />
The reason I say this is that when implementing this into Mathematica I first attempted in a field equation function to use an arbitrary f and the Ricci scalar you calculate from the metric as R. This results in equations that were very different and far more complicated, involving fourth derivatives of B(r) which you would expect as the Ricci scalar has second derivatives and the field equations have second derivatives.
On the other hand replacing R with r 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.
is rewrite it as <br /> g'(r) R_{\mu \nu} - \tfrac12 g(r) g_{\mu \nu} + g_{\mu \nu} \Box g'(r) - \nabla_\mu \nabla_\nu g'(r) = 0<br />
rather than <br /> <br /> \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.<br />
The reason I say this is that when implementing this into Mathematica I first attempted in a field equation function to use an arbitrary f and the Ricci scalar you calculate from the metric as R. This results in equations that were very different and far more complicated, involving fourth derivatives of B(r) which you would expect as the Ricci scalar has second derivatives and the field equations have second derivatives.
On the other hand replacing R with r 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.