# $f(R)$ spherically symmetric field equations derivation mistake?

Abstract:

An f(R) gravitation for galactic environments.

We propose an action-based f(R) modification of Einstein's gravity which admits of a modified Schwarzschild-deSitter metric. In the weak field limit this amounts to adding a small logarithmic correction to the newtonian potential. A test star moving in such a spacetime acquires a constant asymptotic speed at large distances. This speed turns out to be proportional to the fourth root of the mass of the central body in compliance with the Tully-Fisher relation. A variance of MOND's gravity emerges as an inevitable consequence of the proposed formalism

In the paper associated with the above abstract arxiv.org/abs/astro-ph/0603302, the authors have derived field equations for $f(R)$ gravity in the case of a static, spherically symmetric metric (Eq. 3). It appears that they wrote $f(R(r))$ as $f(r)$ for some $f$ where $R$ is the Ricci scalar which only depends on the $r$ coordinate. To me this looks suspicious: the field equations are $$f'(R)R_{\mu \nu} - \tfrac12 f(R) g_{\mu \nu} + g_{\mu \nu} \Box f'(R) - \nabla_\mu \nabla_\nu f'(R) = 0$$ so surely in the replacement $f(R(r)) \to f(r)$ the equation would become
$$\frac{f'(r)}{R'(r)}R_{\mu \nu} - \tfrac12 f(r) g_{\mu \nu} + g_{\mu \nu} \Box \left (\frac{f'(r)}{R'(r)} \right) - \nabla_\mu \nabla_\nu \left (\frac{f'(r)}{R'(r)} \right) = 0$$

however the authors seem to have written it as $$f'(r)R_{\mu \nu} - \tfrac12 f(r) g_{\mu \nu} + g_{\mu \nu} \Box f'(r) - \nabla_\mu \nabla_\nu f'(r) = 0.$$

I do not believe they and others could make a mistake like this so it must be something wrong in my thinking, but I do not know what.

I cannot confirm this is precisely what they have done but I have implemented these equations in Mathematica and assuming the code is correct (which I was able to check against other sources along with reading over the $f(R)$ field equation function multiple times), I am able to obtain equations very similar, though not identical, to the ones they give with the latter prescription I have described.

Tellingly, instead of an arbitrary $f$, a function $f(r) = r$ does not reduce to Einstein's equations. I would appreciate any help in this matter.

• If you multiply everything by R'(r) the only difference is a factor of R'(r) in the second term. And, f(r)=r should not reduce to Einstein's equations because the whole point of the exercise is to modify Einstein's equations. I suspect that d/dr R(r) reduces to 1, but can't say that definitively. – ohwilleke Jul 3 '17 at 22:57
• I cannot see where in the paper the authors used the third equation you wrote. It seems that they go straight from your first equation (which is correct to my knowledge) to the components of the field equations. – Bob Knighton Jul 3 '17 at 23:37
• @BobKnighton well when I tried it in Mathematica. Using the $R$ you obtain from the metric and plugging it into the field equation gives you components which are far more complicated than given. For example, the Ricci scalar has second order derivatives in $B(r)$. This means that you would expect third and fourth order derivatives of it (unless they cancelled miraculously) in the component equations. However if I use $f(r)$ rather than $f(R)$ then I obtain something very similar to their equations. The contracted equation is identical and first two are the same apart from sign differences. – user110503 Jul 3 '17 at 23:43

The third equation can be dismissed immediately by dimensional analysis. Note that the curvature tensor has mass dimension $[R]=2$, and so $f(R)$ also has mass dimension $[f]=2$. However, $r$ has mass dimension $[r]=-1$. Thus, in the third equation, the first term has mass dimension $[f'(r)R_{\mu\nu}]=5$, while the second term has mass dimension $[f(r)g_{\mu\nu}]=2$. Thus, this equation makes no sense from the start.
• I think I understand what you are saying but the way this works does confuse me. So the metric is unitless so the $r^2$ and $B(r)$ terms have to be unitless. Also I don't think $f(R)$ has to have mass units of $2$, just $2d$ where $d$ is some integer. I mean $R^3$ is valid right? So then $B''/B$ would have mass dimension 2 and $h''/h$ would have mass dimension $+2d-6$ mass ($h''$) - $2d$ ($1/h$), so -6? – user110503 Jul 4 '17 at 1:11
• Sorry should have been $-4$. I suppose the action should have the same units so $f(R)$ should always have mass dimensions of 2 as you said. I do think there are problems with their dimensions however. Equation 7 for example has $R$ on the left hand side which has mass dimension 2 and $h''/h$ which has mass dimension $-4$. – user110503 Jul 4 '17 at 10:24