# I $f(R)$ gravity field equation derivation mistake?

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1. Jul 4, 2017

### Dazed&Confused

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 $$f'(R) R_{\mu \nu} - \tfrac12 f(R) g_{\mu \nu} + g_{\mu \nu} \Box f'(R) - \nabla_\mu \nabla_\nu f'(R) = 0$$
is rewrite it as $$g'(r) R_{\mu \nu} - \tfrac12 g(r) g_{\mu \nu} + g_{\mu \nu} \Box g'(r) - \nabla_\mu \nabla_\nu g'(r) = 0$$
rather than $$\frac{g'(r)}{R'(r)}R_{\mu \nu} - \tfrac12 g(r) g_{\mu \nu} + g_{\mu \nu} \Box \left (\frac{g'(r)}{R'(r)} \right) - \nabla_\mu \nabla_\nu \left (\frac{g'(r)}{R'(r)} \right) = 0.$$

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.

2. Jul 4, 2017

### Staff: Mentor

I don't see what you are describing anywhere in the paper. Can you be more specific about which equations in the paper you are asking about?

3. Jul 4, 2017

### Dazed&Confused

So to obtain equations 4 and 7 (or something close to them) on Mathematica I had to use use the second form of the equation I gave. Now, this could be because I implemented it incorrectly on Mathematica, however when I used $R$ and not $r$ my equations were far more complicated as I said, so I'm not sure how a change in the code could result in such a huge difference. And as I have said I have checked it. I'm pretty sure that the calculation of the Ricci tensor, scalar, the Christoffel symbols and so on is correct as I have tested this with the Schwartzchild solution. Therefore it is only likely to be in the field equation (which I have looked over many times).

4. Jul 4, 2017

### Staff: Mentor

The paper's description of how these equations are derived looks sloppy to me, so I'm not sure how to duplicate it. It says equation 4 is a "combination" of two Ricci tensor components, but it's not clear whether it means the two corresponding components of the field equation, which is what would be correct. Similar remarks apply to equation 5. It says equation 6 is the $rr$ component of the field equation, i.e., of equation 2, but that component was already used in equation 4 so it's not clear what additional information equation 6 is supposed to contain. Equation 7 is just the contraction of equation 2, which should be straightforward, but the notation is unclear to me, in particular the primes--are they derivatives with respect to $r$, with respect to $R$, or something else?

This paper does not appear to have been peer-reviewed, so I'm not sure how reliable it is in any case.

5. Jul 4, 2017

### Dazed&Confused

I am able to obtain the contraction. Considering how many terms there are I really do not believe any of this is a coincidence. 4 and 5 are the same except a few terms have the wrong sign. I think equation 6 had an extra term for me.

Now the Ricci scalar I got was $$\frac{A'(r) B'(r)}{2 A(r)^2 B(r)}+\frac{2 A'(r)}{r A(r)^2}-\frac{B''(r)}{A(r) B(r)}+\frac{B'(r)^2}{2 A(r) B(r)^2}-\frac{2 B'(r)}{r A(r) B(r)}-\frac{2}{r^2 A(r)}+\frac{2}{r^2}$$

so I think that differentiating $f$ of this twice with product and chain rules should give many more terms than they find.

6. Jul 4, 2017

### Staff: Mentor

Where are the $e^{- \lambda(r)}$ factors coming from?

7. Jul 4, 2017

### Dazed&Confused

Sorry I used the code for something else and forgot to change the way the metric was written. I have corrected it.

8. Jul 4, 2017

### Staff: Mentor

I don't think the paper is doing this. I think it is just computing the Ricci tensor and writing down field equation components (plus contraction, which doesn't involve taking any extra derivatives).

9. Jul 4, 2017

### Dazed&Confused

But don't you have to do this in the right hand side of their equation 2? And their components had $f'''$ and $f''$ terms; wouldn't they have to come from this differentiation?

10. Jul 4, 2017

### Staff: Mentor

Not to just write down its components or to contract it, no. Those are the only things that the paper appears to be doing. (But see below.)

Do you mean the $h'$ and $h''$ terms? I think those come from contracting the $h_{; \mu \nu}$ term (and from the $h_{; \lambda}{}^{\lambda}$ term, which is already contracted), and then noting that everything is a function of $r$ only, so the only component of the contraction that will be nonzero is $h_{; r}{}^{r}$. (Note that I'm not entirely sure this reasoning is correct; the only nonzero partial derivatives will be with respect to $r$, but that in itself does not guarantee that the only nonzero covariant derivatives will be with respect to $r$; the latter will depend on which connection coefficients are nonzero.) In other words, it's just using the differentiation that already appears in the field equation (2), not putting in any additional differentiation. (It's possible that this is what you meant by "don't you have to do this..." in the first quote above.)

11. Jul 4, 2017

### Dazed&Confused

I don't think I understand what you mean. So the contracted equation is $$f'(R) R - 2 f(R) +3 \Box f'(R)= 0$$
which when using $f(r)$ not $f(R)$ in my code I get their equation 7.

I mean to write the components don't you need to do $\nabla_\mu \nabla_r f'(R) = \nabla_\mu ( f''(R) \partial_r R )$ and so on? So what I think they have done is written this as $\nabla_\mu \partial_r f'(r)$ (ignoring non $r$ partial derivatives).

12. Jul 4, 2017

### Staff: Mentor

I don't have time to check the details of the math. The best I can say is, as I said before, this paper does not appear to be peer-reviewed so I don't know how reliable it is.

13. Jul 6, 2017

### Dazed&Confused

Thank you for your help. I've found similar papers and although I don't know if the authors in this paper did this (my equations still don't completely match), there are ways to get the method to work. If you use
$$f'(R) R - 2 f(R) +3 \Box f'(R)= 0$$
to replace the $f(R)$ term with so that all the $f$s are differentiated, then you can properly consider it a function of $r$ and everything works out.