# $R=0$ solution in field equations

I am dealing with some General Relativity extensions and I am not sure about my knowledge in basic GR since I am having some weird troubles with what I think are basic concepts.

As far as I know, if we have a field equation (we can think about this as, for example, EFE) in vacuum, $$R_{\mu\nu}-\dfrac{1}{2}g_{\mu\nu}R=0$$. In a more general sense, think about it as some equation which contains in the left hand side objects like $$R$$ or $$R_{\mu\nu}$$ and its derivatives, but not higher tensors as Riemann) we can see that the trace of the Einstein equation implies $$R=0$$, and plugging this into the original equation gives us $$R_{\mu\nu}=0$$. Now, is it correct to say that every solution of $$R_{\mu\nu}-\dfrac{1}{2}g_{\mu\nu}R=0$$ must have neccesairly $$R_{\mu\nu}=0$$? I think this is true.

Since the Schwarzschild metric has $$R=0$$, can we conclude that a field equation in the general form I described before verifies Schwarzschild as a solution if Einstein equation $$R_{\mu\nu}-\dfrac{1}{2}g_{\mu\nu}R=0$$ is verified?

I mean, suppose $$D(X,Y)$$ denotes some function that can contain derivatives of its arguments, and then we have the field equation $$D(R,R_{\mu\nu})+R_{\mu\nu}-\dfrac{1}{2}g_{\mu\nu}R=0$$. If Einstein is verified, $$R_{\mu\nu}-\dfrac{1}{2}g_{\mu\nu}R=0$$ and then $$D(R,R_{\mu\nu})=0$$, but since $$D(R,R_{\mu\nu})$$ can only have derivatives of $$R$$ and $$R_{\mu\nu}$$, and Einstein is verified so $$R=0$$ and $$R_{\mu\nu}=0$$, of course its derivatives will be zero and $$D(R,R_{\mu\nu})=0$$, so $$R=0$$ is a solution and since Schwarzschild has $$R=0$$ it must be a solution to this extended theory. Is my reasoning correct? Thanks

• Can you be more explicit about the extendeed theory you're studying? Also your question is kinda not clear. Commented Oct 16, 2020 at 19:39

$$R_{\mu\nu} - \cfrac{1}{2}g_{\mu\nu}R=0\Rightarrow R=0\rightarrow R_{\mu\nu}=0$$

The above statement holds for the absence of matter when one's dealing with Einstein Gravity.

Let's consider $$R^2$$ gravity:

$$RR_{\mu\nu} -\cfrac{1}{4}R^2g_{\mu\nu} + (g_{\mu\nu}\Box - \nabla_{\mu}\nabla_{\nu})R^2=0$$

$$R=0$$ is a solution. This does not imply that $$R_{\mu\nu}=0$$ somehow.

$$R=C\rightarrow R_{\mu\nu} =g_{\mu\nu}\cfrac{C}{4}$$ is also a solution.

Both these solutions satisfy the trace equation: $$\Box R=0$$

I hope this helps.

• Hello! Yes, that's what I am looking at. Suppose we have some field equation derived from the action $S=\dfrac{1}{16\pi G}\int d^4x \sqrt{-g} (R+\alpha R^2)$, which would be Einstein + the field equation you wrote. In the case Einstein holds, $R=0$ and $R_{\mu\nu}=0$, so the total field equation would be automatically satisfied, right? I mean, Schwarzschild ($R=0$) would be the trivial solution to this field equation? Commented Oct 16, 2020 at 20:15
• Yes, in Starobinsky gravity Schwarzchild is a solution, Commented Oct 16, 2020 at 20:18
• If we generalize to higher derivative theories like $f(R)$ or Gauss-Bonnet, this still holds? If Schwarschild is a solution of all this teories, I dont understand why people claims higher derivatives theories of gravity are singularity-free for black holes or Big Bang. Commented Oct 16, 2020 at 20:21
• In f(R) surely it does not hold if the f(R) function is not specified. Gauss Bonnet it does not hold either, the derivatives of Riemann and Ricci are still present. Commented Oct 16, 2020 at 20:23
• I don't understand, if we impose Einstein holds, then Ricci tensor and scalar vanish and then fourth derivative theory without Kretschmann scalar should work. Do you know then how can some $f(R)$ theories be singularity-free? Commented Oct 16, 2020 at 20:27