Physics Asked by Axionlike particles on December 14, 2020
It is well known that Einstein Theory of General Relativity has pathological solutions, called black holes, where the theory fails at some point and gives us infinite values.
In order to solve this problems, there exists UV extensions of General Relativity. These kind of theories try to solve the singularity problem, and some of them are actually singularity-free.
Let’s consider the fourth order gravity, with lagrangian density $R + a R^2 – b R_{munu}R^{munu}$. The field equations in this theory become (in vacuum):
begin{align}
nonumber
& R_{munu}- dfrac{1}{2}g_{munu}R+a left( 2RR_{munu} – dfrac{1}{2}g_{munu}R^2
–
2nabla_mu nabla_nu R
+
2g_{munu} square R right)
& – bleft(
– dfrac{1}{2}g_{munu}R_{alphabeta}R^{alphabeta} – nabla_mu nabla_nu R
+
Box R_{munu}
+
dfrac{1}{2} g_{munu} square R
-2 R_{alphamunubeta}R^{alphabeta}
right) = 0
end{align}
How can we find black hole solutions in this theory? If there are no black hole solutions, can we find at least horizon-like solutions? Since this theory in the limit becomes Einstein GR, we could be able to see how black holes appear from this theory in the limit where we recover GR.
Are there any values $a$ and $b$ such that if we have an Einstein GR solution, this solution is also a solution of the higher derivative field equation above?
One imposes a metric ansatz let's say in four dimensions something like: $$ds^2 = -a(r)dt^2 + b(r)dr^2 + r^2 dOmega^2 $$
Now the tensorial equation you write will become a set of fourth order differential equations. You have to integrate through the equations which seems to be a rather impossible task. Maybe for the specific gauge: $b(r) = 1/a(r)$ one may be lucky, but i personally think it is still pretty hard to obtain a solution. Of course the obtained solution (if exists) should for $alpha = beta =0$ match Schwarzchild Black Hole.
Answered by ApolloRa on December 14, 2020
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