Boston_Guy said:
It would be a theory with a modified Hilbert-Einstein action, where the density is not proportiomal to the Ricci scalar
R, but to some non-linear function of it.
The modified field equations will contain the Ricci curvature in a non-linear way, and might disallow the possibility of a black hole.
By dimensional analysis, one can construct a dimensionless quantity from
c, \hbar,
G, and
R:
<br />
\frac{G \, R}{\hbar \, c^{3}}<br />
Also, we can construct a combination with the dimension of an action density:
<br />
\left[ \frac{S}{x^4} \right] = \mathrm{T}^{-1} \, \mathrm{L}^{-2} \, \mathrm{M} \stackrel{\mathrm{n.u.}}{\rightarrow} \mathrm{L}^{-4}<br />
Because \left[ G \right] = \mathrm{T}^{-2} \, \mathrm{L}^{3} \, \mathrm{M}^{-1} \stackrel{\mathrm{n.u.}}{\rightarrow} \mathrm{L}^{2}, we ought to have \mathcal{L} \propto G^{-2}. Returning the proper powers of \hbar, and
c, we get:
<br />
\mathcal{L} \propto \frac{c^6}{\hbar \, G^2}<br />
Thus, we can write:
<br />
S_{g} = \frac{c^6}{\hbar G^2} \, \int{f\left( \frac{\hbar \, G \, R}{c^3} \right) \, \sqrt{-g} \, d^4 x}<br />
If you require the Planck's constant to drop out of the expression, you ought to have:
<br />
f(x) = A \, x<br />
i.e. the Lagrangian density is proportional to the scalar curvature. This is the ordinary Hilbert-Einstein action. The numerical constant
A is fixed by Newton's Law of universal Gravitation.
We might be tempted to choose a different functional form. For example, let us see if we can get
G, the Universal Gravitational constant, to drop out from the expression. In a sense, this would make the action purely quantum effect. It is easy to see that this happens if:
<br />
f_{2} = B \, x^2<br />
A quadratic function dominates a linear function for large values of the argument, i.e. for:
<br />
R \gg \frac{A}{B} \, L^{-2}_{P}<br />
where L_{P} = \sqrt{G \, \hbar/c^{3}} \sim 10^{-35} \, \mathrm{m} is the Planck length.
There is, however, a problem with choosing a non-linear function. Namely, R contains second derivatives of the metric tensor, but as a 4-gradient. Therefore, they get integrated out when the function is linear. But, one cannot do the same for a nonlinear function, and the equations that we would get would contain derivatives higher than two of the metric tensor.