This is a request for references, mainly for educational aims. In textbooks about general relativity, it is common to present the Riemann and Ricci tensors using the Christoffel symbols. This is easy to understand because it is a straightforward way to perform practical computations and the formulas one obtains are elegant and easy to grasp. Besides, Christoffel symbols are given through the metric and one can do some algebra to get such kind of expressions. But for my aims, I would need references, rather than do the computations, reporting formulas giving the Ricci tensor using the metric explicitly. Also research papers are fine. Can anybody help?
[Physics] Ricci tensor given through the metric
curvaturedifferential-geometrygeneral-relativityresource-recommendationstensor-calculus
Related Solutions
As Prahar says, you just calculate the Ricci scalar $R'$ following the usual definition – the double contraction of the Riemann tensor – which is calculated from the Christoffel symbols and its derivatives where the metric $\gamma'$ expressed as $\exp(2\omega)\gamma$ is substituted everywhere. It will probably not fit one page.
However, there's a faster indirect way to establish the formula.
First, calculate $R'$ for a constant i.e. $(\tau,\sigma)$-independent $\omega$. The distances calculated by the $\gamma'$ metric are just $\exp(\omega)$ times longer than those calculated from $\gamma$. So the Ricci scalar calculated from $\gamma'$, equal to $C/a^2$ where $C$ is a constant and $a$ is some curvature radius, is $\exp(-2\omega)$ times the Ricci scalar calculated from $\gamma$. The only way how to get a formula for $\exp(-2\omega)$ in this simple rescaling by using $\gamma,\gamma'$ only is $$ \exp(-2\omega) = \frac{\sqrt{-\gamma}}{\sqrt{-\gamma'}} $$ Therefore we have $$ R' = R \exp(-2\omega) = R \frac{\sqrt{-\gamma}}{\sqrt{-\gamma'}} $$ for a constant $\omega$. That agrees with your formula in your special case. Just to be sure, if we wrote $\exp(-2\omega)$ to the formula explicitly, instead of the ratio of the two square roots of determinants, it would still be valid because for the two metrics related by the rescaling, the ratio of the two determinants is given by the power of $\exp(\omega)$.
Now, consider a variable $\omega$. It's clear that $R$, the Ricci scalar, only contains up to second derivatives of $\omega$, so the expression for $R'$ can only depend on $\omega$ and the first and second derivatives of it. The dependence on $\omega$ itself has already been clarified because it's fully determined by the case of the constant $\omega$.
Now, the most general relationship for a variable $\omega$ may depend on the derivatives through $\nabla\omega\cdot \nabla \omega$ and $\nabla^2\omega$ because these are the only two covariant expressions one may construct out of the first two derivatives of $\omega$. And the relationship between $R,R'$ has to be given by a nice covariant (tensor structure respecting) formula because the relation between $\gamma,\gamma'$ is also expressed by a nice covariant (tensor structure respecting) formula.
The dependence on $\nabla\omega\cdot \nabla\omega$ actually cannot be there because it is an even function(al) of $\omega$. So you can't say that $R'$ is increased by a multiple of this expression relatively to $R$ because the opposite relationship should give a decrease but this $\nabla\omega\cdot \nabla\omega$ has the same sign in both ways.
That's why the formula only has to be modified by a multiple $\nabla^2\omega$ inserted somewhere. It's really straightforward to convince yourself that up to the ratio of the (square roots of the) determinants, the relationship has to be simply given by $\dots R-2\nabla^2\omega$. For example, you may write down the metric for the sphere in a conformally flat way and verify that if $R=0$ for a flat $\gamma$, $R'=2/a^2$, the desired Ricci scalar for a sphere of radius $a$, may be calculated from the covariant Laplacian of $\omega$.
The simplicity of geometry in lower dimensions is because the Riemann curvature tensor could be expressed in terms of simpler tensor object: scalar curvature and metric (in 2d) or Ricci tensor and metric (in 3d). That fact, of course, does not alter the possibility to write Riemann tensor (as well as Ricci tensor and scalar curvature) as a combination of metric derivatives. But each term in such a combination is not a tensor - only the whole object.
Now, let us recapture, why in lower dimensions we are able to reduce the Riemann tensor to a combination of lower rank tensor objects.
For 3d case, definition of Ricci tensor:
$$ R_{ab} = R_{abcd}g^{ac}$$
contains 6 independent components, exactly the number of independent components in Riemann tensor. So, this equation could be reversed, thus expressing $R_{abcd}$ in terms of $R_{ab}$ and $g_{ab}$.
In 2d case we could similarly start with definition of Ricci scalar:
$$ R = R_{ab} g^{ab} ,$$
and reverse it expressing $R_{ab}$ through $g_{ab}$ and $R$. The next step would be to express Riemann tensor with $g_{ab}$ and $R_{ab}$ (and thus through scalar $R$ only).
Best Answer
As mentioned in the comments calculating algebraic expressions for the Ricci tensor containing the metric, its inverse and its first and second derivatives is straight forward using computer algebra.
The most arbitrary metric $$g_{\alpha\beta}=g_{\beta\alpha}$$ has 10 independent components which are functions of four coordinates $\left\{x_0,x_1,x_2,x_3\right\}$: $$g_{\alpha\beta}(x_0,x_1,x_2,x_3).$$ The metric has an inverse with 10 independent components $$g^{\alpha\beta}=g^{\beta\alpha}.$$
The metric has 40 independent first partial derivatives $$g_{\alpha\beta,\gamma}$$ and 100 independent second partial derivatives (100 and not 160 because of the the symmetry of second derivatives) $$g_{\alpha\beta,\gamma\delta}=g_{\alpha\beta,\delta\gamma}.$$
Using those ingredients ($g_{\alpha\beta}$, $g^{\alpha\beta}$, $g_{\alpha\beta,\gamma}$ and $g_{\alpha\beta,\gamma\delta}$) one can compute 21 components of the Riemann tensor $R_{\alpha\beta\gamma\delta}$. One could eliminate one of those 21 components using the first Bianchi identity.
Just to give one example in this post: $R_{0102}$ has 1510 terms: 4 second derivatives and the rest are contractions of Christoffel symbols:
The Ricci tensor can be constructed from the contraction $$R_{\alpha\beta}=R^\mu_{\,\,\alpha\mu\beta}$$ so it contains the components of the inverse metric an those 21 Riemann tensors:
Writing $R_{\alpha\beta}$ out in terms of $g$ only becomes super messy in case of $R_{01}$ we are talking about 8711 terms. I have no idea how to visualize such an expression here at SE. I have uploaded a PDF (careful it is rather large) of $R_{01}$ here.
I also uploaded .m files containing all 10 independent components of $R_{\alpha\beta}$ Rij.m and the 21 components of $R_{\alpha\beta\gamma\delta}$ Rijkl.m.
As pointed out in the comment of the original question those expressions have only very very limited use. But maybe some conclusions: