Can a conformal flat metric be curved?

[Kurvature]

TL;DR

If the flat (Cartesian) metric is multiplied by a time dependent scale factor could it yield a curvature?

5/18/22
I am an MS in physics.

I need to find out if the following CONFORMAL
METRIC produces zero or nonzero curvature?

I suspect the curvature is zero, but others
have said it's probably not? MAXIMA
sometimes says it is, and other times produces
a Ricci scalar that looks like the FRW scalar
curvature ?

Does someone know the answer to this question
off the top of their head?

If not, could someone possibly plug this metric
into Mathematica and tell me if Ricci scalar
is actually zero or if not, what it actually is ?

|a 0 0 0 |
|0 a 0 0 | = the given spacetime metric
|0 0 a 0 |
|0 0 0 -a|

where: a = a(t) = "the scale factor" is a
simple well behaved function of time.

Thanks in advance, absolutely desperate,
Kurvature

 

[Demystifier]

Penrose diagrams use a conformal transformation to map a curved metric into a flat Minkowski one (often with non-Minkowskian global topology). Hence the inverse conformal transformation maps a flat metric into a curved one. So the answer is - yes.

 

[ergospherical]

The curvature of $g_{ab} = \Omega(x)^2 \eta_{ab}$ is
\begin{align*}
R_{ab} = \frac{1}{\Omega^2}(4\partial_a \Omega \partial_b \Omega - 2\partial_a \partial_b \Omega -(|\nabla \Omega|^2 + \Omega \Delta \Omega)\eta_{ab})
\end{align*}where $\Delta = \partial^a \partial_a$ and $|\nabla \Omega|^2 = \partial^a \Omega \partial_a \Omega$. Such transformations of flat spacetimes have many interesting applications. For example, a particle of charge $q$ coupling to a scalar field $\Phi(x)$ in flat spacetime, as per $q \partial_a \Phi = u^b \nabla_b \left[ (m + q\Phi) u_a \right]$, is equivalently described by a curved spacetime with metric $g_{ab} = (m + q\Phi)^2 \eta_{ab}$ with the curvature generated by some hypothetical matter field whose properties can be explored.

 

[haushofer]

Why do you suspect the curvature to be zero?

 

[martinbn]

If you are looking at $ds^2 = a^2(t)(-dt^2+dx^2+dy^2+dz^2)=-a^2(t)dt^2+a^2(t)dx^2+a^2(t)dy^2+a^2(t)dz^2$
you can change the variables $t'=\int a(t)dt$, the others remain the same, then you get $-dt'^2+a^2(t')dx^2+a^2(t')dy^2+a^2(t')dz^2$ the line element of a typical cosmological solution, which will not be flat unless $a=const$.

Also it is not that hard to caclulate curvature in terms of $a(t)$. Even if you don't use anything that might simplify the calculations, it will not be so bad. All you need is one component of the Riemann tensor that is not zero, you don't have to compute all.

 

[PeterDonis]

MAXIMA
sometimes says it is, and other times produces
a Ricci scalar that looks like the FRW scalar
curvature ?

What specific input are you giving MAXIMA to get these different answers? The same input will always give the same output, so it doesn't make sense to say MAXIMA "sometimes" says one thing and sometimes says another for the same input.