A Can a conformal flat metric be curved?

  • A
  • Thread starter Thread starter Kurvature
  • Start date Start date
  • Tags Tags
    Flat Metric
Kurvature
Messages
22
Reaction score
0
TL;DR Summary
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
 
Physics news on Phys.org
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.
 
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.
 
  • Like
Likes dextercioby, Demystifier and vanhees71
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.
 
Kurvature said:
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.
 
In this video I can see a person walking around lines of curvature on a sphere with an arrow strapped to his waist. His task is to keep the arrow pointed in the same direction How does he do this ? Does he use a reference point like the stars? (that only move very slowly) If that is how he keeps the arrow pointing in the same direction, is that equivalent to saying that he orients the arrow wrt the 3d space that the sphere is embedded in? So ,although one refers to intrinsic curvature...
I started reading a National Geographic article related to the Big Bang. It starts these statements: Gazing up at the stars at night, it’s easy to imagine that space goes on forever. But cosmologists know that the universe actually has limits. First, their best models indicate that space and time had a beginning, a subatomic point called a singularity. This point of intense heat and density rapidly ballooned outward. My first reaction was that this is a layman's approximation to...
So, to calculate a proper time of a worldline in SR using an inertial frame is quite easy. But I struggled a bit using a "rotating frame metric" and now I'm not sure whether I'll do it right. Couls someone point me in the right direction? "What have you tried?" Well, trying to help truly absolute layppl with some variation of a "Circular Twin Paradox" not using an inertial frame of reference for whatevere reason. I thought it would be a bit of a challenge so I made a derivation or...

Similar threads

Replies
4
Views
2K
Replies
40
Views
5K
Replies
14
Views
5K
Replies
7
Views
2K
Replies
13
Views
3K
Replies
11
Views
2K
Replies
2
Views
2K
Back
Top