Is the Space with Ricci Scalar Zero Flat?

[Apashanka]

Given metric ds²=dr²-r²dθ²
Gamma comes as Γ¹₂₂=r,Γ²₁₂=Γ²₂₁=1/r
The Reimann tensor comes as
R₁₁=R²₁₂₁=∂₁Γ²₁₂-Γ^m₁₂Γ²_1m=0,only non zero terms .
Similary R₂₂=R¹₂₁₂=∂₁Γ¹₂₂-Γ^m₂₁Γ¹_m2=0,only non zero terms.
Therefore R(ricci scaler)=0
Is the space flat??

 

[Ibix]

All of the terms of the Riemann tensor are zero so yes, this spacetime (it's got a Lorentzian signature) is flat. Presumably it's just a funny set of coordinates on Minkowski spacetime.

Note that the Ricci scalar being zero is not enough to determine whether or not the spacetime is flat. It's always zero in vacuum, whether the spacetime is flat or not.

 

[phyzguy]

For what it's worth, it is Riemann, not Reimann, and scalar, not scaler.
As to your question, the Ricci scalar will be zero anywhere the stress-energy tensor is zero. For example the space outside a black hole event horizon has a zero Ricci scalar. Is this space flat? (Ibix beat me to it).

 

[Orodruin]

Note that your line element is just the line element for Rindler coordinates on Minkowski space in 1+1 dimensions. Minkowski space is, as should be expected, flat regardless of what coordinates you happen to use.

 

[Ibix]

Rindler coordinates

Knew I should have recognised it...

 

[Apashanka]

Note that your line element is just the line element for Rindler coordinates on Minkowski space in 1+1 dimensions. Minkowski space is, as should be expected, flat regardless of what coordinates you happen to use.

But every flat space doesn't have R to be 0 ,as expected instead of -ve sign in between if +ve sign is there (which reduces to 2-D polar flat) then R is not zero,isn't it?!

 

[Ibix]

But every flat space doesn't have R to be 0 ,as expected instead of -ve sign in between if +ve sign is there (which reduces to 2-D polar flat) then R is not zero,isn't it?!

The definition of a flat space is one for which all components of the Riemann tensor are zero. Thus the Ricci scalar is zero.

I can't follow what you are saying exactly. I think you are claiming that some elements of the Riemann tensor are non-zero in flat space in polar coordinates. This is not correct.

 

[Apashanka]

The definition of a flat space is one for which all components of the Riemann tensor are zero. Thus the Ricci scalar is zero.

I can't follow what you are saying exactly. I think you are claiming that some elements of the Riemann tensor are non-zero in flat space in polar coordinates. This is not correct.

I was misunderstood ,
I got it now

 

[sweet springs]

Given metric ds2=dr2-r2dθ2

An usulal thus plain metric is [tex]ds^2=dr^2+r^2d\theta^2[/tex]. Your case has different sign
[tex]ds^2=dr^2-r^2d\theta^2[/tex].
Would you show me what kind of space your geometry says about?

 

[Ibix]

Would you show me what kind of space your geometry says about?

As noted by @Orodruin in #4, it's Rindler coordinates on flat Minkowski spacetime.

 

[DrGreg]

As noted by @Orodruin in #4, it's Rindler coordinates on flat Minkowski spacetime.

Indeed it is, but an unusual choice of letters. The Rindler metric would usually be written as $$ds^2 = dx^2 - x^2 dt^2$$

 

[sweet springs]

Now I know theta is time. Wrongly I interpreted it angle in xy plane. Thanks.

 

[Orodruin]

Indeed it is, but an unusual choice of letters. The Rindler metric would usually be written as $$ds^2 = dx^2 - x^2 dt^2$$

Alternatively, with the opposite sign convention, it is the Milne model coordinates on the future light-cone of a point in 1+1D Minkowski space:
$$
ds^2 = dt^2 - t^2 dx^2
$$

 

[DrGreg]

Now I know theta is time. Wrongly I interpreted it angle in xy plane. Thanks.

Well, it is "hyperbolic angle" in the XT plane where ##X = x \cosh t## and ##T = x \sinh t## (under the usual convention ##c = 1##).

 

[sweet springs]

Yea, the angle starts from $$-\infty$$ to $$+\infty$$, not from 0 to $$2\pi$$.

 

[Apashanka]

Can anyone please provide any information how is the Ricci tensor arises due to curvature of space??

 

[PeterDonis]

how is the Ricci tensor arises due to curvature of space?

The Ricci tensor doesn't "arise due to" the curvature of spacetime. It describes the curvature of spacetime (more precisely, the part of the curvature of spacetime that is due to the presence of stress-energy at the location where the tensor is evaluated).

Also, as in what I just wrote, it's the curvature of spacetime, not space.