- #1
Giammy85
- 19
- 0
A two-dimensional Rienmannian manifold has a metric given by
ds^2=e^f dr^2 + r^2 dTHETA^2
where f=f(r) is a function of the coordinate r
Eventually I calculated that Ricci scalar is R=-1/r* d(e^-f)/dr
if e^-f=1-r^2 what is this surface?
In this case R comes to be equal to 2
I've read on wikipedia that Ricci scalar of a sphere with radius r is equal to 2/r^2
So, is this surface a sphere of radius r=1?
ds^2=e^f dr^2 + r^2 dTHETA^2
where f=f(r) is a function of the coordinate r
Eventually I calculated that Ricci scalar is R=-1/r* d(e^-f)/dr
if e^-f=1-r^2 what is this surface?
In this case R comes to be equal to 2
I've read on wikipedia that Ricci scalar of a sphere with radius r is equal to 2/r^2
So, is this surface a sphere of radius r=1?
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