Mar 16, 2015, 12:30:03 PM3/16/15
A standard spherically symmetric closed universe is
described by the following well known metric:
dt^2 -d r^2/(1-r^2)-r^2 (d(theta)^2+sin^2(theta)d(phi)^2)
This metric must have an unusually SIMPLE scalar Ricci
Is it simply 1/r^2 ?
Or am I being too simple minded?
Okay, second question. If I multiply the metric by a
"conformal factor" say a(t)^2, then would the ricci scalar
R = 1/a^2 ?
Thanks, George Hammond