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Simple Ricci scalar question
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George Hammond
Mar 16, 2015, 12:30:03 PM3/16/15
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Hi Physics:
A standard spherically symmetric closed universe is
described by the following well known metric:
ds^2 =
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
curvature.
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
be:
R = 1/a^2 ?
Thanks, George Hammond
A standard spherically symmetric closed universe is
described by the following well known metric:
ds^2 =
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
curvature.
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
be:
R = 1/a^2 ?
Thanks, George Hammond
George Hammond
Mar 16, 2015, 12:30:06 PM3/16/15
to
On Sun, 15 Mar 2015 15:12:18 -0400, George Hammond
<george_...@verizon.net> wrote:
NOTE: a-dot = a-double-dot = 0 in the case of interest:
<george_...@verizon.net> wrote:
NOTE: a-dot = a-double-dot = 0 in the case of interest:
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