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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

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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