🚀 go-pugleaf

Dave, although you did not want to say it directly,
you have chosen to reject the idea that the global
metric of spacetime can be approximated by a
Friedmann metric. So when using G(g) = T in cosmology
-whatever the udealization for T- you tell us we are
no longer allowed to assume g Friedmann.
(You haggle about _exactly_ Friedmann. That is irrelevant;
if g is approximatley Friedmann, then G(g) is approximately
G(friedmann). If not, then explain why GR uses so
confidently G(Friedmann) = T(perfect fluid).)

Now that you have rejected this fundamental assumption
of GR cosmology, could you tell me how GR copes with
this loss? It just so happens that nearly all numerical
assertions about global scale are expressed in terms of
Friedmann metrics. Especially the `deceleration parameter q_0'
whose value is the central issue of our argument (I say
it is zero, you say it is not) is defined in terms of
Friedmann metrics.

In answer to your last question, few matter distributions
produce a pumpkin cosmology, especially if the pumpkin
is parabolic and fanning out like a flower. I would
say the only matter distribution resembling a pumpkin
is a pumpkin itself.
Through our correspondence I have learned how to prove
such assertions: because I'm right!

Lest the reader thinks that MTW has dealt with this
subject seriously, consider page 739 for a good laugh.