Re: GR1916 question about g=1?

On Nov 16, 12:43 pm, JanPB <film...@xxxxxxxxx> wrote:
On Nov 16, 12:28 pm, Koobee Wublee wrote:

First of all, you have to understand what g means here. In this case,
g actually refers to the determinant of the metric which is merely a
matrix itself. Because of the mathematical formulation of spacetime
itself, the determinant must always be negative. Want to think about
that one? Thus, it does not make any sense to talk about the
determinant of g being +1.

With the determinant of g being -1, it simplifies the Ricci tensor (a
mere 4-by-4 matrix) drastically. The determinant of the metric
associated with the ordinary spherically symmetric polar coordinate
system does not yield a determinant of -1. So, Schwarzschild quickly
transformed the original metric and original choice of coordinate
system to another metric and another set of coordinate (by keeping the
geometry invariant) to allow for the determinant of the new metric to
be -1. With the Ricci tensor drastically simplified by
Schwarzschild's trick, he was able to arrive at one out of infinite
number of static and spherically symmetric solutions to the field
equations. In Schwarzschild's original solution, the geometry
involved leaves no room for any black holes. Only with Hilbert's
solution towards late 1916 early 1917, that the now called
Schwarzschild metric manifests the black holes.

You write nonsense as usual. You've made too many errors to address
them individually in 5 minutes (which is all the time I have at this
moment :-) )

You are unbelievably ignorant. <shrug>

Start with "determinant of the metric" - there is no such
thing. (You confuse tensors with matrices, that's why you fall into
traps like this one.)

Without the determinant of g, the field equations can never be
derived. <shrug>

http://en.wikipedia.org/wiki/Einstein%E2%80%93Hilbert_action
.

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