Re: GR1916, available online.

On Dec 14, 5:20 pm, Eric Gisse <jowr.pi.nos...@xxxxxxxxxxxxxxxx>
wrote:
On Thu, 13 Dec 2007 22:35:22 -0800 (PST), Koobee Wublee

<koobee.wub...@xxxxxxxxx> wrote:
On Dec 13, 2:54 pm, Eric Gisse wrote:
On Thu, 13 Dec 2007, Koobee Wublee wrote:

Why is the Riemann curvature tensor zero? It is merely a man-made
gage. <shrug>

...because that means spacetime is flat.

Space in Newtonian physics can also be curved just as Riemann had
described it about 150 years ago.

Nope. Space is assumed to be Galilean in Newtonian physics.

"E"ric you're wrong, KW is right.
The developement of tensor analysis sprung
from survey problems, specifically how to
survey land boundary's on a spherical surface.
Get a survey of a large piece of land and
you'll likely find the lengths and angles
vary from a flat surface.

You are encouraged to open a classical mechanics textbook to verify
this statement.



Didn't you know the curvature
tensor describes curvature? Guess not.

The Riemann curvature tensor is derived through geodesic variation
where the covariant derivative involves the Christoffel symbols.

Nope. The Riemann curvature tensor is derived by parallel-transporting
a vector about a closed infintesimal loop.

You are encouraged to open a textbook on GR to verify this statement
as opposed to pulling facts out of your ass as you usually do.

However, there is more than one way to arrange the connection
coefficients that you do not necessarily end up with the Christoffel
symbols. The Riemann curvature tensor is the result of one such
arrangement of connection coefficients. Thus, it is merely a man-made
gage that cannot describes the curvature in only one circumstance and
not in general. <shrug>

It'd help if you actually knew how the curvature tensor was derived.

It'd also help if you knew what you were talking about when you try to
use words like "gauge".



which means a Cartesian
CS is globally applicable, IOW's the metrics
"g_uv" can be transformed to constants every-
where in all time.

This is not true. <shrug>

Actually, it is. The condition is that either the Riemann tensor is
zero or that you are looking at a region of spacetime sufficiently
small such that the first derivatives of the metric components
disappear.

I misunderstood what Mr. Tucker was saying.

That'd be a bad thing if Ken actually knew what he was talking about.
Fortunately, no damage was done.

Hey, I'm the fellow who introduced R^a_bcd=0
into this thread, and I'm interested in anyone
elses PoV about it, and also how to deal with
R^a_bcd =/=0, in spacetime.

In another words, this is what I have been saying all along. The
geometry must be invariant. Depending on the choice of coordinate
system, the metric can only describe this geometry using this
coordinate system. Thus, a non-identity matrix as the metric using
the Euclidean coordinate system would yield a curved space. On the
other hand, an identity matrix as the metric using a non-linear
coordinate system would also yield curved space. What I mean the
coordinate system is dq not q. <shrug>

Your confusion is hilarious. I have repeatedly shown that you have
claimed that you can introduce curvature simply by doing a coordinate
transformation. Then you claim that isn't what you said, and now you
are again saying it explicitly.

When are you going to PROVE anything you have written? Neither of the
things you have written are true - the two simple counterexamples are
flat space in spherical coordinates and a manifold with g_ij = 1 for
all i,j. Neither have curvature and both satisfy your awkwardly-worded
conditions.

Regards
Ken
....

.

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