On May 16, 5:45 am, PD wrote:
On May 16, 1:24 am, Koobee Wublee wrote:

You are making the same mistake again. What you refer to is the
geometry itself not the metric.

Then you are suffering from a basic misalignment on terminology. The
metric *is* the geometry. That's the point.

Consider the Schwarzschild spacetime as described below.

ds^2 = c^2 (1 – 2 U) dt^2 – dr^ / (1 – 2 U) – r^2 dO^2

Where

** U = G M / c^2 / r

** The geometry is ds^2.

** The coordinate is (t, r, O) or (t, r, longitude, latitude).

** The metric is

[c^2 (1 – 2 U), 0, 0, 0]
[0, 1 / (1 – 2 U), 0, 0]
[0, 0, r^2 cos^2(latitude), 0]
[0, 0, 0, r^2]

According to the mathematics of the field equations and thus GR, the
metric is not the geometry, and this is very obvious as explained
above.

If you cannot understand this, you are not fit to teach. If you
understand this, your claim is an utter lie. That would make it
(LYING IS TEACHING) on your part. shrug

Either way, you are not fit to teach. How can I be so point blank?

The equations above represent the
same geometry, yes. They are equivalent. However, the coordinates
are different, and the metrics are different. The metric cannot
adequately describe the geometry despite your voodoo conjectures of
dot products, and the coordinates itself cannot adequately describe
the geometry. It takes both well specified coordinate systems and the
metrics to fully describe the geometries. shrug

We cannot go on without you understand my point of view, and I have
understood yours and pointed the errors in your logic. If you are not
malicious as Eric Gisse is, you need to understand my point of view.
shrug