On Oct 31, 9:07 am, Tom Roberts <tjroberts...@sbcglobal.net> wrote:
> Koobee Wublee wrote:

> [...] it is also what any COMPETENT person _COMPUTES_. That, of
> course, is why it is taught. <shrug>

A better explanation is that the teacher does not understand the
subject very well.

> No. The Riemann curvature tensor is utterly and completely independent
> of coordinates. As is the metric tensor. As is any other tensor. <shrug>

The Riemann curvature tensor is different if I use the rectangular
coordinate or the spherically symmetric polar coordinate.  Thus, it is
coordinate dependent.  You are utterly wrong.

> > The metric along cannot tell
> > how the spacetime is curved.
>
> Not true. Give the metric tensor on an open region of the manifold, one
> can COMPUTE the Riemann curvature tensor throughout that region.

Any curvature tensor is useless if you don't specify what coordinate
system you have.

> As I have said so often: you must LEARN the basics. Essentially
> everything you said in this post is wrong.

It appears to be wrong to you because of your incomplete understanding
of the curvature of space or spacetime.  You need to start with the
basics by learning from my posts here.

> Most especially the part about you being the only person "since
> Riemann" to understand such curvatures.

But that is the truth.  Just by looking at your postings here, they
fully demonstrate a lack of understanding in basic structure of
spacetime.  As I said, it is very lonely to be the only person to have
fully understood the concept of the curvature in space or spacetime
since Riemann.  <shrug>

> > [...] I choose to define
> > Total energy = Kinetic energy - Potential energy
>
> How many legs does a horse have if we call its tail a leg?
> -- FOUR -- calling a tail a leg does not make it one.

You are getting ridiculous.