Ever use cylindrical coordinates?
Brent
On 9/5/2024 2:07 PM, Alan Grayson wrote:
Since the metric tensor is defined on the*flat *tangent space, will
its matrix representation always be diagonal? TY, AG
On Thursday, September 5, 2024 at 12:35:03 PM UTC-6 Brent Meeker wrote:
If you change coordinate systems the vectors and tensors
transform in such a way that the physics is unchanged. That's the
defining property of vectors and tensors and why they are not just
the arrays used to represent them.
Brent
On 9/4/2024 10:42 PM, Alan Grayson wrote:
How is the foregoing consistent with the statement that tensors
are independent of coordinate systems? TY, AG
On Wednesday, September 4, 2024 at 10:39:33 PM UTC-6 Alan Grayson
wrote:
No; you never posted any page numbers in your reference. Had
you done that, I would have immediately studied your specific
reference. But I was about to post something like what you
wrote; namely, that *solving* for the metric tensor *MEANS*
solving for the 16 components defining its function, given a
stress-energy tensor and a coordinate system. The form of the
components must depend on the coordinates. And given a
coordinate system, the components will vary depending on
location in spacetime, and this is what's meant by the
"metric tensor field". I thought the "field" refers to a
*unique* *real numbe*r at each point in spacetime, but it
must refer to the components of the matrix representing the
tensor. Wiki's definition seems misleading since it states
that the metric tensor is a bilinear function of two vectors
on the tangent plane. Those vectors are its arguments, but
sort-of misleading. Is there anything in the foregoing that I
got wrong? AG
On Wednesday, September 4, 2024 at 8:09:07 PM UTC-6 Brent
Meeker wrote:
The metric field is the set of metric tensors, one at
each point. It's not some vector lengths.
When are you gonna read "Relativity DeMystifie". I told
you the page numbers and you can look up more in the index.
Brent
On 9/4/2024 5:59 PM, Alan Grayson wrote:
But my point is therefore that the metric tensor field
is ambiguous !
On Wednesday, September 4, 2024 at 6:40:38 PM UTC-6
Brent Meeker wrote:
It explains this "...and yields *different *real
values for most different pairs."
Brent
On 9/4/2024 1:03 PM, Alan Grayson wrote:
and yields *different *real values for most
different pairs.
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