On 9/4/2024 9:39 PM, 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.
If you have the book you can look up "metric" in the index. There are
multiple entries: 23, 32-45
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
Nope that's it.
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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