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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