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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