I meant   ds^2 = -dt^2 + dx^2 + dy^2 + dz^2

On Friday, September 6, 2024 at 1:53:52 PM UTC-6 Brent Meeker wrote:

> You mean this one?
>
> [image: Schwarzschild Metric]
>
> Brent
>
>
>
>
> On 9/6/2024 5:34 AM, Alan Grayson wrote:
>
> I think we can use the usual metric in relativity, ds^2, with the minus 
> sign in from of dt^2. AG
>
> On Friday, September 6, 2024 at 6:26:14 AM UTC-6 smitra wrote:
>
>> Yes, but then a vector space without a norm. No inner product is 
>> defined, so from a pure math point of view, you are free to define any 
>> arbitrary inner product that satisfies the axioms for it. Alan the 
>> physicist will prefer that inner product that is dictated by the 
>> relevant physics, so this has to take the form of the indefinite Lorentz 
>> inner product in a locally free-falling coordinate system. This means 
>> that you must also consider the metric, so the distance between 
>> infinitesimally separated points. So, labels are arbitrary and hey are 
>> then meaningless, unless you also specify what the distance is between a 
>> point at [ct, x, y, z] and at [c(t+dt), x+dx, y+dy, z+dz]. 
>>
>> Saibal 
>>
>>
>>
>> On 06-09-2024 10:03, Alan Grayson wrote: 
>> > Do the 4-tuple labels on spacetime, (ct, x, y, z), form a vector 
>> > space? I was told on good authority that the answer is negative, but 
>> > now I have grave doubts of that conclusion. AG 
>> > 
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