You mean this one?

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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