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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