"Now ..just LET go" Theory. < Re: What is LET? >:
Hexenmeister wrote: >"Tom Roberts" <> >Bilge wrote:
As soon as you admit absolute simultaneity, even in principle,
you have a galilean invariant theory.
http://www.androcles01.pwp.blueyonder.co.uk/STM/Scoundrels.htm#Bilge
Androcles.
$$ ^.
No more GR.
$$ [ GR's OUT of liNE. ]
GR, a WORLD=point (at an END of it's WORLD-line) ..having DECLARED;
.. [[[ "..NO _PRiOR_ GEOMETRY.!!" ]]] ..
.. [[ "GR .._NOW_, has _NO_ WORLD-line." ]]
.. [[ So, "It's ALL OVER _NOW_ ..just LET go." ]]. ..
Tom (between error-bars) means "simultaneity" _BETWEEN error-bars_
[[..but he ABSOLUTELY _NEVER_ means "vacuum" _BETWEEN error-bars_]].
NOW ..of course, THERE isN'T MUCH LEFT of GR except the DisGRUNTLEs.
It's OVER (finished). ```Brian.
Tom Roberts wrote: > > sal wrote:
"scalar" is often -- perhaps usually -- taken to mean "scalar field on the
manifold" and hence something which is also "invariant", or independent
of choice of coordinates.
Yes. A real-valued field on the manifold (occasionally a complex value
is used).
Time's not, so time's not a scalar, either.
Well, be careful. The time coordinate of a specified observer _is_ a
scalar field on the manifold. Specifically: for observer A, observer A's
coordinates apply a real value t_A(P) to every point P of a given region
of the manifold, and is therefore a scalar field on that region; all
observers will agree that A's time coordinate has the value t_A(P) at
each point P in that region (of course different observers will
naturally use different time coordinates; that's irrelevant).
This has been called an "observer-dependent invariant",
which about sums it up.
In the abstract, "time" without specifying an observer has no meaning in
relativity (or at least no _observable_ meaning).
Tom Roberts
Re: Why TWLS= 0 <> OWLS in any ONE Frame.
Re: GR's OUT of LiNE.
.
