Flat => Curved Spacetime

Hi all...

The Riemann Christoffel tensor (R^a_bcd)
is zero when the metric tensors may be
transformed to constants, globally.

Mathematically that condition is,

R^a_bcd=0, Eq.(1).

As I understand it, Eq.(1) is generally
true in N-dimensional spaces. I'd also
describe that space as "flat" and as
"orthogonal".

In Eq.(CC4) herein,
http://physics.trak4.com/GR_Charge_Couple.pdf
is used an orthogonal quantity "X^2", in 4D
spacetime.
That quantity can be better defined.

Since covariant and contravariant components
can be set equal, in an orthogonal space,
it follows that we may specify a CS condition,

X^2 = x_u x_u = x^u x^u , Eq.(kst1),
{u=0,i , i=1,2,3}.

Expanding in space and time Eq.(kst1),

X^2 = x_i x_i + x_0 x_0 = x^i x^i + x^0 x^0,

needs to be solved.

To do that, let the measures of the spatial lengths
be determined by time (r = ct, per the definition of
the meter by ISU 1983),

x_i == x^0 , x^i == x_0 , Eq.(kst2)

be a solution, whereby the lengths "x_i" are measured
by the time "x^0" in the direction on the "x_i".

Note an expected symmetry of space and time
emerges from Eq.(kst2), in the summation over "i",

x_i x^i = x_0 x^0 , Eq.(kst3),

that provides the invariant,

S^2 = x_u x^u , Eq.(kst4).

that may differ from Eq.(kst1)'s "X^2".
(That S^2 appears in Eq.(CC4) in the ref'd brief).
===================
We now will show Eq(kst1)...Eq(kst4) hold even
in an nonorthogonal spacetime.
=================
We can *fracture* the "orthogonality" by lifting the
requirement of Eq.(1) so that,

R^a_bcd =/= 0, Eq.(1a).

((AE did this differently, he contracted Eq.(1) )).

To solve the *fracture* let,

x^0 = ct , x_0 =Ct , Eq.(kst5),

where c is the speed of light in a vacuum and
C is a different speed of light.
(Physically the "different" speed of light is a
characteristic of the effect of a gravitational field).

Algebraic substitutions of the values in Eq.(kst5)
into Eq.(kst1 to 4) find that system of Eqs holds.

However we have succeeded in "fracturing" the
orthogonality since the components, x^0 =/= x_0 .

Now we can find the difference in the invariants
S^2 from Eq.(kst4) and X^2 from Eq.(kst1), and
return to Eq.(CC4), ie,

S^2 = X^2 + a.b Eq.(CC4),

where the scalar product "a.b" is invariant, and more
generally, that can be re-written as

S^2 = X^2 + h

with "h" being a general invariant.
===================
Employing Eq,(CC4) can set up a couple of vectors,

S(a) = X + a , S(b) = X + b ,

with scalar products,

a.X=0 , b.X =0 , a.b = ab , to find

S^2 = S(a).S(b) = X^2 + ab .

The above shows a means to derive General Relativity
by a straight-forward algebraic procedure.
Regards
Ken S. Tucker
.