Re: GR1916 question about g=1?

On Nov 22, 1:41 pm, Eric Gisse wrote:
On Thu, 22 Nov 2007, Koobee Wublee wrote:

[dq^2] is a 4-by-4 matrix with (dq^i dq^j) being the elements. How in
the world can you possibly construe [dq^2] as a 4-by-1 matrix?

[Learning deficient remarks snipped]

Write the elements of [dq^2].

For those who have learning deficiency, [dq^2] means [(dq)(dq)] as
exactly defined by the following matrix.

[dq^0 dq^0, dq^0 dq^1, dq^0 dq^2, dq^0 dq^3]
[dq^2] = [dq^1 dq^0, dq^1 dq^1, dq^1 dq^2, dq^1 dq^3]
[dq^2 dq^0, dq^2 dq^1, dq^2 dq^2, dq^2 dq^3]
[dq^3 dq^0, dq^3 dq^1, dq^3 dq^2, dq^3 dq^3]

Or a better way to write the above matrix without much confusion of
the indices is the following.

[dq_0 dq_0, dq_0 dq_1, dq_0 dq_2, dq_0 dq_3]
[dq^2] = [dq_1 dq_0, dq_1 dq_1, dq_1 dq_2, dq_1 dq_3]
[dq_2 dq_0, dq_2 dq_1, dq_2 dq_2, dq_2 dq_3]
[dq_3 dq_0, dq_3 dq_1, dq_3 dq_2, dq_3 dq_3]

It follows that [g] is described as follows.

[g_00, g_01, g_02, g_33]
[g] = [g_10, g_11, g_12, g_33]
[g_20, g_21, g_22, g_33]
[g_30, g_31, g_32, g_33]

Thus,

ds^2 = [g] * [dq^2] = g_00 dq_0 dq_0 + g_01 dq_0 dq_1 + ...

Where

** [] * [] = Dot product of two matrices

You can actually transpose either of the matrix, but the result is
always the same because of the symmetry in both [g] and [dq^2].

It is very clear that geometry (ds^2) is the metric [g] times the
coordinate [dq^2]. This very convincingly indicates the metric is
merely a matrix --- no mumble jumble divinity in special properties.
.

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