sci.physics.relativity

Koobee Wublee says...
>
>On Oct 9, 7:45 am, (Daryl McCullough) wrote:
>> Why should anyone take what you have to say seriously,
>
>Why not?

Because you've shown the lack of ability to understand
the most basic facts about tensors and differential
geometry. Why can't you compute areas using non-Cartesian
coordinates?

>> How do you calculate areas in non-Cartesian
>> coordinates?

>If you don't know how to do it

Why should I? I'm just a janitor. You're the one who claims
to understand tensors and Riemannian geometry
better than anyone. Why can't you solve a trivial
problem? It's very mysterious.

But here's the janitor's derivation:

We want to compute the area of a parallelogram with sides
bounded by the displacement vectors A and B. Let's see how
far we can go with our dimly-remembered knowledge of high
school geometry.

The area of a parallelogram with sides A and B is

Area = |A| |B| sin(theta)

where theta is the angle between A and B. But how to
compute the angle theta in terms of A, B, and the metric g?
We can use another fact from high school geometry: The
scalar product of two vectors is given by:

A . B = |A| |B| cos(theta)

Now we use yet another fact from high school geometry:

cos^2(theta) = 1 - sin^2(theta)

So we have

(A . B)^2 = |A|^2 |B|^2 cos^2(theta)
= |A|^2 |B|^2 - |A|^2 |B|^2 sin^2(theta)

So

|A|^2 |B|^2 sin^2(theta) = |A|^2 |B|^2 - (A . B)^2

The left-hand side is just the square of the area. So
we have

Area^2 = |A|^2 |B|^2 - (A . B)^2

we can write |A|^2, |B|^2 and (A . B) in terms of the
metric components g_ij as follows:

|A|^2 = A^i A^j g_ij (summed over i and j)
|B|^2 = B^i B^j g_ij (summed over i and j)
(A . B) = A^i B^j g_ij (summed over i and j)

Putting it altogether (and relabelling the dummy
indices so that we can pull out a common factor of
A^i A^j B^k B^l) we have:

Area^2 = (g_ij g_kl - g_ik g_jl) A^i A^j B^k B^l

So that's how you compute areas in non-Cartesian
coordinates.

--
Daryl McCullough
Ithaca, NY