PWiz said:
The Lorentzian metric changes to a Euclidean one if you use ##ict##. Isn't it better that way? You then have only have one metric tensor to deal with.
No, it's really not the better approach. The approach that you're talking about ignores the distinction between covariant vectors (ones with components A^\mu) and contravariant vectors (ones with components A_\mu). That distinction is extremely important for curved spacetime, but it's also important for flat spacetime, if you're using curvilinear coordinates.
For example, let's just do 3-dimensional vectors in spherical coordinates,nonrelativistically. Then a point in space is described by three numbers, r, \theta, \phi. The path of a particle would similarly be described by three numbers: \frac{dr}{dt}, \frac{d\theta}{dt}, \frac{d\phi}{dt}, which you can think of as components of the velocity vector. But if you're computing the speed v, it's not simply v^2 = \frac{dr}{dt}^2 + \frac{d\theta}{dt}^2 + \frac{d\phi}{dt}^2. Instead, it's more complicated:
v^2 = \frac{dr}{dt}^2 + r^2 \frac{d\theta}{dt}^2 + r^2 sin^2(\theta) \frac{d\phi}{dt}^2. That can be understood as
\sum_{i,j}\ g_{ij} v^i v^j
where g_{rr} = 1, g_{\theta \theta} = r^2, g_{\phi \phi} = r^2 sin^2(\theta).
Of course, it is possible to get a notion of a velocity vector that uses a trivial metric, by letting the components of the velocity vector be:
\frac{dr}{dt}, r \frac{d\theta}{dt}, r sin(\theta) \frac{d\phi}{dt}
That allows the metric to be simple, but in exchange for making the velocity vector more complicated. In a sense, what you'd be doing is incorporating \sqrt{g} into the definition of the velocity vector (which is the same thing that is going on in using ict). That's a convoluted thing to do, and only works in the cases where the metric tensor is diagonal, which isn't always the case.
