Silly question about calculation of $4$-velocity vector

Question

I would like to know if it is possible to state (choose) always that:

$$ u_{\mu} = (-1,0,0,0) \tag{1}.$$

Then, knowing $(1)$ you can raise the index,

$$u^{\mu} = g^{\mu\nu}u_{\nu}, \tag{2}$$

and calculate the components of $u^{\mu}$.

Example:

Calculate the components of four-velocity $u^{\mu}$ for Schwarschild metric:

1. Given $g_{\mu\nu} = Diag\Biggr[(-1+2GM/c^2r),(-1+2GM/c^2r)^{-1},r^2,r^2sin^2\theta\Biggr]$
2. Calculate the inverse metric $g^{\mu\nu} = Diag\Biggr[(1-2GM/c^2r)^{-1},(-1+2GM/c^2r),(r^2)^{-1},(r^2sin^2\theta)^{-1}\Biggr]$
3. State that $u_{\mu} = (-1,0,0,0)$
4. Calculate the components of 4-velocity $u^{\mu}$

$$u^{0} = g^{0\nu}u_{\nu} = g^{00}u_{0} + g^{01}u_{1} + g^{02}u_{2} + g^{03}u_{3} = g^{00}u_{0} = (1-2GM/c^2r)^{-1}(-1) = \frac{-1}{1-2GM/c^2r} $$

$$u^{1} = g^{0\nu}u_{\nu} = g^{10}u_{0} + g^{11}u_{1} + g^{12}u_{2} + g^{13}u_{3} = g^{11}u_{1} = (-1+2GM/c^2r) (0) = 0 $$

$$u^{2} = g^{0\nu}u_{\nu} = g^{20}u_{0} + g^{21}u_{1} + g^{22}u_{2} + g^{23}u_{3} = g^{22}u_{2} = (r^2)^{-1} (0) = 0 $$

$$u^{3} = g^{0\nu}u_{\nu} = g^{30}u_{0} + g^{31}u_{1} + g^{32}u_{2} + g^{33}u_{3} = g^{33}u_{3} = (r^2sin^2\theta)^{-1} (0) = 0 $$

Then,

$$ u^{\mu} = \Biggr( \frac{-1}{1-2GM/c^2r} ,0,0,0\Biggr) \tag{3}$$

Conversely, for the Alcubierre Metric, we have also the $u_{\mu}$ stated as $(1)$, but a four-velocity given by:

$$u^{\mu} = (1,0,0,vf), \tag{4}$$

due to the fact of a non-diagonal metric tensor.

So,

Can I always follow this method in response to the question: "How can I calculate four-velocity components from a given metric..." ?