general relativity - Deriving the 4-momentum of a free particle moving in curved spacetime

Consider a free particle with rest mass $m$ moving along a geodesic in some curved spacetime with metric $g_{\mu\nu}$:

$$S=-m\int d\tau=-m\int\Big(\frac{d\tau}{d\lambda}\Big)d\lambda=\int L\ d\lambda\tag{1}$$

$$L=-m\frac{d\tau}{d\lambda}=-m\Big(-g_{\mu\nu}\frac{dx^\mu}{d\lambda}\frac{dx^\nu}{d\lambda}\Big)^{1/2}\tag{2}$$

The canonical 4-momentum $P_\alpha$ can be derived from the Lagrangian $L$ using the following calculation:

$$\begin{eqnarray*} P_\alpha &=& \frac{\partial L}{\partial(dx^\alpha/d\lambda)}\tag{3} \\ &=& \frac{m}{2}\frac{d\lambda}{d\tau}\Big(g_{\alpha\nu}\frac{dx^\nu}{d\lambda}+g_{\mu\alpha}\frac{dx^\mu}{d\lambda}\Big)\tag{4} \\ &=& m\ g_{\alpha\nu}\frac{dx^\nu}{d\tau}\tag{5} \\ &=& m\ \frac{dx_\alpha}{d\tau} \tag{6} \end{eqnarray*}$$ where we have used the fact that the metric $g_{\mu\nu}$ is symmetric.

Thus, expressed in contravariant form, we have derived an expression for the 4-momentum $P^\alpha$ given by

$$P^\alpha=m\ \frac{dx^\alpha}{d\tau}\tag{7}$$ using a completely general metric $g_{\mu\nu}$.

Is it correct to interpret the components of $P^\alpha$ in the following manner:

$P^0$ is the energy of the particle,

$P^i$ is the 3-momentum of the particle in the $\partial_i$ direction?

In other words is $P^\alpha$ the energy-momentum vector with respect to a local orthonormal basis?