# 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?

• It's correct except that eq. (6) only holds for constant metric. Jan 19, 2020 at 20:31
• But surely a general metric $g_{\alpha\nu}$ can lower the index on $dx^\nu/d\tau$? Jan 20, 2020 at 10:20
• $x_{\alpha}:=g_{\alpha\nu}x^{\nu}$ by definition, so the $\tau$-differentiation would also differentiate the metric. Jan 20, 2020 at 10:23
• Note that $x^\alpha$ is not a vector field (the notation is abused in this case). Hence lowering its index does not actually make any sense. Jan 20, 2020 at 10:49

The covariant component $$-P_0$$ is the conserved total energy (kinetic, potential and rest energy) in stationary spacetimes and if $$x^0=t$$. The transformation from the spatial components of the 4-momentum into the local 3-momentum is best done over the local linear velocity $$v$$ (we use the +--- signature and natural units):
$$P_{\alpha}= \left( \sum_{\beta=1}^{4} \ g_{\alpha \beta} \ \dot{x}^{\beta} \right) - q \ A_{\alpha} = \frac{v_{\alpha}}{\sqrt{1-||v||^2}} \sqrt{-g_{\alpha \alpha}} - (1-||v||^2) \ q \ A_{\alpha}$$
The contravariant momentum is simply $$P^{\alpha}=\dot{x}^{\alpha}$$. If you regard only geodesics for uncharged particles the electromagnetic vector potential $$A$$ and the particle charge $$q$$ can be set to $$0$$.