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

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


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?

  • 2
    $\begingroup$ It's correct except that eq. (6) only holds for constant metric. $\endgroup$
    – Qmechanic
    Jan 19, 2020 at 20:31
  • $\begingroup$ But surely a general metric $g_{\alpha\nu}$ can lower the index on $dx^\nu/d\tau$? $\endgroup$ Jan 20, 2020 at 10:20
  • 2
    $\begingroup$ $x_{\alpha}:=g_{\alpha\nu}x^{\nu}$ by definition, so the $\tau$-differentiation would also differentiate the metric. $\endgroup$
    – Qmechanic
    Jan 20, 2020 at 10:23
  • 2
    $\begingroup$ 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. $\endgroup$
    – TimRias
    Jan 20, 2020 at 10:49

1 Answer 1


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 $0$.


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