Lorentz force in curved spacetime I am trying to derive the equation for Lorentz force mentioned in the following Wikipedia article -
https://en.wikipedia.org/wiki/Maxwell%27s_equations_in_curved_spacetime
viz.,
$$
\frac{d p_{\alpha}}{d t} \, = \, \Gamma^{\beta}_{\alpha \gamma} \, p_{\beta} \, \frac{d x^{\gamma}}{d t} \, +  \, q \, F_{\alpha \gamma} \, \frac{d x^{\gamma}}{d t}. 
$$
Starting from the modified geodesic equation, (based on Wald's Eq. 4.3.2, https://en.wikipedia.org/wiki/Geodesics_in_general_relativity) and using $p^\alpha = m \frac{dx^\alpha}{dt}$, we arrive at,
$$
\frac{d p^\mu}{dt} = -\Gamma^{\mu}_{\alpha \beta} ~p^\alpha\frac{d x^\beta}{dt} + q ~F^{\mu}_{\alpha}\frac{d x^\alpha}{dt}.
$$
Lower $p^\mu$ to $p_\alpha$ by multiplying both sides with $g_{\gamma \mu}$ to get -
$$
\frac{d p_\gamma}{dt} = -g_{\gamma \mu}\Gamma^{\mu}_{\alpha \beta} ~p^\alpha \frac{d x^\beta}{dt} + q~F_{\gamma \alpha}\frac{d x^\alpha}{dt},\\
\mathrm{or , \quad}\frac{d p_\gamma}{dt} = -\Gamma_{\gamma \alpha \beta} ~p^\alpha \frac{d x^\beta}{dt} + q~F_{\gamma \alpha}\frac{d x^\alpha}{dt}.
$$
Relabeling $\gamma$ as $\alpha$ and $\alpha$ as $\gamma$,
$$
\frac{d p_\alpha}{dt} = -\Gamma_{\alpha \gamma \beta} ~p^\gamma  \frac{d x^\beta}{dt} + q~F_{\alpha \gamma}\frac{d x^\gamma}{dt},
$$
which is not the same as the first equation even if we lower $p^\gamma$. What am I missing?
 A: It seems that the following points are inconsistent with a tensorial formulation:
1) The path parameter should be stated $\tau$ (proper time).
2) The covariant derivative is metric compatible (by definition), not the partial derivative.
3) The metric g to raise/lower indices should be applied to a tensor, but the connection $\Gamma$ is not a tensor.
A: You are missing the fact that you cannot lower those indices separately since single components of that equation are NOT tensors. Indeed the correct expression is obtained by the formula (with the Levi-Civita connection):
$$(\nabla_{u}u)_{\mu}=u^{\sigma}\partial_{\sigma}u_{\mu}-\Gamma^{\beta}_{\mu\sigma}u_{\beta}u^{\sigma}=0$$ but in that case you are treating that as acting over the, in your case, covariant vector $u$. (You can do that since you are working in a metric manifold and the metric $g$ allows you to "connect/indentify" vectors and covectors). Be careful that the $u$ in the covariant derivative is a vector and all is applied to a covector $u$. Since $$u^{\sigma}\partial_{\sigma}u_{\mu}=\ddot{x}_{\mu}$$ and the Lorentz contribute goes on the right hand side of the geo eq. then you get the same first equation.
