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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?

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2 Answers 2

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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.

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  • $\begingroup$ Thanks for your answer. I agree that $t$ should be replaced by $\tau$ but what I am not able to follow from your reply is how should one arrive at the first equation from the second (if at all it's possible)? Is there any reference for the first equation like there is for the second (Wald's book)? $\endgroup$ Jan 1, 2018 at 19:03
  • $\begingroup$ The two equations are equivalent, but in order to switch from one another you have to put together the two pieces, partial derivative and connection, to reconstruct the covariant derivative, e.g. on the L.H.S.. Then you can raise/lower the indices and split again the covariant derivative in the two pieces. Be careful that the split has a positive sign in the connection when applied to an obiect with an up index (vector), a negative sign with a down index (dual vector). $\endgroup$ Jan 2, 2018 at 10:13
  • $\begingroup$ Thanks again for being patient with me. As you can see I am not an expert in tensorial notation. Is the third equation already not correct? In other words, I can't go from equation 2 to 3 by multiplying both sides with metric and contracting? What confuses me even more is that the first equation has summation over the top (first) index of the Christoffel symbol whereas the second equation leaves the first index untouched. $\endgroup$ Jan 2, 2018 at 17:57
  • $\begingroup$ The third equation is already not correct as you apply the metric to the partial derivative, not metric compatible. The reason why the summation index in the connection is in one case an up index, while in the other case a down index is because in one case you apply the connection to a dual vector, while in the other case to a vector. To document on the covariant derivative, a reference could be "Spacetime and geometry. An introduction to general relativity" by Sean M. Carroll, Chapter 3 "Curvature", whereof in particular sections 3.1, 3.2, 3.3, 3.4. A few hours, but well expended. $\endgroup$ Jan 2, 2018 at 22:31
  • $\begingroup$ Thank you very much! I can work out the equivalence now. I was trying to manipulate the indices without really understanding what I was doing! $\endgroup$ Jan 3, 2018 at 17:23
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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.

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