So do the whole calculation using A A instead of and then substitute in the formula for A A in terms of g g and only at the end. The Riemann tensor can be constructed from the metric tensor and its first and second derivatives via where the s are Christoffel symbols of the first kind. in terms of derivatives of is much simpler than the one for. As mentioned in the comments calculating algebraic expressions for the Ricci tensor containing the metric, its inverse and its first and second derivatives is straight forward using computer algebra. In short, the expression you provided holds locally for an inertial coordinate system with a timelike coordinate and three space-like coordinates. You can restate this lemma as a rule proven for transforming Christoffel symbols.
Nevertheless, the existence of the four-potential $A_\mu$ is also only ensured locally. On an inertial coordinate system with one time component and three spatial components, the Christoffel symbols vanish and this equation reduces to the usual expression, so your formula is valid, but only locally. $$F_ = \nabla_\mu A_\nu - \nabla_\nu A_\mu$, where $\nabla_\mu$ is the covariant derivative. The electromagnetic field tensor defined with covariant derivatives or with standard partial derivatives is the same simply due to the anti-symmetry of this tensor: Actually, the things are easier than you might think.
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