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$$ \nabla_\mu A^\nu = g^{\nu\alpha} \nabla_\mu A_\alpha = g^{\nu\alpha} ( \partial_\mu A_\alpha - \Gamma^\lambda_{\mu\alpha} A_\lambda ) = g^{\nu\alpha} \partial_\mu ( g_{\alpha\beta} A^\beta ) - g^{\nu\alpha} \Gamma^\lambda_{\mu\alpha} g_{\lambda\beta} A^\beta $$ This simplifies to $$ \nabla_\mu A^\nu = \partial_\mu A^\nu + ( g^{\nu\alpha} \partial_\mu ...


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Explicitly working with the components of $g$ and $\Gamma$ can be messy. Here is a nicer way to derive the coordinate expression for the covariant derivative of a vector (or any tensor) without raising or lowering indices. Consider vector fields $A^\mu, B_\mu$. The covariant derivative of a scalar is the same as the coordinate derivative (in any coordinate ...


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I am reading Schutz on the subject of vectors, and I believe that he says that; given a manifold with two sets of coordinate systems that are related through a linear transformation; a pair of sets of numbers (with the same dimension as the manifold) is a vector if one set of numbers is related to the other set of numbers through the linear transformation.


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I use other symbols in order to prevent confusion in the following. Let a point charge $\:q\:$ moving with position vector $\:\boldsymbol{\xi}\left(t\right)\:$ as in above Figure. Then the volume charge density and the charge current density are expressed via Dirac $\:\delta$-function as follows \begin{align} \rho\left(\mathbf{x},t\right) & ...


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As per, http://en.wikipedia.org/wiki/Four-velocity, we can define four-current density as: $J = \rho_0 U$, where $U$ is the four-velocity. Since it's a scalar times a four-vector, it's another four-vector. $$J = \gamma(v)(\rho_0 c,\rho_0 \vec{v})$$ $$J = (\gamma(v)\rho_0 c,\gamma(v)\rho_0 \vec{v})$$ Now it remains to show that this fits the definition you ...


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It helps to remember that invariant quantities are seen as scalars to the transformation (they have no indices in the target space). In the other hand, covariant quantities are objects that transform in a certain way. Example: Vectors in $R^{2}$, under rotation $R_{ij}$, transform covariantly since $v'_{i}=R_{ij}v_{j}$, but it's length is invariant since ...



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