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Below I'll use Planck units, for which, in particular, $c = \epsilon_{0} = =1$. In fact, the full system of Maxwell's equations provides the statement that the only two vector components of the EM field $\mathbf E, \mathbf B$ are independent (in general, due to a deep symmetry reason, namely that a massless particle has only two polarizations). Next, if we ...

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I think you have all the right pieces to answer the question, here are a few hints that should be of some use. You say that you picked coordinates $\{v^{\mu} \}$. It seems to me that they should instead be called $\{ x^{\mu} \}$, as that is what you're taking partial derivatives with respect to. As you correctly pointed out, you are working with ...

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The key concept needed here is that the Hemholtz decomposition is not necessarily unique. Non-uniqueness can occur because there exist nontrivial vector fields which are both irrotational and divergence-free. For example, the constant 2D velocity field $\vec v = (1,0)$ can be expressed as either $\vec v=-\nabla \phi$ with $\phi(x,y)=-x$, or as $\vec ... 2 "Contraction-orthogonality" of covariant and contravariant basis Contravariant vectors or just "vectors" are defined as elements of the tangent space at a given point. In practice, they are defined with respect to a coordinate-vector basis$\mathbf{e}_{(i)}$, where$\mathbf{e}_{(i)}$is the vector tangent to the$i$-th coordinate line. Then they are given, ... 2 The metric being a rank$(0,2)$tensor transforms under general coordinate transformations$x^\mu \to x'^\mu(x)$as $$g'_{\mu\nu} (x') = \frac{ \partial x^\rho}{ \partial x'^\mu } \frac{ \partial x^\sigma }{ \partial x'^\nu } g_{\rho\sigma} (x)$$ Now set$x'^\mu (x) = x^\mu + \alpha k^\mu(x)$in the above expression and take a limit of small$\alpha$. ... 2 It is pretty much simply a short way to notate both vector field operations by looking at$\nabla$as a vector operator by writing $$\nabla=\left(\frac{\partial}{\partial x},\frac{\partial}{\partial y},\frac{\partial}{\partial z}\right)$$ in$\mathbb{R}^3\$, or equivalently \nabla=\frac{\partial}{\partial ...

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