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$\newcommand{\dv}[2]{\frac{\mathrm{d} #1}{\mathrm{d}#2}}\newcommand{\pdv}[2]{\frac{\partial #1}{\partial #2}} \newcommand{\div}[1]{\nabla \cdot #1} \newcommand{\pdvt}[2]{\frac{\partial^2 #1}{\partial #2^2}} \newcommand{\curl}[1]{\nabla \times \vec{ #1}}$I assume you know how to solve the Maxwell's equations in vacum. In the end you get an expression of the ...


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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 ...


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"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, ...


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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 $\alpha$. ...


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It is pretty much simply a short way to notate both vector field operations by looking at $\nabla$ as a vector operator by writing \begin{equation} \nabla=\left(\frac{\partial}{\partial x},\frac{\partial}{\partial y},\frac{\partial}{\partial z}\right) \end{equation} in $\mathbb{R}^3$, or equivalently \begin{equation} \nabla=\frac{\partial}{\partial ...



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