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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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This is a very formalistic answer, because I have recently reviewed the first chapter of Marion and Thornton in preparation for teaching from the text. It may not be suitable for beginners in the sense of following the math, but the answer that is reached might be helpful even so. I'm going to use the notation $\vec{x}$ (normal weight with arrow overset) ...


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The angular momentum is a vector quantity whose most important properties is that it is conserved and therefore absorbs torques linearly. The angular velocity is a vector quantity whose magnitude specifies a rate at which an object is rotating, and whose direction specifies the axis about which it is rotating. The specific expression you are using is valid ...


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The convention we pick here will interact with the convention we have for matrix multiplication in the following way: If we have matrices $A$ and $B$ and we use the usual convention that the matrix multiplication $AB$ multiplies the rows of $A$ with the columns of $B$ then we have either $$(AB)_{ij}=\sum_kA_{ik}B_{kj}\tag{#}$$ or ...


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Your claim is false (in $\mathbb R^3$). If $\vec{a}$ and $\vec{b}$ are fixed vectors (I assume they are not co-linear and satisfy $|\vec{a}|=|\vec{b}| \neq 0$) there are infinitely many rotations $R \in O(3)$ such that $R\vec{a}=\vec{b}$. One is the rotation $R$ of the angle between $\vec{a}$ and $\vec{b}$ performed around an axis orthogonal to the plane ...


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Short answers Apply the Young calculus (per ACuriousMind's suggestion in the comments). For finding the multiplicity of the trivial representation in a tensor product of representations of $SU(n)$, note that each irreducible representation $D$ of $SU(n)$ has a unique conjugate irreducible representation $\bar D$ such that the Young calculus allows ...


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Horizontal position of indices matters in principle because one might want to raise and lower indices on the Christoffel symbols. If the horizontal position of indices are not observed in a consistent manner, it becomes ambiguous which index was raised or lowered, and so forth, in particular if the connection is not torsionfree. Also note that different ...


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The metric measures lengths in various directions, and also angles between various directions. For example if $\vec{e}_{(1)}$ is the basis vector in the $x^1$-direction, it will have length given by $$ \lVert \vec{e}_{(1)} \rVert^2 = g(\vec{e}_{(1)}, \vec{e}_{(1)}) = g_{11}. $$ If we also have the basis vector $\vec{e}_{(2)}$ in the $x^2$-direction, then the ...


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The metric is an important concept in general relativity. In GR, vectors correspond to weighted directions in spacetime (by "weighted", I mean any scalar multiple of a vector corresponds to the same direction, but weighted differently). The metric tensor can then tell us about the angle between two directions or the magnitude of a given vector, which gives ...


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As Qmechanic pointed out in the comments, you're mixing Einstein and abstract index notation a bit. To make things absolutely clear, we will use early Latin indices for abstract indices $(abc)$ and Greek indices for component indices $(\mu\nu\rho)$ and will always indicate Einstein summation explicitly. First and foremost, an abstract index is nothing more ...



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