# Tag Info

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Replace only one of the $\omega_i$ in the first term of (11.8). That is, replace $\sum\limits_i \omega_i^2$ with $\sum\limits_{i,j}\omega_i\omega_j\delta_{ij}$. Note that what you wrote, $$\left(\sum\limits_i\omega_i x_{\alpha,i}\right)\left(\sum\limits_j\omega_j x_{\alpha,j}\right) = \left(\sum\limits_i\omega_i x_{\alpha,i}\right)^2$$ does not ...

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For doing matrix multiplication, yes, we usually put the dummy indicies together, but this is (assuming one is not working over $\mathbb{H}$) convention really. Don't forget that this is shorthand Einstein convention. When you write $$\mathbf{A} \cdot \mathbf{B} = A_{ij} B_{jk} = \sum_{j=1}^n A_{ij} B_{jk}$$ you are writing a shorthand version in ...

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If you are dealing with spacetime indices (i.e. tensors over the spacetime), then symbols like $\delta^{ab}$ or $\delta_{ab}$ don't make sense. If you lower an index of $\delta^a_b$ you will end up with the metric $g_{ab}$, same for raising an index. This is clear from the definition of $\delta$: $$g_{ab}\delta^b_c=g_{ac}$$ and $$g^{ab}\delta_b^c=g^{ac}$$ ...

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The result is the metric: the effect of the Kronecker delta in your examples is to set $b = c$. The Kronecker delta is really just the identity matrix.

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You're confusing two things. In tensor calculus, the Kronecker delta should be visualized as basically the identity. What it does is relabel an index. Example: $$g_{ab}\delta^b_c=g_{ac}$$ This has nothing whatsoever to do with the Dirac delta function (it's actually a distribution) in this context. In nonrelativistic quantum theory Dirac delta ...

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The co/contra distinction only makes sense when talking about vector fields. Even then the difference only becomes apparent when dealing with curved spaces or at least curvilinear coordinate systems The difference comes from how vectors relate back to the undlying space or manifold on which the fields are defined. Contravariant vectors then are what people ...

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We expect a vector to change in a certain way when we change the scale we use to measure distance. Consider the vector $$\vec{x}=(1, 0, 0)\,\mathrm{m}$$ If we change scale and now measure in centimeters this vector becomes $$\vec{x}=(100, 0 ,0)\,\mathrm{cm}$$ Now consider a vector representing a force: $$\vec{F}=(1,0,0)\,\mathrm{J/m}$$ where I've chosen ...

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The notion of co- and contravariance depends on context: If you wanted to be as clear as possible, you should actually mention with respect to what the components transform co- or contravariantly. In case of the algebraic dual of finite-dimensional vector spaces, the implied context is a change of basis of the vector space. Then, we can look at how the ...

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There are two more points that can be made here. Sorry if I repeat someone. In a way you are right that if you have a vector space and its dual there is no intrinsic way to say which space is the original and which is the dual. This is because there is a canonical isomorphism between a vector space and the dual of its dual. In other words if $V$ is a vector ...

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I) No, it is important to distinguish between covariant and contravariant tensors. OP's link mentions differential geometry. If one has only studied those objects in the context of pseudo-Riemannian manifolds $(M;g)$, which comes equipped with an (invertible) metric $(0,2)$ tensor $g$, then the existence of the musical isomorphism may perhaps unnecessarily ...

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I will say that the standard definition of vectors and one-forms is not the world's cleanest. A modern definition of vectors would say that a vector space is a mapping from the functions on the space to itself that satisfies the Leibniz rule and is linear (alternately, the vector space is the local linear approximation of the space). Then, the set of ...

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This is not really an answer to your question, essentially because there isn't (currently) a question in your post, but it is too long for a comment. Your statement that A co-ordinate transformation is linear map from a vector to itself with a change of basis. is muddled and ultimately incorrect. Take some vector space $V$ and two bases $\beta$ and ...

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First recall that we define a co-vector field $\eta$ from a vector field $v$ via the flat map $\eta~=~v^{\flat}.$ In components we have $\eta_k = g_{ki} v^i.$ Equivalently, we can reconstruct the vector field $v$ via the sharp map $v ~=~\eta^{\sharp}.$ In components we have $v^i = (g^{-1})^{ik} \eta_k.$ Now let us return to OP's question. As Danu writes ...

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Kyle Kanos mentioned Geometric Algebra for Physicists. While geometric algebra is somewhat different in notation from differential forms, the basic concepts are all there, and in many ways, geometric algebra avoids some cumbersome things that differential forms does (I'm thinking of Hodge duality in particular). I think the notation is easier to relate to ...

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If you'd like to quickly obtain an understanding of the basics of differential forms, including their relation to connections, tangent bundles etc. I recommend the first 4 online lectures of the Perimeter Institute from the Gravitational Physics course (13/14, R. Gregory).

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