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## Hot answers tagged tensor-calculus

7 votes
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### Torsion and Compatibility with the Metric

I think what's important to consider here is that the Levi-Civita connection is unique. Once you introduce torsion this is no longer true. Therefore, when you write $g_{ij;k}$ it isn't clear if you're ...
• 615
3 votes

### How to derive $\partial^{\nu}F^{\mu\alpha} + \partial^{\alpha}F^{\nu\mu} + \partial^{\mu}F^{\alpha\nu}=0$ for the Electromagnetic field tensor?

Write down $\partial_\mu G^{\mu\nu} = \frac{1}{2}{\epsilon^{\mu \nu \alpha \beta}\partial_\mu F_{\alpha\beta}}$. Now, fix the free index $\nu$ (to be either of 0, 1, 2, 3), and write down all non-zero ...
• 3,160
3 votes
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### Do I sum over these indices?

Do not sum over $\alpha$ and $\nu$. To clarify, here is the Einstein summation convention in its most accurate form: If a covariant index is followed by the same contravariant index and vice versa, we ...
• 102
3 votes

### Do I sum over these indices?

Do not sum over $\nu,\alpha$. You would only sum over them if they were repeated. The next step would be to plug in the specific formulas for each matrix element and see the cancellation explicitly. ...
• 655
2 votes
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It doesn't seem to be a mistake. We use similar notation in relativity. It means that $$\omega_{xy} = \omega_z,$$ and so on. The general expression, as pointed out in a comment to this answer, is $$\... • 21.3k 2 votes ### Covariant derivative to the metric determinant? Taking a look at the paper and the equations you mentioned in the comments, they're varying I_{W_2} with respect to g_{ab} rather than taking its covariant derivative. To see how you get Eq. 108 ... • 615 1 vote Accepted ### Tensor identity in L&L book 2 These trivialities were developed simply by checking both sides for simple entries from 0,1. The full n dimensional LeviCivita symbol is the coefficient tensor of the n-dimensional signed volume ... • 26 1 vote Accepted ### Trouble understanding tensor notation for relativistic transformations x and x' are coordinates on spacetime. Since each event in spacetime is labelled by a particular value in a given coordinate system, you can change between these two coordinate systems. ... • 21.3k 1 vote ### "Proof" that zero curvature implies \partial_a \Gamma^b_{cd} is symmetric in a and c The claim is not wrong, if you use a matrix function T^a_b(x) to define connection in this way \Gamma^a_{cb}=\partial_cT^a_b, then the change of T^a_b(x) from a point x_0 to point x along a ... 1 vote ### Prove that covariant differentiation obeys the product rule I'm pretty sure that your proof is incorrect; in the last line, you equate$$A_{bc}(\partial_a B^{cd}+\Gamma^d{}_{ea}B^{ce}) = A_{bc}\nabla_a(B^{cd}) and make a similar argument for the other term. ...
• 11

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