Ejercicios Con Tensores Upd Jun 2026

The Levi-Civita symbol ( \epsilon_{ijk} ) (in 3D) is completely antisymmetric: ( \epsilon_{123}=1 ), and it changes sign under any odd permutation of indices. It represents the oriented volume element.

The Riemann tensor ( R^\rho_{\ \sigma\mu\nu} = \partial_\mu \Gamma^\rho_{\nu\sigma} - \partial_\nu \Gamma^\rho_{\mu\sigma} + \Gamma^\rho_{\mu\lambda} \Gamma^\lambda_{\nu\sigma} - \Gamma^\rho_{\nu\lambda} \Gamma^\lambda_{\mu\sigma} ) measures the failure of second covariant derivatives to commute.

Before tackling a physical problem, one must become fluent in the Einstein summation convention (repeated indices imply summation) and the role of the metric tensor. ejercicios con tensores

On a curved manifold (or even in curvilinear coordinates in flat space), ordinary partial derivatives of tensors are not tensors. The covariant derivative ( \nabla_\mu ) corrects this using Christoffel symbols ( \Gamma^\rho_{\mu\nu} ).

Using the identity ( \epsilon_{ijk} \epsilon^{imn} = \delta^m_j \delta^n_k - \delta^n_j \delta^m_k ), prove the following vector calculus identities in Cartesian coordinates (where ( \nabla_i = \partial/\partial x^i )): The Levi-Civita symbol ( \epsilon_{ijk} ) (in 3D)

( \frac{\partial y^i}{\partial W^{kl}} = \sigma'(z^i) \delta^i_k x_l ), where ( z^i = W^{ij}x_j + b^i ) and no sum over i. This exercise reveals how automatic differentiation naturally implements tensor contractions.

While not relativistic, tensors are central to deep learning. Consider a simple neural network layer: output ( y^i = \sigma(W^{ij} x_j + b^i) ), where ( \sigma ) is an activation function. Before tackling a physical problem, one must become

For the 2D metric of a sphere of radius ( R ): ( ds^2 = R^2 d\theta^2 + R^2 \sin^2\theta , d\phi^2 ). Compute all non-zero Christoffel symbols using the formula: ( \Gamma^\sigma_{\mu\nu} = \frac{1}{2} g^{\sigma\rho} (\partial_\mu g_{\nu\rho} + \partial_\nu g_{\mu\rho} - \partial_\rho g_{\mu\nu}) ).