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Considering the covariant tensor $$C = e_{ijk}A^j B^k = A \times B $$ in a 3 spherical-space diagonal metric $$ds^2 = g_{i,i}dx^{i}\cdot dx^i$$

Isn't $C$ the same as "Classical/Newtonian" physics? (Meaning $C_r = A_{\theta} \cdot B_{\phi} - B_{\theta} \cdot A_{\phi} , C_\theta = .... $ )

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  • $\begingroup$ I'm unclear what you are asking. In GR spacetime is 4D and the Levi-Civita symbol has correspondingly 4 indices. $\endgroup$ – jacob1729 Mar 26 at 17:12
  • $\begingroup$ @jacob1729 First of all, Sorry for delayed answer. I didn't get notification (don't know why). I know its 4D but I'm referring to 3+1 General Relativity where you get to have a sub-manifold which refers the the surfaces of $t=C_0$ $\endgroup$ – billy Mar 27 at 11:42
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If $\pi_{ijk}$ is the ordinary Levi-Civita symbol, then the Levi-Civita tensor has components $\epsilon_{ijk}=\sqrt{g}\pi_{ijk}$ (I am assuming the metric is positive definite, but if not then replace $g$ with $|g|$), where $g$ is the determinant of the metric tensor.

Therefore, we have for $C_{i}=\epsilon_{ijk}A^j B^k$ $$ C_1=\sqrt g(A^2B^3-A^3B^2) \\ C_2=\sqrt g(A^3B^1-A^1B^3) \\ C_3=\sqrt g(A^1B^2-A^2B^3). $$

These are the covariant components however. If one wants the contravariant components, then one must raise the indices. If the metric is non-diagonal, this will cause a mixing of the components, but if the metric is diagonal, then we will simply have $C^i=g^{ii}C_i$ (no summation).


Also note that in vector calculus, the components of vectors in say spherical coordinates are usually taken for the orthonormal frame/dreibein $e_r,e_\vartheta,e_\varphi$ rather than the holonomic frame $\partial_r,\partial_\vartheta,\partial_\varphi$. Since the orthonormal basis vector fields are, well, orthonormal, they satisfy $$ e_r\times e_\vartheta=e_\varphi \\ e_\vartheta\times e_\varphi=e_r \\ e_\varphi\times e_r=e_\vartheta, $$ essentially the same relation as the cartesian basis vectors, thus in the orthonormal frame, cross products are calculated the same.

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  • $\begingroup$ Really really thank you! You are right! $\endgroup$ – billy Mar 28 at 11:02

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