anonanon 152k1212 gold badges233233 silver badges398398 bronze badges $endgroup$ 4 $begingroup$ Which is best thanks a bunch! I also posted an analogous problem listed here: math.stackexchange.com/concerns/2135276/… that You may have some insight on? $endgroup$

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RexcirusRexcirus 43544 silver badges1515 bronze badges $endgroup$ two $begingroup$ Your remaining expression would not make Substantially perception: The LHS has cost-free indices $i,j,k$ While your RHS has cost-free indices $i,j,l$. $endgroup$

I am using the Einstein notation, so various indices are summed) or they give two 3-indices symbols with unique signal

Examples of Levi-Civita Homes in 4 Proportions include the symmetric and antisymmetric mother nature in the tensor, its relation towards the metric tensor, and its transformation properties beneath coordinate transformations.

From this behavior, we can easily deduce what $epsilon^ ij4k $ can be without having experiencing the computations Yet again: it is simply the result of exchanging the final two indices of $epsilon^ ijk4 $,

Indeed, there remain a lot of open issues and parts of research associated with Levi-Civita properties in four dimensions.

The Levi-Civita tensor is used in the Einstein field equations to specific the curvature of spacetime in terms of the Power and momentum of issue and radiation.

The Levi-Civita symbol permits the determinant of the sq. matrix, as well as cross product or service of two vectors in 3-dimensional Euclidean Area, for being expressed in Einstein index notation.

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