#### Mentz114

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The tensorial property is the invariance of scalar contractions. For instance a 4-vector is a rank-1 tensor. Suppose we have two 4-vectors [itex]U^\mu, V^\mu[/itex] and we make the contraction [itex]l=U^\mu V_\mu[/itex], where [itex]V_\mu = g_{\mu\alpha}V^\alpha[/itex]. Now apply a coordinate transformation [itex]\Lambda[/itex] to the vectors. The contraction of the transformed vectors is [itex](\Lambda^a_\mu U^\mu)( \Lambda_a^\mu V_\mu)[/itex]. Because [itex]\Lambda[/itex] is not a tensor but a transformation matrix, we can rearrange the last expression to [itex](\Lambda^a_\mu \Lambda_a^\mu) U^\mu V_\mu = U^\mu V_\mu[/itex].I can't vouch for its precision, but in the context of an earlier discussion elswhere it made a lot of sense. Can some of you math whizzes comment on whether, in general, an 'invariant' tensor [a true tensor] is so only in inertial conditions? If so, do we still call it 'invariant' because its so close, or not use that terminology.

What makes U,V tensors is that despite having their components changed by a transformation, their contraction remains the same. The same logic can be applied to any contraction of indexed objects. The objects in question are tensors iff their scalar contractions are invariant.

You must have come across the Ricci scalar which is the self-contraction of the Ricci tensor, [itex]R=R^\mu_\mu[/itex]. Because [itex]R_{\mu\nu}[/itex] is a tensor the contraction is invariant under coordinate tansformation.

(I am not a "math whizz", this is standard textbook stuff.)

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