With uαvβ the components of two vector fields, is
uαvα=u1v1+u2v2+u3v3+u4v4
a scalar invariant under a Lorentz-transformation? And why?
Answer
First note that your expression of uαvα is wrong. Einstein summation convention tell you that you sum if an index appears twice - once up and once down. Thus, in fact
uαvα=u1v1+u2v2+u3v3+u4v4
The quantity uαvβ, on the other hand, is a tensor of rank (2,0), which can be represented by a 4×4 matrix
(u1v1u1v2u1v3u1v4u2v1u2v2u2v3u2v4u3v1u3v2u3v3u3v4u4v1u4v2u4v3u4v4)
so by writing uαvα you refer to its diagonal elements.
Second, you are confusing upper and lower indices. The location of the index determines its transformation properties. Take for example a vector vα, it transform in the following manner
vα′=∂xα′∂xαvα
On the other hand, a co-vector uα transforms differently
uα′=∂xα∂xα′uα
In your case, the quantity uαvβ transform like this
uα′vβ′=∂xα′∂xα∂xβ′∂xβuαvβ
and in particular, the diagonal elements
uα′vα′=∂xα′∂xα∂xα′∂xβuαvβ
In general, it is not invariant under the Lorentz transformations. However, the quantity uαvα has no free indices since you sum over α, and thus it remains invariant under transformations. You can also see it by transforming uα and vα separately
uα′vα′=∂xα′∂xβuβ∂xγ∂xα′vγ=∂xγ∂xβuβvγ=δγβuβvγ=uβvβ
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