Tuesday, 26 July 2016

homework and exercises - Scalar invariance under Lorentz-transformation


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=xαxαvα


On the other hand, a co-vector uα transforms differently



uα=xαxαuα


In your case, the quantity uαvβ transform like this


uαvβ=xαxαxβxβuαvβ


and in particular, the diagonal elements


uαvα=xαxαxαxβuαvβ


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


uαvα=xαxβuβxγxαvγ=xγxβuβvγ=δγβuβvγ=uβvβ


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