|Nov23-12, 12:07 PM||#1|
Question about transverese-traceless gauge in gravity
I have a question about gauges in gravity.
Symmetric tensor field in four dimensions has 10 independet components, when we want to describe massless spin-two field (graviton) we impose harmonic gauge which reduces 10 independet components to 6 and afterwards we use diff invariance and eventually go down from 6 to 2 independet components. And all this happens with harmonic gauge.
What is a transverse-traceless gauge?
Transverese means d_mu (h_mu_nu)=0 and this puts constraints on 4 components out of initial 10 components of symmetric tensor...
Traceless means Tr (h_mu_nu)=0 and this puts constraint on 1 component....
So TT gauge puts constraints 5 constraints on 10 components of symmetric tensor field.... and what to do next how to reduce to 2 components to describe graviton .... how do i cast TT gauge in the same sense as i did for harmonic gauge above... ?
|Nov23-12, 04:24 PM||#2|
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Neitrino, As you say, the first step is to impose the harmonic gauge condition, hμν,ν = 0, or equivalently in momentum space, hμνkν = 0. This can be done covariantly, and reduces the number of independent components from 10 to 6.
There remains the freedom of further gauge transformations, hμν → hμν + e(μkν). These can be used to reduce the number of components from 6 to 2, but not in a Lorentz covariant manner. Choosing a rest frame, one can show it is possible to set hi0 = 0 and hij = 0. Then we also have hijkj = 0. This is what we mean by transverse and traceless, namely transverse to the 3-dimensional k-vector.
|Nov24-12, 04:21 AM||#3|
Thanks Bill_K for you help...
So is it (TT gauge) much the same as Coulomb gauge in vector field theory ?
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