Question about transverese-traceless gauge in gravity

Neitrino

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.
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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... ?

Thank you

Bill_K

Neitrino, As you say, the first step is to impose the harmonic gauge condition, h_μν,^ν = 0, or equivalently in momentum space, h_μνk^ν = 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, h_μν → h_μν + e_(μk_ν). 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 h_i0 = 0 and h_ij = 0. Then we also have h_ijk^j = 0. This is what we mean by transverse and traceless, namely transverse to the 3-dimensional k-vector.

Neitrino

Thanks Bill_K for you help...

So is it (TT gauge) much the same as Coulomb gauge in vector field theory ?


Thanks