Raising and lowering indices in Linearized theory

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zn5252
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Hi there,
in Linearized theory we know that :
gαβ = ηαβ + hαβ
If I multiply out both terms by ηθαηλβ, wouldn't one get :
gθλ = ηθλ + hθλ ? EQ1
But we already know gθλ = ηθλ - hθλ + O(h2) EQ2
How can we reconciliate EQ1 and EQ2 ? Was it an error to have raised the perturbation h with the η ?
Can we also say that gθλ = gθλ ?
I'm a bit confused.
Thanks,
 
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zn5252, There are covariant tensors and contravariant tensors, and it's just a convention that we use the metric tensor to associate one with the other. That is, if Bμ = gμνAν, its a convention to say that A and B are the 'same' tensor. For this to work you must be able to reverse the process and say Aν = gμνBμ, and this requires gμν to be the inverse of gμν, matrixwise, that is, gμνgνσ = δμσ.

In linearized theory, gμν = ημν + hμν. Writing this as a matrix equation, g = η + h, the inverse is g-1 = (η + h)-1 = η - ηhη + ... So again it's a helpful convention to raise and lower indices in the flat space using the flat space metric ημν and its inverse ημν, but the original definition of gμν as the inverse matrix of gμν must be kept.
 
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Remark : The product of the covariant g with its contravariant counterpart is assured to equal the unit tensor to first order In this case. See footnote of Landau & Lifgarbagez page 350.
 
Bill_K said:
zn5252, There are covariant tensors and contravariant tensors, and it's just a convention that we use the metric tensor to associate one with the other. That is, if Bμ = gμνAν, its a convention to say that A and B are the 'same' tensor. For this to work you must be able to reverse the process and say Aν = gμνBμ, and this requires gμν to be the inverse of gμν, matrixwise, that is, gμνgνσ = δμσ.

In linearized theory, gμν = ημν + hμν. Writing this as a matrix equation, g = η + h, the inverse is g-1 = (η + h)-1 = η - ηhη + ... So again it's a helpful convention to raise and lower indices in the flat space using the flat space metric ημν and its inverse ημν, but the original definition of gμν as the inverse matrix of gμν must be kept.

Thanks Bill . I'm just trying to convince the little mathematician in my head ...