Proof of/Reason for SVT Decomposition

  • #1

Main Question or Discussion Point

Using the conventions of these notes (not mine).

For a flat FRW perturbed universe, the metric is can be written in general as:
[tex]ds^2=a^2(\tau)[(1+2A)dt^2-B_idtdx^i-(\delta_{ij}+h_{ij})dx^idx^j][/tex]
I understand intuitively that we can decompose Bi into two parts:
[tex]B_i=B_i^{\perp}+B_i^{\parallel}[/tex]
with
[tex]\nabla \cdot B^{\perp}=\nabla \times B^{\parallel}=0[/tex]
In Fourier space, this means that we decompose the vector into two parts: one parallel to the wavevector k, and one perpendicular.

(And of course, we can write a curl-less vector as a gradient of a scalar.)

He then writes down a similar decomposition for the tensor hij (eq. 4.2.35-4.2.37). What's the reason/justification for the form of that?
 

Answers and Replies

  • #2
WannabeNewton
Science Advisor
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