Scalar Vector Tensor Decomposition

[latentcorpse]

My notes are talking about this scalar vector tensor decopmosition business. Unfortunately, they are not online but they seem to follow this wikipedia article fairly closely:
http://en.wikipedia.org/wiki/Scalar-vector-tensor_decomposition

So our perturbed metric is of the form
[itex]ds^2=a^2(\eta) ( ( 1 + 2 \psi) d \eta^2 - 2 B_i dx^i d \eta - \left( (1-2 \phi) \delta_{ij} + 2 E_{ij} \right) dx^i dx^j)[/itex]
Now apparently the perturbation [itex]E_{ij}[/itex] is symmetric and trace free ([itex]\delta^{ij}E_{ij}=0[/itex])
We can decompose it as follows:
[itex]E_{ij}=\partial_{\langle i} \partial_{j \rangle} E + \partial_{(i} E_{j)}^T + E_{ij}^T[/itex]
where [itex]\partial_{\langle i} \partial_{j \rangle}E=\partial_i \partial_j E - \frac{1}{3} \delta_{ij} \nable^2 E[/itex]

(i) So apparently the first term in the decomposition is the scalar part. Why? It has two indices?

(ii) Again, why is the second term a vector? It also has two indices?

Then he goes on to say that the intial 5 degrees of freedom for [itex]E_{ij}[/itex] (3x3 matrix that's symmetric means 6 d.o.f. and then the constraint of trace free reduces this to 5) get re-distributed as follows:

the scalar gets one degree of freedom.

the vector has two degrees of freedom since it has 3 components and has the constraint that it must be divergence free.

the tensor has two degrees of freedom since [itex]E_{ij}^T[/itex] has five components but [itex]\delta^{ik} \partial_k E_{ij}^T[/itex] is three constraints.

(iii) What is up with the divergence free constraint on the vector bit. He doesn't mention (although he probably meant to) this and I don't see where it has come from?

(iv) I don't understand what [itex]\delta^{ik} \partial_k E_{ij}^T[/itex] is all about or how it leads to three constraints!

Thanks for any help!

 

[BigJDubs]

(i) The scalar part of the decomposition is the first term because it is a scalar quantity. It only has two indices because it is derived from the second-order partial derivatives of the perturbation E_{ij}, which is a second-order tensor. (ii) The second term is a vector because it is derived from the first-order partial derivatives of the perturbation E_{ij}. Since it has three components, it has two degrees of freedom. (iii) The divergence free constraint on the vector bit comes from the fact that the divergence of a vector must be zero for it to be considered divergence free. (iv) \delta^{ik} \partial_k E_{ij}^T is a contraction of the Kronecker delta (\delta^{ik}) and the partial derivative of the tensor part of the perturbation (\partial_k E_{ij}^T). This contraction leads to three constraints because there are three independent components of the partial derivative of the tensor part of the perturbation.