Index Notation of div(a:b) and div(c^transpose d)

[chowdhury]

TL;DR

What is the index notation for divergence of tensor?

What is the index notation of divergence of product of 4th rank tensor and second rank tensor?

What is the index notation of divergence of 3rd rank tensor and vector?

div(a:b) = div(c^transpose. d)
Where a = 4th rank tensor, b is second rank tensor, c is 3rd rank tensor and d is a vector.

 

[anuttarasammyak]

A^{\mu\nu\alpha\beta}B_{\alpha\beta}:=F^{\mu\nu}
or
C^{\mu\nu\alpha}D_{\alpha}:=F^{\mu\nu}
and its divergence is
\frac{\partial F^{\mu\nu}}{\partial x^\nu}=F^{\mu\nu}_{\ \ \ ,\nu}:=G^{\mu}
or in GR with covariant derivative
F^{\mu\nu}_{\ \ \ :\nu}:=G^{\mu}
For all these equations you have to appoint which index and which index should be contracted by dummy indexes. The above shown is an example from many other possible ways.

 

[chowdhury]

Thanks. I am not familiar with the covariant and contra-variant formulations and their manipulations. Can it be written as below? $$ div(a:b) + \frac {\partial^2 G} {\partial t^2} = div(c^{transpose}. d) $$ $$ (a_{ijkl}b_{kl})_{,j} + G_{i,tt}= (c_{ijk}^{transpose} d_{,k}),j $$
$$ (a_{ijkl}b_{kl})_{,j} + G_{i,tt}= (c_{kij} d_{,k}),j $$

 

[anuttarasammyak]

I am not familiar with the symbols ":" and "." used here. Someone will confirm it.

Is d a scalar as you show gradient $d,_k$ ? I am not sure how to interpret "transpose" for 3 indexes entity as $c_{ijk}$. Einstein summation convention is usually for 4-spacetime coordinates, i.e. i=0,1,2,3. It may cause confusion to apply it for i=1,2,3 not including t.

I prefer to note ",t,t" than ",tt" for applying time derivative twice but it would be just a matter of taste.

 

[chowdhury]

1.) ":" means summation over repeated subscripts like $$a_{ijkl} b_{kl}$$ here sum over k and l for allowed.
2.) "." is just a matrix vector multiplication, like $$c_{ijk}d_{k}$$ summ over all allowed k.
3.) It is certainly allowable for index i, j,k,l to include with or without 0, as depending on the problem, here in my case, space, these are from the set of {x,y,z} or {1,2,3}, and 0 does not exist, as I exclusively denote time
4.) I mentioned in my original post d is a vector, for 3D, it is a (3x1) vector, for 2D it is a (2x1) vector.