# Rank 2 covariant tensors and dimesionality

1. Feb 4, 2005

### SIlasX

I've already handed in my (I can only assume) incorrect solution, but I just felt like posting, though I'm not sure if anyone will be able to help.

I have a rank-2 covariant tensor, T sub i,j. This can be written in the form of t sub i,j + alpha*metric tensor*T super k, sub k (I hope my notation makes sense). I am also told that t super i, sub i = 0. I need to find alpha in d-dimensions.

I was thinking that I multiply the whole thing by the inverse metric tensor. That would give me a scalar on the right hand side of the equation and, I believe, would raise one index on t sub i,j so that I could use the given. I'm lost after that.

If it helps, it seems that this is somehow related to the dimensional coefficients of the Ricci tensors and curvature scalars in the Weyl tensor.

2. Feb 4, 2005

### dextercioby

Is that T arbitrary?
$$T_{ij}=t_{ij}+\alpha g_{ij}T^{k} \ _{k}$$ (1)

That means u decomposed the initial tensor into a sum of a tracless tensor (t) and the the trace of the initial tensor (surely,metaphorically speaking,the assertion needs to be understood in terms of space dimentionality).
Take trace of the (1)...You'll find alpha immediately.

Daniel.

3. Feb 4, 2005

### SIlasX

Thanks! :shy: