# Rank and weight of a Tensor

1. Jan 1, 2009

### Jack3145

Given a Riemann Curvature Tensor. How do you know the weight and rank of each:

$$R^{i}_{jki}$$
$$R^{i}_{jik}$$
$$R^{i}_{ijk}$$

Is the Ricci tensor always a zero tensor for diagonal metric tensors?

2. Jan 1, 2009

### shoehorn

The rank of a tensor can be thought of as the number of distinct indices that the tensor has. Thus $R^i_{jkl}$ is a fourth-rank tensor, while the Ricci tensor $R^k_{ikj}=R_{ij}$ is a second-rank tensor. On the other hand, the Ricci scalar $R=R^i_i$ is a scalar quantity and hence a zero-rank tensor.

The weight of a tensor is defined to be the power of $\sqrt{-\det g_{ij}}$ that appear in the tensor.

3. Jan 1, 2009

### Jack3145

What tells the weight?

$$g_{ab}=(1,0,0,0;0,r^{2},0,0;0,0,r^{2}*(sin(\theta))^{2},0;0,0,0,-c^{2}*t^{2})$$
$$(-det(g_{ab}))^{1/2} = r^{2}*sin(\theta)*c*t$$

Last edited: Jan 1, 2009
4. Jan 1, 2009

### Mentz114

No. In fact, it's rarely zero. For instance if you replace 1-2m/r in the Schwarzschild metric with s-2m/r where s is a constant ne to 1, the Ricci tensor gets 2 components.