# Stress-energy tensor diagonalization

1. May 12, 2010

### djy

This question probably applies to symmetric rank-2 tensors in general, but I've been thinking about it specifically in the context of the stress-energy tensor.

For any stress-energy tensor and any metric (with signature -, +, +, +), is it possible to find a coordinate transformation that a) diagonalizes the stress-energy tensor and b) transforms the metric to diag(-1, 1, 1, 1)?

In other words, it seems intuitive to me that, for any stress-energy tensor of a fluid element, one should be able to find an MCRF of the fluid element such that all off-diagonal components of the tensor are zero.

2. May 12, 2010

### cesiumfrog

Regards b: You can always find a field of basis vectors so that the metric (in that basis) takes that form everywhere, but that vector basis won't always correspond to a coordinate field basis. Alternatively, around any point, you can always choose coords so the metric is like Minkowski space when evaluated at exactly that point, but such a metric will not be constant (that choice of coords will give different values of the metric components when evaluated at different points in the neighbourhood) unless the region is free of intrinsic curvature.

3. May 12, 2010

### djy

Thanks -- but I should have specified that my question only concerns one point in spacetime. I agree that one can't find a coordinate basis in curved spacetime where the metric is Minkowskian everywhere.

4. May 12, 2010

### haushofer

About the stress energy tensor I would have to think, but for the metric this is the very definition of the vielbein:

$$g_{\mu\nu}e_{a}^{\mu}e_{b}^{\nu} = \eta_{ab}$$

which is based on the definition of a manifold which says a manifold is locally flat (and in GR thus locally Minkowski).

5. May 12, 2010

### Ich

The stress-energy tensor is diagonal in this basis only for perfect fluids.

6. May 12, 2010

### djy

I think you mean that, for perfect fluids, the tensor is diagonal and $$T^{11} = T^{22} = T^{33}$$.