# Diagonal Stress-Energy Tensor

What does it mean if the stress-energy Tensor ($T_{ij}$) is a diagonal matrix? (i.e. the only nonzero components are density energy $T_{00}$ and pressure $T_{11}$, $T_{22}$, $T_{33}$).

ADD: Where the density energy is constant and pressure is not constant.

• What do you mean by "pressure is not constant"? Do you mean inhomogeneous, anisotropic, or both? (Read this question if you're not familiar with those words.) – Michael Seifert Nov 21 '17 at 14:27
• Also, note that we can always pick the spatial components of $T_{ij}$ to be diagonal; it's a symmetric tensor, so it always has a set of orthogonal principal axes. – Michael Seifert Nov 21 '17 at 14:29
• @Michael Seifert, Sorry, I'm not expert in GR, I mean that the density energy component is a constant, while the stress components are functions of 3 variables (two spatial and one is time) – Alexander Pigazzini Nov 21 '17 at 14:33
• I don't think this is unclear at all. A diagonal stress-energy tensor just means you're using comoving coordinates. If the question is reopened I'll be happy to add an answer expanding on this. – John Rennie Nov 21 '17 at 15:18
• Also related: physics.stackexchange.com/q/184042/25301 – Kyle Kanos Nov 21 '17 at 15:46

To write down the components of the stress-energy tensor you need to choose a coordinate system, and the values of the components will depend on what coordinate system you choose. So when you ask:

the question is really:

And the simplest interpretation is that your coordinates are comoving i.e. the average four velocity of the stuff in the universe in your coordinate system is $(c,0,0,0)$. I say average velocity because in the case of a gas or dust the individual molecules/particles may have non-zero spatial velocities but if these are random they will average to zero on a larger scale.

To understand this I recommend you read my answer to Intuitive understanding of the elements in the stress-energy tensor. This explains how to understand the stress-energy tensor starting with the stress-energy tensor for a point particles:

$$T^{\alpha\beta}({\bf x},t) = \gamma m v^\alpha v^\beta$$

(where $v$ is the coordinate velocity, not the four velocity, and $v^0=c$).

If the velocities are random then all the $v^\alpha v^\beta$ terms will average to zero apart from the diagonal terms so we get a diagonal stress-energy tensor.

There's no special reason why either the energy density or pressure terms have to be constant in either space of time, but not that if they vary in space then the matter/energy distribution is likely to evolve with time into a form where the stress-energy tensor is no longer diagonal i.e. the stuff in your universe will acquire non-zero peculiar velocities.

The obvious example of a universe where the stress-energy tensor is diagonal in comoving coordinates is the FLRW universe, which approximately describes our universe.

• thank you very much for your interesting answer! a clarification.. Can I obtain the Stress-Energy Tensor from Einstein field equation directly? for example if I have an Einstein manifold endowed metric $g$, then I know Ricci-Tensor and Scalar Curvature..Can I calculate the Stress-Energy Tensor from Einstein field-equation directly? – Alexander Pigazzini Nov 22 '17 at 11:26
• @AlexanderPigazzini: yes you can. Typically we are interested in starting with $T$ and working out the metric, but you can do the reverse as well. The only problem is that in most cases you will end up with an unphysical stress-energy tensor e.g. involving negative energy densities or other forms of exotic matter. In fact this is how Alcubierre worked out the (exotic) matter distribution needed for his hyperspatial drive. – John Rennie Nov 22 '17 at 11:30
• ..for example I hypothesized an expanding universe so I did not consider the cosmological constant in the Einstein field equation..the stress-energy tensor that I found is a $4x4$-matrix..I write it here for rows: $(\frac{-\lambda c^4}{8 \pi G}, 0, 0, 0); (0, \frac{-\lambda c^4}{8 \pi G}(r^{10} e^{(\lambda/2)r^2}), 0, 0); (0, 0, \frac{-2 \lambda c^4}{h^2 r^4 \pi G}, 0); (0, 0, 0, \frac{-2 \lambda c^4 t^2}{h^2r^4 \pi G})$ are all constants except $r$ and $t$. – Alexander Pigazzini Nov 22 '17 at 13:41
• ..I realized now that I missed a piece at my previous comment ..the question: "How can I figure out if my Stress-Energy Tensor is a non-physical?" (my Stress-Energy Tensor is the one written for rows in the comment above) – Alexander Pigazzini Nov 22 '17 at 14:31