# A Confused about stress-energy tensor

1. Jun 1, 2016

### JonnyG

I understand the basics of the stress-energy tensor (I think) but I still have a couple questions about it. But first, I'd like to give a quick run down of what I do understand, and I would appreciate if one of you could correct me where I am wrong and also answer my questions afterward.

So consider a collection of matter in spacetime. We can assign each individual particle a four-momentum, but this would be highly impractical. However, for certain collections of matter, like a cloud of gas for example, we can approximate the entire collection with a single four-momentum vector. But this four-momentum vector isn't enough to entirely describe the collection. So we will use a covariant stress-energy tensor. Given a coordinate system, let $\{e_0, e_1, e_2, e_3\}$ be an orthonormal basis for our space. We can define the components of our tensor, in this coordinate system, by its action on the basis vectors:

1) We define $T(e_0, e_0) = T_{00}$ to be the energy density (mass-energy per unit volume) as measured by an observer whose four-velocity is $e_0$.

2) Given $i \neq j$, we define $-T(e_i, e_j)$ to be the momentum density of the matter in the $e_j$ direction, as measured by the observer with a four-velocity of $e_i$.

3) Given $i,k \neq j$, we define $T(e_i, e_k)$ to be the $i$'th component of the momentum in the $e_k$ direction, as measured by the observer with four-velocity $e_j$.

The action of $T$ on arbitrary vectors is thus determined by the above.

Question 1: Why do we require the tensor to be symmetric?\

Question 2: In 2), there is a negative sign infront of the $T(e_i, e_j)$. Why is this?

Just in case you're wondering, I used Wald's explanation of the stress-energy tensor on page 61-62, but replaced his arbitrary vectors with basis vectors as I want to really examine the components of the tensor.

2. Jun 1, 2016

### Staff: Mentor

Only for certain purposes. This approximation treats the cloud of gas as basically a point mass with no internal structure. The situations in which this is useful are very limited.

Yes.

No. In the orthonormal basis, $e_0$ is timelike and $e_1$, $e_2$, $e_3$ are spacelike (that's the only way to have an orthonormal basis in spacetime). So $e_0$ is the only basis vector that can represent a 4-velocity. Thus, $T(e_0, e_j)$ (where $j$ can be 1, 2, or 3) is s the momentum density in the $j$ direction as measured by an observer with 4-velocity $e_0$. (Note also that there is no minus sign in front.)

No. In addition to the issue described above, that only $e_0$ can be a 4-velocity, the basis vector $e_j$ doesn't appear in your $T$ expression at all, so what you're saying here makes no sense.

In fact there is no way to represent momentum using the stress-energy tensor; you can only represent momentum density (as shown above). Quantities like $T(e_i, e_j)$, where both $i$ and $j$ can be 1, 2, or 3, represent stresses: pressure if $i = j$, shear stress if $i \neq j$.

We don't "require" it to be symmetric arbitrarily; we demonstrate that it is symmetric by looking at the physics that it represents.

First, take the case where the second index is 0, i.e., a quantity like $T(e_i, e_0)$, where $i$ can be 1, 2, or 3. This is the energy flux in the $i$ direction as measured by an observer with 4-velocity $e_0$. Saying that the SET is symmetric is saying that this energy flux must be the same as the momentum density $T(e_0, e_i)$. There are various heuristic arguments that one can use to see that this is the case.

Second, take the case where both indexes are 1, 2, or 3, i.e., a quantity like $T(e_i, e_j)$, and $i \neq j$ (since the case where they are equal is on the diagonal and doesn't play a role in whether the tensor is symmetric). This is shear stress, as noted above, and there are, again, various heuristic arguments from the theory of shear stresses for why these must be symmetric, i.e,. $T(e_i, e_j) = T(e_j, e_i)$.

More formally, one can derive the symmetry of the SET, at least with its standard definition, by looking at how it is derived from the Lagrangian, or by looking at the Einstein Field Equation and noting that the Einstein tensor must be symmetric because of how it is derived from the Riemann tensor.

I don't think your equation there is correct; see above.

Last edited: Jun 1, 2016
3. Jun 1, 2016

### JonnyG

@PeterDonis

Thank you for your reply. I checked again and Wald indeed has a negative sign in the equation I mentioned. MTW also has a minus sign in front of that equation as well, on page 131 in box 5.1. Perhaps it is just a convention? I don't know.

4. Jun 2, 2016

### Staff: Mentor

I don't have my copies of either handy at the moment so I can't check to see the context. I'll defer further comment until I can.