Stress-energy tensor and dependence on coordinate system in general relativity

Question

In Wikipedia, components of stress-energy tensor $T^{\alpha \beta}$ is defined as flux of $\alpha$th component of momentum vector across a surface with $x^{\beta}$ coordinate.

What I don't exactly understand is this. Does the choice of (hyper-)surfaces matter? That is, suppose we chose Euclidean coordinate system. Does this mean that surface has to be chosen as Euclidean rectangular "box" (orthogonal to coordinate basis chosen), then we take the limit as area vanishes to zero? I don't believe this is the case, but I am not sure if I am right.

(If surfaces can be non-rectangular-Euclidean-box, then do we need to consider in changes in metric tensor when calculating hypersurface area-or-volume?)

Answer 1

I have never liked the "box" analogy.

Rather than think about $T_{ab}$, I like to think about $T_{a}{}^{b}$, which maps vectors to vectors. Now, imagine probing this tensor with two vectors, a timelike one $u^{a}$, and a spacelike one $s^{a}$.

Then, the vector:

$$u^{a}T_{a}{}^{b}$$

describes the energy-momentum density flowing through spacetime, while the vector

$$s^{a}T_{a}{}^{b}$$

describes the Pressure flux flowing through the spacetime (note that the time component of this will be momentum, but this is reasonable, because

$$\nabla_{b}\left(s^{a}T_{a}{}^{b}\right)=0 \rightarrow {\dot {\vec p}} = {\vec \nabla} \cdot {P_{ij}} $$

(excuse some index abuse in this heuristic discussion), which makes sense from ordinary fluid mechanics.)

The reason why I prefer this interpretation is that it makes the connection with ordinary classical mechanics a lot clearer, and it is much more consistent with how people actually build stress-energy tensors when trying to set up an IVBP for Einstein's equation.

Answer 2

You need to distinguish between the concept of a tensor, and its representation in a particular coordinate system.

The concept of a stress tensor is a (linear) transformation between an infinitesimal element of a surface (defined by its normal direction) and the forces acting on the body if it was cut at that surface. That transformation gives the correct result for every possible surface element at a particular point, whatever its normal direction might be, and is is independent of any choice of coordinate system. In fact it has to be coordinate-system-independent, because the "laws of physics" don't care what coordinate system you use to describe them!

On the other hand the representation of a tensor (for example as an array of numbers) does depend on the choice of coordinate system, and in order to calculate something in a particular physical situation, usually some choices of coordinate system are much easier to work with than others. For example a good choice of coordinate system can make some the numerical coefficients equal to zero, and/or can capture some symmetry of the physical situation in a convenient way.

There is no particular reason why the coordinate system has to represent a "rectangular box" - though it is often convenient to make that choice, since every pair of normalized basis vectors $\vec e_i$ and $\vec e_j$ then satisfy $\vec e_i \cdot \vec e_j = \delta_{ij}$.