Basic understanding of stress tensors in a fluid

Question

So, after having spent the last 9 hours attempting to understand the basic tenets of stress tensors in fluids, I can honestly say that I think I know less now than when I began.

My questions are these: what precisely is a stress tensor in the context of fluid dynamics, and why is it the case that stress tensors are necessarily symmetric in this context?

One of the resources I found referred to the entries of the stress tensor matrix as being the representations of the directional stresses on an infinitesimal cube in a flow, supposing that this cube of fluid is static. This would explain the symmetry of the tensor (since the fluid cube is not rotating, the shear stresses would have to be equal in magnitude at the corners), but it would also indicate to me that the perpendicular forces (the diagonal elements of the tensor) must all be zero, since the particle is also not translating in the fluid. I know this is incorrect, but why?

Disclaimer: I have never actually taken a course in physics. I'm in a graduate mathematics program, and this is my first time encountering any of these concepts.

Answer 1

$T^{ij}$ is nothing more or less than the flow of $i$-momentum across surfaces of constant $j$.¹ As a result, the force exerted across a surface $S$ with unit normal one-form components $n_j$ has components $$ F_{(S)}^i = \int\limits_S T^{ij} n_j \,\mathrm{d}^{d-1}x $$ in $d$ dimensions.

The argument for symmetry is not that the cube is static. The argument is that the cube cannot have infinite angular acceleration as its size shrinks. That is, because we are dealing with a continuous fluid, it should be well behaved as we take our region to be arbitrarily small.

Consider a two-dimensional example of a square covering the area $x_0-L/2 < x < x_0+L/2$, $y_0-L/2 < y < y_0+L/2$. The left surface $x = x_0-L/2$ has $n^\mathrm{left}_j = \delta_j^x$ (sign chosen to correspond to flow of momentum into the square), and so the force on our square due to interactions across the left face has components $F_\mathrm{left}^i \approx L T_\mathrm{left}^{ix}$, where $T_\mathrm{left}^{ix}$ is $T^{ix}$ evaluated at the midpoint $(x_0-L/2,y_0)$. On the opposite surface, $n^\mathrm{right}_j = -\delta_j^x$, so $F_\mathrm{right}^i \approx -L T_\mathrm{right}^{ix}$. Similarly, $F_\mathrm{bottom}^i \approx L T_\mathrm{bottom}^{iy}$ and $F_\mathrm{top}^i \approx -L T_\mathrm{top}^{iy}$.

Torque is a $(d-2)$-form: $\tau = {}^*(\tilde{r} \wedge \tilde{F})$, with $\tilde{r}$ and $\tilde{F}$ the one-forms corresponding to displacement $\vec{r}$ and force $\vec{F}$. In 2D, $\tau = \epsilon_{ij} (r^i F^j - r^j F^i)$. If force components $F_\mathrm{left}^i$ are applied at $(x_0-L/2,y_0)$, then $r_\mathrm{left}^i = -(L/2) \delta^i_x$ and $\tau_\mathrm{left} \approx -(1/2) L^2 T^{yx}$. You can also check $\tau_\mathrm{right} \approx -(1/2) L^2 T^{yx}$ and $\tau_\mathrm{bottom} \approx \tau_\mathrm{top} \approx (1/2) L^2 T^{xy}$.

As we shrink the square down, the midpoints at which we evaluate $T^{ij}$ approach one another and we find the total torque is $\tau \approx L^2 (T_\mathrm{center}^{xy} - T_\mathrm{center}^{yx})$. However, the moment of inertia for a square of surface density $\sigma$ is $\sigma L^4/6$. Thus angular acceleration is $$ \alpha = \lim_{L\to0} \frac{6(T^{xy}-T^{yx})}{\sigma L^2} $$ at any point in the fluid. Thus we must have $T^{xy} = T^{yx}$, in order to avoid $\alpha \to \infty$. Note that this argument holds in higher dimensions, in more general settings than fluids, and for more general geometries/spacetimes.²

The argument does not however hold in the linear acceleration case. For example, the net $x$-force will have terms like $L (T_\mathrm{left}^{xx} - T_\mathrm{right}^{xx})$ and $L (T_\mathrm{bottom}^{xy} - T_\mathrm{top}^{xy})$. Even though mass is $\sigma L^2$, which would seem to imply linear acceleration goes as $1/L$, the fact is the pairs of stress tensor components naturally cancel as they are evaluated at the same point ($T_\mathrm{left}, T_\mathrm{right}, T_\mathrm{bottom}, T_\mathrm{top} \to T_\mathrm{center}$). No constraints are imposed from this consideration.

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¹I avoid saying "in the $j$-direction, since we are really interested in surfaces, and these are characterized by one-forms, not vectors. This is more apparent in non-Cartesian (better still non-diagonal) coordinate systems.

²This symmetry always holds, for any material or field, as long as momentum is conserved. You occasionally see reference to the antisymmetric part of the stress tensor, but this comes from splitting the physics into separate domains, and pretending that momentum is lost when going from one to another (e.g. torques can transfer angular momentum from bulk flow into particle spins, and we choose to treat the latter as some momentum-conservation-violating sink as far as the continuum-modeled fluid is concerned).

Answer 2

I'm not an expert in fluid mechanics, but let me try to answer your question. Let us write the stress tensor as $\tau_{ij}$, with $i,j=x,y,z$. For definiteness, we look at $\tau_{xz}$. The meaning of this component is as follows: consider a directional face element of area $dA$ with normal pointing along $z$-direction, $\hat{z}dA$. Then, $\tau_{xz} dA$ gives you the force acting on this element along $x$-direction (a shear force). This is the definition of stress tensor. I would say, it is not guaranteed that it is symmetric: $\tau_{xz}$ is not always equal to $\tau_{zx}$. But, for isotropic materials, it is indeed symmetric. Actually, one can argue on the basis of thermodynamics (entropy production rate must be positive) that $\tau_{ij}$ may be written as $\tau_{ij} \sim \eta (\frac{\partial v_i}{\partial x_j}+\frac{\partial v_j}{\partial x_i})$, which is evidently symmetric. I'd recommend reading the book by P M Chaikin and T C Lubensky, Principles of Condensed Matter Physics.