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In the field of fluid mechanics, what is the momentum flux tensor? Is there an easy explanation for how it "works"?

Bernhard
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BillyJean
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2 Answers2

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The momentum flux tensor comes from the momentum equation of Navier-Stokes equations: $$ \frac{\partial\left(\rho\mathbf{u}\right)}{\partial t}+\nabla\cdot\mathbf{P}=0\tag{1} $$ Or, using indices (where it is easier to see that $\mathbf{P}$ is a rank-2 tensor): $$ \frac{\partial\left(\rho u_i\right)}{\partial t}+\frac{\partial\Pi_{ij}}{\partial x_j}=0\tag{2} $$ We can split this tensor into three components:

  1. advection of $i$-momentum in the $j$-direction: $\left(\rho u_i\right)u_j$
  2. pressure: $p\delta_{ij}$
  3. stress tensor: $\sigma_{ij}$

The first two are rather straight-forward (but can be elaborated on a little bit more if you need it), but the third is a little more complicated. We generally regard the stress tensor as traceless and symmetric: $$ \sigma_{ij}=\mu\left(\frac{\partial u_j}{\partial x_i}+\frac{\partial u_i}{\partial x_j}-\delta_{ij}\frac{2}{3}\,\frac{\partial u_i}{\partial x_i}\right) $$ where $\mu$ is a (fluid-dependent) constant, referred to as viscosity. This term describes the deformation of the fluid.


Edit

If we absorb bullets 2 and 3 into the single quantity $\tau_{ij}$, then Equation (1) becomes $$ \frac{\partial\left(\rho \mathbf{u}\right)}{\partial t}+\mathbf{u}\cdot\nabla\rho\mathbf{u}+\nabla\cdot\tau=0\tag{3} $$ The reason we do this is because of the subscript $j$ (see Equation (2)) on one of the $u$'s is identical to the gradient operator, $\nabla$. We can further describe the left two terms as a single entity most-often called the material derivative: $$ \frac{D}{Dt}=\frac{\partial}{\partial t}+\mathbf{u}\cdot\nabla\equiv D_t $$ From the Leibniz rule, $$ D_t\rho\mathbf{u}=\mathbf{u}D_t\rho+\rho D_t\mathbf{u} $$ And if we assume that the flow is incompressible, then $D_t\rho=0$ and Equation (3) becomes $$ \rho D_t \mathbf{u}+\nabla\cdot\tau=0\tag{4} $$

Kyle Kanos
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  • The stress tensor is a bit more easy for incompressible flows. – Bernhard Dec 18 '13 at 16:15
  • @Bernhard: thanks for the edit. Yes, it is easier for incompressible flows as the third term is zero. – Kyle Kanos Dec 18 '13 at 16:16
  • @KyleKanos Thanks, can I ask you to elaborate on how you go from the momentum eq. of NS to the index-version? – BillyJean Dec 18 '13 at 17:33
  • @BillyJean: Sure. All vectors can be written as $\mathbf{x}=(x_1,,x_2,,x_3)$ so we can simplify it to $\mathbf{x}=x_i$ Since $\mathbf{u}$ is a vector, then the result of $\nabla\cdot\mathbf{P}$ must also be a vector; since we know that $\nabla=(\partial_x,,\partial_y,,\partial_z)$ is a vector, then $\mathbf{P}$ must be an object that has 2 indices so that contracting it with $\nabla$ makes the resulting object have 1 index. – Kyle Kanos Dec 18 '13 at 17:39
  • This sort of thing should be explained in your standard fluid mechanics textbook (e.g. Kundu & Cohen, Acheson, etc) – Kyle Kanos Dec 18 '13 at 17:40
  • @KyleKanos Which of those books to you recommend for a "newbie"? Or perhaps Batchelors book instead, or Frank Whites? – BillyJean Dec 18 '13 at 18:06
  • White's text is probably a good start. – Kyle Kanos Dec 18 '13 at 18:10
  • @KyleKanos I believe it should be $\partial \Pi_{ij}/\partial x_i$, not $\partial \Pi_{ij}/\partial x_j$. Do you agree? – BillyJean Dec 18 '13 at 19:19
  • @BillyJean: No because there is an implicit summation on $j$, leaving $i$ as the only index. Note that it is $u_i$, not $u_j$ in the time-derivative. – Kyle Kanos Dec 18 '13 at 19:24
  • @KyleKanos Exactly, so the $x$-component of the term $\nabla \cdot \mathbf P$ becomes $\partial_x P_{11} + \partial_y P_{21} + \partial_z P_{31}$. Thus I believe it should read $\partial_i \Pi_{ij}$ – BillyJean Dec 18 '13 at 19:26
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    @BillyJean. Nope, see the wiki page on it. – Kyle Kanos Dec 18 '13 at 19:35
  • @KyleKanos I have read further about this. So I can derive the momentum equation starting from Newtons 2nd, $\rho D_tv = \nabla \cdot \tau$, where $\tau$ is the stress tensor containing #2 and #3 of your answer. But my notes never say anything about #1? Can I ask you to elaborate on this? Thanks – BillyJean Jan 02 '14 at 16:33
  • @BillyJean: See my edit. – Kyle Kanos Jan 02 '14 at 17:02
  • @KyleKanos Thanks, I agree 100% with (3), that is also derived in my book. My I really can't see how to go from (3) to (2)?! – BillyJean Jan 02 '14 at 17:19
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    @BillyJean: You have to realize that $\mathbf{u}\cdot\nabla\rho\mathbf{u}$ is really the "operator" $\left(\mathbf{u}\cdot\nabla\right)$ acting on $\rho\mathbf{u}$. From the partial-time derivative, we have that $\rho\mathbf{u}=\rho u_i$ which means we need a new index on the other $u$ present, which we give $j$. Since it is dotted with $\nabla$, we give that the index $j$ as well; thus we obtain the connection. – Kyle Kanos Jan 02 '14 at 17:32
  • @KyleKanos I got it.. you're good! Surface forces #3 (i.e., non-isotropic) I understand, as well as pressure (isotropic) #2. All good so far. But how does #1 enter, physically? What is it and why is it not taken into account by #3? – BillyJean Jan 02 '14 at 17:39
  • @BillyJean: It is the advection term, the motion of the parcel due to the bulk motion of the entire flow. – Kyle Kanos Jan 02 '14 at 17:49
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You can think about it in analogy with density and current. So let's review the situation with density and current.

Let's suppose I have a fluid whose spatially varying density is described by $\rho =\rho(\vec{x})$. Since the fluid is allowed to flow, $\rho$ may change with time. However, the change of $\rho$ with time is not arbitrary, since the mass cannot teleport, it needs to move continuously. Hence the change of $\rho$ must satisfy a continuity equation: $\partial_t \rho = -\vec{\nabla} \cdot \vec{J}$ where the vector field $\vec{J} = \vec{J}(\vec{x})$ is called the current. (For a fluid it would be defined by $\vec{J} = \rho \vec{u}$, where $\vec{u} = \vec{u}(\vec{x})$ is the velocity of the fluid.) In words, the continuity equation says that the change in density at a point is equal to the amount of matter flowing out of that point.

Now let's talk about the stress tensor. In addition the total mass, we know that the total momentum must also be conserved. We know that each little piece of fluid at position $\vec{x}$ carries some momentum $\vec{p}(\vec{x})$. (For a fluid, we know $\vec{p} = \rho \vec{u}$.) Also, we assume that each little piece of matter interacts only with its neighbors, (i.e., there are no long range forces just as coulomb forces). Thus the momentum density$\vec{p}(\vec{x})$ may change with time, but it must do so continuously. Let's look at the $i$th component of the momentum to keep things simple. This component of the momentum is analogous to the mass density in that it needs to satisfy a continuity equation $\partial_t p_i = \nabla_j \sigma_{ij}$. Here we see that mass current $J_j$ is analogous to the momentum current (or momentum flux as you say) $\sigma_{ij}$. The general equation is the same except you consider $i$ to be a free index. Since we know that momentum is transferred through forces, we interpret $\nabla_j \sigma_{ij}$ as the net force on the piece of the object due to its neighbors. So basically in this way $\sigma_{ij}$ tells you what the local forces (or we usually say stresses) are at each point in the object.

Brian Moths
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