# Why is stress tensor (in this derivation) symmetric?

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1. Aug 25, 2015

### genxium

First by "this derivation" I'm referring to an online tutorial: http://farside.ph.utexas.edu/teaching/336L/Fluidhtml/node9.html

It's said in the above tutorial that the $i-th$ component of the total torque acting on a fluid element is

$\tau_i = \int_V \epsilon_{ijk} \cdot x_{j} \cdot F_{k} \cdot dV + \int_V \epsilon_{ijk} \cdot \sigma_{kj} \cdot dV + \int_V \epsilon_{ijk} \cdot x_{j} \cdot \frac{\partial \sigma_{kl}}{\partial x_{l}} \cdot dV$ -- (*)

where

$\epsilon_{ijk}$ is the permutation tensor (http://mathworld.wolfram.com/PermutationTensor.html),

$F_{i}$ is the i-th component of the "volume force" acting on the fluid element

and $\sigma_{ij}$ is the "stress tensor" such that the i-th component of the total force acting on the fluid element is

$f_{i} = \int_V F_{i} \cdot dV + \oint_{S=\partial V} \sigma_{ij} \cdot dS_{j}$

The tutorial states that (slightly rephrased but tried to keep the same meaning)

from

where "the second term" means $\int_V \epsilon_{ijk} \cdot \sigma_{kj} \cdot dV$ in (*).

It's not obvious to me why "the second term" doesn't approximate $0$ when $V \rightarrow 0$ but instead induces absurdly large angular velocity as it "scales as $V$ and $V \rightarrow 0$".

I understand that it's worth considering $V \ll 1$ thus $V \gg V^{4/3}$ but this is not convincing enough for me to take "the second term" to $0$ and yield the symmetry of $\sigma_{ij}$. For example, I could argue that even if the $x_j$ factor satisfies $x_j \ll 1$, i.e. fluid element very close to point $O$, in the first and third terms, it only makes the first and third terms "small" but NOT necessarily makes the second term "large" especially when $V \rightarrow 0$.

2. Aug 27, 2015

### Orodruin

Staff Emeritus
The first and third term cancel each other due to force equilibrium.

3. Aug 31, 2015

### genxium

Hi @Orodruin, sorry for the late response, I spent sometime reading the following chapters in the same tutorial and found that I still have problems in understanding this question.

Would you please explain or show some reference for

Why is the 1st term having such strong relationship with the 3rd one? In my understanding they're independent to each other (like, having different "origins"? I can't define this well at the moment) and they're just added together in equation (*).

Moreover how is your answer related to that "the 2nd term must be identically zero"?

4. Aug 31, 2015

### Orodruin

Staff Emeritus
Just like for any equilibrium, you need to have force as well as moment equilibrium for any small volume. You are currently working with the moment equilibrium and so you are writing down the moment acting on a small volume. The corresponding argumentation for the force equilibrium will give you $F_i + \partial_j \sigma_{ij} = 0$ and therefore your first and third terms will cancel out.

Requiring moment equilibrium then is giving that the second term is zero. This must be true for any volume and therefore the integrand must be zero.

5. Sep 1, 2015

### genxium

Is "moment equilibrium" required implicitly in equation (*)? I'm afraid this is the essential point by which I'm confused :)

To my understanding by equation (*) I'm looking at the $i-th$ component of torque of a "fluid element" wrt a "might be randomly chosen" axis $O$, and this "fluid element" is NOT EXPLICITLY REQUIRED to be static in this chapter (in fact, static fluid is discussed right in the following chapter: http://farside.ph.utexas.edu/teaching/336L/Fluidhtml/node10.html).

It's reasonable that the "fluid element" is not static but it's also required to match some kind of "equilibrium conditions", e.g. what you mentioned above, however this is not quite obvious to me. Would you mind explaining more explicitly? Is there any constraint on choosing the axis?

Btw, did you mean $F_i + \frac{\partial \sigma_{ij}}{\partial x_i} dV = 0$ by force equilibrium? The units would match in this way and it makes sense to me.