Shear and the stress tensor of a Newtonian fluid

[Apashanka]

Similarly the paper by @buchert and @ehlers
https://arxiv.org/abs/astro-ph/9510056

Here the author has defined $v_{ij}=\frac{\partial v_i}{\partial x_j}=\frac{1}{2}(\frac{\partial v_i}{\partial x_j}+\frac{\partial v_j}{\partial x_i})+\frac{1}{2}((\frac{\partial v_i}{\partial x_j}-\frac{\partial v_j}{\partial x_i})+\frac{1}{3}\theta\delta_{ij}-\frac{1}{3}\theta\delta_{ij}=\sigma_{ij}+\frac{1}{3}\theta\delta_{ij}+\omega_{ij}$ where $\vec \omega=\frac{1}{2}(\vec \nabla×\vec v),v$ is the velocity.
Now the author has called the term $\sigma_{ij}$ the shear,where $\sigma_{ij}=\frac{1}{2}(\frac{\partial v_i}{\partial x_j}+\frac{\partial v_j}{\partial x_i})-\frac{1}{3}\theta\delta_{ij}$ but $\sigma_{ij}$ is the component of total stress tensor isn't it??

 

[Orodruin]

Not in the way the author has defined it. Just because a particular symbol is used for one purpose in some cases does not mean it always denotes that. If it did we would run out of symbols pretty fast. Then you can question the choice since it is pretty reasonable to expect a stress tensor here at some point.

Also note: It is technically not the shear, but the shear strain rate. In fluids stresses arise mainly due to the strain rate, not the strain itself.

 

[Apashanka]

Not in the way the author has defined it. Just because a particular symbol is used for one purpose in some cases does not mean it always denotes that. If it did we would run out of symbols pretty fast. Then you can question the choice since it is pretty reasonable to expect a stress tensor here at some point.

Also note: It is technically not the shear, but the shear strain rate. In fluids stresses arise mainly due to the strain rate, not the strain itself.

The author has mentioned the word shear for $\sigma$.i am just talking of the term not the symbol

 

[Orodruin]

Then no. What the author has given has nothing to do with stresses, it has to do with the strain rate. In order to know the stresses given the strain rate you need the viscosity tensor of the fluid.

 

[Chestermiller]

The author has mentioned the word shear for $\sigma$.i am just talking of the term not the symbol

As the author has defined it, $\sigma$ represents the deviatoric part of the rate of deformation tensor, equal to the actual rate of deformation tensor minus 1/3 its trace. This factors out the volumetric rate of change.

You will also note from the author's equations that the fluid is assumed inviscid.

 

[Apashanka]

As the author has defined it, $\sigma$ represents the deviatoric part of the rate of deformation tensor, equal to the actual rate of deformation tensor minus 1/3 its trace. This factors out the volumetric rate of change.

You will also note from the author's equations that the fluid is assumed inviscid.

The total stress tensor is $\Sigma_{ij}=-p\delta_{ij}+\tau_{ij}$ where $\tau_{ij}$ is the viscous stress tensor given by $\mu[(\frac{\partial}{\partial x_j}v_i+\frac{\partial}{\partial x_i}v_j]$ where $\mu$ is the coefficient of viscocity,Now how to relate this $\Sigma_{ij}$ to $\sigma_{ij}$ defined by the author??

 

[Chestermiller]

The total stress tensor is Σij=−pδij+τijΣij=−pδij+τij where τijτij is the viscous stress tensor given by μ[(∂∂xjvi+∂∂xivj]μ[(∂∂xjvi+∂∂xivj] where μμ is the coefficient of viscocity,Now how to relate this ΣijΣij to σijσij defined by the author??

Now, you are aware that the author is analyzing a fluid with zero viscosity, correct? Also, the author omitted the pressure gradient term in the equation of motion, so he is looking at essentially free-fall of a fluid.

 

[Apashanka]

I have made a schematic diagram and try to relate things(Newtonian fluid) ,
For 1-D motion

The shear stress is $\sigma=\mu\frac{\partial v_y}{\partial z}|_{(x,y)}$
For 2-D motion

Is the shear stress will be $\sigma=\mu(\frac{\partial v_x}{\partial z}+\frac{\partial v_y}{\partial z})|_{(x,y)}$
If not then what will be it in this context??

 

[Chestermiller]

I have made a schematic diagram and try to relate things(Newtonian fluid) ,
For 1-D motion View attachment 243307
The shear stress is $\sigma=\mu\frac{\partial v_y}{\partial z}|_{(x,y)}$
For 2-D motion
View attachment 243308
Is the shear stress will be $\sigma=\mu(\frac{\partial v_x}{\partial z}+\frac{\partial v_y}{\partial z})|_{(x,y)}$
If not then what will be it in this context??

In the author's development, $\mu$ is equal to zero.

 

[Apashanka]

In the author's development, $\mu$ is equal to zero.

Okk but are the terms correct for non zero$\mu$??

 

[Apashanka]

Now, you are aware that the author is analyzing a fluid with zero viscosity, correct? Also, the author omitted the pressure gradient term in the equation of motion, so he is looking at essentially free-fall of a fluid.

That means here the stress tensor is 0,as non-viscous and pressureless fluid,isn't it??

 

[Chestermiller]

According to Bird, Stewart, and Lightfoot, Transport Phenomena, $$\sigma_{ij}=\mu\left(\frac{\partial u_i}{\partial x_j}+\frac{\partial u_j}{\partial x_1}\right)-(\frac{2}{3}\mu-\kappa)\left(\frac{\partial u_x}{\partial x}+\frac{\partial u_y}{\partial y}+\frac{\partial u_z}{\partial z}\right)\delta_{ij}$$where $\kappa$ is the dilatational viscosity. For a mono-atomic gas, $\kappa$ is equal to zero.

 

[Orodruin]

According to Bird, Stewart, and Lightfoot, Transport Phenomena, $$\sigma_{ij}=\mu\left(\frac{\partial u_i}{\partial x_j}+\frac{\partial u_j}{\partial x_i}\right)-(\frac{2}{3}\mu-\kappa)\left(\frac{\partial u_x}{\partial x}+\frac{\partial u_y}{\partial y}+\frac{\partial u_z}{\partial z}\right)\delta_{ij}$$where $\kappa$ is the dilatational viscosity. For a mono-atomic gas, $\kappa$ is equal to zero.

Note that $\sigma$ here is the stress tensor, unlike in the paper linked in the OP, where it is the shear strain rate.

(Also, the 1 in the first term should be an i.)

 

[Chestermiller]

Note that $\sigma$ here is the stress tensor, unlike in the paper linked in the OP, where it is the shear strain rate.

Thanks. Actually, it's the viscous part of the stress tensor. In addition to this, there is an isotropic pressure.

(Also, the 1 in the first term should be an i.)

Thanks. Corrected the typo.

 

[Apashanka]

Note that $\sigma$ here is the stress tensor, unlike in the paper linked in the OP, where it is the shear strain rate.

(Also, the 1 in the first term should be an i.)

Ohh that means for the viscous part is given
$\sigma_{ij}=\mu\left(\frac{\partial u_i}{\partial x_j}+\frac{\partial u_j}{\partial x_1}\right)-(\frac{2}{3}\mu-\kappa)\left(\frac{\partial u_x}{\partial x}+\frac{\partial u_y}{\partial y}+\frac{\partial u_z}{\partial z}\right)\delta_{ij}$ therefore the total stress tensor is
$\sigma_{ij}=\mu\left(\frac{\partial u_i}{\partial x_j}+\frac{\partial u_j}{\partial x_1}\right)-(\frac{2}{3}\mu-\kappa)\left(\frac{\partial u_x}{\partial x}+\frac{\partial u_y}{\partial y}+\frac{\partial u_z}{\partial z}\right)\delta_{ij}+p\delta_{ij}$ ,isn't it??

 

[Orodruin]

Actually, it's the viscous part of the stress tensor. In addition to this, there is an isotropic pressure.

In general, sure.

It is actually a quite nice group theory exercise to show that the relation between the strain rate tensor and the viscous stress tensor can only depend on two constants (the bulk and shear viscosities), but that is probably out of scope here ...

 

[Orodruin]

Ohh that means for the viscous part is given
$\sigma_{ij}=\mu\left(\frac{\partial u_i}{\partial x_j}+\frac{\partial u_j}{\partial x_1}\right)-(\frac{2}{3}\mu-\kappa)\left(\frac{\partial u_x}{\partial x}+\frac{\partial u_y}{\partial y}+\frac{\partial u_z}{\partial z}\right)\delta_{ij}$ therefore the total stress tensor is
$\sigma_{ij}=\mu\left(\frac{\partial u_i}{\partial x_j}+\frac{\partial u_j}{\partial x_1}\right)-(\frac{2}{3}\mu-\kappa)\left(\frac{\partial u_x}{\partial x}+\frac{\partial u_y}{\partial y}+\frac{\partial u_z}{\partial z}\right)\delta_{ij}+p\delta_{ij}$ ,isn't it??

Almost. The force from pressure goes against the surface normal so the pressure term comes with a minus sign.

 

[Apashanka]

Therefore the total stress tensor is
σij=μ(∂ui∂xj+∂uj∂x1)−(23μ−κ)(∂ux∂x+∂uy∂y+∂uz∂z)δij−pδijσij=μ(∂ui∂xj+∂uj∂x1)−(23μ−κ)(∂ux∂x+∂uy∂y+∂uz∂z)δij−pδij ,here according to author p=0p=0 .
But for p=0p=0 this σijσij will have normal (##\sigma_{ii})and tangential components(shear),but the author has called σijσij as shear??

 

[Orodruin]

Again, what the author calls $\sigma$ is not the stress tensor, it is the shear strain rate (technically they used the term "rate of shear", but it is what it is, it is also called deviatoric strain rate).