Frame indifference and stress tensor in Newtonian fluids

[bobinthebox]

During lecture today, we were given the constitutive equation for the Newtonian fluids, i.e. $T= - \pi I + 2 \mu D$ where $D=\frac{L + L^T}{2}$ is the symmetric part of the velocity gradient $L$. Dimensionally speaking, this makes sense to me: indeed the units are the one of a pressure. Also, since the velocity gradient is what we use to model the relative motion between fluid particles, it makes sense for $T$ to have such a form.

However, my professor said that we need the symmetric part of $L$, and not only $L$, because otherwise we loose frame indifference. Unfortunately, he didn't prove this. I'd like to get some feedback on the following argument which should prove the bold sentence.

Assume we have a rotation tensor $Q(t)$ and let $B_t$ and $B_t^{*}$ two configurations related by a rigid motion $x^{*}=Q(t)x + c(t)$. Then we have, for $L^{*}=\nabla_{x^*}v$ the following relation $L^{*}=QLQ^T + \dot{Q}Q^T$. Let's focus on the second addendum of $T$, i.e. $2 \mu D$. If we were only considering $S=2 \mu L$ instead of the symmetric part of $L$, we would obtain $S^{*}=2 \mu L^*=(2 \mu) (QLQ^T + \dot{Q}Q^T) = QSQ^T + 2 \mu \dot{Q}Q^T$ and from here we can see that we are not satisfying the requirement for frame indifference.

As another matter of fact, it's clear that if $S=2 \mu L$, then we don't have symmetry anymore, because of that $\dot{Q}Q^T$ term. However, this is only an algebraic point of view, which has nothing to do with the frame-indifference requirement.

Do you think that this may be what the professor had in mind? Or maybe is there some more physical interpretation?

 

[Frabjous]

For solids, you do this to get rid of rigid body rotations.

 

[bobinthebox]

@caz Could you be more explicit? I'm really sorry but I can't understand what you mean with

get rid of rigid body rotations.

 

[Frabjous]

Rigid body rotation has a velocity gradient and one does not want to generate stresses when a body undergoes rigid body rotation so one wants to define things in constitutive equations so that this does not appear. Defining D thusly accomplishes this. The antisymmetric counterpart is known as the spin tensor, W. L=D+W, so by symmetrizing L you are essentially subtracting out rigid body rotation.

@Chestermiller knows this stuff cold. Hopefully this will conjure him.

 

[Chestermiller]

Rigid body rotation has a velocity gradient and one does not want to generate stresses when a body undergoes rigid body rotation so one wants to define things in constitutive equations so that this does not appear. Defining D thusly accomplishes this. The antisymmetric counterpart is known as the spin tensor, W. L=D+W, so by symmetrizing L you are essentially subtracting out rigid body rotation.

@Chestermiller knows this stuff cold. Hopefully this will conjure him.

No argument with what either of you is saying.

Another perspective on this is that the state of stress in a material has to be independent of what the observer is doing. So, if the observer is rotating, this must not change the constitutive equation which translates the kinematics of the deformation into the stress tensor. If Newton's constitutive equation were based on the velocity gradient tensor, rather than the rate of deformation tensor, one would predict a state of stress that depends on the rotation rate of the observer.

 

[bobinthebox]

Thanks to both of you guys for your helpful answers.

@Chestermiller

No argument with what either of you is saying.

So do you agree with my "formal" argument? I've not been able to find this on the web, nor in the books I have.

If Newton's constitutive equation were based on the velocity gradient tensor, rather than the rate of deformation tensor, one would predict a state of stress that depends on the rotation rate of the observer.

I can't understand why you say "one would predict a state of stress that depends on the rotation rate of the observer." Do you refer to the $\dot{Q}Q^T$ arising from my computations, or is there some physical intuition/principle that I'm missing?

 

[Chestermiller]

Thanks to both of you guys for your helpful answers.

@ChestermillerSo do you agree with my "formal" argument? I've not been able to find this on the web, nor in the books I have.
I can't understand why you say "one would predict a state of stress that depends on the rotation rate of the observer." Do you refer to the $\dot{Q}Q^T$ arising from my computations, or is there some physical intuition/principle that I'm missing?

The physical principle is that the observer can't affect the state of stress.

 

[bobinthebox]

Thanks so much @Chestermiller. For what concerns my "formal" proof on the consequence for frame indifference of taking only the velocity gradient rather than its symmetric part is okay, right?

 

[Chestermiller]

Thanks so much @Chestermiller. For what concerns my "formal" proof on the consequence for frame indifference of taking only the velocity gradient rather than its symmetric part is okay, right?

In my judgment, yes.