Frame indifference and stress tensor in Newtonian fluids

In summary: I will have to think more about it.In summary, 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
  • #1
bobinthebox
29
0
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?
 
Physics news on Phys.org
  • #2
For solids, you do this to get rid of rigid body rotations.
 
  • #3
@caz Could you be more explicit? I'm really sorry but I can't understand what you mean with
caz said:
get rid of rigid body rotations.
 
  • #4
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.
 
Last edited:
  • #5
caz said:
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.
 
  • #6
Thanks to both of you guys for your helpful answers.

@Chestermiller

Chestermiller said:
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.
Chestermiller said:
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?
 
  • #7
bobinthebox said:
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.
 
  • #8
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?
 
  • #9
bobinthebox said:
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.
 

FAQ: Frame indifference and stress tensor in Newtonian fluids

What is frame indifference in Newtonian fluids?

Frame indifference in Newtonian fluids refers to the property that the stress tensor, which describes the distribution of forces within a fluid, is independent of the reference frame used to measure it. This means that the stress tensor will have the same values regardless of whether it is measured in a stationary or moving reference frame.

How does frame indifference relate to the conservation of momentum?

Frame indifference is closely related to the conservation of momentum in Newtonian fluids. This is because the stress tensor is used to calculate the rate of change of momentum within a fluid, and the fact that it is frame indifferent ensures that the conservation of momentum holds true in all reference frames.

Can frame indifference be violated in certain situations?

Yes, frame indifference can be violated in non-Newtonian fluids, where the stress tensor is not independent of the reference frame. This can occur in fluids with complex rheological properties, such as viscoelastic fluids, where the stress tensor may depend on the history of deformation.

How is the stress tensor calculated in Newtonian fluids?

The stress tensor in Newtonian fluids is calculated using the constitutive equation, which relates the stress tensor to the rate of deformation of the fluid. This equation is based on the assumption that the fluid behaves in a linear, viscous manner, and is given by the product of the dynamic viscosity and the rate of strain tensor.

What are some practical applications of frame indifference and the stress tensor?

Frame indifference and the stress tensor have many practical applications in fields such as fluid mechanics, engineering, and geophysics. They are used to model and analyze the behavior of fluids in various systems, such as in the design of pipelines, pumps, and turbines, as well as in the study of atmospheric and oceanic flows.

Back
Top