Fluid pressure at an interface

[joshmccraney]

Suppose we have an incompressible, viscous sessile drop subject to a time-dependent pressure field $p$ on a substrate. Let $\mu$ be dynamic viscosity, $u$ be the fluid velocity field, $\kappa_{1/2}$ curvatures of the fluid surface, $\sigma$ surface tension, $\hat n$ normals to the equilibrium surface, and $\eta$ the disturbed interface.

Disturbances to the equilibrium surface generate pressure gradients, and thereby flows. A pressure balance at the interfacial surface yields $$p-\mu \hat n \cdot(\nabla \otimes u) \cdot \hat n = - \sigma( \Delta_\Gamma \eta + (\kappa_1^2+\kappa_2^2)\eta)$$

The RHS is flow from the capillary pressure (Young-Laplace equation). The LHS is inertial pressure (first term) and viscous pressure (second term). I do not understand where the viscous pressure entered the pressure balance.

After googling I found this site: http://web.mit.edu/1.63/www/Lec-notes/Surfacetension/Lecture2.pdf

where equation (3) looks like the LHS, and if we look at their definition of $T$ we see there is a transpose velocity component (not shown in the pressure balance above, and the implication $\hat n \cdot -p I \cdot \hat n = -p$)? Can someone help me understand this? Thanks!

 

[Chestermiller]

I agree with the MIT analysis. What happened to the $(\nabla \otimes u)^T$ term in your equation for the stress tensor for a Newtonian fluid? Why is it missing?

 

[joshmccraney]

I agree with the MIT analysis. What happened to the $(\nabla \otimes u)^T$ term in your equation for the stress tensor for a Newtonian fluid? Why is it missing?

It's from this paper, equation (4): https://pubs.rsc.org/en/content/articlepdf/2016/sm/c6sm01928e

Author does not mention it. Do you have an idea?

 

[Chestermiller]

It's from this paper, equation (4): https://pubs.rsc.org/en/content/articlepdf/2016/sm/c6sm01928e

Author does not mention it. Do you have an idea?

Sorry. I don't have access to that article...retired and all.

 

[joshmccraney]

Sorry. I don't have access to that article...retired and all.

Attached is the Mathematical formulation. Everything before this is introduction, so not useful for my question.

 

[Chestermiller]

Isn't n dot (a tensor) dot n the same as n dot (the transpose of the tensor) dot n? If so, then all they are missing is a factor of 2.

 

[joshmccraney]

Isn't n dot (a tensor) dot n the same as n dot (the transpose of the tensor) dot n? If so, then all they are missing is a factor of 2.

So you think what they wrote is in fact wrong?

 

[joshmccraney]

Isn't n dot (a tensor) dot n the same as n dot (the transpose of the tensor) dot n? If so, then all they are missing is a factor of 2.

Also, what would be a good way to know if $\hat n \cdot \nabla u \cdot \hat n = \hat n \cdot (\nabla u)^T \cdot \hat n$? Like, where did your intuition come from?

 

[Chestermiller]

Also, what would be a good way to know if $\hat n \cdot \nabla u \cdot \hat n = \hat n \cdot (\nabla u)^T \cdot \hat n$? Like, where did your intuition come from?

Just evaluate it in Cartesian Coordinates and see whether it is correct.

 

[joshmccraney]

Just evaluate it in Cartesian Coordinates and see whether it is correct.

Thinking about it and reviewing notes, perhaps we don't need to show it's true for cartesian. We know $$ \nabla u = D + \Omega : \\
D \equiv \frac{1}{2} (\nabla u + \nabla u ^T),
\Omega \equiv \frac{1}{2} (\nabla u - \nabla u ^T)$$
and we know $D=D^T$ and $\Omega = -\Omega^T$. Then

$$n \cdot \nabla u \cdot n = n \cdot (D + \Omega) \cdot n$$

where we note $$n\cdot\Omega\cdot n = n_i \Omega_{ij} n_j = -n_i \Omega_{ji} n_j$$ so then $-\Omega = \Omega \implies \Omega = 0.$ Also we can show since $D = D^T$ that $\nabla u ^T = \nabla u$. I think this proves what we seek to show, right?

 

[Chestermiller]

The transpose of the velocity gradient tensor is not equal to the velocity gradient tensor itself.

 

[joshmccraney]

The transpose of the velocity gradient tensor is not equal to the velocity gradient tensor itself.

Can you elaborate please? I thought $n \cdot \nabla u \cdot n = n \cdot ( \nabla u)^T \cdot n$.

 

[Chestermiller]

Can you elaborate please? I thought $n \cdot \nabla u \cdot n = n \cdot ( \nabla u)^T \cdot n$.

That is correct, but it doesn’t guarantee that the gradient of the velocity and its transpose are equal.

 

[joshmccraney]

That is correct, but it doesn’t guarantee that the gradient of the velocity and its transpose are equal.

I must be missing something. Isn't it enough to show that $(n \cdot \nabla u \cdot n + n \cdot (\nabla u)^T \cdot n)/2 = n \cdot \nabla u \cdot n$?

 

[Chestermiller]

You seem to be trying to prove that $\nabla u=(\nabla u)^T$. Just because $(n \cdot \nabla u \cdot n + n \cdot (\nabla u)^T \cdot n)/2 = n \cdot \nabla u \cdot n$ is true doesn't necessarily mean that $\nabla u=(\nabla u)^T$. As a matter of fact, the latter is not generally correct.

 

[joshmccraney]

You seem to be trying to prove that $\nabla u=(\nabla u)^T$. Just because $(n \cdot \nabla u \cdot n + n \cdot (\nabla u)^T \cdot n)/2 = n \cdot \nabla u \cdot n$ is true doesn't necessarily mean that $\nabla u=(\nabla u)^T$. As a matter of fact, the latter is not generally correct.

Gotcha. But it is true that $(n \cdot \nabla u \cdot n + n \cdot (\nabla u)^T \cdot n)/2 = n \cdot \nabla u \cdot n$, right?

Also, when I take the inner product $$M_{ij} = \int (n \cdot \nabla \nabla \phi_i \cdot n) \phi_j : u = \nabla \phi$$ it turns out $M_{ij} \neq M_{ji}$. I have reason to think $M$ should be symmetric. What do you think?

 

[Chestermiller]

Gotcha. But it is true that $(n \cdot \nabla u \cdot n + n \cdot (\nabla u)^T \cdot n)/2 = n \cdot \nabla u \cdot n$, right?

Yes. You can show this using Cartesian coordinates.

Also, when I take the inner product $$M_{ij} = \int (n \cdot \nabla \nabla \phi_i \cdot n) \phi_j : u = \nabla \phi$$ it turns out $M_{ij} \neq M_{ji}$. I have reason to think $M$ should be symmetric. What do you think?

I think I don't understand this notation.

 

[joshmccraney]

Yes. You can show this using Cartesian coordinates.

Yes, I did this for the problem I'm working on and it's true.

I think I don't understand this notation.

Sorry, let me ask the question in a better way: is it true that the notation here are always equivalent $\nabla \otimes \vec u = \nabla \vec u$?

 

[Chestermiller]

Yes, I did this for the problem I'm working on and it's true.


Sorry, let me ask the question in a better way: is it true that the notation here are always equivalent $\nabla \otimes \vec u = \nabla \vec u$?

That's my understanding

 

[joshmccraney]

Thanks!