# Fluid pressure at an interface

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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!

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Mentor
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?

Gold Member
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?

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

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Mentor
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.

Gold Member
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?

Gold Member
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?

Mentor
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.

Gold Member
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?

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

Gold Member
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##.

Mentor
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.

Gold Member
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 ##?

Mentor
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.

Gold Member
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?

Mentor
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.

Gold Member
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##?

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Mentor
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
Gold Member
Thanks!