I When to use which dimensionless number

[Chestermiller]

We are looking for the unit vector in the same direction as a super theta. You can’t just divide a vector by its magnitude and use the same symbol for it. Plus, if you are dividing a super theta by its own magnitude to get the unit vector, you need to do the algebra correctly and you need to use a new symbol for it.

 

[member 428835]

In post 37 you say:

Good. Next, determine the magnitude of $\hat{a}^{\theta}$ in terms of the g's. Then, divide $\hat{a}^{\theta}$ by its magnitude to get the in-plane unit vector $\hat{u}^{\theta}$ in the direction perpendicular to $\hat{a}^z$.

and in post 47 you show how to compute the magnitude of $\hat{a}^{\theta}$. What I tried doing in post 48 was dividing $\hat{a}^{\theta}$ by it's magnitude, making it a unit vector.

Also, did I use the same symbol? Thought I used your suggestion $\hat{u}^{\theta}$.

 

[member 428835]

Yea, I dropped a term in the algebra. We should have
$$
\hat u^\theta
=\frac{g_{zz}\hat{a}_\theta - g_{\theta z}\hat{a}_z} {\sqrt{g_{zz}}\sqrt{g_{zz}g_{\theta\theta}-g_{\theta z}^2}}.$$

 

[Chestermiller]

Yea, I dropped a term in the algebra. We should have
$$
\hat u^\theta
=\frac{g_{zz}\hat{a}_\theta - g_{\theta z}\hat{a}_z} {\sqrt{g_{zz}}\sqrt{g_{zz}g_{\theta\theta}-g_{\theta z}^2}}.$$

Right. Now, by extension, please write down what $\hat{u}^z$ would be.

 

[member 428835]

Right. Now, by extension, please write down what $\hat{u}^z$ would be.

Without showing any of the legwork since it's the same as your approach, we have $$\hat{u}^z = \frac{g_{\theta\theta} \hat a_z-g_{\theta z}\hat a_\theta}{\sqrt{g_{\theta \theta}}\sqrt{g_{zz}g_{\theta\theta}-g_{\theta z}^2}}.$$
Do you agree?

 

[Chestermiller]

Without showing any of the legwork since it's the same as your approach, we have $$\hat{u}^z = \frac{g_{\theta\theta} \hat a_z-g_{\theta z}\hat a_\theta}{\sqrt{g_{\theta \theta}}\sqrt{g_{zz}g_{\theta\theta}-g_{\theta z}^2}}.$$
Do you agree?

Yes. I'll be back later to continue.

 

[Chestermiller]

To refresh your memory, we are going to be performing a force balance over a "window" in the free surface formed by the intersection of planes of constant $\theta$ and z with our free surface. These intersections form grid lines on our free surface. The window will be situated between z and z + dz, and between $\theta$ and $\theta+d\theta$. In terms of $g_{zz}$, $g_{\theta z}$, $g_{\theta \theta}$, dz, and $d\theta$, what are the lengths of the four sides of our window?

 

[member 428835]

To refresh your memory, we are going to be performing a force balance over a "window" in the free surface formed by the intersection of planes of constant $\theta$ and z with our free surface. These intersections form grid lines on our free surface. The window will be situated between z and z + dz, and between $\theta$ and $\theta+d\theta$. In terms of $g_{zz}$, $g_{\theta z}$, $g_{\theta \theta}$, dz, and $d\theta$, what are the lengths of the four sides of our window?

I'm not quite sure what to write. I'm trying to compare what we have to a simpler problem. Geometrically, if we were in cylindrical coordinates the side lengths would be $r\Delta\theta,\Delta r,\Delta r,(r+\Delta r)\Delta \theta$. This makes me think the lengths should be something like $a \Delta z, b \Delta z, c \Delta \theta, d \Delta \theta$. The $g$-subs (as oppose to -supers) were dot products of the in plane vectors $a$-subs (not necessarily orthonormal). I'm stuck; any hints?

To make sure I understand what we've done, we started by finding two vectors that lie tangent in the plane and call them $a$-sub. Then we carefully aligned them with our selected coordinate system and called them $a$-super. Then we normalized them and called them $u$. Right?

 

[Chestermiller]

I'm not quite sure what to write. I'm trying to compare what we have to a simpler problem. Geometrically, if we were in cylindrical coordinates the side lengths would be $r\Delta\theta,\Delta r,\Delta r,(r+\Delta r)\Delta \theta$. This makes me think the lengths should be something like $a \Delta z, b \Delta z, c \Delta \theta, d \Delta \theta$. The $g$-subs (as oppose to -supers) were dot products of the in plane vectors $a$-subs (not necessarily orthonormal). I'm stuck; any hints?

To make sure I understand what we've done, we started by finding two vectors that lie tangent in the plane and call them $a$-sub. Then we carefully aligned them with our selected coordinate system and called them $a$-super. Then we normalized them and called them $u$. Right?

We found two vectors in the tangent plane that align with the curved grid lines of constant theta and constant z within the surface. These vectors, when multiplied by the corresponding differentials in the coordinates represent actual differential position vectors within the surface in terms of both their spatial length and their direction. Then we found the a-super vectors normal to the grid lines, with the characteristic that, when dotted with the subs, the dot product is equal to 1 or zero. We then used this to find the equations for unit vectors in the plane of the surface that are normal to the grid lines.

Regarding our little window in the surface, the sides of the window have differential position vectors along their edges (from corner to corner) of $\mathbf{a}_{\theta}d\theta$ and $\mathbf{a}_{z}dz$. So, what are the lengths of these edges?

 

[member 428835]

Regarding our little window in the surface, the sides of the window have differential position vectors along their edges (from corner to corner) of $\mathbf{a}_{\theta}d\theta$ and $\mathbf{a}_{z}dz$. So, what are the lengths of these edges?

Would the sides be $\sqrt{g_{\theta \theta}}|_zd\theta$, $\sqrt{g_{\theta \theta}}|_{z+dz}d\theta$, $\sqrt{g_{zz}}|_\theta dz$, $\sqrt{g_{zz}}|_{\theta+d\theta}dz$?

 

[Chestermiller]

Would the sides be $\sqrt{g_{\theta \theta}}|_zd\theta$, $\sqrt{g_{\theta \theta}}|_{z+dz}d\theta$, $\sqrt{g_{zz}}|_\theta dz$, $\sqrt{g_{zz}}|_{\theta+d\theta}dz$?

Josh! You're the man! Excellent.

Now we are ready to do a force balance on our little window. In terms of the unit normal u's, what are the in plane tensile forces due to surface tension on the 4 edges of our window (recall that surface tension acts normal to the sides of our window)? (There is also a pressure force acting perpendicular to our window that we'll get to soon)

 

[member 428835]

Josh! You're the man! Excellent.

Thanks!

Now we are ready to do a force balance on our little window. In terms of the unit normal u's, what are the in plane tensile forces due to surface tension on the 4 edges of our window (recall that surface tension acts normal to the sides of our window)? (There is also a pressure force acting perpendicular to our window that we'll get to soon)

I think they would be $$\sigma \sqrt{g_{\theta \theta}}|_zd\theta\, \hat u^z\\
\sigma\sqrt{g_{\theta \theta}}|_{z+dz}d\theta\, \hat u^z\\
\sigma\sqrt{g_{zz}}|_\theta dz\, \hat u^\theta \\
\sigma\sqrt{g_{zz}}|_{\theta+d\theta}dz\, \hat u^\theta$$

 

[Chestermiller]

Thanks!I think they would be $$\sigma \sqrt{g_{\theta \theta}}|_zd\theta\, \hat u^z\\
\sigma\sqrt{g_{\theta \theta}}|_{z+dz}d\theta\, \hat u^z\\
\sigma\sqrt{g_{zz}}|_\theta dz\, \hat u^\theta \\
\sigma\sqrt{g_{zz}}|_{\theta+d\theta}dz\, \hat u^\theta$$

Good, except that the ones at evaluated z and theta should be in the negative u directions. Now substitute your equations for the u's into these equations. What do you get?

Given the equations for the differential position vectors along the edges of the window, can you write a vector equation for the area of the window, including the normal vector to the window?

 

[member 428835]

Good, except that the ones at evaluated z and theta should be in the negative u directions. Now substitute your equations for the u's into these equations. What do you get?

$$\left.-\sigma \sqrt{g_{\theta \theta}}d\theta\, \frac{g_{\theta\theta} \hat a_z-g_{\theta z}\hat a_\theta}{\sqrt{g_{\theta \theta}}\sqrt{g_{zz}g_{\theta\theta}-g_{\theta z}^2}}\right|_z = \left.-\sigma d\theta\, \frac{g_{\theta\theta} \hat a_z-g_{\theta z}\hat a_\theta}{\sqrt{g_{zz}g_{\theta\theta}-g_{\theta z}^2}}\right|_z\\

\left.\sigma\sqrt{g_{\theta \theta}}d\theta\, \frac{g_{\theta\theta} \hat a_z-g_{\theta z}\hat a_\theta}{\sqrt{g_{\theta \theta}}\sqrt{g_{zz}g_{\theta\theta}-g_{\theta z}^2}}\right|_{z+dz} = \left.\sigma d\theta\, \frac{g_{\theta\theta} \hat a_z-g_{\theta z}\hat a_\theta}{\sqrt{g_{zz}g_{\theta\theta}-g_{\theta z}^2}}\right|_{z+dz}\\

-\left.\sigma\sqrt{g_{zz}} dz\, \frac{g_{zz}\hat{a}_\theta - g_{\theta z}\hat{a}_z} {\sqrt{g_{zz}}\sqrt{g_{zz}g_{\theta\theta}-g_{\theta z}^2}} \right|_{\theta} = -\left.\sigma dz\, \frac{g_{zz}\hat{a}_\theta - g_{\theta z}\hat{a}_z} {\sqrt{g_{zz}g_{\theta\theta}-g_{\theta z}^2}} \right|_{\theta}\\

\left.\sigma\sqrt{g_{zz}} dz\, \frac{g_{zz}\hat{a}_\theta - g_{\theta z}\hat{a}_z} {\sqrt{g_{zz}}\sqrt{g_{zz}g_{\theta\theta}-g_{\theta z}^2}} \right|_{\theta+d\theta} = \left.\sigma dz\, \frac{g_{zz}\hat{a}_\theta - g_{\theta z}\hat{a}_z} {\sqrt{g_{zz}g_{\theta\theta}-g_{\theta z}^2}} \right|_{\theta+d\theta}$$

Given the equations for the differential position vectors along the edges of the window, can you write a vector equation for the area of the window, including the normal vector to the window?

I think this would be
$$||\sqrt{g_{\theta \theta}}|_zd\theta\, \hat u^z \times \sqrt{g_{zz}}|_\theta dz\, \hat u^\theta|| \hat u^z \times \hat u^\theta = \sqrt{g_{\theta \theta}}|_zd\theta\cdot \sqrt{g_{zz}}|_\theta dz \, \hat u^z \times \hat u^\theta$$

 

[Chestermiller]

I think this would be
$$||\sqrt{g_{\theta \theta}}|_zd\theta\, \hat u^z \times \sqrt{g_{zz}}|_\theta dz\, \hat u^\theta|| \hat u^z \times \hat u^\theta = \sqrt{g_{\theta \theta}}|_zd\theta\cdot \sqrt{g_{zz}}|_\theta dz \, \hat u^z \times \hat u^\theta$$

It's much simpler than this. The area of the would be $\hat{a}_{\theta}\times \hat{a}_z d\theta dz$. So the pressure force acting on the window from inside to outside would be $p\hat{a}_{\theta}\times \hat{a}_z d\theta dz$, where p is the gauge pressure inside the fluid at the interface.

 

[member 428835]

It's much simpler than this. The area of the would be $\hat{a}_{\theta}\times \hat{a}_z d\theta dz$. So the pressure force acting on the window from inside to outside would be $p\hat{a}_{\theta}\times \hat{a}_z d\theta dz$, where p is the gauge pressure inside the fluid at the interface.

Of course, makes sense! I'm with you. So we've accounted for surface tension and pressure.

 

[Chestermiller]

Of course, makes sense! I'm with you. So we've accounted for surface tension and pressure.

Are you saying there are other forces acting on our window?

 

[member 428835]

Are you saying there are other forces acting on our window?

Well, we're ignoring gravity. Viscous forces though from the side walls, what do you think?

 

[Chestermiller]

Well, we're ignoring gravity. Viscous forces though from the side walls, what do you think?

I think that the interface has no mass, so that the gravitational force on the window is zero. The viscous contribution to the normal stress at the interface, when lumped in with the pressure (the shear stress is, of course, zero) might contribute a little, and we should consider that later (although I don't think it would be important). It should be neglected for now.

 

[member 428835]

I'm with you, and agree with everything you've said so far!

 

[Chestermiller]

I'm with you, and agree with everything you've said so far!

OK. Please write down the force balance on the window, and divide by dtheta and dz. What do you get?

 

[member 428835]

OK. Please write down the force balance on the window, and divide by dtheta and dz. What do you get?

$$
\left.\sigma d\theta\, \frac{g_{\theta\theta} \hat a_z-g_{\theta z}\hat a_\theta}{\sqrt{g_{zz}g_{\theta\theta}-g_{\theta z}^2}}\right|_{z+dz} \left.-\sigma d\theta\, \frac{g_{\theta\theta} \hat a_z-g_{\theta z}\hat a_\theta}{\sqrt{g_{zz}g_{\theta\theta}-g_{\theta z}^2}}\right|_z+
\left.\sigma dz\, \frac{g_{zz}\hat{a}_\theta - g_{\theta z}\hat{a}_z} {\sqrt{g_{zz}g_{\theta\theta}-g_{\theta z}^2}} \right|_{\theta+d\theta}-\left.\sigma dz\, \frac{g_{zz}\hat{a}_\theta - g_{\theta z}\hat{a}_z} {\sqrt{g_{zz}g_{\theta\theta}-g_{\theta z}^2}} \right|_{\theta}+
p\hat{a}_{\theta}\times \hat{a}_z d\theta dz = \vec 0\\

\frac{\partial}{\partial z}\left(\sigma \frac{g_{\theta\theta} \hat a_z-g_{\theta z}\hat a_\theta}{\sqrt{g_{zz}g_{\theta\theta}-g_{\theta z}^2}}\right)+
\frac{\partial}{\partial \theta}\left(\sigma \frac{g_{zz}\hat{a}_\theta - g_{\theta z}\hat{a}_z} {\sqrt{g_{zz}g_{\theta\theta}-g_{\theta z}^2}} \right)+
p\hat{a}_{\theta}\times \hat{a}_z = \vec 0
$$

where the rhs is 0 since the membrane does not have any mass. Something doesn't seem right though: the pressure term is the only term that is normal to the surface. If my work above is correct, then there would not be anything to balance it, implying $p=0$. What do you think?

 

[Chestermiller]

$$
\left.\sigma d\theta\, \frac{g_{\theta\theta} \hat a_z-g_{\theta z}\hat a_\theta}{\sqrt{g_{zz}g_{\theta\theta}-g_{\theta z}^2}}\right|_{z+dz} \left.-\sigma d\theta\, \frac{g_{\theta\theta} \hat a_z-g_{\theta z}\hat a_\theta}{\sqrt{g_{zz}g_{\theta\theta}-g_{\theta z}^2}}\right|_z+
\left.\sigma dz\, \frac{g_{zz}\hat{a}_\theta - g_{\theta z}\hat{a}_z} {\sqrt{g_{zz}g_{\theta\theta}-g_{\theta z}^2}} \right|_{\theta+d\theta}-\left.\sigma dz\, \frac{g_{zz}\hat{a}_\theta - g_{\theta z}\hat{a}_z} {\sqrt{g_{zz}g_{\theta\theta}-g_{\theta z}^2}} \right|_{\theta}+
p\hat{a}_{\theta}\times \hat{a}_z d\theta dz = \vec 0\\

\frac{\partial}{\partial z}\left(\sigma \frac{g_{\theta\theta} \hat a_z-g_{\theta z}\hat a_\theta}{\sqrt{g_{zz}g_{\theta\theta}-g_{\theta z}^2}}\right)+
\frac{\partial}{\partial \theta}\left(\sigma \frac{g_{zz}\hat{a}_\theta - g_{\theta z}\hat{a}_z} {\sqrt{g_{zz}g_{\theta\theta}-g_{\theta z}^2}} \right)+
p\hat{a}_{\theta}\times \hat{a}_z = \vec 0
$$

where the rhs is 0 since the membrane does not have any mass. Something doesn't seem right though: the pressure term is the only term that is normal to the surface. If my work above is correct, then there would not be anything to balance it, implying $p=0$. What do you think?

You can factor out the $\sigma$ from the derivatives. Regarding your question, the spatial derivatives of the coordinate basis vectors $\hat{a}_{\theta}$ and $\hat{a}_z$ along the surface have components normal to the surface. This is what the pressure has to balance.

In anticipation of what we do next, what is the magnitude of the vector $\hat{a}_{\theta}\times \hat{a}_z$?

 

[member 428835]

From your previous post we have $$\hat{a}_\theta=\frac{\partial R}{\partial \theta}\hat{r}+R\hat{\theta}\tag{3a}$$and$$\hat{a}_z=\frac{\partial R}{\partial z}\hat{r}+\hat{z}\tag{3b}\implies \\
\hat{a}_\theta \times \hat{a}_z = \left( \frac{\partial R}{\partial \theta}\hat{r}+R\hat{\theta} \right) \times \left( \frac{\partial R}{\partial z}\hat{r}+\hat{z} \right)\\
=-R \frac{\partial R}{\partial z}\hat z-\frac{\partial R}{\partial \theta}\hat \theta+R\hat r \implies\\
||\hat{a}_\theta \times \hat{a}_z|| = \sqrt{\left( R \frac{\partial R}{\partial z} \right)^2+\left( \frac{\partial R}{\partial \theta} \right)^2+R^2}$$
Geometrically this would be the area of a window along the surface. Is this what you're looking for?

 

[Chestermiller]

From your previous post we have $$\hat{a}_\theta=\frac{\partial R}{\partial \theta}\hat{r}+R\hat{\theta}\tag{3a}$$and$$\hat{a}_z=\frac{\partial R}{\partial z}\hat{r}+\hat{z}\tag{3b}\implies \\
\hat{a}_\theta \times \hat{a}_z = \left( \frac{\partial R}{\partial \theta}\hat{r}+R\hat{\theta} \right) \times \left( \frac{\partial R}{\partial z}\hat{r}+\hat{z} \right)\\
=-R \frac{\partial R}{\partial z}\hat z-\frac{\partial R}{\partial \theta}\hat \theta+R\hat r \implies\\
||\hat{a}_\theta \times \hat{a}_z|| = \sqrt{\left( R \frac{\partial R}{\partial z} \right)^2+\left( \frac{\partial R}{\partial \theta} \right)^2+R^2}$$
Geometrically this would be the area of a window along the surface. Is this what you're looking for?

It's also equal to $\sqrt{g_{zz}g_{\theta \theta}-g_{\theta z}^2}$. Can you show this?

 

[member 428835]

It's also equal to $\sqrt{g_{zz}g_{\theta \theta}-g_{\theta z}^2}$. Can you show this?

I don't think so. I know we've seen that show up a bunch in the denominators. It also looks like a Jacobian, which relates transformed area. How would you show it, or do you have any suggestions on how to start me out?

 

[Chestermiller]

I don't think so. I know we've seen that show up a bunch in the denominators. It also looks like a Jacobian, which relates transformed area. How would you show it, or do you have any suggestions on how to start me out?

Here's a hint. The cross product of two vectors is equal to the magnitudes of the two vectors times the sine of the angle between them. The cosine of the angle between two vectors is equal to their dot product divided by the magnitudes of the two vectors. sin^2=1-cos^2

 

[member 428835]

Here's a hint. The cross product of two vectors is equal to the magnitudes of the two vectors times the sine of the angle between them. The cosine of the angle between two vectors is equal to their dot product divided by the magnitudes of the two vectors. sin^2=1-cos^2

Beautiful. We have $$|\hat a_\theta \times \hat a_z| = |\hat a_\theta|| \hat a_z|\sin\theta\\
= |\hat a_\theta|| \hat a_z|\sqrt{1-\cos^2\theta}\\
=|\hat a_\theta|| \hat a_z|\sqrt{1-\left(\frac{\hat a_\theta\cdot\hat a_z}{|\hat a_\theta|| \hat a_z|}\right)^2}\\
=\sqrt{(|\hat a_\theta|| \hat a_z|)^2-(\hat a_\theta\cdot \hat a_z)^2}\\
=\sqrt{\sqrt{\hat a_\theta\cdot\hat a_\theta}^2\sqrt{\hat a_z\cdot\hat a_z}^2-(\hat a_\theta\cdot \hat a_z)^2}\\
=\sqrt{g_{\theta\theta}g_{zz}-g_{\theta z}^2}$$

 

[Chestermiller]

OK. So, now going back to post #72, we have:
$$\sigma \frac{\partial}{\partial z}\left(\frac{g_{\theta\theta} \hat a_z-g_{\theta z}\hat a_\theta}{\sqrt{g_{zz}g_{\theta\theta}-g_{\theta z}^2}}\right)+
\sigma \frac{\partial}{\partial \theta}\left( \frac{g_{zz}\hat{a}_\theta - g_{\theta z}\hat{a}_z} {\sqrt{g_{zz}g_{\theta\theta}-g_{\theta z}^2}} \right)+
p\hat{N}= \hat{0}$$where $\hat{N}=\hat{a}_{\theta}\times \hat{a}_z$. The next step is to determine the pressure p by dotting the equation with $\hat{N}$ to obtain:
$$\sigma \hat{N}\centerdot \frac{\partial}{\partial z}\left(\frac{g_{\theta\theta} \hat a_z-g_{\theta z}\hat a_\theta}{\sqrt{g_{zz}g_{\theta\theta}-g_{\theta z}^2}}\right)+
\sigma \hat{N}\centerdot \frac{\partial}{\partial \theta}\left( \frac{g_{zz}\hat{a}_\theta - g_{\theta z}\hat{a}_z} {\sqrt{g_{zz}g_{\theta\theta}-g_{\theta z}^2}} \right)+
p(g_{zz}g_{\theta\theta}-g_{\theta z}^2)= 0$$
See if you can show that this is the same as:
$$\sigma \left(\frac{g_{\theta\theta} \hat a_z-g_{\theta z}\hat a_\theta}{\sqrt{g_{zz}g_{\theta\theta}-g_{\theta z}^2}}\right)\centerdot \frac{\partial \hat{N}}{\partial z}+
\sigma \left( \frac{g_{zz}\hat{a}_\theta - g_{\theta z}\hat{a}_z} {\sqrt{g_{zz}g_{\theta\theta}-g_{\theta z}^2}} \right)\centerdot \frac{\partial \hat{N}}{\partial \theta}+
p(g_{zz}g_{\theta\theta}-g_{\theta z}^2)= 0$$

 

[member 428835]

See if you can show that this is the same as:
$$\sigma \left(\frac{g_{\theta\theta} \hat a_z-g_{\theta z}\hat a_\theta}{\sqrt{g_{zz}g_{\theta\theta}-g_{\theta z}^2}}\right)\centerdot \frac{\partial \hat{N}}{\partial z}+
\sigma \left( \frac{g_{zz}\hat{a}_\theta - g_{\theta z}\hat{a}_z} {\sqrt{g_{zz}g_{\theta\theta}-g_{\theta z}^2}} \right)\centerdot \frac{\partial \hat{N}}{\partial \theta}+
p(g_{zz}g_{\theta\theta}-g_{\theta z}^2)= 0$$

I have no clue how you did this. What I'm thinking, though it doesn't seem applicable here, is if you rotate a vector $\vec a$ by a given amount and dot it with $\vec b$ it will be the same as if you rotate $\vec b$ by the given amount and dot it with $\vec a$. But the derivatives aren't rotations. Honestly, to me this looks like dark magic.

 

[Chestermiller]

What is the dot product of N with the expressions in parenthesis?

 

[member 428835]

From the second to last equation in post 79, I'll consider the far left term (dropping the constant $\sigma$). The product rule implies
$$\frac{\partial \hat{N}}{\partial z} \centerdot \left(\frac{g_{\theta\theta} \hat a_z-g_{\theta z}\hat a_\theta}{\sqrt{g_{zz}g_{\theta\theta}-g_{\theta z}^2}}\right)=\frac{\partial}{\partial z}\left[\hat{N}\centerdot\left(\frac{g_{\theta\theta} \hat a_z-g_{\theta z}\hat a_\theta}{\sqrt{g_{zz}g_{\theta\theta}-g_{\theta z}^2}}\right)\right] -\hat{N}\centerdot \frac{\partial}{\partial z}\left(\frac{g_{\theta\theta} \hat a_z-g_{\theta z}\hat a_\theta}{\sqrt{g_{zz}g_{\theta\theta}-g_{\theta z}^2}}\right)$$
By the definition of $\hat N$ we have
$$\hat{N}\centerdot\left(\frac{g_{\theta\theta} \hat a_z-g_{\theta z}\hat a_\theta}{\sqrt{g_{zz}g_{\theta\theta}-g_{\theta z}^2}}\right) = (\hat a_\theta \times \hat a_z)\centerdot\left(\frac{g_{\theta\theta} \hat a_z-g_{\theta z}\hat a_\theta}{\sqrt{g_{zz}g_{\theta\theta}-g_{\theta z}^2}}\right)\\
=\left(\frac{g_{\theta\theta} (\hat a_\theta \times \hat a_z)\centerdot \hat a_z-g_{\theta z}(\hat a_\theta \times \hat a_z)\centerdot\hat a_\theta}{\sqrt{g_{zz}g_{\theta\theta}-g_{\theta z}^2}}\right).
$$
Notice $(\hat a_\theta \times \hat a_z)\centerdot \hat a_z = 0$ and $(\hat a_\theta \times \hat a_z)\centerdot\hat a_\theta = 0$ since the dot product of two orthogonal vectors is zero. Thus the product inside the $z$ derivative from the first equation is zero, and we are left with

$$\frac{\partial \hat{N}}{\partial z} \centerdot \left(\frac{g_{\theta\theta} \hat a_z-g_{\theta z}\hat a_\theta}{\sqrt{g_{zz}g_{\theta\theta}-g_{\theta z}^2}}\right)= -\hat{N}\centerdot \frac{\partial}{\partial z}\left(\frac{g_{\theta\theta} \hat a_z-g_{\theta z}\hat a_\theta}{\sqrt{g_{zz}g_{\theta\theta}-g_{\theta z}^2}}\right)
$$
Then I agree with your final equation in post 79 except that I think the two $\sigma$ terms are off by a sign. What do you think?

 

[Chestermiller]

From the second to last equation in post 79, I'll consider the far left term (dropping the constant $\sigma$). The product rule implies
$$\frac{\partial \hat{N}}{\partial z} \centerdot \left(\frac{g_{\theta\theta} \hat a_z-g_{\theta z}\hat a_\theta}{\sqrt{g_{zz}g_{\theta\theta}-g_{\theta z}^2}}\right)=\frac{\partial}{\partial z}\left[\hat{N}\centerdot\left(\frac{g_{\theta\theta} \hat a_z-g_{\theta z}\hat a_\theta}{\sqrt{g_{zz}g_{\theta\theta}-g_{\theta z}^2}}\right)\right] -\hat{N}\centerdot \frac{\partial}{\partial z}\left(\frac{g_{\theta\theta} \hat a_z-g_{\theta z}\hat a_\theta}{\sqrt{g_{zz}g_{\theta\theta}-g_{\theta z}^2}}\right)$$
By the definition of $\hat N$ we have
$$\hat{N}\centerdot\left(\frac{g_{\theta\theta} \hat a_z-g_{\theta z}\hat a_\theta}{\sqrt{g_{zz}g_{\theta\theta}-g_{\theta z}^2}}\right) = (\hat a_\theta \times \hat a_z)\centerdot\left(\frac{g_{\theta\theta} \hat a_z-g_{\theta z}\hat a_\theta}{\sqrt{g_{zz}g_{\theta\theta}-g_{\theta z}^2}}\right)\\
=\left(\frac{g_{\theta\theta} (\hat a_\theta \times \hat a_z)\centerdot \hat a_z-g_{\theta z}(\hat a_\theta \times \hat a_z)\centerdot\hat a_\theta}{\sqrt{g_{zz}g_{\theta\theta}-g_{\theta z}^2}}\right).
$$
Notice $(\hat a_\theta \times \hat a_z)\centerdot \hat a_z = 0$ and $(\hat a_\theta \times \hat a_z)\centerdot\hat a_\theta = 0$ since the dot product of two orthogonal vectors is zero. Thus the product inside the $z$ derivative from the first equation is zero, and we are left with

$$\frac{\partial \hat{N}}{\partial z} \centerdot \left(\frac{g_{\theta\theta} \hat a_z-g_{\theta z}\hat a_\theta}{\sqrt{g_{zz}g_{\theta\theta}-g_{\theta z}^2}}\right)= -\hat{N}\centerdot \frac{\partial}{\partial z}\left(\frac{g_{\theta\theta} \hat a_z-g_{\theta z}\hat a_\theta}{\sqrt{g_{zz}g_{\theta\theta}-g_{\theta z}^2}}\right)
$$
Then I agree with your final equation in post 79 except that I think the two $\sigma$ terms are off by a sign. What do you think?

I don't think so, but we can check that later. So, from all this, if you do all the arithmetic, what equation do you get for the pressure p? We can compare notes.

 

[member 428835]

I get (using the negative result I got) $$p = \frac{\sigma}{\left(g_{zz}g_{\theta\theta}-g_{\theta z}^2\right)^{3/2}} \left[ \left(g_{\theta\theta} \hat a_z-g_{\theta z}\hat a_\theta\right)\centerdot \frac{\partial \hat{N}}{\partial z}+
\left( g_{zz}\hat{a}_\theta - g_{\theta z}\hat{a}_z \right)\centerdot \frac{\partial \hat{N}}{\partial \theta}\right]
$$
Nothing special here except simple algebra.

 

[Chestermiller]

I get (using the negative result I got) $$p = \frac{\sigma}{\left(g_{zz}g_{\theta\theta}-g_{\theta z}^2\right)^{3/2}} \left[ \left(g_{\theta\theta} \hat a_z-g_{\theta z}\hat a_\theta\right)\centerdot \frac{\partial \hat{N}}{\partial z}+
\left( g_{zz}\hat{a}_\theta - g_{\theta z}\hat{a}_z \right)\centerdot \frac{\partial \hat{N}}{\partial \theta}\right]
$$
Nothing special here except simple algebra.

I meant to flesh out N, and do the dot products.

 

[member 428835]

I meant to flesh out N, and do the dot products.

hahahahhaha I thought so! To be clear, we're talking about taking the coordinates back into $z$ and $r$, right? I'll wait to crunch the numbers until you confirm. I know sometimes you have tricks up your sleeve.

 

[Chestermiller]

hahahahhaha I thought so! To be clear, we're talking about taking the coordinates back into $z$ and $r$, right? I'll wait to crunch the numbers until you confirm. I know sometimes you have tricks up your sleeve.

I'll try to write out my result so we can compare.

 

[Chestermiller]

OK, here's what I got for the pressure (if I did the arithmetic correctly). I corrected the sign error that you pointed out. Please check my results:

$$p=\frac{1}{R\left[1+\left[\frac{\partial R}{\partial z}\right)^2+\left(\frac{1}{R}\frac{\partial R}{\partial \theta}\right)^2\right]^{1/2}}-\frac{\left[1+\left(\frac{1}{R}\frac{\partial R}{\partial \theta}\right)^2\right]\frac{\partial^2 R}{\partial z^2}-2\frac{\partial R}{\partial z}\left(\frac{1}{R}\frac{\partial R}{\partial \theta}\right)\frac{1}{R}\frac{\partial ^2R}{\partial \theta \partial z}+\left[1+\left(\frac{\partial R}{\partial z}\right)\right]^2\frac{1}{R^2}\frac{\partial^2R}{\partial \theta^2}-\frac{1}{R}\left(\frac{1}{R}\frac{\partial R}{\partial \theta}\right)^2}{\left[1+\left[\frac{\partial R}{\partial z}\right)^2+\left(\frac{1}{R}\frac{\partial R}{\partial \theta}\right)^2\right]^{3/2}}$$

 

[member 428835]

OK, here's what I got for the pressure (if I did the arithmetic correctly). I corrected the sign error that you pointed out. Please check my results:

$$p=\frac{1}{R\left[1+\left[\frac{\partial R}{\partial z}\right)^2+\left(\frac{1}{R}\frac{\partial R}{\partial \theta}\right)^2\right]^{1/2}}-\frac{\left[1+\left(\frac{1}{R}\frac{\partial R}{\partial \theta}\right)^2\right]\frac{\partial^2 R}{\partial z^2}-2\frac{\partial R}{\partial z}\left(\frac{1}{R}\frac{\partial R}{\partial \theta}\right)\frac{1}{R}\frac{\partial ^2R}{\partial \theta \partial z}+\left[1+\left(\frac{\partial R}{\partial z}\right)\right]^2\frac{1}{R^2}\frac{\partial^2R}{\partial \theta^2}-\frac{1}{R}\left(\frac{1}{R}\frac{\partial R}{\partial \theta}\right)^2}{\left[1+\left[\frac{\partial R}{\partial z}\right)^2+\left(\frac{1}{R}\frac{\partial R}{\partial \theta}\right)^2\right]^{3/2}}$$

Hmmmm I'm not getting exactly this. Did you compute this result in Mathematica? If so I can send you my notebook. I plugged your result and mine in Mathematica and I am not getting zero when fully simplifying the difference of them both.

If you have Mathematica, let me know and I'll send you my notebook. If you don't have Mathematica, can we agree on the following components:

where $N_z = \partial_z\hat N$ and $N_\theta = \partial_\theta\hat N$. Note vectors are denoted as $\{a,b,c\} \equiv a\hat r + b\hat \theta + c\hat z$. Everything else is self explanatory I think.

 

[Chestermiller]

Hmmmm I'm not getting exactly this. Did you compute this result in Mathematica? If so I can send you my notebook. I plugged your result and mine in Mathematica and I am not getting zero when fully simplifying the difference of them both.

If you have Mathematica, let me know and I'll send you my notebook. If you don't have Mathematica, can we agree on the following components:View attachment 220353
where $N_z = \partial_z\hat N$ and $N_\theta = \partial_\theta\hat N$. Note vectors are denoted as $\{a,b,c\} \equiv a\hat r + b\hat \theta + c\hat z$. Everything else is self explanatory I think.

I confirm all those components.

 

[member 428835]

I confirm all those components.

Ok, then here is what follows:

Notice I used the FullSimplify command, so ideally this is concise. It's not the same as yours (I input that in already). My results based on post 84

 

[Chestermiller]

Ok, then here is what follows:View attachment 220354
Notice I used the FullSimplify command, so ideally this is concise. It's not the same as yours (I input that in already). My results based on post 84

As best I can tell, this is equivalent to what I had.

 

[member 428835]

As best I can tell, this is equivalent to what I had.

Shoot, maybe I typed it into Mathematica wrong. So, if both are correct, then we're good?

 

[Chestermiller]

Shoot, maybe I typed it into Mathematica wrong. So, if both are correct, then we're good?

Yes. But, I'd also like to include the viscous part of the stress in the boundary condition at the free surface. In terms of the components of the stress tensor (in cylindrical coordinates) and our equation for the normal to the free surface, what is the traction vector at the free surface?

 

[member 428835]

Yes. But, I'd also like to include the viscous part of the stress in the boundary condition at the free surface. In terms of the components of the stress tensor (in cylindrical coordinates) and our equation for the normal to the free surface, what is the traction vector at the free surface?

If the stress tensor is $\sigma$ then are you asking for $\sigma \cdot \hat N$? Shouldn't we instead take $\sigma \cdot \hat u_z$ and $\sigma \cdot \hat u_\theta$ since we're concerned with finding shear? Evidently the stress tensor $\sigma$ has components

where we can obviously omit the $\nabla\cdot\hat V$ terms (incompressible). So the matrix appears as
$$
\sigma =
\begin{bmatrix}
\tau_{rr} & \tau_{r\theta} & \tau_{rz}\\
\tau_{r\theta} & \tau_{\theta\theta} & \tau_{\theta z}\\
\tau_{rz} & \tau_{\theta z} & \tau_{zz}
\end{bmatrix}
$$
and $\hat u_z,\hat u_\theta$ have bee previously defined. What do you think?

 

[Chestermiller]

If the stress tensor is $\sigma$ then are you asking for $\sigma \cdot \hat N$? Shouldn't we instead take $\sigma \cdot \hat u_z$ and $\sigma \cdot \hat u_\theta$ since we're concerned with finding shear? Evidently the stress tensor $\sigma$ has components View attachment 220423
where we can obviously omit the $\nabla\cdot\hat V$ terms (incompressible). So the matrix appears as
$$
\sigma =
\begin{bmatrix}
\tau_{rr} & \tau_{r\theta} & \tau_{rz}\\
\tau_{r\theta} & \tau_{\theta\theta} & \tau_{\theta z}\\
\tau_{rz} & \tau_{\theta z} & \tau_{zz}
\end{bmatrix}
$$
and $\hat u_z,\hat u_\theta$ have bee previously defined. What do you think?

I'd like to see (in terms of the tau's),$$\hat{\sigma}\centerdot \hat{n}$$The above is the viscous stress (traction) vector at the interface
$$(\hat{\sigma}\centerdot \hat{n})\centerdot \hat{n}$$The above is the normal viscous stress component at the interface
$$\hat{\sigma}\centerdot \hat{n}-((\hat{\sigma}\centerdot \hat{n})\centerdot \hat{n})\hat{n}$$The above is the viscous shear stress at the interface (which must be zero)

 

[member 428835]

I'd like to see (in terms of the tau's),$$\hat{\sigma}\centerdot \hat{n}$$The above is the viscous stress (traction) vector at the interface
$$(\hat{\sigma}\centerdot \hat{n})\centerdot \hat{n}$$The above is the normal viscous stress component at the interface
$$\hat{\sigma}\centerdot \hat{n}-((\hat{\sigma}\centerdot \hat{n})\centerdot \hat{n})\hat{n}$$The above is the viscous shear stress at the interface (which must be zero)

Gotcha. So
$$
\sigma \cdot \hat N =
\begin{bmatrix}
\tau_{rr} & \tau_{r\theta} & \tau_{rz}\\
\tau_{r\theta} & \tau_{\theta\theta} & \tau_{\theta z}\\
\tau_{rz} & \tau_{\theta z} & \tau_{zz}
\end{bmatrix}
\cdot
\begin{bmatrix}
R\\
-R_\theta\\
-R_zR
\end{bmatrix}
$$
where below is $\sigma \cdot \hat N$

and below here is $((\sigma\cdot\hat N)\cdot\hat N)\hat N$

 

[member 428835]

How does this look Chet?

 

[Chestermiller]

I don't understand the things that are bolded and the things that are not bolded. I get the following for the stress vector:

$$\hat{\sigma}\centerdot \hat{n}=(\sigma_{rr}n_r+\sigma_{r\theta}n_{\theta}+\sigma_{rz}n_z)\hat{r}+(\sigma_{r\theta}n_r+\sigma_{\theta \theta}n_{\theta}+\sigma_{\theta z}n_z)\hat{\theta}+(\sigma_{rz}n_r+\sigma_{\theta z}n_{\theta}+\sigma_{zz}n_z)\hat{z}$$
And, for the normal component of fluid stress at the interface, I get:
$$\hat{n}\centerdot \hat{\sigma}\centerdot \hat{n}=\sigma_{rr}(n_r)^2+\sigma_{\theta \theta}(n_{\theta})^2+\sigma_{zz}(n_z)^2+2\sigma_{r\theta}n_rn_\theta+2\sigma_{rz}n_rn_z+2\sigma_{\theta z}n_{\theta}n_z$$

 

[member 428835]

I don't understand the things that are bolded and the things that are not bolded.

Yea sorry, it's just how Mathematica outputs the symbols, but there is no significance with bold and unbold letters.

I get the following for the stress vector:

$$\hat{\sigma}\centerdot \hat{n}=(\sigma_{rr}n_r+\sigma_{r\theta}n_{\theta}+\sigma_{rz}n_z)\hat{r}+(\sigma_{r\theta}n_r+\sigma_{\theta \theta}n_{\theta}+\sigma_{\theta z}n_z)\hat{\theta}+(\sigma_{rz}n_r+\sigma_{\theta z}n_{\theta}+\sigma_{zz}n_z)\hat{z}$$

Awesome, that's what I get too (sorry, in the future I'll write everything in PF rather than snippets of Mathematica). It seems from post 97 that what I have as $\tau_{ij}$ you have as $\sigma_{ij}$ and you write $\hat n = \langle n_r, n_\theta, n_z\rangle$ where I wrote $\hat n = \langle R, -R_\theta, -R R_z \rangle$. So we get the same results.

And, for the normal component of fluid stress at the interface, I get:
$$\hat{n}\centerdot \hat{\sigma}\centerdot \hat{n}=\sigma_{rr}(n_r)^2+\sigma_{\theta \theta}(n_{\theta})^2+\sigma_{zz}(n_z)^2+2\sigma_{r\theta}n_rn_\theta+2\sigma_{rz}n_rn_z+2\sigma_{\theta z}n_{\theta}n_z$$

Ok, I'm getting $$
R^2(\sigma_{rr}-2\sigma_{rz}R_z+\sigma_{zz}R_z^2)-2R(\sigma_{r\theta}-\sigma_{\theta z}R_z)R_\theta+\sigma_{\theta\theta}R_\theta^2
$$
where I've adopted your notation using $\sigma_{ij}$ instead of $\tau_{ij}$, where the normal vector is written explicitly. I think we're getting the same thing here.