When to use which dimensionless number

[member 428835]

Now its your turn. Please solve Eqns. 4 for the coefficients A and B in terms of $g_{\theta \theta}$, $g_{\theta z}$, and $g_{zz}$, and then write out the equation for $\hat{a}^{\theta}$ in terms of these.

Ahhh this is related to covariant and contravariant vectors, right? I think I've done this before but it was about a year ago. At any rate, we have $$A = \frac{g_{zz}}{-g_{\theta z}^2+g_{zz}g_{\theta\theta}}\\
B = \frac{g_{\theta z}}{g_{\theta z}^2-g_{zz}g_{\theta\theta}}\implies\\
\hat{a}^\theta = \frac{g_{zz}}{-g_{\theta z}^2+g_{zz}g_{\theta\theta}}\hat{a}_\theta+\frac{g_{\theta z}}{g_{\theta z}^2-g_{zz}g_{\theta\theta}}\hat{a}_z.$$

Sorry my response took so long. I should now be able to respond much faster (very busy week for me).

 

[Chestermiller]

Ahhh this is related to covariant and contravariant vectors, right? I think I've done this before but it was about a year ago. At any rate, we have $$A = \frac{g_{zz}}{-g_{\theta z}^2+g_{zz}g_{\theta\theta}}\\
B = \frac{g_{\theta z}}{g_{\theta z}^2-g_{zz}g_{\theta\theta}}\implies\\
\hat{a}^\theta = \frac{g_{zz}}{-g_{\theta z}^2+g_{zz}g_{\theta\theta}}\hat{a}_\theta+\frac{g_{\theta z}}{g_{\theta z}^2-g_{zz}g_{\theta\theta}}\hat{a}_z.$$

Sorry my response took so long. I should now be able to respond much faster (very busy week for me).

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 $\hat{a}^z$.

 

[member 428835]

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 $\hat{a}^z$.

$$
\hat{u}^\theta = \frac{\frac{g_{zz}}{-g_{\theta z}^2+g_{zz}g_{\theta\theta}}\hat{a}_\theta+\frac{g_{\theta z}}{g_{\theta z}^2-g_{zz}g_{\theta\theta}}\hat{a}_z}{\sqrt{\frac{g_{zz}^2+g_{\theta z}^2}{(g_{zz}g_{\theta\theta}-g_{\theta z}^2)^2}}}\\
= \left(\frac{g_{zz}}{-g_{\theta z}^2+g_{zz}g_{\theta\theta}}\hat{a}_\theta+\frac{g_{\theta z}}{g_{\theta z}^2-g_{zz}g_{\theta\theta}}\hat{a}_z \right)\sqrt{\frac{(g_{zz}g_{\theta\theta}-g_{\theta z}^2)^2}{g_{zz}^2+g_{\theta z}^2}}\\
=\frac{g_{zz}\hat{a}_\theta+g_{\theta z}\hat{a}_z}{\sqrt{g_{zz}^2+g_{\theta z}^2}}$$
though I loosely let the absolute values cancel the denominators in the last step, so perhaps one term is off by a sign?

 

[Chestermiller]

$$
\hat{u}^\theta = \frac{\frac{g_{zz}}{-g_{\theta z}^2+g_{zz}g_{\theta\theta}}\hat{a}_\theta+\frac{g_{\theta z}}{g_{\theta z}^2-g_{zz}g_{\theta\theta}}\hat{a}_z}{\sqrt{\frac{g_{zz}^2+g_{\theta z}^2}{(g_{zz}g_{\theta\theta}-g_{\theta z}^2)^2}}}\\
= \left(\frac{g_{zz}}{-g_{\theta z}^2+g_{zz}g_{\theta\theta}}\hat{a}_\theta+\frac{g_{\theta z}}{g_{\theta z}^2-g_{zz}g_{\theta\theta}}\hat{a}_z \right)\sqrt{\frac{(g_{zz}g_{\theta\theta}-g_{\theta z}^2)^2}{g_{zz}^2+g_{\theta z}^2}}\\
=\frac{g_{zz}\hat{a}_\theta+g_{\theta z}\hat{a}_z}{\sqrt{g_{zz}^2+g_{\theta z}^2}}$$
though I loosely let the absolute values cancel the denominators in the last step, so perhaps one term is off by a sign?

That's not what I get. What do you get for the magnitude?

 

[member 428835]

That's not what I get. What do you get for the magnitude?

$$\sqrt{\frac{g_{zz}^2+g_{\theta z}^2}{(g_{zz}g_{\theta\theta}-g_{\theta z}^2)^2}}$$

 

[Chestermiller]

$$\sqrt{\frac{g_{zz}^2+g_{\theta z}^2}{(g_{zz}g_{\theta\theta}-g_{\theta z}^2)^2}}$$

I get $$\frac{\sqrt{g_{zz}}}{\sqrt{g_{zz}g_{\theta \theta}-g_{\theta z}^2}}$$

 

[member 428835]

I get $$\frac{\sqrt{g_{zz}}}{\sqrt{g_{zz}g_{\theta \theta}-g_{\theta z}^2}}$$

Ok, so we have:
$$\hat{a}^\theta = \frac{g_{zz}}{-g_{\theta z}^2+g_{zz}g_{\theta\theta}}\hat{a}_\theta+\frac{g_{\theta z}}{g_{\theta z}^2-g_{zz}g_{\theta\theta}}\hat{a}_z \implies\\
|\hat{a}^\theta| = \sqrt{\left(\frac{g_{zz}}{-g_{\theta z}^2+g_{zz}g_{\theta\theta}}\right)^2+\left(\frac{g_{\theta z}}{g_{\theta z}^2-g_{zz}g_{\theta\theta}}\right)^2}\\
=\sqrt{\frac{g_{zz}^2}{(-g_{\theta z}^2+g_{zz}g_{\theta\theta})^2}+\frac{g_{\theta z}^2}{(g_{\theta z}^2-g_{zz}g_{\theta\theta})^2}}\\
=\sqrt{\frac{g_{zz}^2+g_{\theta z}^2} {(-g_{\theta z}^2+g_{zz}g_{\theta\theta})^2}}$$
What do you think?

 

[Chestermiller]

Ok, so we have:
$$\hat{a}^\theta = \frac{g_{zz}}{-g_{\theta z}^2+g_{zz}g_{\theta\theta}}\hat{a}_\theta+\frac{g_{\theta z}}{g_{\theta z}^2-g_{zz}g_{\theta\theta}}\hat{a}_z \implies\\
|\hat{a}^\theta| = \sqrt{\left(\frac{g_{zz}}{-g_{\theta z}^2+g_{zz}g_{\theta\theta}}\right)^2+\left(\frac{g_{\theta z}}{g_{\theta z}^2-g_{zz}g_{\theta\theta}}\right)^2}\\
=\sqrt{\frac{g_{zz}^2}{(-g_{\theta z}^2+g_{zz}g_{\theta\theta})^2}+\frac{g_{\theta z}^2}{(g_{\theta z}^2-g_{zz}g_{\theta\theta})^2}}\\
=\sqrt{\frac{g_{zz}^2+g_{\theta z}^2} {(-g_{\theta z}^2+g_{zz}g_{\theta\theta})^2}}$$
What do you think?

I think that $\hat{a}_{\theta}$ and $\hat{a}_{z}$ are not orthogonal, nor are they unit vectors.

 

[member 428835]

I think that $\hat{a}_{\theta}$ and $\hat{a}_{z}$ are not orthogonal, nor are they unit vectors.

Riiiiiiight, as you said bevore $$\hat{a}_\theta=\frac{\partial R}{\partial \theta}\hat{r}+R\hat{\theta}\tag{3a}$$and$$\hat{a}_z=\frac{\partial R}{\partial z}\hat{r}+\hat{z}\tag{3b}$$
My bad. So I would take the norm by transforming back to these definitions. Somehow I have a feeling you have a more general approach? Could you share it and I'll do the footwork? I'd like to see how to take a vector norm with non-orthonormal basis vectors.

 

[Chestermiller]

Riiiiiiight, as you said bevore $$\hat{a}_\theta=\frac{\partial R}{\partial \theta}\hat{r}+R\hat{\theta}\tag{3a}$$and$$\hat{a}_z=\frac{\partial R}{\partial z}\hat{r}+\hat{z}\tag{3b}$$
My bad. So I would take the norm by transforming back to these definitions. Somehow I have a feeling you have a more general approach? Could you share it and I'll do the footwork? I'd like to see how to take a vector norm with non-orthonormal basis vectors.

Just use Eqns. 5 in post #35.

 

[member 428835]

Just use Eqns. 5 in post #35.

I tried thinking about how equations 5 help, but I can't see how. At any rate, we have
$$\hat{a}_\theta=\frac{\partial R}{\partial \theta}\hat{r}+R\hat{\theta}\tag{3a}$$
$$\hat{a}_z=\frac{\partial R}{\partial z}\hat{r}+\hat{z}\tag{3b}$$
$$\hat{a}^\theta = \frac{g_{zz}}{-g_{\theta z}^2+g_{zz}g_{\theta\theta}}\hat{a}_\theta+\frac{g_{\theta z}}{g_{\theta z}^2-g_{zz}g_{\theta\theta}}\hat{a}_z \\
=\frac{-g_{zz}}{g_{\theta z}^2-g_{zz}g_{\theta\theta}} \left( \frac{\partial R}{\partial \theta}\hat{r}+R\hat{\theta} \right)+\frac{g_{\theta z}}{g_{\theta z}^2-g_{zz}g_{\theta\theta}}\left( \frac{\partial R}{\partial z}\hat{r}+\hat{z} \right)\\
= \frac{g_{\theta z} \partial_zR -g_{zz} \partial_\theta R}{g_{\theta z}^2-g_{zz}g_{\theta\theta}} \hat{r} +
\frac{-g_{zz} R}{g_{\theta z}^2-g_{zz}g_{\theta\theta}} \hat{\theta}
+ \frac{g_{\theta z} }{g_{\theta z}^2-g_{zz}g_{\theta\theta}} \hat{z}$$
where from here I could compute the norm as a regular problem. Would you mind highlighting the procedure you referenced, involving equations 5? Again, I'm sorry for the amount of time it's taken for me to reply!

 

[Chestermiller]

$$\hat{a}^{\theta}=\frac{g_{zz}\hat{a}_{\theta}-g_{\theta z}\hat{a}_z}{g_{zz}g_{\theta \theta}-g_{\theta z}^2}$$
$$\hat{a}^{\theta}\centerdot \hat{a}^{\theta}=\frac{(g_{zz}\hat{a}_{\theta}-g_{\theta z}\hat{a}_z)\centerdot (g_{zz}\hat{a}_{\theta}-g_{\theta z}\hat{a}_z)}{(g_{zz}g_{\theta \theta}-g_{\theta z}^2)^2}$$
$$\hat{a}^{\theta}\centerdot \hat{a}^{\theta}=\frac{g_{zz}^2(\hat{a}_{\theta}\centerdot \hat{a}_{\theta})-2g_{\theta z}g_{zz}(\hat{a}_z\centerdot \hat{a}_{\theta})+g_{\theta z}^2(\hat{a}_z\centerdot \hat{a}_z)}{(g_{zz}g_{\theta \theta}-g_{\theta z}^2)^2}$$
$$\hat{a}^{\theta}\centerdot \hat{a}^{\theta}=\frac{g_{zz}^2g_{\theta \theta}-2g_{\theta z}^2g_{zz}+g_{\theta z}^2g_{zz}}{(g_{zz}g_{\theta \theta}-g_{\theta z}^2)^2}$$
$$\hat{a}^{\theta}\centerdot \hat{a}^{\theta}=\frac{g_{zz}(g_{zz}g_{\theta \theta}-g_{\theta z}^2)}{(g_{zz}g_{\theta \theta}-g_{\theta z}^2)^2}$$
$$\hat{a}^{\theta}\centerdot \hat{a}^{\theta}=\frac{g_{zz}}{g_{zz}g_{\theta \theta}-g_{\theta z}^2}$$

 

[member 428835]

Riiiiiight, the magnitude of any vector $\vec v =\sqrt{ \vec v \cdot \vec v}$ since $\vec v \cdot \vec v = |v| |v| \cos \theta = |v|^2$. Shoot, this makes perfect sense now! Okay, so I agree the magnitude is $$\frac{\sqrt{g_{zz}}}{\sqrt{g_{zz}g_{\theta \theta}-g_{\theta z}^2}}.$$ From the above we know
$$\hat{a}^\theta = \frac{g_{zz}}{-g_{\theta z}^2+g_{zz}g_{\theta\theta}}\hat{a}_\theta+\frac{g_{\theta z}}{g_{\theta z}^2-g_{zz}g_{\theta\theta}}\hat{a}_z\implies\\
\hat u^\theta = \sqrt{g_{zz}}\hat{a}_\theta - \frac{g_{\theta z}}{\sqrt{g_{zz}}}\hat{a}_z\\
=\frac{g_{zz}\hat{a}_\theta - g_{\theta z}\hat{a}_z} {\sqrt{g_{zz}}}.$$
How does that look?

 

[Chestermiller]

Riiiiiight, the magnitude of any vector $\vec v =\sqrt{ \vec v \cdot \vec v}$ since $\vec v \cdot \vec v = |v| |v| \cos \theta = |v|^2$. Shoot, this makes perfect sense now! Okay, so I agree the magnitude is $$\frac{\sqrt{g_{zz}}}{\sqrt{g_{zz}g_{\theta \theta}-g_{\theta z}^2}}.$$ From the above we know
$$\hat{a}^\theta = \frac{g_{zz}}{-g_{\theta z}^2+g_{zz}g_{\theta\theta}}\hat{a}_\theta+\frac{g_{\theta z}}{g_{\theta z}^2-g_{zz}g_{\theta\theta}}\hat{a}_z\implies\\
\hat u^\theta = \sqrt{g_{zz}}\hat{a}_\theta - \frac{g_{\theta z}}{\sqrt{g_{zz}}}\hat{a}_z\\
=\frac{g_{zz}\hat{a}_\theta - g_{\theta z}\hat{a}_z} {\sqrt{g_{zz}}}.$$
How does that look?

I think you already had the equation for $\hat{a}^{\theta}$. If you're trying to represent the unit vector in the same direction as $\hat{a}^{\theta}$ (which this is not), then at you're missing a factor of $\sqrt{g_{zz}g_{\theta \theta}-g_{\theta z}^2}$ in the denominator.

 

[member 428835]

I think you already had the equation for $\hat{a}^{\theta}$. If you're trying to represent the unit vector in the same direction as $\hat{a}^{\theta}$ (which this is not), then at you're missing a factor of $\sqrt{g_{zz}g_{\theta \theta}-g_{\theta z}^2}$ in the denominator.

You previously posted

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 $\hat{a}^z$.

We found ##$\hat{a}^{\theta}$ and its magnitude, right? What I showed was the division. Am I missing something (perhaps messed up the bookkeeping)?

 

[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!