I When to use which dimensionless number

[Chestermiller]

I've seen all this in papers, but sometimes you do things differently. If you want to see how others have done it I am happy to send you the publication but I wouldn't mind reworking this with you. Again, you've done things differently in the past and I always learn something.

OK Josh. In this spirit, this is how I would obtain the Young-Laplace equation for the free surface:
Using cylindrical coordinates, let $R(t, \theta, z)$ represent the instantaneous radial coordinate of the free surface as a function of t, $\theta$, and z. This fully describes the shape of the free surface at any time t. Let $\mathbf{r}$ represent a position vector drawn from the origin (lower left hand corner of trough) to an arbitrary point $[R(t,\theta,z),\ \theta,\ z]$ on the free surface at time t. We have:$$\mathbf{r}=R(\theta, z)\hat{r}+z\hat{z}\tag{1}$$where, for convenience, we have suppressed the dependence on t. From Eqn. 1, an arbitrary differential position vector within the "plane" of the free surface is then given by:
$$\mathbf{dr}=\left(\frac{\partial R}{\partial \theta}d\theta+\frac{\partial R}{\partial z}dz\right)\hat{r}+R\hat{\theta}d\theta+\hat{z}dz\tag{2}$$This equation can be rearranged to give:
$$\mathbf{dr}=\hat{a}_{\theta}d\theta+\hat{a}_zdz\tag{3}$$with $$\hat{a}_\theta=\frac{\partial R}{\partial \theta}\hat{r}+R\hat{\theta}\tag{3a}$$and$$\hat{a}_z=\frac{\partial R}{\partial z}\hat{r}+\hat{z}\tag{3b}$$where $\hat{a}_{\theta}$ and $\hat{a}_z$ are "coordinate basis vectors" for the free surface, and are tangent to the surface (i.e., lie within the "plane" of the surface) at all locations.

I think I'll stop here for now and give you a chance to digest these equations, particularly Eqns. 3, which are very fundamental.

I still don't think I fully understand the problem that you are trying to solve, and I don't see where there are any natural length scales for the problem, unless the end at z = L is a natural barrier. However, for now, I'm content to focus on the free surface. Is the angle of the trough usually taken to be pretty small?

 

[member 428835]

OK Josh. In this spirit, this is how I would obtain the Young-Laplace equation for the free surface:

I agree, Young-Laplace seems correct. I say this because surface tension dominates. However, in other flows, say the canonical flow down an inclined plane, we don't use Young-Laplace. So when do you know to use this equation?

Using cylindrical coordinates, let $R(t, \theta, z)$ represent the instantaneous radial coordinate of the free surface as a function of t, $\theta$, and z. This fully describes the shape of the free surface at any time t. Let $\mathbf{r}$ represent a position vector drawn from the origin (lower left hand corner of trough) to an arbitrary point $[R(t,\theta,z),\ \theta,\ z]$ on the free surface at time t. We have:$$\mathbf{r}=R(\theta, z)\hat{r}+z\hat{z}\tag{1}$$where, for convenience, we have suppressed the dependence on t. From Eqn. 1, an arbitrary differential position vector within the "plane" of the free surface is then given by:
$$\mathbf{dr}=\left(\frac{\partial R}{\partial \theta}d\theta+\frac{\partial R}{\partial z}dz\right)\hat{r}+R\hat{\theta}d\theta+\hat{z}dz\tag{2}$$This equation can be rearranged to give:
$$\mathbf{dr}=\hat{a}_{\theta}d\theta+\hat{a}_zdz\tag{3}$$with $$\hat{a}_\theta=\frac{\partial R}{\partial \theta}\hat{r}+R\hat{\theta}\tag{3a}$$and$$\hat{a}_z=\frac{\partial R}{\partial z}\hat{r}+\hat{z}\tag{3b}$$where $\hat{a}_{\theta}$ and $\hat{a}_z$ are "coordinate basis vectors" for the free surface, and are tangent to the surface (i.e., lie within the "plane" of the surface) at all locations.

I think I'll stop here for now and give you a chance to digest these equations, particularly Eqns. 3, which are very fundamental.

Ok, I understand why you've selected this coordinate system, how to compute $\mathbf{dr}$, and that it lies along the plane of the surface. Could you elaborate on the basis vectors, specifically how you selected them (or perhaps how the equation suggests them)?

I still don't think I fully understand the problem that you are trying to solve, and I don't see where there are any natural length scales for the problem, unless the end at z = L is a natural barrier. However, for now, I'm content to focus on the free surface. Is the angle of the trough usually taken to be pretty small?

That's a good thing you don't understand the problem, because so far we haven't prescribed any boundary conditions, so as is we're just analyzing a general flow, right? But, what if we had draining at $z=\pm L\implies \mathbf{r}=0$ there. Then we have a characteristic length scale and height scale, right? And yea, typically $R\ll L$ and $\alpha \sim 30^\circ$. But you can make whatever assumptions you want, I'm just here to learn from you.

 

[Chestermiller]

I agree, Young-Laplace seems correct. I say this because surface tension dominates. However, in other flows, say the canonical flow down an inclined plane, we don't use Young-Laplace. So when do you know to use this equation?

Whenever the free surface has significant curvature, Young-Laplace needs to be used.

Ok, I understand why you've selected this coordinate system, how to compute $\mathbf{dr}$, and that it lies along the plane of the surface. Could you elaborate on the basis vectors, specifically how you selected them (or perhaps how the equation suggests them)?

This all comes from Gauss' analysis of surfaces (2D manifolds). He realized that, if you have two arbitrary coordinates within a surface in 3D space (describing lines of, say, constant x1 and x2) you could represent differential position vectors joining closely neighboring points within the surface by an equation of the general form $d\vec{r}=\vec{a_{1}}(x_1,x_2)dx_1+\vec{a_2}(x_1,x_2)dx_2$ (with the a1 vector pointing locally along the lines of constant x2, and the a2 vectors pointing locally along the lines of constant x1). You could then use this to establish all the metrical properties of the 2D surface. You have basically set up a curvilinear coordinate system within the surface. The basis vectors automatically suggest themselves once you specify the spatial coordinates used to grid-up the surface.

So, are you ready to move to the next step now?

 

[member 428835]

So, are you ready to move to the next step now?

Makes a lot of sense! Yes, I'm ready to see what's next!

 

[Chestermiller]

Makes a lot of sense! Yes, I'm ready to see what's next!

What we are going to do next is focus on a tiny window of curved free surface between the grid lines $\theta$ and $\theta + d\theta$ in the $\theta$ in-plane direction, and the grid lines z and z + dz in the z in-plane direction. The opposite edges of this tiny window are not quite parallel to one another, nor are they quite perpendicular to their adjacent edges. We are going to do a force balance on the tiny window to derive the relationship between the normal force exerted by the pressure differential across the window and the surface tension forces at the edges of the window.

To do this force balance, we are going to need to derive equations for the two unit normal in-plane vectors to the grid lines of constant $\theta$ and z within the surface. We need these because the surface tension forces at the edges of the window act perpendicular to the edges.

To implement this derivation, we first define two in-plane "reciprocal coordinate basis vectors" $\hat{a}^{\theta}$ and $\hat{a}^z$ within the surface as follows:

$\hat{a}^{\theta}\centerdot \hat{a}_z=0$ and $\hat{a}^{\theta}\centerdot \hat{a}_{\theta}=1$
and
$\hat{a}^{z}\centerdot \hat{a}_{\theta}=0$ and $\hat{a}^{z}\centerdot \hat{a}_{z}=1$

So $\hat{a}^{\theta}$ is perpendicular to the lines of constant z, and its orientation and magnitude are such that it is at a typically small angle to the lines of constant $\theta$, and the product of its magnitude with that of $\hat{a}_{\theta}$ times the cosine of the angle between them is equal to unity. A similar geometric interpretation applies to $\hat{a}^z$.

In this derivation, what we are going to do is express $\hat{a}^{\theta}$ and $\hat{a}^z$ each as a linear combination of $\hat{a}_{\theta}$ and $\hat{a}_z$. For $\hat{a}^{\theta}$, we first write:$$\hat{a}^{\theta}=A\hat{a}_{\theta}+B\hat{a}_z$$where A and B are coefficients (to be determined). If we dot $\hat{a}^{\theta}$ with $\hat{a}_z$ and $\hat{a}_{\theta}$, we obtain:
$$\hat{a}^{\theta}\centerdot \hat{a_z}=g_{\theta z}A+g_{zz}B=0\tag{4a}$$and$$\hat{a}^{\theta}\centerdot \hat{a}_{\theta}=g_{\theta \theta}A+g_{\theta z}B=1\tag{4b}$$where the three "metrical coefficients" for the surface, $g_{\theta \theta}$, $g_{\theta z}$, and $g_{zz}$, are defined by:
$$g_{\theta \theta}=\hat{a}_{\theta}\centerdot \hat{a}_{\theta}\tag{5a}$$
$$g_{\theta z}=\hat{a}_{\theta}\centerdot \hat{a}_{z}\tag{5b}$$and $$g_{zz}=\hat{a}_{z}\centerdot \hat{a}_{z}\tag{5c}$$
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.

 

[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 $\sqrt{g_{\theta \theta}}|_zd\theta$, $\sqrt{g_{\theta \theta}}|_{z+dz}d\theta$, $\sqrt{g_{zz}}|_\theta dz$, $\sqrt{g_{zz}}|_{\theta+d\theta}dz$?