I When to use which dimensionless number

[member 428835]

Hi PF!

I've been reading about low gravity capillary driven flows, and no authors use Reynolds number when measuring importance of inertia in capillary driven flows. Instead most use the Ohnesorge number. Can someone explain why this is?

Thanks!

 

[BvU]

Did you google ?

 

[member 428835]

Did you google ?

I did. The Ohnesorge number is evidently like a Reynolds number but with a capillary length scale. What would cause someone to think of this?

 

[BvU]

Significance of surface tension in capillary flow

 

[member 428835]

Significance of surface tension in capillary flow

But there doesn't seem to be anything intrinsically wrong with Reynolds number. Sure, it doesn't measure surface tension, but why isn't it really ever used for low Bond number capillary flows?

 

[BvU]

@Chestermiller : you know of someone in PF with specific expertise here ? Unfortunately for Josh, mine is just hearsay .

 

[Chestermiller]

@Chestermiller : you know of someone in PF with specific expertise here ? Unfortunately for Josh, mine is just hearsay .

Very often in systems like this, we start out with the differential equations and use a systematic approach developed by Hellums and Churchill to reduce the equations to dimensionless form. The key dimensionless groups automatically emerge from this methodology (with some small leeway on selection of the groups). As first year graduate students at Michigan, we were taught this simple methodology by Stu Churchill back in 1963 (at that time, he was chairman of the ChE Dept.). It has served me well over the years.

 

[BvU]

Thanks, Chet !

Also consulted with my roommate (from my point of view an expert...) who knows all these numbers, but he didn't come much further than: we use it in droplet formation to find correlations.

My own googling didn't get me far, but I did like the overview lemma.

As a physicist, when I 'm in a bad mood, I am inclined to think that these chemical engineers make up these dimensionless numbers as much as they can for two reasons: 1. as a claim to fame and immortality and 2. to make parity plots (preferably log-log) where at least three of these numbers appear with fractional exponents. But only when I'm in a bad mood, of course. And my roommate is a mathematician.

 

[member 428835]

hahahahaha this is a funny point! Ok then, maybe it's simply from scaling the equations! Thanks for your guys' help!

 

[boneh3ad]

Well, the Ohnesorge number is related to the Weber number and the Reynolds number, so under certain conditions ($We\approx 1$) it is effectively equivalent to Reynolds number.

The greater point here is that the choice of dimensionless group depends on the phenomena being studied at the time. The point made by @Chestermiller is important in that these dimensionless groups fall out of the governing equations (most importantly the Navier-Stokes equations) when you make them dimensionless based on scales that are important to the problem you are studying. However, what those scales are is up to you to define and will change from one class of problems to the next. Alternatively, one can carry out a dimensional analysis of the problem (e.g. by the Buckingham Pi Theorem) and those groups will fall out based on the various combinations of important parameters to the problem.

The bottom line is that different physics are dominant for different problems, and your choice of dimensionless groupings must reflect that.

 

[Chestermiller]

Josh,

Why don't you provide the dimensional equations for your specific problem, and I'll illustrate the procedure.

Chet

 

[member 428835]

Josh,

Why don't you provide the dimensional equations for your specific problem, and I'll illustrate the procedure.

Chet

The problem is fluid flowing down a wedge, though I think I understand the scaling, if that's what you're referring to? If $z$ is the direction of flow, $y$ is the coordinate going into the wedge, and $x$ goes from the corner upwards, $\alpha$ the half-corner angle of the wedge, and $L$ a characteristic length (in $z$ direction) and $H$ characteristic height (in $x$ direction) then scales look like
$$z \sim L\\
x \sim H\\
y \sim H\tan\alpha.$$
In this problem the Bond number is very small, and gravity can be neglected. Then Young-Laplace governs pressure (we assume negligible surface stress, since air is above the fluid) and the transverse curvature is assumed much greater than axial, reducing pressure equation to:
$$\Delta P = \sigma\frac{1}{R}\implies\\
P \sim \frac{\sigma}{fH}$$
where $f = (\cos\theta/\sin\alpha-1)^{-1}$ is a geometric function such that $fH=R$ where $R$ is the radius of curvature in the $x-y$ plane and $\theta$ is static contact angle of fluid. Then scaling the $z$ component of Navier Stokes yields to order $\epsilon \equiv H/L$:
$$P_z = \sin^2\alpha\partial^2_xw+\cos^2\alpha\partial^2_yw.$$
It makes perfect sense to me that the Ohnesorge number emerges with the inertial terms (not included here for simplicity). However, to get a Reynolds number you'd have the Weber number like boneh3ad said. It's just interesting to me that this fluids problem doesn't really require a Reynolds number. What do you both think?

 

[Chestermiller]

I think I'd still like to see the formulation of the differential equations and boundary conditions.

 

[member 428835]

I think I'd still like to see the formulation of the differential equations and boundary conditions.

Before I continue, I forgot a very important scale: $w$. If we assume pressure balances viscosity, then dimensional constant density NS yields in $z$ component
$$\partial_zP = \mu \nabla^2 w\implies\\
\frac{\sigma}{H f L} \sim \mu\left(\frac{1}{H^2}+\frac{1}{H^2\tan^2\alpha}+\frac{1}{L^2}\right)w\implies\\
\frac{\sigma}{H f L} \sim \mu\left(\frac{\tan^2\alpha + 1}{H^2\tan^2\alpha}\right)w\implies\implies\\
\frac{\sigma}{H f L} \sim \mu\left(\frac{\sec^2\alpha}{H^2\tan^2\alpha}\right)w\implies\\
\frac{\sigma}{H f L} \sim \mu\left(\frac{1}{H^2\sin^2\alpha}\right)w\implies\\
w \sim \frac{\sigma H\sin^2\alpha}{ f L \mu}\implies\\
w \sim \frac{\epsilon \sigma \sin^2\alpha}{ f \mu}.
$$
Notice I neglected $1/L^2$ term since $\epsilon = H^2/L^2\ll 1$ (slender column). Also, I would guess time scales as $t \sim L/w$, though I'm not sure why I'm inclined to use $w$ rather than $u,v$. Most of the flow is $z$ directed; is that good enough reason? Anyways, the formulation of the differential equations, assuming we start with dimensional NS and constant density, would be (for $z$ component, though I could do the other two if you'd like as well)

$$
\rho\frac{D w}{Dt} = \partial_zP+\mu \nabla^2w\implies\\
\frac{\rho W}{L/W}\frac{D w}{Dt} = \frac{\sigma}{HfL}\partial_zP+\frac{\mu W}{H^2\sin^2\alpha} \nabla^2w\implies\\
\frac{\rho W^2}{L}\frac{HfL}{\sigma}\frac{D w}{Dt} = \partial_zP+\frac{HfL}{\sigma}\frac{\mu W}{H^2\sin^2\alpha} \nabla^2w
$$

but we already balanced pressure and viscosity, so we know both terms on the RHS are now $O(1)$, thus we have

$$\frac{\rho W^2}{L}\frac{HfL}{\sigma}\frac{D w}{Dt} = \partial_zP+\nabla^2w\implies\\
\left(\frac{\epsilon \sigma \sin^2\alpha}{ f \mu}\right)^2\frac{\rho }{1}\frac{Hf}{\sigma}\frac{D w}{Dt} = \partial_zP+\nabla^2w\implies\\
\left(\frac{\epsilon^2 \sin^4\alpha}{ f }\right) \frac{\sigma \rho H}{\mu^2}\frac{D w}{Dt} = \partial_zP+\nabla^2w\implies\\
\left(\frac{\epsilon^2 \sin^4\alpha}{ f }\right) \frac{1}{Oh^2}\frac{D w}{Dt} = \partial_zP+\nabla^2w\implies\\$$
and $Oh \equiv \mu/\sqrt{\rho \sigma H}$. How does this look?

 

[member 428835]

And boundary conditions...oh gosh. So zero slip on the wedge, zero stress along the free surface. Hmmmm physically aren't those it?

 

[Chestermiller]

The problem is fluid flowing down a wedge, though I think I understand the scaling, if that's what you're referring to? If $z$ is the direction of flow, $y$ is the coordinate going into the wedge, and $x$ goes from the corner upwards, $\alpha$ the half-corner angle of the wedge, and $L$ a characteristic length (in $z$ direction) and $H$ characteristic height (in $x$ direction) then scales look like
$$z \sim L\\
x \sim H\\
y \sim H\tan\alpha.$$
In this problem the Bond number is very small, and gravity can be neglected. Then Young-Laplace governs pressure (we assume negligible surface stress, since air is above the fluid) and the transverse curvature is assumed much greater than axial, reducing pressure equation to:
$$\Delta P = \sigma\frac{1}{R}\implies\\
P \sim \frac{\sigma}{fH}$$
where $f = (\cos\theta/\sin\alpha-1)^{-1}$ is a geometric function such that $fH=R$ where $R$ is the radius of curvature in the $x-y$ plane and $\theta$ is static contact angle of fluid. Then scaling the $z$ component of Navier Stokes yields to order $\epsilon \equiv H/L$:
$$P_z = \sin^2\alpha\partial^2_xw+\cos^2\alpha\partial^2_yw.$$
It makes perfect sense to me that the Ohnesorge number emerges with the inertial terms (not included here for simplicity). However, to get a Reynolds number you'd have the Weber number like boneh3ad said. It's just interesting to me that this fluids problem doesn't really require a Reynolds number. What do you both think?

I don't understand your description of the physical problem. Can you please provide a diagram?

 

[Andy Resnick]

The problem is fluid flowing down a wedge,
<snip>
In this problem the Bond number is very small, and gravity can be neglected. <snip>

But what is driving the flow? An (external) applied pressure gradient? Marangoni effect? Something else?

Edit: hang on, I re-read your OP. Capillary driven flow. Got it.

Edit #2: Here's a reference you may find useful, especially sections 2.2 and 2.3: https://ac.els-cdn.com/S03019322060...t=1512499461_7ce3a4db15d949752b591ed73f002381

 

[member 428835]

I don't understand your description of the physical problem. Can you please provide a diagram?

Here is a schematic of the flow.

 

[Chestermiller]

But what is driving the flow? An (external) applied pressure gradient? Marangoni effect? Something else?

Edit: hang on, I re-read your OP. Capillary driven flow. Got it.

Hi Andy.

If you understand the geometry, maybe you can provide a diagram before Josh does.

 

[Andy Resnick]

Here is a schematic of the flow.

Hi Andy.

If you understand the geometry, maybe you can provide a diagram before Josh does.

This is a 'standard' problem- but not a simple one. There's at least one book and 1 NASA flight experiment that is devoted to this specific problem. Can't immediately find a PDF breaking down the governing equations, but I'm sure the OP can find a suitable reference.

 

[member 428835]

This is a 'standard' problem- but not a simple one. There's at least one book and 1 NASA flight experiment that is devoted to this specific problem. Can't immediately find a PDF breaking down the governing equations, but I'm sure the OP can find a suitable reference.

Yes, you're right. This is a canonical problem in capillary driven corner flows. There are several papers describing what I posted, though perhaps slightly more dense that what I wrote? And the NASA flight experiment you refer to is the topic of my grant proposal due Dec 15. In fact, my previous adviser at a different university than that I attend now was the PI for the flight experiment you refer to: CFE ICF flight experiments.

Hahahahhaa small world. So, as you may wonder, how did this question initially surface about dimensionless numbers? My current adviser asked me for Reynolds numbers and Ohnesorge numbers for the very flight experiments you reference (CFE ICF). I was confused because I've never considered Reynolds numbers for that flow since the scaling didn't automatically provide any. I was curious what people on these forums thought

 

[Chestermiller]

Here is a schematic of the flow.

I'm sorry Josh. I can't make sense out of this figure. I'm more confused than ever. Can you please try again?

 

[member 428835]

I'm sorry Josh. I can't make sense out of this figure. I'm more confused than ever. Can you please try again?

Here's a sketch I drew a while ago. I probably should've used this earlier. The corner angle measures $2\alpha$ (not shown).

 

[Chestermiller]

Here's a sketch I drew a while ago. I probably should've used this earlier. The corner angle measures $2\alpha$ (not shown).

This is much better. As I understand it, you have a trough with liquid flowing down the trough. It is a transient situation, with an advancing wetting front. You are introducing fluid at the left. Does this capture it?

 

[member 428835]

This is much better. As I understand it, you have a trough with liquid flowing down the trough. It is a transient situation, with an advancing wetting front. You are introducing fluid at the left. Does this capture it?

Your understanding is correct! Yes, it's transient, and some of the flows have an advancing front, and others are pinned. I think the scales work out; what do you think? I suppose you could specify a flow-rate condition at a specified height at an end, but I don't think that would change the analysis would it?

 

[Chestermiller]

Your understanding is correct! Yes, it's transient, and some of the flows have an advancing front, and others are pinned. I think the scales work out; what do you think? I suppose you could specify a flow-rate condition at a specified height at an end, but I don't think that would change the analysis would it?

I don't know yet. I'm just starting to think about this problem.

 

[Chestermiller]

Is the flow supposed to be a uniform velocity profile at the inlet? Is that what is specified? I assume this is set up in cylindrical coordinates, correct? My first step would be to parameterize the free surface shape, and get surface tangents, unit normal, and curvature.

 

[member 428835]

Is the flow supposed to be a uniform velocity profile at the inlet? Is that what is specified? I assume this is set up in cylindrical coordinates, correct? My first step would be to parameterize the free surface shape, and get surface tangents, unit normal, and curvature.

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.

 

[Andy Resnick]

Yes, you're right. This is a canonical problem in capillary driven corner flows. There are several papers describing what I posted, though perhaps slightly more dense that what I wrote? And the NASA flight experiment you refer to is the topic of my grant proposal due Dec 15. In fact, my previous adviser at a different university than that I attend now was the PI for the flight experiment you refer to: CFE ICF flight experiments.

Hahahahhaa small world. So, as you may wonder, how did this question initially surface about dimensionless numbers? My current adviser asked me for Reynolds numbers and Ohnesorge numbers for the very flight experiments you reference (CFE ICF). I was confused because I've never considered Reynolds numbers for that flow since the scaling didn't automatically provide any. I was curious what people on these forums thought

I'm familiar with these experiments :)

Yeah, it's tricky because there's a lot going on. For the CFE-ICF experiments concerned with two-phase bubbly flow, I would expect the capillary number to be most important, and the Oh and We numbers to be less relevant. To be sure, two-phase flow regimes are segregated using terms involving the Reynolds number:

http://cdn.intechopen.com/pdfs/40637/InTech-Two_phase_flow.pdf

If the liquid phase does not completely wet the solid, then the real issue is contact line motion- I don't think that problem can be scaled to generate a dimensionless number (similarity parameter).

Maybe you've already seen this:
https://www.cambridge.org/core/jour...erior-corner/8C88D0772C50153457B155B8B1DDB029

 

[member 428835]

I'm familiar with these experiments :)

Yeah, it's tricky because there's a lot going on. For the CFE-ICF experiments concerned with two-phase bubbly flow, I would expect the capillary number to be most important, and the Oh and We numbers to be less relevant. To be sure, two-phase flow regimes are segregated using terms involving the Reynolds number:

http://cdn.intechopen.com/pdfs/40637/InTech-Two_phase_flow.pdf

That's a good link for dimensionless numbers.

If the liquid phase does not completely wet the solid, then the real issue is contact line motion- I don't think that problem can be scaled to generate a dimensionless number (similarity parameter).

Maybe you've already seen this:
https://www.cambridge.org/core/jour...erior-corner/8C88D0772C50153457B155B8B1DDB029

Oh yes, when I was Mark's (Weislogel) student he gave me his thesis to familiarize myself with the subject, the bulk of which is in this publication. I've read this paper once or twice ;)

 

[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)?