# Vector field, cylindrical coordinates

Gold Member

## Homework Statement

Describe the following vector field: $$\bold v (\bold x)=\frac{\bold a \times \bold x}{(\bold a \times \bold x)(\bold a \times \bold x)}$$ with $$\bold a = \text{constant}$$.
Calculate its divergence and curl. In what region is there a potential for $$\bold v$$? Calculate it.
Hint: Use cylindrical coordinates in which $$\bold a$$ is an axis.

None given.

## The Attempt at a Solution

I don't really know how to use the hint. I've calculated $$\bold a \times \bold x$$ to be worth $$(a_2 x_3-x_2a_3) \hat i -(a_1x_3-a_3x_1) \hat j + (a_1x_2-a_2x_1)\hat k$$.

I wrote $$a_1=\rho \cos \varphi$$, $$a_2=\rho \sin \varphi$$ and $$a_3=a_3$$. I'm stuck here, I have no idea about how to continue. I think I should write $$\bold x$$ in cylindrical coordinates but I don't know how that would help and there's no relation with $$\bold a$$.

Homework Helper
Hi fluidistic! I don't really know how to use the hint. I've calculated $$\bold a \times \bold x$$ to be worth $$(a_2 x_3-x_2a_3) \hat i -(a_1x_3-a_3x_1) \hat j + (a_1x_2-a_2x_1)\hat k$$.

No, don't use coordinates, just draw a cylinder, and use the fact that a x x is perpendicular to both a and x. Then describe that result in cylindrical coordinates. Gold Member
Hi fluidistic! No, don't use coordinates, just draw a cylinder, and use the fact that a x x is perpendicular to both a and x. Then describe that result in cylindrical coordinates. Thanks for the help, I really appreciate it!
I've drew a cylinder, sketched $$\bold a$$ growing with the z axis, I sketched an arbitrary $$\bold x$$ and their cross product is another vector going into the page for the $$\bold x$$ vector I chose.
So for example $$\bold a = (a_1,0,a_3)$$ and $$\bold x =(x_1,0,0)$$ expressed in cylindrical coordinates. I could guess the cross product to be of the form $$(h_1,\frac{\pi}{2},0)$$. Obviously I'm still doing something wrong. I didn't get the point yet.

Homework Helper
So for example $$\bold a = (a_1,0,a_3)$$ and $$\bold x =(x_1,0,0)$$ expressed in cylindrical coordinates. I could guess the cross product to be of the form $$(h_1,\frac{\pi}{2},0)$$. Obviously I'm still doing something wrong. I didn't get the point yet.

That's a bit weird. Say a = aez, x = (r, θ, 0).

What is the magnitude of a x x ?

And what is its direction?​

Gold Member
That's a bit weird. Say a = aez, x = (r, θ, 0).

What is the magnitude of a x x ?

And what is its direction?​

Magnitude: $$ar\sin \theta$$.
Direction: lies in the plane z=0 and orthogonal to $$\bold x$$. Well I realize there are still 2 choices. I use my right hand to determine the direction.
For example if you choose $$\theta =\frac{\pi}{2}$$, the direction of the vector resulting of the cross product points in the same direction than $$-\hat i$$.

Also I didn't understand why you chose $$\bold x \perp \bold a$$.

Homework Helper
oops!

Sorry, I got confused. I meant x = (r, 0, z).

Gold Member

Sorry, I got confused. I meant x = (r, 0, z).

In this case both vectors lie in the x-z plane, right? The magnitude of their cross product would be $$ar\sin \theta$$ where $$\theta$$ is the smallest angle between both vector. Its direction would be in the $$\hat j$$ direction.
I still don't see where we're going... Thanks a lot for your help.

Homework Helper
No, in cylindrical coordinates, it's r along and z up, so the magnitude of the cross product will just be ar. (and i'm off to bed :zzz: see you in the morning )​

Gold Member
No, in cylindrical coordinates, it's r along and z up, so the magnitude of the cross product will just be ar. (and i'm off to bed :zzz: see you in the morning )​
Hey good night and good morning! I'm still stuck to understand.
We have $$\bold x =(r,0,z)$$. When I look at the picture at http://mathworld.wolfram.com/CylindricalCoordinates.html, I understand that if I have an r, I have an x-y component. Also, if I have a z component then I understand that $$\bold x$$ lies in the x-z plane.
As for the coordinates of $$\bold a$$, I'd write them as $$(0,\theta,a)$$ where $$\theta$$ can be any angle.
Because of the z component of $$\bold x$$, $$\bold a$$ and $$\bold x$$ are not orthogonal, hence $$\sin \theta \neq 1$$ and thus I do not understand why the modulus of $$\bold a \times \bold v =ar$$.
I think I'm wrong somewhere, I'll need someone to tell me where.

Also you wrote "r is along ...". Along what? Did you forget a word? Or it's my English once again? ahahah!

Staff Emeritus
Homework Helper
Draw the xz axes on your paper, making the z-axis the vertical axis. Now you want to draw two vectors tail to tail starting at the origin. Vector a will point along the z-axis, and vector x makes an angle θ with the z-axis. The horizontal component of x lies in the xy plane, so its length is r. Now you know that $|\vec{a}\times\vec{x}|=|\vec{a}||\vec{x}|\sin\theta$, and you should be able to see from the picture that $r=|\vec{x}|\sin\theta$. Therefore, you get $|\vec{a}\times\vec{x}|=ar$.

Gold Member
Draw the xz axes on your paper, making the z-axis the vertical axis. Now you want to draw two vectors tail to tail starting at the origin. Vector a will point along the z-axis, and vector x makes an angle θ with the z-axis. The horizontal component of x lies in the xy plane, so its length is r. Now you know that $|\vec{a}\times\vec{x}|=|\vec{a}||\vec{x}|\sin\theta$, and you should be able to see from the picture that $r=|\vec{x}|\sin\theta$. Therefore, you get $|\vec{a}\times\vec{x}|=ar$.

Hey vela!
Thanks to you I've found it! I will try to continue the rest alone. I've a terrible wisdom teeth pain, because they are growing. Further, it's a bit late and I feel very tired. Tomorrow I'll post my problems or say what I've done.
Thanks to both.

Homework Helper
Hey fluidistic! (just got up :zzz: …)

vela has sorted it out, so I'll just add …
Also you wrote "r is along ...". Along what? Did you forget a word? Or it's my English once again? ahahah!

ah, "along" (like "up" and "down") can either a preposition or an adverb.

As a preposition, you're right, we must say "along something", but as an adverb, we can say "move along, nothing to see" (a quote from The Simpsons ), just as we can say "go up!" or "look down!" I've a terrible wisdom teeth pain, because they are growing.

It'll only get worse … go to the dentist as soon as possible, and have them out. Gold Member
Hey fluidistic! (just got up :zzz: …)

vela has sorted it out, so I'll just add …

ah, "along" (like "up" and "down") can either a preposition or an adverb.

As a preposition, you're right, we must say "along something", but as an adverb, we can say "move along, nothing to see" (a quote from The Simpsons ), just as we can say "go up!" or "look down!" It'll only get worse … go to the dentist as soon as possible, and have them out. Ok thanks for the English lesson. I feel much less pain than yesterday, though I'm not free of pain. I have no time to go to dentist, let alone being operated.

So back to the problem, am I right writing $$v(\bold x)=\frac{1}{ar}$$ (I'm not sure if I should carry an absolute value somewhere), where r represent the projection of $$\bold x$$ into the x-y plane?
To describe this field I'd say that when $$\bold x$$ tends to be parallel to the z axis, $$v(\bold x)$$ tends to infinity. While when $$\bold x$$ is far away from the z axis, $$v(\bold x)$$ is close to 0.
Now to take the divergence of this field, I think I should take the divergence for cylindrical coordinates... Right? Same for the curl.
About the potential, I remember the formula $$\vec E =- \vec \nabla \phi$$. So $$\phi=-\int _1^2 \vec E d\vec s$$. So I'd replace $$\vec E$$ by $$\bold x=\frac{1}{ar}$$ which is defined for $$r\neq 0$$, or in other words there exist a potential function for all the field except the z-axis (or the axis of $$\bold a$$). I'm not sure I'm right on all this.

Homework Helper
… am I right writing $$v(\bold x)=\frac{1}{ar}$$ (I'm not sure if I should carry an absolute value somewhere), where r represent the projection of $$\bold x$$ into the x-y plane?
To describe this field I'd say that when $$\bold x$$ tends to be parallel to the z axis, $$v(\bold x)$$ tends to infinity. While when $$\bold x$$ is far away from the z axis, $$v(\bold x)$$ is close to 0.

Yes, the magnitude of v(x) is 1/ar (both a and r are positive, so absolute value is irrelevant ), and its direction is … ?

Note that 1/ar is independent of z and φ, so the magnitude of v(x) will depend only on the cylindrical shell that x is on.

Gold Member
Yes, the magnitude of v(x) is 1/ar (both a and r are positive, so absolute value is irrelevant ), and its direction is … ?

Note that 1/ar is independent of z and φ, so the magnitude of v(x) will depend only on the cylindrical shell that x is on.

Hmm ok. Direction: same as $$\hat j$$? I just use the right hand rule with a and x.

I have a question, why didn't they give $$v(\bold x)=\frac{1}{\bold a \times \bold x}$$? Isn't the same as the expression they gave?

Homework Helper
Hmm ok. Direction: same as $$\hat j$$? I just use the right hand rule with a and x.

Or, in words, it's always "horizontal" and tangent to the cylinder.
I have a question, why didn't they give $$v(\bold x)=\frac{1}{\bold a \times \bold x}$$? Isn't the same as the expression they gave?

Nooo … 1/vector doesn't mean anything, does it? Gold Member
Or, in words, it's always "horizontal" and tangent to the cylinder.
I didn't realize the cylinder was along the y axis. I thought it was along the z axis. So is $$\bold a$$ the radius of the cylinder? Or $$\bold x$$? Or none of these 2 vectors? Now I guess "r" is.

tiny-tim said:
Nooo … 1/vector doesn't mean anything, does it? Oops you're right!

How would I continue? For the divergence of the field, $$\vec \nabla \frac{1}{ar} \hat k$$, where $$\hat k$$ is the unit vector of the cylindrical coordinates of the axis in which the cylinder is infinite. I took r as the radius of it.
Is this right?

Edit: If so, then the divergence of the field is worth $$-\frac{1}{ar^2}$$.

Last edited:
Staff Emeritus
Homework Helper
I didn't realize the cylinder was along the y axis. I thought it was along the z axis. So is $$\bold a$$ the radius of the cylinder? Or $$\bold x$$? Or none of these 2 vectors? Now I guess "r" is.
So far you have that the magnitude of the cross product depends only on the magnitude of a and the radial distance r from the z axis. Now what about its direction?

Draw the xy plane and a circle of radius r centered on the origin. The z axis will be coming out of the page, so $\vec{a}\times\vec{x}$ will lie in the plane of the paper. Draw a vector x that goes from the origin to a point on the circle. Figure out what direction $\vec{a}\times\vec{x}$ is and draw that vector with its tail at the tip of x. Do this for a couple of points on the circle, and you should see a pattern.

Gold Member
So far you have that the magnitude of the cross product depends only on the magnitude of a and the radial distance r from the z axis. Now what about its direction?

Draw the xy plane and a circle of radius r centered on the origin. The z axis will be coming out of the page
Until here everything's fine.

vela said:
, so $\vec{a}\times\vec{x}$ will lie in the plane of the paper.
I don't see why. Where are a and x?

vela said:
Draw a vector x that goes from the origin to a point on the circle. Figure out what direction $\vec{a}\times\vec{x}$ is and draw that vector with its tail at the tip of x. Do this for a couple of points on the circle, and you should see a pattern.
Ok for the rest, but I need the direction of a. z-axis I guess?

Staff Emeritus
Homework Helper
Until here everything's fine.

I don't see why. Where are a and x?

Ok for the rest, but I need the direction of a. z-axis I guess?
Yes, a points along the z-axis. The vector x goes from the origin to some point on the circle of radius r. The vector v = axx is the vector assigned to that point in space.

Gold Member
Yes, a points along the z-axis. The vector x goes from the origin to some point on the circle of radius r. The vector v = axx is the vector assigned to that point in space.

Ah thanks. I'm not sure I get it. What I get is $$\bold a \times \bold x$$ tangential to the circle at each point of it. So it would be so over the whole cylinder.
I will ask a stupid question... but why is the origin of $$\bold a \times \bold x$$ the point assigned by $$\bold x$$?

Staff Emeritus
Ah thanks. I'm not sure I get it. What I get is $$\bold a \times \bold x$$ tangential to the circle at each point of it. So it would be so over the whole cylinder.