Vector field, cylindrical coordinates

In summary: 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... with vector a?
  • #1

fluidistic

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Homework Statement


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


Homework Equations


None given.


The Attempt at a Solution


I don't really know how to use the hint. I've calculated [tex]\bold a \times \bold x[/tex] to be worth [tex](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[/tex].

I wrote [tex]a_1=\rho \cos \varphi[/tex], [tex]a_2=\rho \sin \varphi[/tex] and [tex]a_3=a_3[/tex]. I'm stuck here, I have no idea about how to continue. I think I should write [tex]\bold x[/tex] in cylindrical coordinates but I don't know how that would help and there's no relation with [tex]\bold a[/tex].
 
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  • #2
Hi fluidistic! :smile:
fluidistic said:
I don't really know how to use the hint. I've calculated [tex]\bold a \times \bold x[/tex] to be worth [tex](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[/tex].

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. :wink:
 
  • #3
tiny-tim said:
Hi fluidistic! :smile:


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. :wink:

Thanks for the help, I really appreciate it!
I've drew a cylinder, sketched [tex]\bold a[/tex] growing with the z axis, I sketched an arbitrary [tex]\bold x[/tex] and their cross product is another vector going into the page for the [tex]\bold x[/tex] vector I chose.
So for example [tex]\bold a = (a_1,0,a_3)[/tex] and [tex]\bold x =(x_1,0,0)[/tex] expressed in cylindrical coordinates. I could guess the cross product to be of the form [tex](h_1,\frac{\pi}{2},0)[/tex]. Obviously I'm still doing something wrong. :frown: I didn't get the point yet.
 
  • #4
fluidistic said:
So for example [tex]\bold a = (a_1,0,a_3)[/tex] and [tex]\bold x =(x_1,0,0)[/tex] expressed in cylindrical coordinates. I could guess the cross product to be of the form [tex](h_1,\frac{\pi}{2},0)[/tex]. Obviously I'm still doing something wrong. :frown: I didn't get the point yet.

That's a bit weird. :confused:

Say a = aez, x = (r, θ, 0).

What is the magnitude of a x x ?

And what is its direction?​
 
  • #5
tiny-tim said:
That's a bit weird. :confused:

Say a = aez, x = (r, θ, 0).

What is the magnitude of a x x ?

And what is its direction?​

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

Also I didn't understand why you chose [tex]\bold x \perp \bold a[/tex].
 
  • #6
oops!

Sorry, I got confused. :redface:

I meant x = (r, 0, z).
 
  • #7


tiny-tim said:
Sorry, I got confused. :redface:

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 [tex]ar\sin \theta[/tex] where [tex]\theta[/tex] is the smallest angle between both vector. Its direction would be in the [tex]\hat j[/tex] direction.
I still don't see where we're going... Thanks a lot for your help.
 
  • #8
No, in cylindrical coordinates, it's r along and z up, so the magnitude of the cross product will just be ar. :wink:

(and I'm off to bed :zzz: see you in the morning :smile:)​
 
  • #9
tiny-tim said:
No, in cylindrical coordinates, it's r along and z up, so the magnitude of the cross product will just be ar. :wink:

(and I'm off to bed :zzz: see you in the morning :smile:)​
Hey good night and good morning! I'm still stuck to understand.
We have [tex]\bold x =(r,0,z)[/tex]. 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 [tex]\bold x[/tex] lies in the x-z plane.
As for the coordinates of [tex]\bold a[/tex], I'd write them as [tex](0,\theta,a)[/tex] where [tex]\theta[/tex] can be any angle.
Because of the z component of [tex]\bold x[/tex], [tex]\bold a[/tex] and [tex]\bold x[/tex] are not orthogonal, hence [tex]\sin \theta \neq 1[/tex] and thus I do not understand why the modulus of [tex]\bold a \times \bold v =ar[/tex].
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!
 
  • #10
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 [itex]|\vec{a}\times\vec{x}|=|\vec{a}||\vec{x}|\sin\theta[/itex], and you should be able to see from the picture that [itex]r=|\vec{x}|\sin\theta[/itex]. Therefore, you get [itex]|\vec{a}\times\vec{x}|=ar[/itex].
 
  • #11
vela said:
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 [itex]|\vec{a}\times\vec{x}|=|\vec{a}||\vec{x}|\sin\theta[/itex], and you should be able to see from the picture that [itex]r=|\vec{x}|\sin\theta[/itex]. Therefore, you get [itex]|\vec{a}\times\vec{x}|=ar[/itex].

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.
 
  • #12
Hey fluidistic! :smile:

(just got up :zzz: …)

vela :smile: has sorted it out, so I'll just add …
fluidistic said:
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 :wink:), just as we can say "go up!" or "look down!" :smile:
fluidistic said:
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. :redface:
 
  • #13
tiny-tim said:
Hey fluidistic! :smile:

(just got up :zzz: …)

vela :smile: 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 :wink:), just as we can say "go up!" or "look down!" :smile:


It'll only get worse … go to the dentist as soon as possible, and have them out. :redface:
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 [tex]v(\bold x)=\frac{1}{ar}[/tex] (I'm not sure if I should carry an absolute value somewhere), where r represent the projection of [tex]\bold x[/tex] into the x-y plane?
To describe this field I'd say that when [tex]\bold x[/tex] tends to be parallel to the z axis, [tex]v(\bold x)[/tex] tends to infinity. While when [tex]\bold x[/tex] is far away from the z axis, [tex]v(\bold x)[/tex] 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 [tex]\vec E =- \vec \nabla \phi[/tex]. So [tex]\phi=-\int _1^2 \vec E d\vec s[/tex]. So I'd replace [tex]\vec E[/tex] by [tex]\bold x=\frac{1}{ar}[/tex] which is defined for [tex]r\neq 0[/tex], or in other words there exist a potential function for all the field except the z-axis (or the axis of [tex]\bold a[/tex]). I'm not sure I'm right on all this.
 
  • #14
fluidistic said:
… am I right writing [tex]v(\bold x)=\frac{1}{ar}[/tex] (I'm not sure if I should carry an absolute value somewhere), where r represent the projection of [tex]\bold x[/tex] into the x-y plane?
To describe this field I'd say that when [tex]\bold x[/tex] tends to be parallel to the z axis, [tex]v(\bold x)[/tex] tends to infinity. While when [tex]\bold x[/tex] is far away from the z axis, [tex]v(\bold x)[/tex] is close to 0.

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

Your description is unclear.

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.
 
  • #15
tiny-tim said:
Yes, the magnitude of v(x) is 1/ar (both a and r are positive, so absolute value is irrelevant :wink:), and its direction is … ?

Your description is unclear.

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 [tex]\hat j[/tex]? I just use the right hand rule with a and x.

I have a question, why didn't they give [tex]v(\bold x)=\frac{1}{\bold a \times \bold x}[/tex]? Isn't the same as the expression they gave?
 
  • #16
fluidistic said:
Hmm ok. Direction: same as [tex]\hat j[/tex]? 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 [tex]v(\bold x)=\frac{1}{\bold a \times \bold x}[/tex]? Isn't the same as the expression they gave?

Nooo … 1/vector doesn't mean anything, does it? :wink:
 
  • #17
tiny-tim said:
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 [tex]\bold a[/tex] the radius of the cylinder? Or [tex]\bold x[/tex]? Or none of these 2 vectors? Now I guess "r" is.
tiny-tim said:
Nooo … 1/vector doesn't mean anything, does it? :wink:
Oops you're right!

How would I continue? For the divergence of the field, [tex]\vec \nabla \frac{1}{ar} \hat k[/tex], where [tex]\hat k[/tex] 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 [tex]-\frac{1}{ar^2}[/tex].
 
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  • #18
fluidistic said:
I didn't realize the cylinder was along the y axis. I thought it was along the z axis. So is [tex]\bold a[/tex] the radius of the cylinder? Or [tex]\bold x[/tex]? 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 [itex]\vec{a}\times\vec{x}[/itex] 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 [itex]\vec{a}\times\vec{x}[/itex] 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.
 
  • #19
vela said:
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 [itex]\vec{a}\times\vec{x}[/itex] 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 [itex]\vec{a}\times\vec{x}[/itex] 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?
 
  • #20
fluidistic said:
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.
 
  • #21
vela said:
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 [tex]\bold a \times \bold x[/tex] 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 [tex]\bold a \times \bold x[/tex] the point assigned by [tex]\bold x[/tex]?
 
  • #22
That's what a vector field is. It's a vector v(x) assigned to each point x in space.
 
  • #23
fluidistic said:
Ah thanks. I'm not sure I get it. What I get is [tex]\bold a \times \bold x[/tex] tangential to the circle at each point of it. So it would be so over the whole cylinder.
Right. So if you look again at the figure on http://mathworld.wolfram.com/CylindricalCoordinates.html, how would you express the vector axx at point x in terms of the cylindrical basis vectors?
 
  • #24
vela said:
Right. So if you look again at the figure on http://mathworld.wolfram.com/CylindricalCoordinates.html, how would you express the vector axx at point x in terms of the cylindrical basis vectors?
A positive multiple of [tex]\hat \theta[/tex]. And thanks for the clarification.
 

1. What is a vector field in cylindrical coordinates?

A vector field in cylindrical coordinates is a mathematical representation of a vector quantity that varies at every point in a three-dimensional space defined by cylindrical coordinates. It is usually denoted as F(r, θ, z) and can be visualized as arrows pointing in different directions and magnitudes at each point in the cylindrical coordinate system.

2. How is a vector field represented in cylindrical coordinates?

A vector field in cylindrical coordinates can be represented using either Cartesian or polar coordinates. In polar coordinates, the vector field is described in terms of a magnitude and direction at each point, while in Cartesian coordinates, it is represented by three components: Fr, Fθ, and Fz.

3. What are the applications of vector fields in cylindrical coordinates?

Vector fields in cylindrical coordinates are used in many fields of science and engineering, including fluid dynamics, electromagnetism, and heat transfer. They are also used in computer graphics and animation to simulate realistic movements of objects in three-dimensional space.

4. How can one visualize a vector field in cylindrical coordinates?

A vector field in cylindrical coordinates can be visualized by plotting arrows at different points in the coordinate system, with the length and direction of each arrow representing the magnitude and direction of the vector at that point. This can be done using computer software or by hand using graph paper.

5. How is the divergence and curl of a vector field calculated in cylindrical coordinates?

The divergence and curl of a vector field in cylindrical coordinates can be calculated using the standard formulas for polar coordinates. The divergence can be calculated as ∇·F = 1/r ∂(rFr)/∂r + 1/r ∂Fθ/∂θ + ∂Fz/∂z, while the curl can be calculated as ∇×F = (1/r) ∂Fz/∂θ - ∂Fθ/∂z + ∂(rFθ)/∂r.

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