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Vector field, cylindrical coordinates

  1. Mar 12, 2010 #1

    fluidistic

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    1. The problem statement, all variables and given/known data
    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.


    2. Relevant equations
    None given.


    3. 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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  3. Mar 12, 2010 #2

    tiny-tim

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    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:
     
  4. Mar 12, 2010 #3

    fluidistic

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    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.
     
  5. Mar 12, 2010 #4

    tiny-tim

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    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?​
     
  6. Mar 12, 2010 #5

    fluidistic

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    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].
     
  7. Mar 12, 2010 #6

    tiny-tim

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    oops!

    Sorry, I got confused. :redface:

    I meant x = (r, 0, z).
     
  8. Mar 12, 2010 #7

    fluidistic

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    Re: oops!

    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.
     
  9. Mar 12, 2010 #8

    tiny-tim

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    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:)​
     
  10. Mar 12, 2010 #9

    fluidistic

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    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!
     
  11. Mar 12, 2010 #10

    vela

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    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].
     
  12. Mar 12, 2010 #11

    fluidistic

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    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.
     
  13. Mar 13, 2010 #12

    tiny-tim

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    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:
     
  14. Mar 13, 2010 #13

    fluidistic

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    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.
     
  15. Mar 13, 2010 #14

    tiny-tim

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    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.
     
  16. Mar 13, 2010 #15

    fluidistic

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    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?
     
  17. Mar 13, 2010 #16

    tiny-tim

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    Or, in words, it's always "horizontal" and tangent to the cylinder.
    Nooo … 1/vector doesn't mean anything, does it? :wink:
     
  18. Mar 13, 2010 #17

    fluidistic

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    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.


    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].
     
    Last edited: Mar 13, 2010
  19. Mar 13, 2010 #18

    vela

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    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.
     
  20. Mar 13, 2010 #19

    fluidistic

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    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?
     
  21. Mar 13, 2010 #20

    vela

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    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.
     
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