1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Problem to do with curls

  1. Jan 20, 2009 #1
    If r is the position vector in three dimensions, a is a constant vector and [tex]\nabla[/tex] is the gradient operator, evaluate

    [tex]\nabla[/tex] x (a x r).

    I don't understand what a constant vector is, it can't be the same as a scalar function. I've searched for a while and couldn't find anything that made sense, so any help would be vey much appreciated.
     
  2. jcsd
  3. Jan 20, 2009 #2

    gabbagabbahey

    User Avatar
    Homework Helper
    Gold Member

    A constant vector is a vector whose length and direction do not depend on position.

    For example; v=i+2j+687k qualifies as a constant vector. Its components are not functions of position.

    In contrast, the vector v=3x2j is not a constant vector since its lentgh depends on position (specifically where you are along the x-axis)

    Since constant vectors do not vary with position, their curl, divergence and gradient are all zero.
     
  4. Jan 20, 2009 #3
    but if the constant vector is first "crossed" with the position vector... shouldn't it give me some values...
     
  5. Jan 20, 2009 #4
    ohhhh but they cancel out... ahhh i see. Your right. Thank you gabbagabbahey :)
     
  6. Jan 20, 2009 #5

    gabbagabbahey

    User Avatar
    Homework Helper
    Gold Member

    Careful...

    [tex]\vec{\nabla}\times\vec{a}=\vec{\nabla}\cdot\vec{a}=\vec{\nabla}(\vec{a})=0[/tex]

    But that doesn't necessarily mean

    [tex]\vec{\nabla}\times(\vec{a}\times\vec{r})=0[/tex]

    You should find the following vector identity useful:

    [tex]\vec{\nabla}\times(\vec{A}\times\vec{B})=(\vec{B}\cdot\vec{\nabla})\vec{A}-(\vec{A}\cdot\vec{\nabla})\vec{B}+\vec{A}(\vec{\nabla}\cdot\vec{B})-\vec{B}(\vec{\nabla}\cdot\vec{A})[/tex]
     
  7. Jan 20, 2009 #6
    hmmm okay, if A is a constant vector and div A is equal to 0 then the above formula reduces to

    [tex]\vec{\nabla}\times(\vec{A}\times\vec{B})=(2(\vec{B}\ \cdot \vec{\nabla})\vec{A})[/tex]

    right?

    is

    [tex](\vec{B}\ \cdot \vec{\nabla}) = (\vec{\nabla}\ \cdot\vec{B}) [/tex]
     
  8. Jan 20, 2009 #7

    gabbagabbahey

    User Avatar
    Homework Helper
    Gold Member

    No,

    [tex](\vec{B}\ \cdot \vec{\nabla}) \neq (\vec{\nabla}\ \cdot\vec{B}) [/tex]

    In Cartesian coordinates,

    [tex](\vec{B}\ \cdot \vec{\nabla})= B_x\frac{\partial}{\partial x}+B_y\frac{\partial}{\partial y}+B_z\frac{\partial}{\partial z} [/tex]

    which is a differential operator that operates on whatever vector is on the right of the operator.

    Of course, any differential operator that operates on a constant vector will be taking some derivative of some constant and therefor gives a result of zero.

    Therefor, [tex](\vec{r}\cdot \vec{\nabla})\vec{a}=0[/tex]

    And

    [tex]\vec{\nabla}\times(\vec{a}\times\vec{r})=-(\vec{a}\cdot \vec{\nabla})\vec{r}+\vec{a}(\vec{\nabla}\cdot\vec{r})[/tex]
     
  9. Jan 20, 2009 #8

    gabbagabbahey

    User Avatar
    Homework Helper
    Gold Member

    You can then use the fact that [tex]\vec{r}=x\hat{x}+y\hat{y}+z\hat{z}[/tex] to simplify further.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Problem to do with curls
  1. Curl problem (Replies: 2)

  2. Curl problem (Replies: 3)

Loading...