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Curl of the position vector

  1. Jan 9, 2013 #1
    1. The problem statement, all variables and given/known data

    compute the curl of:

    [tex]\vec{r}[/tex]
    and
    [tex]\frac{\vec{r}}{r^3}[/tex]


    2. Relevant equations

    [tex]\vec{r}=x\hat{x}+y\hat{y}+z\hat{z}[/tex]

    [tex]r^3=(x^2+y^2+z^2)^\frac{3}{2}[/tex]

    3. The attempt at a solution

    I figured out that the curl of [tex]\vec{r} = 0[/tex] as my book says it should be....

    however...I also need to prove the second vector field up there also is equal to 0...Is it as simple as saying this?:

    since [tex]∇\times\vec{r}=0 [/tex]... therefore, [tex]∇\times\vec{r}(\frac{1}{r^3})=0[/tex] as well. ?????

    My book states that we should be able to see that these equal zero before doing the calculations to save us time
     
    Last edited: Jan 9, 2013
  2. jcsd
  3. Jan 9, 2013 #2

    Dick

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    It's simple to say that. To prove it you need a vector calculus identity about the curl of a vector times a function. Do you know one? And yes, if you have that, there is no need do the whole calculation.
     
  4. Jan 9, 2013 #3
    We have not learned one yet, the last problem I did was a proof of [tex]∇\times∇ψ(\vec{r})=0[/tex] and that was the problem directly before this, so no identities like you mentioned thus far.
     
  5. Jan 9, 2013 #4

    Dick

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    Ok, that works too. What's the gradient of 1/r?
     
    Last edited: Jan 9, 2013
  6. Jan 9, 2013 #5
    gradient of 1/r is [tex]-\frac{\vec{r}}{r^3}[/tex]
     
  7. Jan 9, 2013 #6

    Dick

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    Ok, so? The curl of that must then be zero. That's what you proved in the previous exercise, right? How does that relate to your problem?
     
    Last edited: Jan 9, 2013
  8. Jan 9, 2013 #7
    Cool thanks for that connection i totally missed haha...does this look sufficient?

    [url=http://postimage.org/][PLAIN]http://s13.postimage.org/opdekff9z/photo_8.jpg[/url] upload photos[/PLAIN]
     
  9. Jan 9, 2013 #8

    Dick

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    Well, no. You proved that the curl of any gradient vector is zero in the previous exercise. You then showed that the vector r over r^3 is the gradient of -1/r. So the curl of vector r over r^3 is...??? The curl of ANY gradient is zero. That's where the skipping of some calculation comes in. You don't have to repeat the previous proof.
     
    Last edited: Jan 9, 2013
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