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Stokes Theorem

  1. Jul 10, 2009 #1
    For stokes theorem, can someone tell me why [tex]\hat{a} \bullet \vec{ds} = ds[/tex]? My notes say it's because they are parallel, but I'm not sure what that means.

    Also to get things clear, Stokes theorem is the generalized equation of Green's theorem. The purpose of Stokes theorem is to provide a means of calculating the "flux" [not the curl- which is the tendency for a point to rotate], or the amount of given substance moving through a surface [in our case a 3dimensional].


    THanks,



    JL
     
  2. jcsd
  3. Jul 10, 2009 #2

    Cyosis

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    Are you familiar with inner products. For vectors [itex]\vec{a},\vec{b}[/itex] the inner product, or dot product in this case, is given by [itex]\vec{a} \cdot \vec{b}=|\vec{a}|| \vec{b}|\cos\theta[/itex], with [itex]\theta[/itex] the angle between the two vectors. What is the angle between parallel vectors?
     
  4. Jul 10, 2009 #3
    Yep, you can do a dot product between two vectors, or you can dot a vector to get the projection onto an axis.
     
  5. Jul 10, 2009 #4

    Cyosis

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    Then do you understand now why [itex]\hat{a} \cdot \vec{ds} = ds[/itex] when they are parallel?
     
  6. Jul 10, 2009 #5
    No haha, that's my question, but I'll try to think about it more.
     
  7. Jul 10, 2009 #6

    Cyosis

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    It's a dot product. Write this dot product in the form I listed in post 2 then ask yourself what is the angle between two parallel vectors?
     
  8. Jul 10, 2009 #7
    Ok, so it is:

    [tex] \hat{a} _{n} \cdot \vec{ds}=| \hat{a} _n| |\vec{ds}|\cos \theta[/tex]
     
  9. Jul 10, 2009 #8

    Cyosis

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    Yep that is correct. Now for my second question, what's the value of [itex]\theta[/itex] if those vectors are parallel? Secondly what is the length of [itex]\hat{a}[/itex]? If you don't know try to explain to me what [itex]\hat{a}[/itex] is.
     
  10. Jul 10, 2009 #9
    [itex]\hat{a}[/itex] is the unit normal vector, thus it has been "normalized" (to a length of 1). If we set [tex]\theta = 0[/tex] then we get [tex]\vec{ds} = |ds| = ds[/tex]
     
  11. Jul 10, 2009 #10
    But what if [tex]\theta[/tex] is not zero, then this is false. So it is not parallel in all conditions- this is part of the reason I am horrible with proofs.
     
  12. Jul 10, 2009 #11
    Ohh, never mind. What I needed was to show that particular condition (which you helped me discover) was true- not whether it was always true. By having that condition being true, I can use this in the derivation of Stokes theorem.


    Thanks,


    JL
     
  13. Jul 10, 2009 #12
    This would have helped a while back ago, \vec{ds}, \hat{a} _n are by definition parallel.
     
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