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Simple question about cross product properties

  1. May 15, 2009 #1
    Hello, this is pretty straightforward. I need to take cross product:
    kq(x/(x^2+y^2),y/(x^2+y^2)) x (d/dx,d/dy)

    since kq is a scalar can i just leave it outside the calculation until the very end and for now just calculate the cross product of the two and once i get a definite answer, multiply that answer by kq. Basically can i do this:

    ((x/(x^2+y^2),y/(x^2+y^2)) x (d/dx,d/dy)) dot kq
     
  2. jcsd
  3. May 15, 2009 #2

    gabbagabbahey

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    Yes, that's fine. (However, writing "dot" kq is confusing since the dot product is undefined between a scalar and a vector...I would write it as [itex]kq\left((x/(x^2+y^2),y/(x^2+y^2))\times (d/dx,d/dy)\right)[/itex] instead)
     
  4. May 15, 2009 #3
    yeah thank you. would you know if it would help to convert to polar coordinates when calculating the flux?
     
  5. May 15, 2009 #4

    Mark44

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    Am I missing something here? It looks like you are crossing two vectors in R2. The cross product I know about is defined only for two vectors in R3.
     
  6. May 15, 2009 #5

    gabbagabbahey

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    It depends on which surface you are trying to calculate the flux through.

    Also, are you actually trying to calculate the curl of (x/(x^2+y^2),y/(x^2+y^2))? If so, you need to realize that [itex]\vec{nabla}\time\vec{F}[/itex] is very different from [itex]\vec{F}\times\vec{\nabla}[/itex]! The former is the curl of F, which will be a vector. While the latter is a vector differential operator which operates on some vector.
     
  7. May 15, 2009 #6
    well surface is x^2+y^2=a^2. and for curl ill add a 0 vector for z and the del operator will be (d/dx, d/dy, d/dz) making it R3. and i actually have to prove its conservative away from the region. so i will take [tex]\nabla[/tex] X F in that order yeah i realize i have it swapped before hand. I dont really know what to do about the flux though. Thank you so much for help by the way

    Edit-im gettng something wierd for the k vector of cross product= (y/(x^2+y^2)) d/dx - (x/(x^2+y^2)) d/dy how does this equal 0? NVM, i see it after quotient rule
     
    Last edited: May 15, 2009
  8. May 15, 2009 #7
    alrite so i said flux=[tex]\int[/tex][tex]\int[/tex][tex]\nabla[/tex]dotF dA.
    since i just proved that dot product = 0. im left with

    [tex]\int[/tex][tex]\int[/tex] (0) dx dy over the domain: x^2 + Y^2=a^2
    sooo, what now
     
  9. May 15, 2009 #8

    gabbagabbahey

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    [tex]\int_{\mathcal{S}}\vec{\nabla}\cdot\vec{F}dA[/tex] Does not represent the flux of F through [itex]\mathcal{S}[/itex]....It doesn't even represent the flux of div(F) since the flux of a scalar is not well defined (to the best of my knowledge anyways).

    I think perhaps you should post the entire original problem word for word so that we can see exactly what you are supposed to compute...
     
  10. May 15, 2009 #9
    The 2D vector F = kq((x)/(x^2+y^2), (y)/(x^2+y^2))

    Show that this vector eld is both conservative, and divergence free
    away from the origin (0; 0). Done dot product=0 and cross product=0

    Find the
    Find the flux of this vector field over the circle C given by x2+y2 = a2 also says that flux is integral of F.Nds but in my book it says that this is equal to
    double integral of the dot product of del operator FdA

    and yeah youre right that integral doesnt but the double integral with the Domain being C of that does represent the flux
     
    Last edited: May 15, 2009
  11. May 15, 2009 #10
    and this Use the Divergence theorem to show that the flux will be the same
    even if the curve C is not a circle, but a reasonably nice curve enclosing the origin
    (closed, orientable, simple (i.e. doesn't cross itself), simply connected, enclosing the
    origin). But this only makes things worse so...
     
  12. May 15, 2009 #11

    gabbagabbahey

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    No, I'm sure your book doesn't say that. What your book probably says is:

    [tex]\oint_{\mathcal{S}}\vec{F}\cdot\hat{N}dS=\int_{\mathcal{V}}(\vec{\nabla}\cdot\vec{F})dV[/tex]

    Where [itex]\mathcal{S}[/itex] is the closed surface that bounds the volume [itex]\mathcal{V}[/itex]

    right? (this is known as the divergence theorem, which is sometimes called "Gauss' theorem" or "Green's theorem")

    But(!) this doesn't help you here since the surface [itex]x^2+y^2\leq a^2[/itex] is just a disk of radius [itex]a[/itex]; which is an open surface.

    Instead you have to calculate the flux [tex]\int_{\mathcal{S}}\vec{F}\cdot\hat{N}dS[/tex] directly.

    Begin by expressing the unit normal to the surface in Cartesian unit vectors....does F have any component parallel to the normal n this case? If not, then F.NdS=0 and so the flux is zero.
     
  13. May 15, 2009 #12
    oooo alright i see this now thanks!!!!
     
  14. May 15, 2009 #13

    gabbagabbahey

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    OH!:bugeye:

    You should not have left this part of the problem description out of your previous post!

    After reading this, it seems that they want you to use the 2D version of the divergence theorem. The version I posted above is the 3D version.

    And the "flux" they want you to calculate is also the 2D version of flux; so instead of calculating [tex]\int_{\mathcal{S}} \vec{F}\cdot\hat{k}dxdy[/tex] over the disk, they actually want you to calculate [tex]\int_{\mathcal{C}} \vec{F}\cdot\hat{r}rd\theta[/tex] over the circle (I'm using [itex]r[/itex] and [itex]\theta[/itex] to represent polar coordinates, and [itex]\hat{r}[/itex] to represent the radial unit vector)

    And that is what you are supposed to apply the divergence theorem to.

    The point of the problem is to show that div(F) is not actually zero everywhere, and is in fact infinite at the origin.
     
  15. May 15, 2009 #14
    but i can't calculate the flux at the origin. as the vector is undefined...so how do u take dot product of this [tex]\nabla[/tex]dot F when F doesnt even exist at the origin
     
  16. May 15, 2009 #15

    gabbagabbahey

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    You can't calculate [itex]\vec{\nabla}\cdot\vec{F}[/itex] directly (at the origin). But you can calculate

    [tex]\int_{\mathcal{C}} \vec{F}\cdot\hat{N}ds[/tex]

    You will get the same non-zero answer no matter what curve [itex]\mathcal{C}[/itex] you use (as long as it encloses the origin exactly once); and from that you can conclude that [itex]\vec{\nabla}\cdot\vec{F}[/itex] cannot be zero at the origin.

    Then, if you have been taught about the dirac delta function, you should be able to figure out what [itex]\vec{\nabla}\cdot\vec{F}[/itex] actually is.
     
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