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Vector Calculus

  1. Mar 21, 2009 #1
    1. The problem statement, all variables and givenknown data
    Let r be a position vector from the origin (r=xi+yj+zk), whose magnitude is r, and let f(r) be a scalar function of r. Sketch the field lines of f(r)r

    2. Relevant equations
    1 [tex]\nabla[/tex]x([tex]\nabla[/tex][tex]\Psi[/tex])=0
    2 [tex]\nabla[/tex].([tex]\nabla[/tex]xv)=0
    3 [tex]\nabla[/tex]x([tex]\nabla[/tex]xv)=[tex]\nabla[/tex]([tex]\nabla[/tex].v)-[tex]\nabla[/tex][tex]\^{}2[/tex]v
    4 [tex]\nabla[/tex].([tex]\Psi[/tex]v)=[tex]\Psi[/tex][tex]\nabla[/tex].v+v.[tex]\nabla[/tex][tex]\Psi[/tex]


    3. The attempt at a solution
    I can't get started on this question. I have no idea how you can draw a sketch of the field lines when the scalar function is unknown. My intuition says you should be able to use some of those identities but I need a push in the right direction. I hope someone can give me that.
     
  2. jcsd
  3. Mar 21, 2009 #2
    More relevant equations (the site was crashing when I tried to put them all into one post):
    5 [tex]\nabla[/tex]x([tex]\Psi[/tex]v)=[tex]\Psi[/tex][tex]\nabla[/tex]xv+([tex]\nabla[/tex][tex]\Psi[/tex])xv
    6 [tex]\nabla[/tex].(v.w)=w.([tex]\nabla[/tex]xv)-v.([tex]\nabla[/tex]xw)
    7 [tex]\nabla[/tex]x(vxw)=v([tex]\nabla[/tex].w)-w([tex]\nabla[/tex].v)+(w.[tex]\nabla[/tex])v-(v.[tex]\nabla[/tex])w
     
  4. Mar 21, 2009 #3

    gabbagabbahey

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    Why not start by computing the divergence and curl of f(r)r....
     
  5. Mar 21, 2009 #4
    Ok, I did that using the correct formulae.
    Div:
    3f(r)+x[tex]\partial[/tex]f(r)/[tex]\partial[/tex]x+y[tex]\partial[/tex]f(r)/[tex]\partial[/tex]y+z[tex]\partial[/tex]f(r)/[tex]\partial[/tex]z
    Curl:
    (z[tex]\partial[/tex]f(r)/[tex]\partial[/tex]y-y[tex]\partial[/tex]f(r)/[tex]\partial[/tex]z)i+(x[tex]\partial[/tex]f(r)/[tex]\partial[/tex]z-z[tex]\partial[/tex]f(r)/[tex]\partial[/tex]x)j+(y[tex]\partial[/tex]f(r)/[tex]\partial[/tex]x-x[tex]\partial[/tex]f(r)/[tex]\partial[/tex]y)k

    I really can't see how that helps since I don't know f(r), and f(r) could be positive or negative. I know that the divergence of a vector field is the (flux of an infinitessimal box/unit volume) placed at a point in the field but I don't know how that helps. I don't know what curl is exactly so I don't know how that helps me visualise the field.
     
  6. Mar 21, 2009 #5

    gabbagabbahey

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    Yuck!:yuck:...since f is a function only of r, use spherical coordinates to find the div and curl instead......remember, [tex]\frac{\partial f(r)}{\partial{r}}=f'(r)[/tex]
     
  7. Mar 21, 2009 #6
    When I switch to spherical polars, r becomes r=r[tex]\widehat{r}[/tex] and grad changes appropriately.
    For the div I get:
    Div = f(r) + rf'(r)
    and for the curl I get zero.
    I still cant see what this means. The div could be positive or negative depending on the actual function.
     
  8. Mar 21, 2009 #7

    gabbagabbahey

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    The curl is zero, which tells you that the field lines don't 'rotate'....You are correct that Div (f(r)r) can be pos/neg or zero, so why not sketch and label all 3 cases?
     
  9. Mar 21, 2009 #8
    Ok I see. Thanks very much for your consistent help.
     
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