1. Limited time only! Sign up for a free 30min personal 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!

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


    User Avatar
    Homework Helper
    Gold Member

    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.

    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


    User Avatar
    Homework Helper
    Gold Member

    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


    User Avatar
    Homework Helper
    Gold Member

    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.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Discussions: Vector Calculus
  1. Calculus on vectors (Replies: 14)

  2. Vector calculus (Replies: 12)

  3. Vector calculus -.- (Replies: 5)

  4. Vector calculus (Replies: 6)

  5. Vector Calculus (Replies: 1)