Sketching Field Lines of f(r)r: A Guiding Intuition

[MrB3nn]

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

Homework Equations

1 \nablax(\nabla\Psi)=0
2 \nabla.(\nablaxv)=0
3 \nablax(\nablaxv)=\nabla(\nabla.v)-\nabla\^{}2v
4 \nabla.(\Psiv)=\Psi\nabla.v+v.\nabla\Psi

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.

 

[MrB3nn]

More relevant equations (the site was crashing when I tried to put them all into one post):
5 \nablax(\Psiv)=\Psi\nablaxv+(\nabla\Psi)xv
6 \nabla.(v.w)=w.(\nablaxv)-v.(\nablaxw)
7 \nablax(vxw)=v(\nabla.w)-w(\nabla.v)+(w.\nabla)v-(v.\nabla)w

 

[gabbagabbahey]

Why not start by computing the divergence and curl of f(r)r...

 

[MrB3nn]

Ok, I did that using the correct formulae.
Div:
3f(r)+x\partialf(r)/\partialx+y\partialf(r)/\partialy+z\partialf(r)/\partialz
Curl:
(z\partialf(r)/\partialy-y\partialf(r)/\partialz)i+(x\partialf(r)/\partialz-z\partialf(r)/\partialx)j+(y\partialf(r)/\partialx-x\partialf(r)/\partialy)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.

 

[gabbagabbahey]

Yuck!...since f is a function only of r, use spherical coordinates to find the div and curl instead...remember, \frac{\partial f(r)}{\partial{r}}=f'(r)

 

[MrB3nn]

When I switch to spherical polars, r becomes r=r\widehat{r} and grad changes appropriately.
For the div I get:
Div = f(r) + rf'(r)
and for the curl I get zero.
I still can't see what this means. The div could be positive or negative depending on the actual function.

 

[gabbagabbahey]

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?

 

[MrB3nn]

Ok I see. Thanks very much for your consistent help.