Expert: Paul Klarreich - 2/12/2006

Question
hi again, sorry, here is the question in full:

Show that a spherically symmetric vector field F of the form F = f(r)R is conservative by showing that there exists a potential phi such that F = grad phi

(key: F = bold f and R = bold r hat)
I think r = sqrt(x^2 + y^2 + z^2) making it distance to any point and then R = unit vector field = P/r where P = xi + yj + zk (position vector field)

im having great difficulty with this although im sure its relatively easy! i have a feeling integration is involved to find phi? thanks very much!

Answer
Hi, James,

You wrote:

Question: hi again, sorry, here is the question in full:

Show that a spherically symmetric vector field F of the form F = f(r)R is conservative by showing that there exists a potential phi such that F = grad phi

(key: F = bold f and R = bold r hat)
I think r = sqrt(x^2 + y^2 + z^2) making it distance to any point and then R = unit vector field = P/r where P = xi + yj + zk (position vector field)

im having great difficulty with this although im sure its relatively easy! i have a feeling integration is involved to find phi? thanks very much!
----------------------------------------------------------
I am afraid this is beyond my recollections of vector fields, and I really don't know what I'm doing here.  HOWEVER, that hasn't always stopped me in the past, so I will see what I can do.  If what I'm writing is garbage, you know what to do with it.

WARNING: THE MATERIAL BELOW MAY CONTAIN FRACTIONS AND OTHER MATERIAL INAPPROPRIATE FOR CERTAIN COMPUTING SYSTEMS.  BE SURE TO VIEW IT IN A FIXED-SIZE FONT, SUCH AS COURIER.

Given some function G(r).  What is its gradient?  This would be the vector field with components which we now compute:  
[Sorry I can't make partial derivative characters -- you'll have to accept d's.]


 d        d        d
< -- G(r), -- G(r), -- G(r) >
 dx       dy       dz  

What is the derivative with respect to x?

Using the chain rule,

d          DG dr
-- G(r) =  -- --
dx         Dr dx

-- the capital D is the old-fashioned derivative, the small one is the partial.

Since r = sqrt(x^2 + y^2 + z^2),

dr                               x        x
-- = 1/2 (...)^(-1/2) (2x) = --------- = ---
dx                           sqrt(...)    r

So we have:

d         x G'(r)
-- G(r) = -------
dx          r


Likewise for the y- and z-components, we get the gradient to be:

G'(r)
----- <x,y,z>
 r

So all we have to do now is consider  f(r) to be  G'(r)/r.  Then G'(r) = r f(r), and compute  G(r) by integrating.  Even if we don't know how to integrate it, we can always write:

       {t=r
G(r) =  |   t f(t) dt
       }t=0

and this is your phi(r).

Does any of this make sense?  I wish I knew.  Good luck.