Separation Vector: Showing $\nabla(\frac{1}{||\vec{r}||})$

  • Thread starter Thread starter Yeldar
  • Start date Start date
  • Tags Tags
    Vector
Yeldar
Messages
6
Reaction score
0
Separation Vector

Let [itex]\vec{r}[/itex] be the separation vector from a fixed point [itex](\acute{x},\acute{y},\acute{z})[/itex] to the source point [itex](x,y,z)[/itex].

Show that:

[tex]\nabla(\frac{1}{||\vec{r}||}) = \frac {-\hat{r}} {||\vec{r}||^2}[/tex]

Now, I've attempted this comeing from the approach that [itex]||\vec{r}|| = (\vec{r} \cdot \vec{r})^\frac {1} {2}[/itex] but it dosent seem to get me anywhere, am I missing something blatently obvious?

Thanks.
 
Last edited:
Go back to the definition of the gradient in spherical coordinates.
 
Wouldnt that just complicate things further?


In Spherical Coordinates:

[tex]\displaystyle{ \nabla = \hat{r} \frac {\partial{}{}} {\partial{}{r}} + \frac {1}{r} \hat{\phi}\frac {\partial{}{}} {\partial{}{\phi}} + \frac {1}{r sin \phi} \hat{\theta}\frac {\partial{}{}} {\partial{}{\theta}} }[/tex]



I just don't see how that could simplify things?
 
Okay, nevermind on this...

Went with a totally different appraoch and things worked out nicely without having to go into spherical coordinates.


Thanks again.
 

Similar threads

  • · Replies 12 ·
Replies
12
Views
2K
  • · Replies 21 ·
Replies
21
Views
2K
Replies
3
Views
3K
Replies
1
Views
3K
Replies
26
Views
3K
Replies
19
Views
5K
  • · Replies 11 ·
Replies
11
Views
2K
Replies
3
Views
2K
Replies
1
Views
2K
Replies
12
Views
2K