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

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The discussion focuses on calculating the gradient of the function \( \frac{1}{||\vec{r}||} \), where \( \vec{r} \) is the separation vector from a fixed point to a source point. The key result is that \( \nabla(\frac{1}{||\vec{r}||}) = \frac{-\hat{r}}{||\vec{r}||^2} \). One participant initially attempts to derive this using the definition of \( ||\vec{r}|| \) but finds it unproductive. Another suggests using spherical coordinates for the gradient, although this approach is met with skepticism regarding its complexity. Ultimately, a different method yields successful results without resorting to spherical coordinates.
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Separation Vector

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

Show that:

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

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

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


In Spherical Coordinates:

\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}} }



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.
 
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