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Divergence of vector r.

Hi, can someone please explain this to me.
I have the vector field
v=r=r^r. \mathbf{v}=\mathbf{r}=\dfrac{ \hat{r}}{r}.

Taking the divergence using Spherical Polar Coordinates gives me
∇⋅v=1r2, \boldsymbol{\nabla}\cdot\mathbf{v}=\dfrac{1}{r^2},

however, if I convert this to cartesian coordinates I get the divergence to be 3. What is going on here?
Reply 1
Avatar for Mark13
Mark13
Original post by KeyFingot
Hi, can someone please explain this to me.
I have the vector field
v=r=r^r. \mathbf{v}=\mathbf{r}=\dfrac{ \hat{r}}{r}.

Taking the divergence using Spherical Polar Coordinates gives me
∇⋅v=1r2, \boldsymbol{\nabla}\cdot\mathbf{v}=\dfrac{1}{r^2},

however, if I convert this to cartesian coordinates I get the divergence to be 3. What is going on here?


You need to check your working for the cartesian case - I think you may have made the mistake of thinking that \hat{r}/r = (x,y,z) in cartesian coordinates, which isn't the case (but would give a divergence of 3).
Reply 2
Avatar for KeyFingot
KeyFingot
OP
Original post by Mark13
You need to check your working for the cartesian case - I think you may have made the mistake of thinking that \hat{r}/r = (x,y,z) in cartesian coordinates, which isn't the case (but would give a divergence of 3).


Oh I see now, cheers

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