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I need to find the divergence of

$$\frac{\vec{r}}{r^3}$$

I think this is the way to solve (but I would like someone to check)

r = {x,y,z}
r^3 = (x^2+y^2+z^2)^(3/2)

For r, let us says Ax = x/((x^2+y^2+z^2)^(3/2))

WLOG, y and z are the same.

Can I just do a 

dAx/dx + dAy/dy + dAz/dz

to show that divergence = 0?

lcv
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Jackson Hart
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2 Answers2

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First, note that $\vec{r} = (x,y,z)$ and $r = \|\vec{r}\| = (x^2+y^2+z^2)^{1/2}$.

Thus, the vector field is $\dfrac{\vec{r}}{r^3} = \left(\dfrac{x}{(x^2+y^2+z^2)^{3/2}},\dfrac{y}{(x^2+y^2+z^2)^{3/2}},\dfrac{z}{(x^2+y^2+z^2)^{3/2}}\right)$.

So, the divergence of this vector field is:

$\nabla \cdot \dfrac{\vec{r}}{r^3} = \dfrac{\partial}{\partial x}\left[\dfrac{x}{(x^2+y^2+z^2)^{3/2}}\right]+\dfrac{\partial}{\partial y}\left[\dfrac{y}{(x^2+y^2+z^2)^{3/2}}\right]+\dfrac{\partial}{\partial z}\left[\dfrac{z}{(x^2+y^2+z^2)^{3/2}}\right]$

You need to show that this expression is $0$. This just involves using the quotient rule carefully.

JimmyK4542
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    It might be outside the scope of what the OP was interested in, but it's worth emphasizing that the vanishing of the divergence is valid everywhere except the origin. One gets a $\delta$-function there, as may be confirmed via the divergence theorem. – Semiclassical Sep 07 '14 at 02:07
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Here's another way to do it which I find considerably less computationally intensive:

Not that for any (sufficiently differentiable) vector field $\vec X$ and function $f$ we have

$\nabla \cdot (f \vec X) = \nabla f \cdot \vec X + f \nabla \vec X, \tag{1}$

a standard identity which may be found at en.m.wikipedia.org/wiki/Vector_calculus_identities; it is also very easy to prove applying the Leibniz product rule for derivatives to the partials occurring in (1). If we take $\vec X = \vec r = (x, y, z)$ and $f(r) = r^{-3}$, by (1) we have

$\nabla \cdot (r^{-3} \vec r) = \nabla (r^{-3}) \cdot \vec r + r^{-3} \nabla \cdot \vec r. \tag{2}$

Now it is easy to see that

$\nabla \cdot \vec r = 3; \tag{3}$

furthermore,

$\nabla (r^{-3}) \cdot \vec r = \nabla_{\vec r} (r^{-3}) = \vec r [r^{-3}]. \tag{4}$

Next,

$\vec r = r \vec e_r, \tag{5}$

where $\vec e_r$ is the unit vector field in the radial direction. As such,

$\vec r [r^{-3}] = r \vec e_r [r^{-3}] = r \dfrac{\partial [r^{-3}]}{\partial r} = r(-3 r^{-4}) = -3r^{-3}; \tag{6}$

using (3)-(6) in (2) yields

$\nabla \cdot (r^{-3} \vec r) = 0, \tag{7}$

as per request.

Hope this helps. Cheers,

and as always,

Fiat Lux!!!

Robert Lewis
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