# Taking the gradient of 1/r (solid sherical harmonics?)

1. Aug 1, 2009

### hanson

Hi all, just an very elementary question, arising from the first study of generating harmonic solutions.

How to get the gradient twice for
$1/r$ in spherical coordinates?

Last edited: Aug 1, 2009
2. Aug 2, 2009

### tiny-tim

Hi hanson!

What's the difficulty?

1/r obviously doesn't depend on θ or φ, so just use ∂/∂r.

3. Aug 2, 2009

### hanson

Hello. Yea,
I think I can get the gradient of 1/r once, using the ∂/∂r, and that will gives me a vector.

What confuses me is taking it another time, which will gives me a tensor. However, the answer I got is different from what is supposed to be.

4. Aug 2, 2009

### arildno

This is SO wrong!

Let us do this properly (sort of!):

First:
$$\nabla\frac{1}{r}=-\frac{1}{r^{2}}\vec{i}_{r}$$

Now, $$\nabla^{2}\frac{1}{r}=(\vec{i}_{r}\frac{\partial}{\partial{r}}+\vec{i}_{\theta}\frac{1}{r\sin\phi}\frac{\partial}{\partial\theta}+\vec{i}_{\phi}\frac{\partial}{r\partial\phi})\cdot(-\frac{1}{r^{2}}\vec{i}_{r})$$

Now, you are to apply each of these operators PRIOR to perform the dot products!

Since the vector [itex]\vec{i}_{r}[/tex] depends upon both the angular variables, differentiating that vector will yield components PARALLELL to the direction of the angular variable.
Thus, non-zero contributions to the full Laplacian of 1/r will appear from these differentiations.

Last edited: Aug 2, 2009
5. Aug 2, 2009

### arildno

Okay, I'm generous today, so we use the identities:
$$\frac{\partial\vec{i}_{r}}{\partial\phi}=\vec{i}_{\phi}$$
$$\frac{\partial\vec{i}_{r}}{\partial\theta}=\sin\phi\vec{i}_{\theta}$$

Thus, we get:
$$\nabla^{2}\frac{1}{r}=\frac{2}{r^{3}}-\frac{1}{r^{3}}-\frac{1}{r^{3}}=0$$

As you should get.

6. Aug 2, 2009

### arildno

In general, when dealing with a function f(r), where r is the radial vector in spherical coordinates, we have the simple relation:
$$\nabla^{2}f=\frac{1}{r^{2}}\frac{d}{dr}(r^{2}\frac{df}{dr})$$

7. Aug 2, 2009

### hanson

Hi, arildno. Thanks for the detail reply.
But what if it is not a dot product between $$\nabla$$ and $$\nabla \frac{1}{r}$$? If it is a dot product, then we will get a Laplacian. However, if it is like a "dyadic product" (I am not sure if this is the right term to use, but seems to be), then we should get a tensor out of this, right? And the result should be something like $$\frac{\vec{\vec{I}}}{r^3} + 3 \frac{\vec{r}\vec{r}}{r^5}$$, where $$\vec{r} = r\vec{i}_{r}$$ and $$\vec{\vec{I}}$$ is the identity matrix. I am not sure how to get this. I kind of got the second term, but not the first one. Could you please help?

p.s. I am sorry about the missing "arrows" above the identity matrix and the "r"s. They cannot display properly, since I am not familiar with latex typing here....

Last edited: Aug 2, 2009
8. Aug 2, 2009

### arildno

OKay, I was a bit unsure if it was the Laplacian or the dyadic tensor you were after; it is, however, fairly trivial to set up in this case (remember that the Laplacian is the sum of the diagonal terms)!

You get the tensor:
$$f^{,,}\vec{i}_{r}\vec{i}_{r}+\frac{f^{,}}{r}\vec{i}_{\theta}\vec{i}_{\theta}+\frac{f^{,}}{r}\vec{i}_{\phi}\vec{i}_{\phi}$$

9. Aug 2, 2009

### arildno

Note that this is rewritable as:
$$\frac{3}{r^{3}}\vec{i}_{r}\vec{i}_{r}-\frac{I}{r^{3}}=\frac{3}{r^{5}}\vec{r}\vec{r}-\frac{I}{r^{3}}$$
where I is the identity matrix.

10. Aug 2, 2009

### hanson

Oh...the "f' here refers to a general function? So, you get this formula basically by performing similar steps below, right?

I have done this is the most boring and lengthy way

In spherical coordinates,
$$\nabla = \vec{i}_r \frac{\partial}{\partial r}+ \vec{i}_{\phi} \frac{1}{r} \frac{\partial}{\partial \phi}+\vec{i}_\theta \frac{1}{r sin \phi} \frac{\partial}{\partial \theta}$$

So,
$$\nabla \left(\frac{1}{r}\right) = \vec{i}_r \frac{-1}{r^2}$$

which is

$$\nabla(\nabla(\frac{1}{r})) = \vec{i}_r \vec{i}_r \frac{2}{r^3} + \vec{i}_\phi \vec{i}_\phi \frac{-1}{r^3}+\vec{i}_\theta \vec{i}_\theta \frac{-1}{r^3}$$

And, it is rewritten as
$$\vec{i}_r \vec{i}_r \frac{3}{r^3} +\vec{i}_r \vec{i}_r \frac{-1}{r^3} + \vec{i}_\phi \vec{i}_\phi \frac{-1}{r^3}+\vec{i}_\theta \vec{i}_\theta \frac{-1}{r^3},$$

which is
$$\vec{i}_r \vec{i}_r \frac{3}{r^3} - \frac{\vec{\vec{I}}}{r^3}$$

p.s. I edited this 100 times but the latex still doesn't show up properly...i want to give up...

But the above steps are basically correct, right?

Last edited: Aug 2, 2009
11. Aug 3, 2009

### arildno

YEs, f was a general scalar function, only dependent upon the radial variable.

What you have done above is correct, and you Latex seems just fine.

If you want a similar tensorial form for a general f, you might rewrite post 8 as:
$$r\frac{d}{dr}(\frac{f^{,}}{r})\vec{i}_{r}\vec{i}_{r}+\frac{f^{,}}{r}I$$
where I is, again, the identity matrix.

Last edited: Aug 3, 2009
12. Aug 3, 2009

### hanson

Thanks arildno!

13. Aug 4, 2009

### tiny-tim

divergence theorem

You can also find ∇2 of a spherically symmetric f(r) using the divergence theorem

V2f(r)dV = ∫S f(r).dS = ∫S df/dr er.dS

So if V is the volume between two concentric spheres of radius r and r + dr (so that S is the surface of the two spheres, and er.dS = ±dS),

then 4πr22f(r) dr = d/dr(4πr2 df/dr) dr,

so ∇2f(r) = (1/r2) d/dr(r2 df/dr)