Griffiths E&M 3.33 write e-field of dipole moment in coordinate free form

[naele]

Homework Statement

Show that the electric field of a "pure" dipole can be written in the coordinate-free form

<br /> E_{dip}(r)=\frac{1}{4\pi\epsilon_0}\frac{1}{r^3}[3(\vec p\cdot \hat r)\hat r-\vec p].

Homework Equations

Starting from
E_{dip}(r)=\frac{p}{4\pi\epsilon_0r^3}(2\cos \hat r+\sin\theta \hat \theta)

The Attempt at a Solution

The equation immediately above assumes a spherical coordinate system such that p is oriented along z. We can therefore write
\vec p=p\hat z
\hat z = \cos\theta \hat r - \sin\theta \hat \theta \implies \vec p=p\cos\theta\hat r-p\sin\theta\hat\theta
From equation 3.102 in the book we know that \hat r\cdot \vec p=p\cos\theta

Try as I might I don't know how to show, geometrically or via manipulation, that p\sin\theta\hat \theta=(\vec p \cdot \hat \theta)\hat \theta. From there it's easy to get to the desired result.

 

[jdwood983]

I find this problem is easier to see in reverse than forwards (that is, working from the solution to get the starting point)

<br /> 3(\mathbf{p}\cdot\hat{\mathbf{r}})\hat{\mathbf{r}}-\mathbf{p}=3p\cos\theta\hat{\mathbf{r}}-p\cos\theta\hat{\mathbf{r}}+p\sin\theta\hat{\boldsymbol{\theta}}=2p\cos\theta\hat{\mathbf{r}}+p\sin\theta\hat{\boldsymbol{\theta}}<br />

which looks a lot like what you have in your relevant equations

 

[naele]

I find this problem is easier to see in reverse than forwards (that is, working from the solution to get the starting point)

<br /> 3(\mathbf{p}\cdot\hat{\mathbf{r}})\hat{\mathbf{r}}-\mathbf{p}=3p\cos\theta\hat{\mathbf{r}}-p\cos\theta\hat{\mathbf{r}}+p\sin\theta\hat{\boldsymbol{\theta}}=2p\cos\theta\hat{\mathbf{r}}+p\sin\theta\hat{\boldsymbol{\theta}}<br />

which looks a lot like what you have in your relevant equations

Well I guess that solves the problem. Seems a little anticlimactic. My professor recommended we draw a picture and show how things cancel out by dotting vectors together.

 

[Jolb]

I was able to do it by using pictures...backwards. Here's how to derive the given equation from the desired equation.

Let theta be the polar angle and p point toward theta=0.

First draw the p vector and the r hat vector intersecting at some angle theta. Project p onto the r axis.
\overrightarrow{p}\cdot \widehat{r}=pcos(\Theta )

Then draw the p vector (parallel to the original p vector) at some point at vector r from the origin (where the dipole is actually located). Resolve the p vector onto the (r, theta) coordinates.
<br /> \widehat{p}=cos(\Theta )\widehat{r}+sin(\Theta )\left ( -\widehat{\Theta } \right )<br />
Plugging those into the problem statement equation and doing some algebra gives you the equation you were supposed to start with. I'm sure you could just work through it backwards.

 

[jdwood983]

Well I guess that solves the problem. Seems a little anticlimactic. My professor recommended we draw a picture and show how things cancel out by dotting vectors together.

It is a little anticlimactic doing it backwards. I spent about 30 minutes trying to do the geometry forwards before I decided to do it backwards. When I saw the answer, my first thought was "That was it?"

As Jolb said, you can do it graphically if you draw the appropriate vectors, but I personally think it's easier to do it mathematically.