# Laplace equation charge reflection

1. Sep 16, 2012

### merrypark3

1. The problem statement, all variables and given/known data

Show that

If $\phi$(x,y,z) is a solution of Laplace's equation, show that
$\frac{1}{r}\phi (\frac{x}{r^2} ,\frac{y}{r^2} , \frac{z}{r^2} )$ is also a solution

2. Relevant equations

3. The attempt at a solution

let $\psi= \frac{1}{r} \phi (\frac{x}{r^2} ,\frac{y}{r^2} , \frac{z}{r^2} )$ is a solution.

Then in the spherical coordinate,

$\psi=\frac{1}{r} \phi ( \frac{1}{r} , \theta , \varphi )$

So input $\psi$ to the spherical laplace equation.

$\frac{1}{r^2}\frac{∂}{∂r} (r^2 \frac{∂\psi}{∂r}) = \frac {2}{r^4} \frac{∂\phi}{∂r} - \frac{1}{r^3} \frac{∂^2 \phi}{∂r^2}$

The derivation with other angles, same to the original one, except 1/r times factor.

But the r part for the original one is

$\frac{1}{r^2}\frac{∂}{∂r} (r^2 \frac{∂\phi}{∂r}) = \frac {2}{r} \frac{∂\phi}{∂r} + \frac{∂^2 \phi}{∂r^2}$

What's wrong with me?
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Sep 17, 2012

### gabbagabbahey

Here you are assuming what you are trying to prove. That is not usually a good idea.

Instead, just compute the Laplacian of $\psi= \frac{1}{r} \phi (\frac{x}{r^2} ,\frac{y}{r^2} , \frac{z}{r^2} )$ and show that it is zero if $\nabla^2 \phi(x,y,z)=0$

No. What are $x$, $y$ and $z$ in terms of spherical coordinates $r$, $\theta$ and $\varphi$? What does that make $\frac{x}{r^2}$, $\frac{y}{r^2}$, and $\frac{z}{r^2}$?

3. Sep 17, 2012

### merrypark3

$\psi= \frac{1}{r} \phi (\frac{x}{r^2} ,\frac{y}{r^2} , \frac{z}{r^2} )= \frac{1}{r} \phi(\frac{sin \theta cos\varphi}{r}, \frac{sin \theta sin\varphi}{r},\frac{cos \theta}{r})$

$\frac{∂}{∂r} \phi(\frac{sin \theta cos\varphi}{r}, \frac{sin \theta sin\varphi}{r},\frac{cos \theta}{r})= -( \frac{ sin\theta cos\varphi}{r^2} \frac{∂\phi}{∂(\frac{sin \theta cos\varphi}{r})}+\frac{ sin\theta sin\varphi}{r^2} \frac{∂\phi}{∂(\frac{sin \theta sin\varphi}{r})}+\frac{ cos\theta}{r^2}\frac{∂\phi}{∂(\frac{sin \theta sin\varphi}{r})} )$

Then how can I calculate the terms like $\frac{∂\phi}{∂(\frac{sin \theta cos\varphi}{r})}$?

4. Sep 17, 2012

### gabbagabbahey

good.

Just use a substitution like $\bar{x} \equiv \frac{\sin \theta \cos\varphi}{r}$, $\bar{y} \equiv \frac{\sin \theta \sin\varphi}{r}$, and $\bar{z} \equiv \frac{\cos \theta}{r}$ to make it easier to write:

$$\frac{\partial \phi\left( \frac{x}{r^2} ,\frac{y}{r^2} , \frac{z}{r^2} \right)}{\partial (\frac{\sin \theta \cos\varphi}{r})} = \frac{ \partial \phi( \bar{x}, \bar{y}, \bar{z})}{ \partial \bar{x}}$$?

In the end, you should get something like $\nabla^2 \psi = \text{some factor} * \left[ \frac{ \partial^2 \phi( \bar{x}, \bar{y}, \bar{z})}{ \partial^2 \bar{x}} + \frac{ \partial^2 \phi( \bar{x}, \bar{y}, \bar{z})}{ \partial^2 \bar{y}} + \frac{ \partial^2 \phi( \bar{x}, \bar{y}, \bar{z})}{ \partial^2 \bar{z}} \right]$

Aside: Changing the size on all your posts makes them difficult to read (for my eyes anyways), please stop it.

5. Apr 6, 2013

### steelclam

I am stuck on this too

Hi!

I was just trying to solve this problem through this same approach (make the substitutions $\overline{x}=\sin(\theta)\cos(\phi)/r$, etc...). But it turns out to be *very* complicated. Do you have finished this one? Can you show me more explicit steps? Thanks a lot!!