Retarded potentials

1. Mar 20, 2008

stunner5000pt

1. The problem statement, all variables and given/known data
Griffith's problem 10.8
Show that retarded potentials satisfy the Lorentz condition. Hint proceed as follows
a) Show that
$$\nabla\cdot\left(\frac{J}{R}\right)=\frac{1}{R}\left(\nabla\cdot\vec{J}\right)+\frac{1}{R}\left(\nabla '\cdot\vec{J}\right)-\nabla '\cdot\left(\frac{J}{R}\right)$$
b) Show that $$\nabla\cdot\vec{J}=-\frac{1}{c}\frac{\partial\vec{J}}{\partial t_{r}}\cdot(\nabla R)$$
c) Note that $$\vec{J}=\vec{J}\left(\vec{r'},t_{r}\right)$$
$$\nabla '\cdot J=-\frac{\partial \rho}{\partial t}-\frac{1}{c}\frac{\partial J}{\partial t_{r}}\cdot (\nabla ' R)$$

where $$\vec{R}=\vec{r}-\vec{r'}$$

2. The attempt at a solution

I managed to do the first and second parts but its the third part that i am unable to prove.
Ok so i know that $$\vec{J}=\vec{J}\left(\vec{r'},t_{r}\right)=\vec{J}\left(\vec{r'},t-\frac{\vec{r}-\vec{r'}}{c}\right)$$

To make it simpler for me to understand lets do it for one dimension.
$$\frac{\partial J_{x}}{\partial x'} = \frac{\partial J_{x}}{\partial t_{r}}\frac{\partial t_{r}}{\partial x'}$$
But $$\frac{\partial t_{r}}{\partial x'}=\frac{1}{c}\frac{\partial R}{\partial x'}$$
so $$\frac{\partial J_{x}}{\partial x'} = \frac{1}{c}\frac{\partial J_{x}}{\partial t_{r}}\frac{\partial R}{\partial x'}$$

THat explains the second term which i need to get in the proof. But how do i get the first term?

Also is it supposed to be $$\nabla\cdot J=-\frac{\partial \rho}{\partial t}$$
or is it supposed to be $$\nabla'\cdot J=-\frac{\partial \rho}{\partial t}$$

2. Mar 21, 2008

Reshma

Use the product rules:
$$\vec \nabla \cdot \left(\frac{\vec J}{R}\right) = {1\over R}\left(\vec{\nabla} \cdot \vec J\right) + \vec J \cdot \vec{\nabla} \left({1\over R}\right)$$

$$\vec{\nabla}' \cdot \left(\frac{\vec J}{R}\right) = {1\over R}\left(\vec{\nabla}' \cdot \vec J\right) + \vec J \cdot \vec{\nabla}' \left({1\over R}\right)$$

Where:

$$\vec R = \vec r -\vec{r}'$$

$$\vec \nabla ({1\over R}) = -\vec{\nabla}' ({1\over R})$$

Last edited: Mar 21, 2008
3. Mar 21, 2008

stunner5000pt

i cna use that part for the first two
but i cannot get the second part to work

while i was asleep i thought of something though

does this work? I have clearly forgotten how to apply chain rule....

$$\frac{\partial J_{x}}{\partial x'}=\frac{\partial J_{x}}{\partial x'}+\frac{\partial J_{x}}{\partial t_{r}}\frac{\partial t_{r}}{\partial x'}\frac{\partial x}{\partial x'}$$

is this correct??

4. Mar 22, 2008

Reshma

I will show you the steps for $\vec \nabla \cdot \vec J$

$$\frac{\partial t_r}{\partial x} = -{1\over c}\frac{\partial R}{\partial x}$$

So,
$$\vec \nabla \cdot \vec J = \frac{\partial J_x}{\partial x} + \frac{\partial J_y}{\partial y} + \frac{\partial J_z}{\partial z}$$

$$= \frac{\partial J_x}{\partial t_r}\frac{\partial t_r}{\partial x} + \frac{\partial J_y}{\partial t_r}\frac{\partial t_r}{\partial y} + \frac{\partial J_z}{\partial t_r}\frac{\partial t_r}{\partial z}$$

$$= -{1\over c} \frac{\partial \vec {J}}{\partial t_r}\cdot (\vec \nabla R)$$

Can you prove the same for Div J'?

Last edited: Mar 22, 2008
5. Mar 22, 2008

stunner5000pt

I got it now thanks a lot