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Retarded potentials

  1. Mar 20, 2008 #1
    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
    [tex] \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) [/tex]
    b) Show that [tex] \nabla\cdot\vec{J}=-\frac{1}{c}\frac{\partial\vec{J}}{\partial t_{r}}\cdot(\nabla R)[/tex]
    c) Note that [tex] \vec{J}=\vec{J}\left(\vec{r'},t_{r}\right)[/tex]
    [tex] \nabla '\cdot J=-\frac{\partial \rho}{\partial t}-\frac{1}{c}\frac{\partial J}{\partial t_{r}}\cdot (\nabla ' R) [/tex]

    where [tex] \vec{R}=\vec{r}-\vec{r'} [/tex]

    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 [tex] \vec{J}=\vec{J}\left(\vec{r'},t_{r}\right)=\vec{J}\left(\vec{r'},t-\frac{\vec{r}-\vec{r'}}{c}\right)[/tex]

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

    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 [tex] \nabla\cdot J=-\frac{\partial \rho}{\partial t}[/tex]
    or is it supposed to be [tex] \nabla'\cdot J=-\frac{\partial \rho}{\partial t}[/tex]

    Thanks for your help!
     
  2. jcsd
  3. Mar 21, 2008 #2
    Use the product rules:
    [tex] \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)[/tex]

    [tex] \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)[/tex]

    Where:

    [tex]\vec R = \vec r -\vec{r}'[/tex]

    [tex]\vec \nabla ({1\over R}) = -\vec{\nabla}' ({1\over R})[/tex]
     
    Last edited: Mar 21, 2008
  4. Mar 21, 2008 #3
    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....

    [tex] \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'}[/tex]

    is this correct??
     
  5. Mar 22, 2008 #4
    I will show you the steps for [itex]\vec \nabla \cdot \vec J[/itex]

    [tex]\frac{\partial t_r}{\partial x} = -{1\over c}\frac{\partial R}{\partial x}[/tex]

    So,
    [tex]\vec \nabla \cdot \vec J = \frac{\partial J_x}{\partial x} + \frac{\partial J_y}{\partial y} + \frac{\partial J_z}{\partial z}[/tex]

    [tex]= \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}[/tex]

    [tex] = -{1\over c} \frac{\partial \vec {J}}{\partial t_r}\cdot (\vec \nabla R)[/tex]

    Can you prove the same for Div J'?
     
    Last edited: Mar 22, 2008
  6. Mar 22, 2008 #5
    I got it now thanks a lot
     
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