Proving Lorentz Condition for Retarded Potentials in Griffith 10.8

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Homework Help Overview

The discussion revolves around Griffith's problem 10.8, which involves proving that retarded potentials satisfy the Lorentz condition. The problem is situated within the context of electromagnetic theory, specifically dealing with the mathematical formulation of current density and its divergence in relation to retarded potentials.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the application of the product rule in vector calculus to manipulate expressions involving the divergence of current density. There are attempts to clarify the relationship between spatial and temporal derivatives, particularly in the context of retarded time.

Discussion Status

Some participants have made progress on the initial parts of the problem but express difficulty with the final part. There is an ongoing exploration of the chain rule and its application to the problem, with various expressions being proposed and questioned. Guidance has been offered regarding the use of the product rule and the relationships between derivatives.

Contextual Notes

Participants are navigating through the complexities of the problem, with some uncertainty about the correct application of derivatives and the definitions of terms involved. There is mention of potential confusion regarding the notation used for divergence.

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Homework Statement


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 let's 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!
 
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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:
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??
 
stunner5000pt said:
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??

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:
Reshma said:
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'?

I got it now thanks a lot
 

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