# Feynman Expression for the field of a point charge

1. Apr 12, 2015

### Telemachus

Hi there. I'm reading Jackson's Classical Electrodynamics.

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

In chapter 6, the equation for the electric field of a moving point charge is derived.

I could follow the mathematics to get the electric field for the moving charge, which is given in equation 6.57 in Jackson.

$\displaystyle \vec E(\vec x,t) = \frac{q}{4\pi \epsilon_0} \left ( \left [ \frac{\hat R}{kR^2} \right ]_{ret}+\frac{\partial}{c \partial t} \left [ \frac{\hat R}{kR} \right ]_{ret} - \frac{\partial}{c^2 \partial t} \left [ \frac{\vec v}{kR} \right ]_{ret} \right )$

Where $\vec R=\vec x-\vec x'$ is the vector from the source point $\vec x'$ to the observation point $\vec x$, $R=\left |\vec x-\vec x' \right |$, $\hat R=\frac{\vec R}{R}$.

The charge density has been used to get this formula, and the current density, which are given by: $\rho (x',t')=q\delta (\vec x'-\vec r_0(t'))$, $\vec J(\vec x',t')=\rho \vec v(t')$, and $\vec r_0(t')$ is the vector that points to the point charge.

$\displaystyle k=1- \frac{v(t´)}{c} \cdot \hat R$ and ret means that whats inside the square brackets must be evaluated at the retarded time $t'=t-\frac{R}{c}$

2. Relevant equations

The Feynman expression for the field reads

$\displaystyle \vec E(\vec x,t) = \frac{q}{4\pi \epsilon_0} \left ( \left [ \frac{\hat R}{R^2} \right ]_{ret}+\frac{ [R]_{ret} \partial }{c \partial t} \left [ \frac{\hat R}{R^2} \right ]_{ret} + \frac{\partial}{c^2 \partial t} \left [ \hat R \right]_{ret} \right )$

I'm having trouble to get this last expresion.

3. The attempt at a solution

In principle I should have:

$\displaystyle \left [ \frac{ \hat R }{ kR^2 } \right ]_{ret}=\left [ \frac{ \hat R }{ R^2 } \right ]_{ret}$

But I don't see how to get this. It looks like:

$\displaystyle \left [ \frac{1}{k} \right ]_{ret}=1$, but I couldn't get this result. The other expresions are even more complicated, because of the time derivatives, but I wanted to start with this one, which looks simpler.

Last edited: Apr 12, 2015
2. Apr 12, 2015

### Telemachus

I think I have realized why that factor has to be one inside the retarded brackets. It is actually a retardation factor, so I think thats the reason. Anyway, I would like to find a rigorous demonstration.

3. Apr 12, 2015

### TSny

This is a fairly tedious exercise.

$\left [ \frac{1}{k} \right ]_{ret} \neq 1$. So, the first term of equation (6.57) does not reduce to the first term of the Feynman formula.
In the printing of the 3rd edition of Jackson that I have access to, the equation is (6.58) rather than (6.57).

The book The Classical Electromagnetic Field by L. Eyges outlines the derivation but leaves most of the algebra to the reader. The key steps are:

(1) Show that $\frac{dt}{dt'}= 1- \left [\vec{v} \cdot \hat{R}/c \right ]_{ret} = \left [ k \right ]_{ret}$

(2) Show that $\left [ \frac{1}{k} \right ]_{ret} = \frac{dt'}{dt} = 1 - \frac{d}{dt} \left [\frac{R}{c} \right ]_{ret}$

(3) Show that $\vec{v}_{ret} = - \left [ k \right ]_{ret} \frac{d}{dt} \left [\vec{R} \right ]_{ret}$

Use these relations to reduce (6.57) to the Feynman formula. When I tried it I got lots of terms. But quite a few of the terms canceled to give the result.

4. Apr 13, 2015

### Telemachus

Thank you (you're right, its eq. 6.58).

5. Apr 13, 2015

### Telemachus

Ok. I've found this.

I'll show you the steps, because I had trouble in part 2.

$\displaystyle t'=t-\frac{\left |\vec x - \vec r_0' \right |}{c}$ (for simplicity the primed ' denotes that must be evaluated at the retarded time)

Then: $\displaystyle \frac{\partial t'}{\partial t}=1-\frac{1}{c} \frac{\partial \left |\vec x - \vec r_0' \right |}{\partial t'} \frac{\partial t'}{\partial t}$

Then, using that:

$\displaystyle \frac{\partial \left |\vec x - \vec r_0' \right |}{\partial t'}=\frac{\partial \left |\vec x - \vec r_0' \right |}{\partial \vec r_0'}\frac{\partial \vec r_0'}{\partial t'}=-\hat R \cdot \vec v(t')$

Where I have used: $\displaystyle \frac{\partial \left |\vec x - \vec r_0' \right |}{\partial \vec r_0'}=-\hat R$ and $\vec v(t')=\displaystyle \frac{\partial \vec r_0'}{\partial t'}$

So I get the desired result: $\displaystyle \frac{\partial t'}{\partial t}=\frac{1}{1-\frac{\vec v(t')}{c}\cdot \hat R}$

Now for this I've used that $t=t'+\frac{\left |\vec x - \vec r_0' \right |}{c}$

Then: $\displaystyle \frac{\partial t'}{\partial t}=1+\frac{1}{c} \frac{\partial \left |\vec x - \vec r_0' \right |}{\partial t} \frac{\partial t'}{\partial t}$

But now: $\displaystyle \frac{\partial \left |\vec x - \vec r_0' \right |}{\partial t}=\frac{\partial \left |\vec x - \vec r_0' \right |}{\partial \vec r_0'}\frac{\partial \vec r_0'}{\partial t}=-\hat R \cdot \frac{\partial \vec r_0'}{\partial t}$

I'm not sure what $\displaystyle\frac{\partial \vec r_0'}{\partial t}$ represents. Is it just the velocity at the current time? I don't know if it has a physical interpretation, now I'm sure its not the current velocity, because it is the (current) time derivative for the retarded position.

Ok, that step wasn't necesary and I just can rewrite it this way.

$\displaystyle \frac{\partial t'}{\partial t}=1+\frac{\partial}{\partial t}\left [\frac{R}{c} \right ]_{ret} \frac{\partial t'}{\partial t}$
And get the desired result (I have the partial derivative instead of the total derivative, is it the same in this case?)

Last edited: Apr 13, 2015
6. Apr 13, 2015

### TSny

OK for (1).

For (2), I don't see how you are getting $\displaystyle \frac{\partial t'}{\partial t}=1+\frac{\partial}{\partial t}\left [\frac{R}{c} \right ]_{ret} \frac{\partial t'}{\partial t}$.

We have $t'=t - \frac{\left |\vec x - \vec r_0' \right |}{c} = t - \left [\frac{R}{c} \right ]_{ret}$. It seems to me that the result for (2) follows straightaway from this.
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$\displaystyle\frac{\partial \vec r_0'}{\partial t}$ represents the rate at which the retarded position of the particle changes with respect to the present time. If you are located at the point where you are calculating the field and you could see the particle, then you would of course see the particle at its retarded position. So, $\displaystyle\frac{\partial \vec r_0'}{\partial t}$ represents the rate at which you see the position of the particle changing. It's an "apparent velocity".

7. Apr 13, 2015

### Telemachus

You are right, I see the mistake I've made. I'm getting confused with the variables, even the fist one doesn't agree with the definition you gave for t' I think.

I have $t=t'+ \displaystyle \left [\frac{R}{c} \right ]_{ret}$

Then
$\displaystyle \frac{\partial t'}{\partial t}=1+\frac{\partial}{\partial t}\left [ \frac{R}{c} \right ]_{ret}=1+\frac{\partial |x-r_0'|}{\partial r_0'}\frac{\partial r_0'}{\partial t'}=1-\hat R \cdot \vec v(t')$.

For (3) I've tried:

$\left [ \vec R \right ]_{ret}=\vec x- \vec r_0'$
Then:

$\displaystyle \frac{d}{dt}\left [ \vec R \right]_{ret}=\frac{d \vec R}{dt'}\frac{dt'}{dt}=-\frac{d \vec r_0'}{dt'}\frac{dt'}{dt}=-\vec v_{ret} \left (1-\frac{d}{dt} \left [ \frac{R}{c} \right ]_{ret} \right)$

I think I've made some mistake here, but I don't know how to get that thing right.

I have also tried using the distance instead of the vector:

$\displaystyle \frac{d}{dt}\left [ R \right]_{ret}=\frac{d R}{dt'}\frac{dt'}{dt}=-\hat R \cdot v(t') \frac{dt'}{dt}=-\hat R \cdot v(t') \left (1-\frac{d}{dt} \left [ \frac{ R}{c} \right ]_{ret} \right)$

Then

$\displaystyle \frac{d}{dt}\left [ R \right]_{ret} \left (1-\frac{\hat R \cdot v(t')}{c} \right )=-\hat R \cdot \vec v(t')=[k]_{ret} \frac{d}{dt}\left [ R \right]_{ret}$

$\therefore [\hat R \cdot \vec v]_{ret}=-[k]_{ret} \frac{d}{dt}\left [ R \right]_{ret}$

Its different in that I have the dot product with the versor $\hat R$

I couldn't get the result in vector form.

Last edited: Apr 13, 2015
8. Apr 14, 2015

### TSny

$t'$ is the retarded time: $t' = t - \frac{1}{c} \left [ R \right]_{ret}$

I'm not sure if you are trying to get (1) or (2), here. For (1), I still think your derivation in post #5 is correct. You found $\frac{\partial t'}{\partial t} = 1/k'$ where $k' = \left [ k \right ]_{ret} = 1- \frac{1}{c} \left [ \vec{v} \cdot \hat{R}\right ]_{ret}$. You can just flip both sides to get $\frac{\partial t}{\partial t'} = \left [ k \right ]_{ret}$ which is relation (1).

For (3), note that $\left [ \vec{v} \right ]_{ret} = \frac{\partial \vec{r}_0(t')}{\partial t'} =-\frac{\partial (\vec{x} - \vec{r}_0(t'))}{\partial t'} = - \frac{\partial \vec{R}(t')}{\partial t'} = -\frac{\partial t'}{\partial t} \frac{\partial}{\partial t} \left [ \vec{R} \right ]_{ret}$.

9. Apr 15, 2015

### Telemachus

Thanks. Now how do I plug in all of these terms into the equation for $\vec E$ to get the Feynman expression? don't do it, just give me some hints so I can work it out. It still looks quiet cumbersome, and I don't want to complicate it more than it is by making silly things :p

Last edited: Apr 15, 2015
10. Apr 15, 2015

### TSny

Here is where it does get cumbersome. I didn't see any tricks. Just substitute the expressions and then take the time derivatives in the second and third terms. (Note that k cancels out in the third term before taking the time derivative.)

11. Apr 15, 2015

### Telemachus

Is this ok? I'll put an image so I don't have to write all that in Latex.

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12. Apr 15, 2015

### TSny

Yes, that will be fine.

13. Apr 15, 2015

### Telemachus

The thing in the middle is the ugly one. I have the first and the third term, but I don't see how all that cancels out to get the second term.

14. Apr 15, 2015

### TSny

Looks pretty good so far. I'm not sure about your last two terms. It seems to me that the last term should not be there yet. It will be produced by the next to last term. However, in the next to last term it is a bit hard for me to tell if you have a vector symbol over the R at the end of that term.

15. Apr 15, 2015

### TSny

For me, the middle expression of E generated 5 terms and the last expression also generated 5 terms.

16. Apr 15, 2015

### Telemachus

Yes, I just absorbed the R in the denominator into the $[ \vec R ]_{ret}$. So thats wrong, I was tempted because that gave the term I was looking for :p

Thank you verymuch for your assistance TSny. I'll try to work it out from here, and I'll let you know if I have any trouble.

17. Apr 15, 2015

### Telemachus

Lets see if I'm working the third term propperly:

$\frac{\partial}{c^2 \partial t} \left [ \frac{\vec v}{kR} \right ]_{ret} =\frac{\partial}{c^2 \partial t} \left ( \left [ \frac{1}{kR} \right ]_{ret} [ \vec v ]_{ret} \right )=-\frac{\partial}{c^2 \partial t} \left ( \left [ \frac{1}{kR} \right ]_{ret} [ \vec k ]_{ret} \frac{\partial}{\partial t} [ \vec R ]_{ret} \right )=-\frac{\partial}{c^2 \partial t} \left ( \left [ \frac{1}{R} \right ]_{ret} \frac{\partial}{\partial t} [ \vec R ]_{ret} \right )$

And:
$-\frac{\partial}{c^2 \partial t} \left ( \left [ \frac{1}{R} \right ]_{ret} \frac{\partial}{\partial t} [ \vec R ]_{ret} \right )=-\frac{\partial}{c^2 \partial t} \left ( \left [ \frac{1}{R} \right ]_{ret} \right ) \frac{\partial}{\partial t} [ \vec R ]_{ret}- \left [ \frac{1}{R} \right ]_{ret} \frac{\partial^2}{c^2 \partial t^2} [ \vec R ]_{ret}$

18. Apr 15, 2015

### TSny

OK. Looks good. Note $\vec{R} = R \hat{R}$.

19. Apr 15, 2015

### Telemachus

I think I'm almost done, I have to do something with the middle term to get it in the form given by Feynmann, but I don't know what precisely. I show you what I did in the pictures. The two terms in the middle in the last picture should give the time derivative of the electric field, but I don't see how to condense those terms.

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20. Apr 15, 2015

### TSny

Everything looks OK to me, except in your final expression for E I believe the second terms has the wrong sign. [EDIT: Your sign is correct!]

Last edited: Apr 15, 2015