Feynman Expression for the field of a point charge

[Telemachus]

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

1. Homework Statement

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 what's inside the square brackets must be evaluated at the retarded time $t'=t-\frac{R}{c}$

2. Homework 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.

 

[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 that's the reason. Anyway, I would like to find a rigorous demonstration.

 

[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.

 

[Telemachus]

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

 

[Telemachus]

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

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: $\displaystyle \frac{\partial t'}{\partial t}=\frac{1}{1-\frac{\vec v(t')}{c}\cdot \hat R}$

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

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?)

 

[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".

 

[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.

 

[TSny]

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.

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

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

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

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

 

[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

 

[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.)

 

[Telemachus]

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

 

[TSny]

Yes, that will be fine.

 

[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.

 

[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.

 

[TSny]

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.

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

 

[Telemachus]

Yes, I just absorbed the R in the denominator into the $[ \vec R ]_{ret}$. So that's 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.

 

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

 

[TSny]

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

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

 

[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.

 

[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!]

 

[Telemachus]

The sign came from $\left [ \frac{1}{k} \right ]_{ret} = \frac{dt'}{dt} = 1 - \frac{d}{dt} \left [\frac{R}{c} \right ]_{ret}$, when I replaced in the first term

$[ \frac{\hat R}{kR^2} ]_{ret}=[ \frac{1}{k} ]_{ret} [ \frac{\hat R}{R^2} ]_{ret}=(1 - \frac{d}{dt} [\frac{R}{c}]_{ret}) [ \frac{\hat R}{R^2} ]_{ret}$

Now, according to the expression I get for $\vec E$ and to the Feynman expression I should have

$[R]_{ret} \frac{\partial }{c \partial t} \left [ \frac{\hat R}{R^2} \right ]_{ret} =\frac{1}{c} \left ( [ \frac{\hat R}{R^2} ]_{ret} \frac{d}{dt} [R]_{ret}+ \frac{\partial}{\partial t} [ \frac{\hat R}{R} ]_{ret} \right )$

Which is what is left to demonstrate if all the previous is ok.

 

[TSny]

Sorry. I think your sign is OK. I believe you have the correct result. You just need to show that your middle two terms of E are equivalent to the single middle term of Feynman.

 

[TSny]

$[R]_{ret} \frac{\partial }{c \partial t} \left [ \frac{\hat R}{R^2} \right ]_{ret} =\frac{1}{c} \left ( [ \frac{\hat R}{R^2} ]_{ret} \frac{d}{dt} [R]_{ret}+ \frac{\partial}{\partial t} [ \frac{\hat R}{R} ]_{ret} \right )$

I think the first term on the right should be negative, which is what you had originally.

 

[Telemachus]

Yes, you're right. How can I demonstrate that? that would be the last step.

 

[TSny]

You could expand the single term of Feynman and show it gives your two terms.

Note $[R]_{ret} \frac{\partial }{c \partial t} \left [ \frac{\hat R}{R^2} \right ]_{ret} = [R]_{ret} \frac{\partial }{c \partial t} \left [\frac{1}{R} \cdot \frac{\hat R}{R} \right ]_{ret}$

 

[Telemachus]

I think I have something wrong, because the two terms I obtain from Feynman expression look different to the terms I obtained. I'll show you what I get in the pic. Now I see that the second term is fine. And the other can be rewritten to obtain what I was looking for :)

Thank you verymuch TSny. You helped me a lot to get what I was looking for.

 

[TSny]

OK. Glad it worked out.