- #1
Telemachus
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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.
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