Can anyone studying John Jackson's Classical Electrodynamics help? I don't know how eqn. 2.5 was obtained.
My method was to substitute eq. 2.4 back to 2.1, then do the derivative. But it did not work. I used Mathematica, still it won't give me the right equation.
Can someone give me an idea?
Thanks
How to get eqn. 2.5 from John Jackson's electrodynamics
Sorry that this is a little late - I am new to the forum!
With a bit of ugly algebra you are correct.
Plug (2.4) into (2.1). Then take the derivative with respect to x. After you take the derivative let x = a. Then multiply both sides by a negative epsilon.
Make sure that you realize that x and y are vectors seperated by angle gamma such that
|x-y| = [x^2 - 2xycos(gamma) +y^2]^(1/2)
and that
|x-y'| = [x^2 - 2xy'cos(gamma) +y'^2]^(1/2)
I did this one out by hand - took a few pages but it simplified to Jackson's expression.
Plug (2.4) into (2.1). Then take the derivative with respect to x. After you take the derivative let x = a. Then multiply both sides by a negative epsilon.
Make sure that you realize that x and y are vectors seperated by angle gamma such that
|x-y| = [x^2 - 2xycos(gamma) +y^2]^(1/2)
and that
|x-y'| = [x^2 - 2xy'cos(gamma) +y'^2]^(1/2)
I did this one out by hand - took a few pages but it simplified to Jackson's expression.