How to get eqn. 2.5 from John Jackson's electrodynamics

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Joined: Sat Oct 07, 2006 5:59 am

How to get eqn. 2.5 from John Jackson's electrodynamics

Post by yyxt » Mon May 28, 2007 7:19 pm

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?

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Joined: Wed Jul 25, 2007 10:28 pm

Sorry that this is a little late - I am new to the forum!

Post by A.Sphere » Wed Jul 25, 2007 10:39 pm

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.

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