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Re: "4/3 Problem" Resolution (fwd), comment on



Here is a comparison between Feynman's derivation of B=vxE/c^2 and my
derivation of -div(A)=v.E/c^2 that better illustrates the similarities
between the two.


Feynman's derivation of B=vxE/c^2:
----------------------------------

From Feynman's "Lectures on Physics", vol. 2, pg. 26-3, where Feynman
derives B=vxE/c^2 for a charge moving with constant speed v in the
x-direction,

"For the z-component [of B],

B_z = d(A_y)/dx - d(A_x)/dy

Since A_y is zero we have just one derivative to get. Notice, however, that
A_x is just v(phi) [he has set c=1], and d/dy of v(phi) is just -v*E_y. So

B_z = v*E_y

Similarly,

B_y = d(A_x)/dz - d(A_z)/dx = +v*d(phi)/dz

and

B_y = -v*E_z

Finally, B_x is zero, since A_y and A_z are both zero. We can write the
magnetic field simply as

B = vxE [c=1]"

or with c put back in,

B = vxE/c^2


My derivation of -div(A)=v.E/c^2:
---------------------------------

For an electron moving with constant velocity v,

-div(A) = -(d(A_x)/dx + d(A_y)/dy + d(A_z)/dz) (*)

A_x is just v_x(phi) (with c=1), A_y is v_y(phi), and A_z is v_z(phi), and
since d/dx of v_x(phi) is -v_x*E_x, d/dy of v_y(phi) is -v_y*E_y, and d/dz
of v_z(phi) is -v_z*E_z, you can write (*) as

-div(A) = (v_x*E_x + v_y*E_y + v_z*E_z)

or, with c put back in,

-div(A) = v.E/c^2

--
Dave Rutherford
"New Transformation Equations and the Electric Field Four-vector"
http://www.softcom.net/users/der555