- #1
jmc8197
- 9
- 0
I've tried a simple derivation of the Lienard-Wiecher potential for 2 discrete charges separated by dr, but end up with a result which isn't the same as the theoretically correct version:
Take two charges q1 @r1, q2 @r2 with r1 > r2, dr = r1 - r2, parallel
and both traveling at velocity v away from and along r1, r2. The
observation point is at r = 0. The potential of q1 that arrives at q2
is from a retarded time t' = dr/(v + c). This is when q1 was at r2 +
dr - vt' = r2 + dr - v dr/(v + c) = r2 + dr/(1 + B) where B = v/c. So
the potential at the observation point can be replaced by an
equivalent static q1 and q2 separated by dr/(1 + B), equivalent to
increasing the charge density by (1 + B) at r.
The total potential is then q2/r2 + q1/(r2 + dr/(1 + B)) which for
small dr and q1 = q2 = q can be written using the first two terms of a taylor expansion:
q/r2 + q/r2 (1 - dr/(1 + B)r2)
= 2q/r2 - q dr/(1 + B)r2^2 )
I'm not sure how to proceed further, since I was hoping to end up with
an expression 2q/(1 + B) r2. Can anyone see where I've gone wrong?
Take two charges q1 @r1, q2 @r2 with r1 > r2, dr = r1 - r2, parallel
and both traveling at velocity v away from and along r1, r2. The
observation point is at r = 0. The potential of q1 that arrives at q2
is from a retarded time t' = dr/(v + c). This is when q1 was at r2 +
dr - vt' = r2 + dr - v dr/(v + c) = r2 + dr/(1 + B) where B = v/c. So
the potential at the observation point can be replaced by an
equivalent static q1 and q2 separated by dr/(1 + B), equivalent to
increasing the charge density by (1 + B) at r.
The total potential is then q2/r2 + q1/(r2 + dr/(1 + B)) which for
small dr and q1 = q2 = q can be written using the first two terms of a taylor expansion:
q/r2 + q/r2 (1 - dr/(1 + B)r2)
= 2q/r2 - q dr/(1 + B)r2^2 )
I'm not sure how to proceed further, since I was hoping to end up with
an expression 2q/(1 + B) r2. Can anyone see where I've gone wrong?