Classical Electrodynamics |
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Page 86
... results can be obtained from this expansion. If we let x → 0, only the m = 0 term
survives, and we obtain the integral representation: –––3 s cos kz Ko(kp) dk (
3.150) Vpo -- z* T Jo If we replace p” in (3.150) by R' = p” + p” – 2pp'cos ...
... results can be obtained from this expansion. If we let x → 0, only the m = 0 term
survives, and we obtain the integral representation: –––3 s cos kz Ko(kp) dk (
3.150) Vpo -- z* T Jo If we replace p” in (3.150) by R' = p” + p” – 2pp'cos ...
Page
12.4 Laboratory angle 0, of particle 3 versus center of —lf O momentum angle 6'
for 2 - 1 O # 7r and o. o. 1. 6'— it is a straightforward, although tedious, matter to
obtain the result: 2 2 * — m.? (E1 + mosmo, + corror-ro) 2 * — * — 2\ 2 + p cos 9.
12.4 Laboratory angle 0, of particle 3 versus center of —lf O momentum angle 6'
for 2 - 1 O # 7r and o. o. 1. 6'— it is a straightforward, although tedious, matter to
obtain the result: 2 2 * — m.? (E1 + mosmo, + corror-ro) 2 * — * — 2\ 2 + p cos 9.
Page
Then we obtain o . e!% #s gas-on") als (14.127) 7. 2_% The integral is a Dirac
delta function. Then dI(o) e^e”B” sin” 0 d() where 6 is measured relative to the
velocity v. The presence of the delta function guarantees that the radiation is
emitted ...
Then we obtain o . e!% #s gas-on") als (14.127) 7. 2_% The integral is a Dirac
delta function. Then dI(o) e^e”B” sin” 0 d() where 6 is measured relative to the
velocity v. The presence of the delta function guarantees that the radiation is
emitted ...
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Contents
Introduction to Electrostatics | 1 |
Nš 3 | 3 |
Greens theorem | 14 |
Copyright | |
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