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...forms r(t) = 1 2π ∫ +∞ −∞ dω exp(−iωt) r(ω) ; F ext(t) = 1 2π Using the expression (14) we arrive at the relation r(ω) = { m0(−iω) 2 + e2 c 2 ℓ ∫ ∞ 0 ∫ +∞ −∞ dT [exp(iωT) − 1+ dω exp(−iωt) F ext(ω) . =-=(24)-=- + (−iωT) exp(iωT)] 1 T d dT ( J1(cT/ℓ) T )} −1 F ext(ω) . (25) Standard techniques, such as the convolution theorem, allows us to write the inhomogeneous solution in the form ˙v(t) = ∫ +∞ dt −∞ ′ G(t...

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...ssed in the last section. II. EVALUATION OF THE SELF-FORCE OF A POINT CHARGED PARTICLE The Lagrangian L(ℓ ) leads to the following linear partial differential equations: ( 1 − ℓ 2 □ ) □Aα = − 4π c jα =-=(8)-=- where we used the Lorentz gauge ∂α Aα = 0. To determine these potentials, it is useful to find the retarded Green function for the equation ( 1 − ℓ 2 □ ) □G(x − x ′ , t − t ′ , ℓ ) = − 4π δ(x − x ′ )...

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...which is subjected to the causality condition that G = 0 for t < t ′ . In that way the solution of equation (8) will be: Aα (x, t, ℓ ) = 1 ∫ c d 3 x ′ dt ′ G (x − x ′ , t − t ′ , ℓ )jα (x ′ , t ′ ) . =-=(10)-=- Following the procedure described in reference 21, we arrive at G(R, T, ℓ ) = c θ(T − R/c) ℓ √ c2T 2 (√ ) c2T 2 − R2 J1 − R2 ℓ , (11) 6where R = |x − x ′ |, T = t − t ′ may be shown to be: ∫ F s (t)...

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...ges in the motion of the particle which occur during short time intervals of order r0/c are negligible. One then obtains the Abraham-Lorentz equation of motion: (m0 + mel) ˙v − 2 3 e 2 c 3 ¨v = F ext =-=(3)-=- where we have added a mechanical nonelectromagnetic mass m0. As remarked by Feynman 8 , one would be in trouble only if the energy changes were also infinite. Unfortunately, this is the case: even if...

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... they were able to show that a particle subjected to an external force F ext obeys the following equation of motion: 4 3 U 2 ˙v − c2 3 e2 2e2 ¨v + c3 3c3 ∞∑ n=2 (−1) n n!c n γn d n+1 v dt n+1 = F ext =-=(1)-=- where ˙v is the acceleration of the particle and U represents its electrostatic energy: U = 1 ∫ 2 d 3 ∫ x d 3 x ′ ρ(x)ρ(x′ ) |x − x ′ | . (2) The constants γn are proportional to r n−1 0 and characte...

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...bility that includes a cut-off ℓ leads to a Lagrangian containing second order derivatives of the electromagnetic potentials Aα = (A, iφ): L(ℓ ) = − 1 16π Fαβ Fαβ − 2 ℓ 8π ∂β Fαβ ∂γ Fαγ + 1 c jα Aα , =-=(7)-=- where Fαβ = ∂α Aβ − ∂β Aα is the usual electromagnetic field tensor and jα = (j, icρ) is the conserved four-current. At distances much larger than the cut-off, the fields described by equation (7) be...

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...cs for a point particle by 21 mel = e2 2ℓ c 2 , (21) it is possible to express p in terms of the cut-off ℓ and the classical radius of the electron r0 = e 2 /mc 2 as: p = 3m0ℓc 2 e 2 3ℓ = −3 + 2 r0 . =-=(22)-=- This shows that p is necessarily larger than −3/2, for both ℓ and r0 are positive constants. Therefore, we are led to the conclusion that x must be restricted to positive values. Consequently, if p <...

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...hes. Then, the possible solutions of equation (15) in the absence of external forces are determined by the condition that m0 η 2 + e2 c2 ∫ ∞ dT [exp(−ηT) − 1 + ηT exp(−ηT)] ℓ 1 [ ] d J1(c T/ℓ ) = 0 . =-=(16)-=- T dT T 0 The T-integration may be performed, 24 ( η 2 + c2 ℓ 2 ) 1/2 ( 2η 2 − c2 ℓ 2 giving ) 7 = 2η 3 − c3 ℓ 3 − 3m0 c 3 e 2 η 2 . (17)Squaring both sides and noticing that η = 0 is a doubly degene...

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...(t) = [ ˙v(0) − 1 mτ ∫ t 0 dt ′ exp(−t ′ /τ) F ext(t ′ ] ) exp(t/τ) , (5) unless we impose, following Dirac 9 , the very peculiar initial condition: ˙v(0) = 1 mτ ∫ ∞ dt 0 ′ exp(−t ′ /τ) F ext(t ′ ) . =-=(6)-=- 3But in this case, the acceleration ˙v(t) would depend on the force F ext(t + t ′ ) at times greater than t. This non-causal effect, which is more pronunciated during times t ′ of order τ is called ...

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... = 1 ∫ c d 3 x ′ dt ′ G (x − x ′ , t − t ′ , ℓ )jα (x ′ , t ′ ) . (10) Following the procedure described in reference 21, we arrive at G(R, T, ℓ ) = c θ(T − R/c) ℓ √ c2T 2 (√ ) c2T 2 − R2 J1 − R2 ℓ , =-=(11)-=- 6where R = |x − x ′ |, T = t − t ′ may be shown to be: ∫ F s (t) = − and J1 is the Bessel function of order one. The self-force d 3 [ xρ(x, t) ∇φ(x, t) + 1 c ∂A ∂t ] (x, t) . (12) Now, instead of pe...

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... (iii) p < 0. There is a continuous set of solutions x = x(p). To see that, consider the inverse relation p = p(x) which, according to (19), is given by p = 2x − 1 x 2 + ( 1 + x 2) 1/2 ( ) 1 − 2 x2 . =-=(20)-=- When x ≫ 1, p approaches zero as −1/x 2 , while for x ≪ 1, p behaves approximately as −3/2 + 2x. A plot of the graph of p versus x helps us to grasp these features (see figure 1). 80 2 4 x 6 8 10 -0...

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...7)Squaring both sides and noticing that η = 0 is a doubly degenerate solution, we may rewrite (17) as the cubic equation: m0 η 3 − 3 m 4 2 0 c3 e2 η 2 + 1 3 e2 1 η − ℓ3 4 c e2 1 m0 c − ℓ 4 2 3 = 0 . =-=(18)-=- ℓ 3 The solutions of (17) are a subset of those determined by the cubic equation (18). The solutions of such an equation are well known and it can be verified that its complex conjugate roots do not ...