Single-Particle Electrodynamics - Assassination Science

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The case of (6.33) with m = 3 is more subtle, again due to the fact that the radial r d integral is, due to its inverse-cube nature, infinite. The term proportional to n i dn j d in (6.33) avoids this fate (due to the fact that it contains an extra power of r d /2ε in all terms), but the term proportional to δ ij does not. However, it will be found that, in fact, terms involving δ ij will all cancel in our final expressions, before they need to be integrated; let us therefore simply write its integral, unperformed, as I ij Performing, then, the integral of the n i dn j d part explicitly, we find η 0 ( 4 3 πε3 ) −2 ∫ ∫ d 3 r r≤ε r ′ ≤ε 3 . ∫ d 3 r ′ r −3 r2 s n i sns j f(n d ) = −η 1 d d 2 n d n i dn j d f(n d) + I ij 3 , (6.38) Again, the result (6.38) may be equivalently incorporated into our calculations by making the simple substitution rd −3 r2 s n i sn j s −→ I ij 3 − rd −1 ni dn j d , (6.39) and then performing the r d integration as if there had been no r s -dependence at all. Unfortunately, it is not so simple to devise such a straightforward crosscheck of (6.39) as was performed earlier for (6.35), due to the presence of the divergent integral I ij 3 . Instead, the author has thoroughly verified this result by directly comparing the term-by-term integrations of rd −3 with those of rd 0 (which latter have been verified above), which are of course in the ratio (n + 1)/8ε 3 (n − 2) if the overall integrand (including 4πrd 2 factor) has dependence rd n . (This can be seen quickly by integrating the polynomials from first principles, and noting that the extra factor of r 3 d contributes (2ε) 3 .) 251
6.4.6 Angular outer integrals We now turn to the question of performing integrals over the angular coördinates n d of r d , viz., integrals of the form ∫ d 2 n d f(n d ). (6.40) Clearly, since the volume of integration of r d is a simple sphere, we may separate the r d and n d integrals completely by employing spherical coördinates. (This was not possible for the r s integrals above, since the volume V s is not, in general, spherical.) Obviously, if the f(n d ) in (6.40) contains an odd number of factors of n d , the integral will vanish identically, by symmetry. We are thus left with integrals of the form ∫ d 2 n d n i dn j d , ∫ d 2 n d n i dn j d nk dn l d, ∫ d 2 n d n i dn j d nk dn l dn m d n n d, (6.41) and so on. We shall at this point employ clairvoyance for the third and final time, and predict that the integrals (6.41) exhaust those that we shall need in the radiation reaction calculations. Now, consideration of three-space covariance alone requires the integrals (6.41) to be able to be written solely in terms of the three-covariant quantity δ ij —there being no other covariant quantity available after the integration has been performed. This, together with due consideration of the symmetry of the expressions (6.41), already tells us that the answers must be of the form ∫ d 2 n d n i dn j d = β 2 δ ij , ∫ d 2 n d n i dn j { d nk dn l d = β 4 δ ij δ kl + δ ik δ jl + δ il δ jk} , ∫ d 2 n d n i dn j { d nk dn l dn m d n n d = β 6 δ ij δ kl δ mn + δ ij δ km δ ln + δ ij δ kn δ lm 252
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Single-Particle Electrodynamics Mr.
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Copyright Declaration I understand
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I apologise to Dr. Geoff Taylor for
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Contents 1 Overview of this Thesis
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3.2.2 Einsteinian rigidity . . . .
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6 Radiation Reaction 220 6.1 Introd
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A Notation and Conventions 314 A.1
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A.8.18 MCLF and CACS components . .
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E.2 Quantum field theory . . . . .
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G.7 test3int: Testing of 3-d integr
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lood pressure: such inane things ar
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1.2 What do I need to read? If the
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Section 2.7: Classical limit of qua
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Section 5.2: History of the retarde
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Section 6.7: Inverse-cube integrals
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Section A.7: Matrices Notation and
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Appendix C: Supplementary Proofs Se
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Section E.2: Quantum field theory T
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esults, not the establishment of ne
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e forgiven by their respective auth
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tron. The proton and neutron clearl
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most appropriate in the task of rem
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(see Section A.8): f ≡ d τ p. (2
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T 0i ≡ T i0 = E×B, T ij = 1 2 (E
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densities and the conservation laws
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2.4 Lagrangian mechanics An alterna
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canonical momentum: b a ≡ ∂ ˙q
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to be the appropriate degrees of fr
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task. Firstly, we need to select ap
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canonical energy, canonical momentu
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Let us return to the electric charg
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should also be carefully noted agai
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For an elementary particle, the mas
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2.6.7 Unit four-spin It is often ne
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the case. The reason, pointed out b
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period it appeared that the magnitu
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tionally reflect the difference bet
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has precessed by an amount ˙σ dt.
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is in terms of the quantities of Sp
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2.6.10 The FitzGerald three-spin In
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such systems. On the other hand, fo
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equations, one then finds, for eith
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i.e., the torque. Clearly, the corr
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The component ∆L 0 is our warning
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vanishes, for v = 0: P | v=0 = 0; (
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wish of them—not least of which b
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The author will, in the remaining c
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(see equation (A.58) of Section A.8
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6(c 1·c 3 ) + 4c 2 2 = 0, 8(c 1·c
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and reverting t(τ) from (2.91), on
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Chapter 3 Relativistically Rigid Bo
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than considering the rest frame of
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We also note Pearle’s proof of hi
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properties of the constituent r are
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where we set u(τ) = r (3.7) becaus
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+ 1 { [1 ] .... + (r· ˙v) v + 2 [
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applied than the former.) We shall
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which varies with time. We deem tha
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desirable properties—but unfortun
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the magnitude d of d describes the
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shall compute the net force on the
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we shall make the analysis relativi
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Now, let us consider such a spin-ha
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adreact of Appendix G; the expressi
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4.2.2 The magnetic-charge dipole Ma
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ut its employment of moving constit
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point of the loop at angle θ to th
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Now, the power into each positive c
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or, on reärranging the terms, −(
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Again noting the relations (4.23) a
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its discovery. Mathematically, the
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direction: E ≡ jE y . Now, let us
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(i.e., 2πε/v orb with v = 1, the
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a sufficient amount of inertia, the
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The fact that the Penfield-Haus eff
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tum of ±µ×E. Firstly, one can sh
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concepts precisely and correctly, a
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on the structure of spacetime, one
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4.3.4 Relativistic Lagrangian deriv
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we know that the operator equations
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invariance requirements. They sough
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That (4.73) is an amazing result is
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is an important example. Clearly, t
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let us therefore add a Pauli term (
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On the other hand, for a Pauli mome
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moment actually cancels with the Di
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Chapter 5 The Retarded Fields I hav
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to particles carrying dipole moment
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workers. In 1969, Cohn [53] unfortu
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manifestly-covariant field expressi
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as a mathematical aid—it not bein
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integrate over the proper time τ,
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Equation (5.28) is, in the above no
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this is Jackson’s equation (14.13
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moments of a particle. There are a
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It can be noted that equation (5.51
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the overall Lorentz-gauge four-pote
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Applying (5.61) to the delta functi
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The results above are for a particl
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(One should carefully note that the
Page 201 and 202: under differentiation, the latter e
Page 203 and 204: Because the field expressions diver
Page 205 and 206: expression (5.81) in the direction
Page 207 and 208: We now note that the non-zero elect
Page 209 and 210: clearly, for any ε, this parametri
Page 211 and 212: For a point electric dipole moment,
Page 213 and 214: where the convenient constant η 1
Page 215 and 216: the former generates solely an elec
Page 217 and 218: The position of the mechanical cent
Page 219 and 220: when the particle is in arbitrary m
Page 221 and 222: Chapter 6 Radiation Reaction Accord
Page 223 and 224: perhaps, fortunately,—such a “f
Page 225 and 226: Lorentz performed this calculation
Page 227 and 228: 6.2 Previous analyses To the author
Page 229 and 230: other than pedagogy, in the past fi
Page 231 and 232: immediate practicality, since then
Page 233 and 234: 6.3 Infinitesimal rigid bodies We n
Page 235 and 236: enefit that, by Gauss’s law, ther
Page 237 and 238: particular times depend on the acce
Page 239 and 240: ground our understanding of the qua
Page 241 and 242: whence the reverse transformation i
Page 243 and 244: We define η 0 ≡ 1 ( ) 4 −2 ∫
Page 245 and 246: and ending values of each loop are
Page 247 and 248: = 1 8( 4 3 π(2ε)3 ) 2 , (6.23) wh
Page 249 and 250: = 4 3 π(2ε)5 { 1 5 − 3 4 α +
Page 251: where by the notation ∫ d 2 n d w
Page 255 and 256: 6.4.7 Is there an easier way? It mi
Page 257 and 258: 6.5.3 Final expression for the reta
Page 259 and 260: Using the fact that t ret ≡ −R,
Page 261 and 262: 6.5.9 Modified retarded normal vect
Page 263 and 264: 6.6 Divergence of the point particl
Page 265 and 266: “godlike” parameter, i.e., one
Page 267 and 268: Using the chain rule on (6.74), we
Page 269 and 270: Our first use of (6.84) is to compu
Page 271 and 272: The significance of this result is
Page 273 and 274: lead to delta-function contribution
Page 275 and 276: (6.99) then gives { (a·b) − 5(n
Page 277 and 278: 6.7.1 The bare inverse-cube integra
Page 279 and 280: 6.7.2 Inverse-cube integral with tw
Page 281 and 282: we immediately find that the r d -i
Page 283 and 284: 6.8.4 Duality symmetry consideratio
Page 285 and 286: in order that Maxwell’s equations
Page 287 and 288: This asserted result of the author
Page 289 and 290: obtain a non-vanishing integral ove
Page 291 and 292: − 1 2 ˜µ2 η 1 ( ˙v· ¨σ)σ
Page 293 and 294: If one examines the author’s calc
Page 295 and 296: performed the electric charge radia
Page 297 and 298: and the spin renormalisation term f
Page 299 and 300: motion of the particle. This has, i
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6.10.2 The Ternov-Bagrov-Khapaev ef
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where the characteristic polarisati
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detail, for arbitrary values of g,
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Substituting (6.151) into (6.152),
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importantly, the hyperbolic tangent
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twice, then it clearly will not do
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Appendix A Notation and Conventions
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Filename bibliog.tex claspcle.tex c
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Filename algebra.h algebra1.c algeb
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costella, british or american must
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fonts of other resolutions, or font
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typographically, the mathematical n
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quantity which has been designated
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noted that the inline fraction symb
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The d’Alembertian operator is den
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It should be noted that quantum mec
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A.3.15 Pointlike objects If an obje
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as a mathematically G-covariant qua
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The adjective enumerated will be us
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Each of the i m take values from th
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distributed factor; for example, A
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A.7.2 Square matrices Matrices with
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A.8.2 Lorentz tensors Rank-zero, ra
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AB denotes the four-tensor that is
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and G are four-tensors, then ε(A,
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The use of this signature of metric
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ackets; e.g., C α | MCLF ≡ [C α
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The first way is to simply measure
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is referred to as the parity operat
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three-vector conversion. In any cas
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A.9.9 Magnitude of a three-vector T
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A.9.13 Cross-products An alternativ
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B.2.2 The field strength tensor The
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B.2.9 Explicit electromagnetic comp
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Also, (a×b)·(c×d) ≡ (a·c)(b·
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{ (σ·∇)rd m r s (n s·A)n d = r
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For two electric charges q 1 and q
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where z is the position of the char
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Appendix D Retarded Fields Verifica
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(−2) ≡ − e { 3 a U mρ 2 2 ρ
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F µν + 1 2ȧUM [να U α R µ]
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Comparing (D.11) and (D.12), we see
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efore neutral particles were discov
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Appendix E The Interaction Lagrangi
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G µ ≡ G 1 k/k µ + G 2 k/B µ +
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u 2 σ µν u 1 −→ ˜Σ µν ,
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powers of ∂ 2 acting on the field
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International Journal of Modern Phy
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we shall not claim any predictive c
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least, if one wants to derive the e
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and electric charge interaction ter
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way that the “Lagrangian-based”
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Θ ≡ ζ 2 (E·B) , ∂ ′ ≡
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prospective solution of the combine
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Appendix G Computer Algebra Mrs. Du
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) dependence to extract the three-g
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mately chosen, both for simplicity,
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appendical reference. In following
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Program Filename Description kinema
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G.4 kinemats: Kinematical quantitie
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The reason for us naming it after F
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[ ˙ Σ 0 ] = γ 2 (v· ˙σ) + γ
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− 25γ 10 (γ + 1) −2 ˙v 2 (v
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+ 124γ 11 (v· ˙v) 2 (v·¨v), (
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+ 2γ 5 (γ + 1) −2 ( ˙v·σ)¨v
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( Σ ... ) 0∣ ∣ ∣v=0 = 0, (
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+ 4γ 8 (v· ˙v) 2 (v·σ ′ ), (
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fields. These products are labelled
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+ 3γ 4 (γ + 1) −1 (v· ˙σ)¨v
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G.5.2 Covariant field expressions S
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B ′d 1 = κ 2 ¨σ ′ ×n + κ 3
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Expanding out these expressions exp
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G.6 radreact: Radiation reaction G.
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Firstly, iterating the differential
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we define the redshift factor λ
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G.6.8 Sum and difference variables
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− 1 16 r d(r d· ˙v) 3 − 1 8 r
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− 1 120 r4 d (n d·.... v)n d + 1
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+ 1 4 r2 d (n d· ˙v) 2 ˙v − 1
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+ 1 8 r2 d r s ˙v 2 (n s· ˙v)n d
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− 1 48 r3 d r s (n d· ˙v)(n s·
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Using (G.28) and (G.55) in (G.56),
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+ 1 8 r4 d (n d· ˙v) 2 ¨σ − 1
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+ 3 2 r dr s ˙v 2 (n s· ˙v) ˙σ
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E q 2 = r −2 d n d + 1 2 r−1 d
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− r d (n d· ˙v)( ˙v· ¨σ)n d
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+ 39 4 r s(n d· ˙v)(n d·¨v)(n d
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+ 3rd −1 r s(n d· ˙v)(n d·σ)(
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+ 1 4 r s(n s·... v) ˙σ + 15 8 r
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+ 1 8 r−2 d r3 s (n d· ˙v)(n s
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− 3 4 r−2 d r2 s (n d· ˙σ)(n
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− 9 16 r s(n d·σ)(n s·¨v)¨v
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B q 1 = −rd −1 n d× ˙v + n d
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− 39 8 r d(n d· ˙v) 2 (n d·σ)
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− 1 2 r−1 d r s(n s·¨v)n d ×
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+ 3 4 r s(n d·σ)(n s· ˙v) ˙v×
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− 5 12 r d(n d·¨v)¨v×σ − 1
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If we add the various E a n and B a
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+ 15 16 r d ˙v 2 (n d· ¨σ)n d +
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+ 3 16 r−2 d r3 s (n d· ˙v)(n d
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+ 3 8 r s(n d·¨v)(n d·σ)(n s·
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+ 1 24 r(n·¨v)(¨v·σ)n − 5 24
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λE q 2 = r −2 d n d + 1 2 r−1
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+ 27 4 (n d· ˙v) 3 (n d·σ)n d
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+ 1 2 r−1 d r s(n d·σ)(n s·¨v
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− 1 2 r−1 d r s(n s· ˙v)n d
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+ 1 8 r−1 d r s(n s· ˙v)( ˙v·
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λ(σ·∇)E d 1 = −rd −2 ¨v
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− 11rd −1 (n d· ˙v)(n d·σ)(
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+ 1 6 (¨v· ¨σ)σ − 1 2 (... v
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− 3rd −2 r2 s (n d· ˙v)(n d·
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d·... − 3rd −1 (n d·σ)(n σ)
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− 12rd −2 r s(n d·¨v)(n d·σ
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+ 101 4 (n d·¨v)(n d·σ)( ˙v·
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+ 7rd −2 (n d·σ)(¨v·σ)n d +
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− 57 2 r−1 d (n d· ˙v) 2 (n d
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+ (n d· ˙v)(n d·... σ)σ + 3(n
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− rd −2 r2 s (n s· ˙v)(n s·
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+ 1 8 r−1 d (n d· ˙v) 2 ( ˙v·
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P (n) ad (r d, r s ) = ˙σ·E a n(
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F (2) dq = 1 15 η 1 ˙v 2 σ − 1
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F (M) µd = −3η ′ 3σ× ˙σ,
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N F (1) dd = 1 10 η 1( ˙v·σ) ˙
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% ID string: Test 3-d integrations.
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+rd^{-3}(n.a)(n.b) +rd^{-3}(n.c)(n.
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+(n.q)(n.r)(ns.s)(ns.t)ns +(ns.u)v
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+\f{1}{105}(i.o)(j.m)(k.l)a +\f{2}{
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+rd^{-2}rs^2(n.u)(n.v)(ns.w)(ns.x)y
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+\f{1}{30}(o.p)a*m +\fourthirds a
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+\f{1}{105}(w.y)(x.z)a’*a +\f{1}{
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Bibliography [1] M. Abraham, Ann. P
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[39] H. J. Bhabha, Proc. Roy. Soc.
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[80] J. R. Ellis, J. Math. Phys. 7
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[124] M. Kolsrud and E. Leer, Phys.
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[167] M. Pavsic (1987): see [34]. [
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[212] I. M. Ternov, V. G. Bagrov an
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It doesn’t matter whether you suc
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