Single-Particle Electrodynamics - Assassination Science

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For the complete integral (6.110), we therefore have η 0 ( 4 3 πε3 ) −2 ∫ ∫ d 3 r r≤ε d 3 r ′ ˜r d −3 n˜r d in˜rd j = δ ij 3 r ′ ≤ε { −8η ′ 3 + 3η ′′ 3 } . (6.113) We can now see why the apparent vanishing of an n d integration cannot be trusted, for the inverse-cube fields; for example, from (6.109) and (6.113), we have, for the integral of the dipolar field expression over the interior of the sphere, ( 4 3 πε3 ) −1 ∫ ∫ d 3 r r≤ε d 3 r ′ 3(n d·a)n d − a 4πr 3 r ′ ≤ε ( ) 4 = −η 3a ′ 3 πε3 = − 1 a, (6.114) 3 the correct result (5.86) of Chapter 5, despite the fact that the n d integral would appear to otherwise vanish. Note that, as advertised, the divergent quantity η ′′ 3 has disappeared from this final result (which will, incidentally, be a contributor to our final equations of motion). 6.7.3 Inverse-cube integral with four normals Finally, we consider the inverse-cube integral involving four factors of n˜rd : η 0 ( 4 3 πε3 ) −2 ∫ ∫ d 3 r r≤ε d 3 r ′ ˜r −3 r ′ ≤ε d n˜r d in˜rd jn˜rd kn˜rd l. Clearly, our analysis above for the two-factor case carries through identically, except that the n d -integration yields, of course, 1 { } δij δ kl + δ ik δ jl + δ il δ jk , 15 and that there is now yet another extra factor of u 2 in our integrand. By noting, from (6.111), that u 6 1 − u 2 ≡ u4 1 − u 2 − u4 ≡ u2 1 − u 2 − u2 − u 4 , 279
we immediately find that the r d -integral’s contribution is the same as before, with the addition this time of a term −u 4 in the integrand; thus, η 0 ( 4 3 πε3 ) −2 ∫ ∫ d 3 r r≤ε d 3 r ′ ˜r −3 r ′ ≤ε d n˜r d in˜rd j = 1 { } { δij δ kl + δ ik δ jl + δ il δ jk − 43 } 15 5 η′ 3 + 3η 3 ′′ . (6.115) 6.8 Computation of the self-interactions We can now use the expressions of the previous sections, as well as the power, force and torque equations of motion of Chapter 4, to obtain the radiation reaction equations of motion themselves. 6.8.1 Non-radiative equations of motion The equations of motion in Chapter 4, ignoring radiation reaction, were obtained for an arbitrary velocity of the particle. Since we have, in this chapter, placed the “receiving” body at rest, we need only consider these equations for the simplified situation v = 0. dipole and magnetic dipole, we then find P q = 0, P d = d ˙σ·E, P µ = µ ˙σ·B, F q = qE, F d = d(σ·∇)E + d( ˙v·σ)E + d ˙σ×B, For a point charge, electric F µ = µ(σ·∇)B + µ( ˙v·σ)B − µ ˙σ×E + µσ×J, N q = 0, N d = dσ×E, N µ = µσ×B. (6.116) 280
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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
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under differentiation, the latter e
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Because the field expressions diver
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expression (5.81) in the direction
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We now note that the non-zero elect
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clearly, for any ε, this parametri
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For a point electric dipole moment,
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where the convenient constant η 1
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the former generates solely an elec
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The position of the mechanical cent
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when the particle is in arbitrary m
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Chapter 6 Radiation Reaction Accord
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perhaps, fortunately,—such a “f
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Lorentz performed this calculation
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6.2 Previous analyses To the author
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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 and 252: where by the notation ∫ d 2 n d w
Page 253 and 254: 6.4.6 Angular outer integrals We no
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: 6.7.2 Inverse-cube integral with tw
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
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Page 303 and 304: 6.10.2 The Ternov-Bagrov-Khapaev ef
Page 305 and 306: where the characteristic polarisati
Page 307 and 308: detail, for arbitrary values of g,
Page 309 and 310: Substituting (6.151) into (6.152),
Page 311 and 312: importantly, the hyperbolic tangent
Page 313 and 314: twice, then it clearly will not do
Page 315 and 316: Appendix A Notation and Conventions
Page 317 and 318: Filename bibliog.tex claspcle.tex c
Page 319 and 320: Filename algebra.h algebra1.c algeb
Page 321 and 322: costella, british or american must
Page 323 and 324: fonts of other resolutions, or font
Page 325 and 326: typographically, the mathematical n
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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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