FINITE SELF MASS OF THE ELECTRON II

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1 FINITE SELF MASS OF THE ELECTRON II ABSTRACT The charge of the electron is postulated to be distributed or extended in space. The differential of the electron charge is set equal to a function of electron charge coordinates multiplied by a four-dimensional differential volume element. The four-dimensional integral of this function is required to equal the electron charge in all Lorentz frames. The self-mass of such an electron is found to be finite. I. LOGARITHMIC DIVERGENCE OF THE SELF-MASS This section will show that the self-mass of a point electron diverges logarithmically. The electron self-mass is associated with the following sequence of events: a point electron propagates from spacetime point y to spacetime point w where it emits a photon, the electron then propagates to spacetime point z where it absorbs the same photon, finally the electron then propagates to spacetime point x. The propagator for this process, which is second order in the charge e, is Date: August, 25.

2 2 FINITE SELF MASS OF THE ELECTRON II S F2 x y = S F x z ieγ µ S F z w ieγ µ S F w yd F z wd 4 z d 4 w where γ µ γ µ µ =,, 2, 3 both represent the Dirac matrices, S F x z, S F z w, S F w y represent electron propagators. The electron propagators are given by S F x z = S F z w = S F w y = exp ip x zs F p d4 p 2π 4, 2 exp ik z ws F k d4 k 2π 4, 3 exp ip w ys F p d4 p 2π 4, 4 where we choose the following representations in momentum space for the electron propagators: S F p = i /p + m, 5 p 2 m 2 S F p = i /p + m, 6 p 2 m 2

3 FINITE SELF MASS OF THE ELECTRON II 3 S F k = i/k + m 2 k 2 + m 2 k k 2 + m 2 + iɛ k + k 2 + m 2 iɛ 7 where p, p, k are four-momentum vectors, /p = γ µ p µ, /p = γ µ p µ, /k = γ µ k µ, m is called the bare electron mass. The observable or physical mass is given by m = m +δm where δm is the self-mass. The photon propagator is given by D F z w = i exp iq z wd F q d4 q 2π 4. 8 Choose the following representation for the photon propagator in momentum space: D F q = 2 q q q + iɛ q + q iɛ 9 where q is a four-momentum vector. It should be noted that the Eq. 7 Eq. 9 are not the usual choice for the propagators. Introduce the following abbreviations: P = S F pexp+ip y d4 p 2π 4,

4 4 FINITE SELF MASS OF THE ELECTRON II P = S F p exp ip x d4 p 2π 4. Then using γ µ /k + mγ µ = 2/k + 4m. S F2 x y = ie 2 P expip z i 2/k + 4m 2 exp ik z w k 2 + m 2 d 4 k 2π 4 k k 2 + m 2 + iɛ k + k 2 + m 2 iɛ i d4 q exp iq z w 2π 4 2 q q q + iɛ q + q iɛ d 4 z d 4 w exp ip w P = P expip z I I 2 exp ip w dz dw I 3 P. 2 where dk I = 2π exp ik z w 2/k + 4m 2 k 2 + m 2 k k 2 + m 2 + iɛ + k + k 2 + m 2 iɛ, 3 dq I 2 = 2π exp iq z w 2 q q q + iɛ q + q iɛ, 4

5 FINITE SELF MASS OF THE ELECTRON II 5 I 3 = exp[ik + q p z]d 3 z exp[i k q + p w]d 3 w = 2π 3 δ 3 k + q p 2π 3 δ 3 k q + p = 2π 3 δ 3 p p2π 3 δ 3 k q + p. 5 Here δ denotes the delta function. Use contour integration find I = ihz w exp i k 2 + m 2 z w 2γ k 2 + m 2 2γj k j + 4m 2 k 2 + m 2 ihw z exp+i k 2 + m 2 z w +2γ k 2 + m 2 2γ j k j + 4m 2 k 2 + m 2, 6 where H is the unit step function, the index j =, 2, 3. Again use contour integration find I 2 = i exp i q z w Hz w 2 q i exp i q w z Hw z. 7 2 q The product I I 2 contains four terms. Since Hz w Hw z =, two of the terms are zero. So

6 6 FINITE SELF MASS OF THE ELECTRON II 2γ I I 2 = i 2 k 2 + m 2 2γ j k j + 4m 2 q 2 k 2 + m 2 Hz w exp i k 2 + m 2 + q z w +2γ + i 2 k 2 + m 2 2γ j k j + 4m 2 q 2 k 2 + m 2 Hw z exp i k 2 + m 2 + q w z. 8 Introduce I = exp i k 2 + m 2 + q z w Hz w exp[ip z p w ]dz dw, 9 I 2 = exp +i k 2 + m 2 + q z w Hw z exp[ip z p w ]dz dw. 2 In Eq. 9, change variables from z to t = z w, so I = Ht exp i k 2 + m 2 + q p t dt exp[ip p w ]dw. 2 By the theory of generalized functions, 2 I = 2π δp p I S where

7 FINITE SELF MASS OF THE ELECTRON II 7 i I S = k k2 + m 2 + q + πδ 2 + m 2 p + q p. 22 In Eq. 2, change variables from w to t = w z, so I 2 = Ht exp i k 2 + m 2 + q + p t dt exp[ip p z ]dz. 23 By the theory of generalized functions, 2 I 2 = 2π δp p I S2 where i I S2 = k k2 + m 2 + q + p + πδ 2 + m 2 + q + p. 24 A delta function vanishes when its argument is not zero. The argument of the delta functions in Eqs is never zero, so both delta functions are zero may be dropped. Then S F2 x y = e 2 P I 2γ k 2 + m 2 2γj k j + 4m + I 2+2γ k 2 + m 2 2γ j k j + 4m d 3 k d 3 q I 3 P 2π 3 2π 3 = e 2 P d 3 k 2π 3 d 3 q 2π 3 2 q 2 k 2 + m 2 2π 3 δ 3 p p2π 3 δ 3 k q + p 2π δp p [I S 2γ k 2 + m 2 2γ j k j + 4m + ] I S2 +2γ k 2 + m 2 2γ j k j + 4m P. 25

8 8 FINITE SELF MASS OF THE ELECTRON II Use d 4 p 2π 4S Fp exp ip x2π 4 δ 4 p p = S F pexp ip x, 26 find S F2 x y = ie 2 S F pexp ip x 2π3 δ 3 +k + q p 2 q 2 d 3 k d 3 q k 2 + m 2 2π 3 2π 3 [I S 2γ k 2 + m 2 2γ j k j + 4m + ] I S2 +2γ k 2 + m 2 2γ j k j + 4m S F pexpip y d4 p 2π 3 = S F pexp ip x iσ 2 ps F pexp +ip y d4 p 2π 4 27 where Σ 2 p = e 2 d 3 k d 3 q 2π 3 2π 3 2π3 δ 3 k + q p 2 q 2 k 2 + m 2 [I S 2γ k 2 + m 2 2γj k j + 4m + ] I S2 +2γ k 2 + m 2 2γ j k j + 4m. 28 Σ 2 /p = m is identified approximately as δm the self-mass.

9 FINITE SELF MASS OF THE ELECTRON II 9 It is convenient to continue the calculation in the frame where the electron is initially at rest, so p = k = q. Then Σ 2 /p = m,p = = e 2 d 3 q 2π 3 2 q 2 q 2 + m 2 [ 2γ q 2 + m 2 + 2γj q j + 4m q2 + m 2 + q m + 2γ q 2 + m 2 + 2γj q j + 4m q2 + m 2 + q + m ]. 29 The terms in the numerator, which are linear in q, q 2, q 3, integrate to zero. Put the terms in brackets over a common denominator. Then [ Σ 2 /p = m,p = = e2 d 3 4m q2 + m 2 + 8m q + q q 2 + m 2 ] 2π 3 2 q 2 q 2 + m 2 2 q q + q 2 + m 2 = e2 m [ d q 4π 2 q + 2 ] + 3 q 2 + m 2 q2 + m 2 Thus Σ 2 /p = m,p = = δm d q / q log q as q, the self-mass of the point electron is logarithmically divergent. II. EXTENDED ELECTRON COORDINATES In the rest frame of an electron charge distribution, let x µ r = x r, x r, x 2 r, x 3 r denote a spacetime charge point. Let x µ r = x r, x r, x 2 r, x 3 r denote the center of the charge distribution. The charge distribution of the electron is assumed to have a well-defined center, which is identified as the argument of the wave function. The shape of the charge distribution

10 FINITE SELF MASS OF THE ELECTRON II depends on the motion of the charge, is assumed to be unaffected by any interaction. Sometimes the superscript will be omitted, we will write x r = x r, x r, x 2 r, x 3 r x r = x r, x r, x 2 r, x 3 r. Introduce x r = x r x r or equivalently x µ r = x µ r x µ r. In a frame of reference in which the charge distribution moves with a constant speed β in the +x 3 direction, let x m = x m, x m, x 2 m, x 3 m denote a spacetime charge point, let x m = x m, x m, x2 m, x3 m denote the center of the charge distribution. Introduce x m = x m x m. A Lorentz transformation yields x r = x m, x 2 r = x 2 m, x 3 r = γ x 3 m β x m, x r = γ x m β x 3 m where γ = / β 2. In the rest frame, the electron charge e is equal to ρ r x r δ x r d4 x r where ρ r x r is the charge density in the rest frame. 3 So an element of charge in the rest frame is given by de x r = ρ r x r δ x rd 4 x r. 3 In the moving frame described above, the element of charge is de x m = ρ r x m δγ x m β x3 m d4 x m Consider a uniformly charged spherical surface of radius a, which is at rest, is centered at x = x r, x 2 r, x 3 r. The charge density is given by

11 FINITE SELF MASS OF THE ELECTRON II ρ r x r = e 4πa 2δ x r 2 + x 2 r 2 + x 3 r 2 a 33 When the charge moves in the +x 3 direction with a speed β, the charge density is given by ρ r L x m = e 4πa 2δ x m 2 + x 2 m 2 + γ 2 x 3 m β x m 2 a 34 III. FINITE SELF-MASS OF THE EXTENDED ELECTRON Take the extended electron to be initially at rest. It will be assumed that when an element of electron charge emits a photon of fourmomentum q, then the electron recoils with a speed β = k /k. When an element of electron charge absorbs a photon of four-momentum q, then the electron acquires a four-momentum p = p. The photons are emitted absorbed at charge points, so replace D F z w by D F z w. Each charge point propagates between emission absorption of the photon with momentum k, so S F z w is replaced by S F z w. In addition, the first e in Eq. is replaced by the four-dimensional integral of de z, the second e is replaced by the four-dimensional integral of de w. Then Eq. is replaced by S F2e x y = S F x z ide zγ µ S F z w ide wγ µ S F w yd F z w d 4 z d 4 w, 35

12 2 FINITE SELF MASS OF THE ELECTRON II Eq. 2 is replaced by S F2e x y = ie 2 P expip z i 2/k + 4m 2 exp ik z w k 2 + m 2 d 4 k 2π 4 k k 2 + m 2 + iɛ k + k 2 + m 2 iɛ i d4 q exp iq z w 2π 4 2 q q q + iɛ q + q iɛ de z de w d 4 z d 4 w exp ip w P 36 e e Change variables from z to z from w to w. get S F2e x y = ie 2 P expip z i 2/k + 4m 2 exp ik z w k 2 + m 2 d 4 k 2π 4 k k 2 + m 2 + iɛ k + k 2 + m 2 iɛ i d4 q exp iq z w 2π 4 2 q q q + iɛ q + q iɛ exp ip z de z de w exp+ip w d 4 z d 4 w exp ip w P e e 37

13 In place of Eq. 8 for I I 2 is FINITE SELF MASS OF THE ELECTRON II 3 2γ I I 2 e = i 2 k 2 + m 2 2γ j k j + 4m 2 q 2 k 2 + m 2 Hz w exp i k 2 + m 2 + q z w +2γ + i 2 k 2 + m 2 2γ j k j + 4m 2 q 2 k 2 + m 2 Hw z exp i k 2 + m 2 + q w z, 38 which is to be multiplied by exp ip zde zexp+ip wde w. For the purpose of illustration, pick the electron charge to be uniformly distributed on a spherical surface of radius a in the electron rest frame. For z w >, the electron is initially at rest, so exp+ip w r de w r e = expip w r ip w r 4πa 2 δ w r 2 + w 2 r 2 + w 3 r 2 aδ w rd 4 w r =. 39

14 4 FINITE SELF MASS OF THE ELECTRON II After emission of the photon, the electron travels with a speed β = k /k. Choose k to be in the direction of z 3. Then exp ip z m de z m e = exp ip z m + ip z m 4πa 2 δ z m 2 + z 2 m 2 + γ 2 z 3 m β z m 2 aδγ z m β z 3 md 4 z m = exp ip β z m 3 4πa 2δ z m 2 + z m z m 3 2 /γ 2 a d3 z m = γ exp im βγẑm 3 4πa 2δ zˆ m ad 3 ẑ m = expi q a exp i q a 2i q a = sin q a q a 4 where βγ = k / k 2 k 2 = q /m, z m = ẑm, z m 2 = ẑm, 2 z m/γ 3 = ẑm, 3 p = m. For w z >, the electron is finally at rest. After emission of the photon, the electron travels backward in time with a speed β = k /k. So exp ip z r de z r e = exp ip z r + ip z r 4πa 2 δ z r 2 + z 2 r 2 + z 3 r 2 aδ z rd 4 z r =, 4

15 FINITE SELF MASS OF THE ELECTRON II 5 exp+ip w m de w m e = exp+ip w m ip w m 4πa 2 δ w m 2 + w 2 m 2 + γ 2 w 3 m β w m 2 aδγ w m β w3 m d4 w m = exp+ip β w m 3 4πa 2δ w m 2 + w m w m 3 2 /γ 2 a d3 w m = γ exp im βγŵm 3 4πa 2δ wˆ m ad 3 ŵ m = expi q a exp i q a 2i q a = sin q a q a 42. So Eq. 3 is replaced by Σ 2e /p = m,p = = e 2 m 4π 2 e2 2π 3 [ d 3 4m q2 + m 2 q + 8m q + q 2 + m 2 ] 2 q 2 q 2 + m 2 2 q q + q 2 + m 2 sin q a = q a [ d q q + 2 ] sin q a q 2 + m 2 q2 + m 2 q a The integr is finite for small q since sin q a/ q a = at q =. Take the absolute value of the integr note that for large q d q [ + q q + q 2 + m ] q2 + m 2 d q q Thus the integral converges absolutely, the self-mass is finite for the assumed charge distribution. This suggests, but does not prove that the self-mass will be finite for a charge distributed over a finite

16 6 FINITE SELF MASS OF THE ELECTRON II region of space. The integral for the self-mass will be evaluated in the appendix. APPENDIX Write Eq. 43 in the form Σ 2e /p = m,p = = e2 m 4π 2 [ d q q + 2 ] + q 2 + m 2 q2 + m 2 expi q a exp i q a. 45 2i q a Multiply the first term in the brackets by q + q 2 + m 2 / q + q 2 + m 2. Then multiply q 2 + m 2 /m2 by q 2 + m 2 / q 2 + m 2. The result is Σ 2e /p = m,p = = e2 m [ q d q 4π 2 2ia m 2 q2 + m 2 + q + ] expi q a exp i q a. 46 q 2 + m 2 m 2 Change variables in the terms multiplied by expi q a by replacing q by q, find Σ 2e /p = m,p = = e2 m 4π 2 2ia I 2 + I 3 + I 4 47 where I 2 = + d q exp i q a q m 2 q2 + m 2, 48

17 FINITE SELF MASS OF THE ELECTRON II 7 + I 3 = d q exp i q a q q 2 + m 2, 49 + I 4 = d q exp i q a sgnq m 2 = +2i m 2 a.5 5 Set q = m q, find I 2 = d q exp i q m a q m q 2 +, 5 I 3 = d q exp i q a m q q 2 +, 52 To integrate Eqns. 5 52, the following integrals will be used with the convolution: I = d q exp i q m a = 2K m a, 6 53 q 2 + I 2 = d q exp i q m a q = 2πiδ m a, 5 54 I 3 = d q exp i q m a q = πi sgnm a Here K is the modified Bessel function of order zero, which is given by K x = πi 2 [J ix + in ix]. 7 56

18 8 FINITE SELF MASS OF THE ELECTRON II Here J is the Bessel function of order zero, N is the Neuman function of order zero. Use the convolution 6 to write I 2 = 2πm 2K α2πiδ ma αdα = 2i m K αδα madα, 57 I 3 = 2πm 2K α πisgnma αdα = +i ma 2K αdα. m 58 K can be written as a series. For small argument, K takes the form K x ln x x 2 γ 2 4 where γ = is Euler s constant. Then so K x x x 2 I 2 = 2i K 2i [ ma m m m a m a 2 ln x + γ, ln x + γ, ln x 2 ] + γ, 6 2 So I 3 2i m a ln m a + m a γ 62 m 2 2 I 2 + I 3 + I 4 2i [ 3m a m 2 ln m a 2 5 ] + γ. 63 4

19 FINITE SELF MASS OF THE ELECTRON II 9 finally Σ 2e /p = m,p = e2 m [ m a 3ln 3γ π2π 2 2] ACKNOWLEDGMENTS I can not thank Ben enough for his continued help suoport. References [] M. Peskin D. Schroeder, An Introduction to Quantum Field Theory Perseus Books, Reading MA. 995, pp [2] D.S. Jones, Generalized Functions McGraw-Hill, Berkshire, Engl 966, p. 99. [3] S. Weinberg, Gravitation Cosmology: Principles Applications of the General Theory of Relativity John Wiley & Sons, Inc. New York, 972, pp [4] Scattering Beyond. [5] D.S. Jones, Generalized Functions McGraw-Hill, Berkshire, Engl 966, p [6] D.S. Jones, Generalized Functions McGraw-Hill, Berkshire, Engl 966, p. 58. [7] George Arfken, Mathematical Methods for Physicists-3rd edition Academic Press, Boston 985, p. 62 [8] George Arfken, Mathematical Methods for Physicists-3rd edition Academic Press, Boston 985, p. 575 p. 598.

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