Spin Precession in Electromagnetic Field

Transcript

1 Spin Preession in Eletromagneti Field Eunil Won, v T-BMT Equation This note is an expliit alulation of the spin preession known as T-BMT equation. The method used in this note is based on arxiv: and Jakson, Classial Eletrodynamis 3rd edition with SI unit. In the rest frame of a partile, the time evolution of spin 3-vetor s in eletri E and magneti B fiel an be written as = µs B + E 1 where µ and d are magneti and eletri dipole moments, respetively : magneti dipole moment, EDM: eletri dipole moment. Now, let us disuss how it looks like when it is written in relativistially ovariant form. In a referene frame, 1 let us define the momentum and spin 4-vetor of a relativisti partile as S µ and P µ, respetively. In the frame that the partile is at rest, we have s µ = 0, s, p ν = m, 0 where is the speed of light. Note that s is the spin 3-vetor in the rest frame of the partile. Then, Lorentz invariane required that S µ P µ = s µ p µ = 0, S µ S µ = s µ s µ = s s = s 3 1 In this note, apital-lettered 4-vetors are defined in the lab frame, and lowerased 4-vetors are defined in the rest frame of the partile. Also, we use sign onvention of S µ = S 0, S, S µ = S 0, S. 1

2 are satisfied. Also, if we use veloity 4-vetor U µ =, v, S µ U µ = s µ u µ = 0, S 0 = β S 4 is satisfied. β = v/, = 1/ 1 β. Now, the spin 4-vetor S µ an be obtained by Lorentz transformation on s µ just like the ase for the 4-momentum. Aording to the notation in Jakson p. 55, Lorentz transformation of spae-time an be expressed as x 0 = x 0 β x, x = x + 1 β β xβ βx 0, 5 when two frames move with the relative veloity β. To apply them to our spin ase, let us assume that -system is the rest frame of the partile. Then with S 0 = β S, s = S + 1 β β Sβ βs 0 1 = S + β β Sβ = S β Sβ 6 is obtained. Sine we are interested in the inverse of them, let s use S 0 = s 0 + β s, S 0 = S 0 = β s, so β S = S 0 = β s an be obtained. Now, the reverse transformation is S = s + β Sβ = s + β sβ. 7 Now let us onstrut the relativisti equation of motion for spin. The left-hand side of suh equation should be the Lorentz ovariant form of µ /dτ. The right-hand side may ontain eletromagneti field, spin, or veloity 4-vetor. If we an neglet terms like µ B, we may write the spin equation as see arguments developed in Jakson, p.

3 563. In priniple we an have terms like U µ ν F νλ S λ? µ dτ = 1F µν S ν + U µ F νλ U ν S λ + 3 F µν S ν + 4 U µ F νλ U ν S λ 8 where i i = 1,, 3, 4 are onstants to be determined later and F µν = F µν = 0 E x / E y / E z / E x / 0 B z B y, E y / B z 0 B x E z / B y B x 0 0 B x B y B z B x 0 E z / E y /. 9 B y E z / 0 E x / B z E y / E x / 0 Here F µν is the field-strength tensor Jakson p. 556 and F µν = 1 ϵµνρσ F ρσ. Now, the spatial omponents of Eq. 8 at the partile rest frame are i = 1 F ij s j + 3 F ij s j 10 and using s µ = 0, s, s µ = 0, s, one an write them as F 1j S j = B z s y B y s z = s B x F j S j = B z s x + B x s z = s B y F 3j S j = B y s x B x s y = s B z F 1j S j = E z s y + E y s z / = E s x / F j S j = E z s x E x s z / = E s y / F 3j S j = E y s x + E x s y / = E s z /. 11 Therefore, i = 1s B i + 3 E s i / 1 3

4 is satisfied. This should be onsistent with Eq. 1 so we onlude that 1 = µ, 3 = d 13 must be satisfied. On ther other hand, the value of an be obtained from the relativisti equation of motion m du µ dτ = ef µν U ν and S µ U µ = 0. If we differentiate S µ U µ = 0 with respet to the proper time, µ U µ dτ = S du µ µ dτ = S µ e m F µν U ν = e m F µν U µ S ν 14 is satisfied where we use anti-symmetri nature of F µν with indies µ, ν at the last step. Now, ontrat the veloity 4-vetor U µ to Eq. 8 and use U µ U µ =, one gets U µ µ dτ = µf µν U µ S ν + U µ U µ F νλ U ν S λ df µν U µ S ν + 4 U µ U µ F νλ U ν S λ = µ + F µν U µ S ν + d + 4 F µν U µ S ν = e m F µν U µ S ν 15 so µ + e F µν U µ S ν + d + 4 F µν U µ S ν = 0 16 m must be satisfied. Therefore, = µ e, 4 = d m 17 are found. The spin 4-vetor equation of motion is now µ dτ = µf µν S ν µ e U µ F νλ U ν S λ df µν S ν 1 m U µ F νλ U ν S λ 18 and Eq. 18 is known as the relativisti Thomas-Bargmann-Mihel-Telegdi T-BMT equation, or the spin 4-vetor equation of motion. Now let us look at the spatial omponents of µ /dτ. Before we do this, let us define 4

5 µ µ e/m and rewrite the T-BMT equation as µ dτ = µf µν S ν µ U µ F νλ U ν S λ df µν S ν 1 U µ F νλ U ν S λ. 19 To further simplify the disussion, let us write the and EDM parts separately as = +. 0 EDM Then 1 = µf 1ν S ν µ U 1 F νλ U ν S λ 1 and the first term F 1ν S ν beomes F 1ν S ν = F 10 S 0 + F 1 S + F 13 S 3 = E x /β S + B z S y + B y S z = E x /β S + S B x so in general F iν S ν = β S Ei + S Bi 3 5

6 is satisfied. Now, the seond term U 1 F νλ U ν S λ beomes U 1 F νλ U ν S λ = U 1 F 0λ U 0 S λ + F 1λ U 1 S λ + F λ U S λ + F 3λ U 3 S λ = U 1 F 01 U 0 S 1 + F 0 U 0 S + F 03 U 0 S 3 + U 1 F 10 U 1 S 0 + F 1 U 1 S + F 13 U 1 S 3 + U 1 F 0 U S 0 + F 1 U S 1 + F 3 U S 3 + U 1 F 30 U 3 S 0 + F 31 U 3 S 1 + F 3 U 3 S = U 1 E x /S x + E y /S y + E z /S z + U 1 E x /v x S 0 + B z v x S y + B y v x S z + U 1 E y /v y S 0 + B z v y S x + B x v y S z + U 1 E z /v z S 0 + B y v z S x + B x v z S y = U 1 E S β S β S B. 4 So, U i F νλ U ν S λ = U i E S β S β S B = β i E S β S β S B. 5 Therefore, = { µ S B + S β E } + µ β { } S E + β S + β S B is satisfied. Now EDM term an be obtained without suh ompliated steps by noting that the following transformation F µν F µν 6 6

7 an be made by E/ B and B E/. Therefore F iν S ν = F iν S ν E = S B, E B E + S βb i 7 and U i F νλ U ν S λ = U i F νλ U ν S λ E B, B E = β i B S β Bβ S β S E 8 are satisfied. Therefore = d S βb S E EDM + β } { S B β S E + β Bβ S an be written. So, EDM = { µ S B + S β E } { + µ β S E + } β S + β S B = d S βb S E + β{ S B β S E } + β Bβ S is satisfied idential to arxiv: Eq. 15. The above expressions an be rewritten with µ = e 4m g, µ = µ e m = e g = e 4m g m. 9 7

8 If we do that, = e { g S B + S β E } 4m { + g β S E + } β S + β S B 30 is made. On the other hand, the Lorentz fore equation tells us and β dβ = Now, dβ = e β B + E m β e β β B + m β = e m 31 1 β = e 3 m. β dβ = 1 dβ = 1 dβ β = β dβ = e 3 m 3 and give us = = s = S β Sβ 33 β β β Sdβ 3 β dβ β Sβ β β β Sdβ 1 e m dβ S β dβ S β β Sβ. 34 We want to write above equations with physial quantities defined at the rest frame of 8

9 the partile. To do that, first we put Eq. 30 and others in the above equation. Then, = e m g { S B + S β E } g { β S E + 1 e m g { S B + S β E } } β S + β S B g { β S E + β S + β S B} ββ 1 e β B + E m β Sβ 1 e β S β B + E β m 1 e m β Sβ 35 and is quite ompliated. Now let us use β S = β s and the relation between S and s. First, let s ollet terms with the eletri field only. Then, E term only = e m s βe g 1 + e S E g g β β + m 1 + e m β g β s g 1 g β = e m g + e m β + 1 s β E β s { g S E β + 1 }. 36 9

10 Now, let s use S = s + /β sβ to rewrite above. Then, E terms only = e m g + e m β = e g m + 1 s β E β s + 1 s Now let s ollet the magneti field terms. Then, B terms only = e m S B g { g s E β β s + 1 } β E β e g m β S B g 1 g β + e m β B S 1 β e 1 β Sβ B m = e g m s + g β Sβ β + e m β sβ B s + B β Sβ B β = e g m s B + e g β sβ B m + e g β s Bβ 38 m is obtained. Finally, using a b d = a db + b d a + da b, β 10

11 we an further simplify the above as, B terms only Therefore, the term beomes = e m + = e g m s B + e g β sβ B m + e g β s B + β sb β + B ββ S m = e g m + g β s B + e g β Bβ s m = e g + 1 s B + e g β Bβ s. m m g + 1 g s B + g β Bβ s + 1 E s β. 39 Note that there is a typo of sign in Eq. 18 of arxiv: v3. Now, the EDM term beomes EDM = = d + d EDM β β EDM S βb S E + β } { S B β S E + β Bβ S S βb β S E β + β } { S B β S E + β Bβ S β

12 Again, olleting terms with magneti-field-only gives us EDM, B terms only = d 1 S βb + S Bβ β + β Bβ Sβ = d = d + β 1 S βb + S Bβ β Bβ Sβ 1 s βb + s + β Bβ sβ s ββ Bβ = B β. 41 1

13 There are in total four terms with the eletri field, so EDM, E terms only = d 1 S E + β S E β 1 β S β E 1 = d 1 = d S E + β S E 1 = d + β + = d s + + = d β S E + S β E β E β β S β S E β E E + S β β + β β S S E + S β E β s ββ E + s β E β E β β s + s ββ s E + is satisfied. Simplifying the above gives us EDM = d s E + β s 4 β s s B β

14 If we summarize what we did so far, it beomes EDM = e m + = d g + 1 g s B + g + 1 E s β, s E + β Bβ s β s s B β. 44 If we hange the onstant term d in the EDM part as d = eη/4m, then d = eη/m. The spin equation of motion is expressed as = ω s s 45 where ω s is the angular veloity of the spin preession. So in our ase, ω s = e m + η g + 1 B E g β + β B β Bβ g + 1 β E is satisfied. We would like to point out again that the spin vetor s is defined at the partile rest frame and E and B are defined in the laboratory frame. The last step is to alulate the spin preession angular veloity with respet to the partile momentum diretion. To do that, one has to alulate the angular veloity of the partile and subtrat it from the spin preession angular veloity vetor. For that, let s define N β/β = p/p and differentiate it with respet to time. Then, we get 46 dn = β β β β 3 β β = β β β β 3 β β = ω p N 47 Note that formally d = η e s where s is spin of muon. And for muon, s = ħ. So, d = η m η m/s e MeV s η e m MeV e ħ = m 14

15 in general. Now, if we apply it to the partile in the eletromagneti field, we get dn = β β β β β β3 e = β B + E βm β = e B N + 1 E m β βn N E 1 β N = e B N + 1 { E m β N N E } = e B + 1 m β eβ β 3 β β B + m β N E + βn N E N E N. 48 So, ω p = e m 1 β N E B 49 is obtained. Therefore the spin preession angular veloity with respet to the partile 15

16 motion ω a = ω s ω p = e g + 1 m + η E e m 1 β N E B B = e g B m + η E = e m + η E g B g β + β B g β Bβ β Bβ β + β B g β Bβ β + β B is satisfied. The usual experimental onditions require g + 1 g β g 1 1 β E β E β E 50 B β = 0, E β = 0 51 and then ω a = e m g g B 1 β E + η E 1 + β B 5 is finally obtained. This equation is ommonly used in g-/edm experiments. For the muon, one defines a µ g / and then ω a = e m a µ B a µ 1 β E + η E 1 + β B. 53 This must be the master formula used in the muon g-/edm experiments. 16

17 Interpretation of the Results.1 When EDM is Zero When there is only B field that satisfies β B = 0 with no EDM, ω s beomes ω s = e a µ + 1 B 54 m aording to our alulation. The first term is the muon spin preession in its rest frame and the seond one is Thomas s preession beause we have an relativistially aelerating frame. In this ase, ω p beomes ω p = e 1 m B 55. Therefore, ω a = ω s ω p = e m a µb 56 where the relativisti effet is aneled out. If we have an eletri field E that satisfies = 0, we get ω a = e m a µ B a µ 1 β E The reason of having terms with E field is due to the fat that the partile now sees the B field from moving E field with respet to the partile known as the motional B field and is a relativisti effet.. When is Zero For the EDM term when no B field is presented and = 0, we get ω a = e η E m 58 17

18 and that is expeted. Now if there is B field with β B = 0, one gets ω a = e m η E + β B. 59 Again, we observe a motional E field due to the motion of the partile. You may ask why the motional fiel appear differently in and EDM? A short, perhaps an irresponsible answer is: we have shown why it is by an expliit alulation. But another, may be better answer is due to the fat that there is different treatment of E and B fiel embedded in the Maxwell equations. Do you have better answer? 18