Solutions Ph 236a Week 2

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1 Solutions Ph 236a Week 2 Page 1 of 13 Solutions Ph 236a Week 2 Kevin Bakett, Jonas Lippune, and Mak Scheel Octobe 6, 2015 Contents Poblem Pat (a Pat (b Pat (c Poblem Pat (a Pat (b Pat (c Pat (d Poblem Pat (a Pat (b Pat (c Pat (d Pat (e Poblem Pat (a Pat (b Pat (c Pat (d Poblem Pat (a Pat (b

2 Solutions Ph 236a Week 2 Page 2 of 13 Pat (a Poblem 1 Fo a paticle in its est fame, we know that its 4-velocity is given by u = (1, 0 and we ae given in the poblem statement that its spin vecto is S = (0, s. Thus Pat (b S u = 0, d dτ ( S u = 0 (1.1 Fist, let us examine the deivative spatial components of S while in the paticles est fame so that d s dτ = A 1F i js j (1.2 Whee i, j ae spatial indices that un ove 1, 2, 3. To see what happens, let us pick a spin component and expand (fo example, the z-diection ds z dτ = A 1F z js j = A 1 (s x B y s y B x (1.3 But this is just the fom we expect of the usual 3-vecto coss poduct. So using that esult with equation (1 fom the poblem set, we get To get A 2, we stat with d s dτ = A 1 s B, A 1 = ge 2m d dτ ( S u = 0 S d u dτ = u d S dτ (1.4 (1.5 Note, that d u dτ is an acceleation, which is foce divided by mass. Since the electomagnetic foce is given by qf α β uβ, we can wite S d u dτ = q m S αf α βu β = A 1 S α F α βu β + A 2 (1.6 whee we used the fact that A 2 u u = A 2. Now we can plug in the esult of A 1 fom above and solve fo A 2 and we get A 1 = ge 2m, A 2 = F α βs α u β e m (g 1 (1.7 2

3 Solutions Ph 236a Week 2 Page 3 of 13 Pat (c Let Ẽ = 0 and s = (s 0, S. Thus the paticle s 4-velocity is given by u = γ(1, ṽ. Then S u = 0 = γ( s 0 + s ṽ s ṽ = s 0 (1.8 Now note that since Ẽ = 0 F α β = 0 if eithe α o β = 0. So then ds 0 dτ = A 1F 0 βs β + A 2 u 0 = γa 2 (1.9 and fom that, in the lab fame with time coodinate t, the time ate of change of the helicity is, d dt ( s ṽ = ds0 dt = ds dτ dτ dt = A 2 = F α βs α u β e m (g 2 1 = γf i js i v j e m (g 2 1 = γv j e ( s j B m (g 1 ( In the case whee g = 2, the quantity ( g 2 1 = 0 which implies the time ate of change of the helicity is 0, so the helicity is constant. Poblem 2 Pat (a Stat by evaluating v = P( v = P αβ v β = g αβ v β + u α u β v β = v α + u α u β v β (2.1 To show that it is othogonal to u, v u = (v α + u α u β v β u α = v α v α + u α u α u β v β = v α u α (1 + u α u α = 0 (2.2 since u is a 4-velocity. Thus v is othogonal to u.

4 Solutions Ph 236a Week 2 Page 4 of 13 Pat (b We want to show that v = P( v and we know fom above that v = v α + u α u β v β so then P( v = (g αβ + u α u β (v β + u β u γ v γ = g αβ v β + g αβ u β u γ v γ + u α u β v β + u α u β u β u γ v γ = v α + u α u γ v γ + u α u β v β u α u γ v γ = v α + u α u β v β = v (2.3 Thus we have shown that P is unique. Pat (c Since we ae given a non-null vecto q we can define q 2 = q q and we don t need to be afaid of dividing by q 2. Now define ou pojection tenso fo q as P q = g αβ 1 q 2 q αq β (2.4 It can be shown using the same aguments as in pats (a and (b that P q is indeed the pojection tenso and that it is unique. Pat (d Fo a null vecto k, constuct its pojection tenso as P k = ck α k β fo an abitay eal numbe c. Thus, fo an abitay vecto v, we have k P k ( v = k α ck α k β v β = ck β v β (0 = 0 (2.5 Thus P k is indeed the pojection opeato, howeve it is not unique because c is abitay. Poblem 3 Pat (a Using the usual spheical coodinates, the Catesian coodinates ae given by x = sin θ cos φ, y = sin θ sin φ, z = cos θ. (3.1

5 Solutions Ph 236a Week 2 Page 5 of 13 Since e is the tangent vecto of the cuves paametized by with constant θ and φ, we find e = = xi x i = sin θ cos φ e x + sin θ sin φ e y + cos θ e z, (3.2 whee i = 1, 2, 3 and we used x 1 = x, x 2 = y, and x 3 = z. Similaly, we find Pat (b e θ = θ = xi θ e φ = φ = xi φ x i = cos θ cos φ e x + cos θ sin φ e y sin θ e z, x i = sin θ sin φ e x + sin θ cos φ e y. (3.3 Recall that the spheical coodinates ae given by = x 2 + y 2 + z 2, ( z θ = accos, x2 + y 2 + z 2 φ = { accot(x/y if y 0 accot(x/y + π if y < 0, (3.4 whee accos etuns a value in [0, π] and accot etuns a value in [0, π] (when extending its domain to include ±. We find that d = i dx i = x dx i 1 ( i = 2 2x dx + 2y dy + 2z dz x 2 + y 2 + z 2 = sin θ cos φ dx + sin θ sin φ dy + cos θ dz, (3.5

6 Solutions Ph 236a Week 2 Page 6 of 13 and dθ = i θ dx i = = = 1 1 z 2 x 2 + y 2 + z 2 ( 2xz dx 2(x 2 + y 2 + z 2 3/2 2yz dy 2(x 2 + y 2 + z 2 + dz 3/2 x2 + y 2 + z 2 x 2 + y 2 + z 2 xz dx zy dy + (x 2 + y 2 dz x 2 + y 2 (x 2 + y 2 + z 2 3/2 2z 2 dz 2(x 2 + y 2 + z 2 3/2 xz zy dx + dy x2 + y 2 (x 2 + y 2 + z 2 x2 + y 2 (x 2 + y 2 + z 2 and finally x2 + y 2 x 2 + y 2 + z 2 dz = 2 sin θ cos φ cos θ dx sin θ sin φ cos θ sin θ 3 sin θ cos θ cos φ cos θ sin φ = dx + dy sin θ dφ = i φ x i 1 = 1 + x 2 /y 2 = Pat (c 1 2 sin 2 θ ( dx y x dy y 2 = ( sin θ sin φ dx + sin θ cos φ dy dy sin θ 2 dz dz, (3.6 1 ( x 2 + y 2 y dx + x dy = sin φ sin θ dx + cos φ sin θ dy. (3.7 We have found in pat (a that and in pat (b we found that e = sin θ cos φ e x + sin θ sin φ e y + cos θ e z, e θ = cos θ cos φ e x + cos θ sin φ e y sin θ e z, e φ = sin θ sin φ e x + sin θ cos φ e y, (3.8 d = sin θ cos φ dx + sin θ sin φ dy + cos θ dz, dθ = cos θ cos φ dx + cos θ sin φ dy sin θ dφ = sin φ sin θ dx + cos φ sin θ dy. (3.9 dz,

7 Solutions Ph 236a Week 2 Page 7 of 13 Recall that ( e x, e y, e z ae basis vectos and thus they ae linealy independent. It can be shown (but is not necessay that ( e, e θ, e φ ae also linealy independent. And since these ae 3 vectos in a 3-dimensional space, it follows that they fom a basis. Similaly, it can be shown that ( d, dθ, dφ ae linealy independent and since the space of 1-foms is also 3-dimensional, it follows that they fom a basis too. Thus it emains to be shown that ( d, dθ, dφ ae dual to ( e, e θ, e φ. Since ( dx, dy, dz ae dual to ( e x, e y, e z, we have that dx i, e j = δ i j, (3.10 whee dx 1 = dx, dx 2 = dy, dx 3 = dz, e 1 = e x, e 2 = e y, and e 3 = e z. We find and and finally d, e = sin 2 θ cos 2 φ + sin 2 θ sin 2 φ + cos 2 θ = 1, d, e θ = sin θ cos θ cos 2 φ + sin θ cos θ sin 2 φ sin θ cos θ = 0, d, e φ = sin 2 θ sin φ cos φ + sin 2 θ sin φ cos φ = 0, (3.11 dθ, e = sin θ cos θ cos2 φ + sin θ cos θ sin2 φ sin θ cos θ = 0, dθ, e θ = cos 2 θ cos 2 φ + cos 2 θ sin 2 φ + sin 2 θ = 1, dθ, e φ = sin θ cos θ sin φ cos φ + sin θ cos θ sin φ cos φ = 0, (3.12 Thus we have shown that dφ, sin φ cos φ sin φ cos φ e = + = 0, dφ, cos θ sin φ cos φ cos θ sin φ cos φ e θ = + = 0, sin θ sin θ dφ, e φ = sin 2 φ + cos 2 φ = 1. (3.13 dxī, e j = δī j, (3.14 whee dx 1 = d, dx 2 = dθ, dx 3 = dφ, e 1 = e, e 2 = e θ, and e 3 = e φ, and so we have shown that ( d, dθ, dφ ae dual to ( e, e θ, e φ. Pat (d Recall that the components of the metic tenso ae given by g ī j = e ī e j. (3.15

8 Solutions Ph 236a Week 2 Page 8 of 13 Using (3.8 we find g 1 1 = e e = sin 2 θ cos 2 φ + sin 2 θ sin 2 φ + cos 2 θ = 1, g 1 2 = g 2 1 = e e θ = sin θ cos θ cos 2 φ + sin θ cos θ sin 2 φ = 0, sin θ cos θ g 1 3 = g 3 1 = e e φ = sin 2 θ sin φ cos φ + sin 2 θ sin φ cos φ = 0, g 2 2 = e θ e θ = 2 cos 2 θ cos 2 φ + 2 cos 2 θ sin 2 φ + 2 sin 2 θ = 2, g 2 3 = g 3 2 = 2 sin θ cos θ sin φ cos φ + 2 sin θ cos θ sin φ cos φ = 0, g 3 3 = 2 sin 2 θ sin 2 φ + 2 sin 2 θ cos 2 φ = 2 sin 2 θ. (3.16 Hence the components of the metic tenso in spheical coodinates ae g ī j = ( sin 2 θ The full metic tenso in all its gloy is Pat (e g = g ī j dxī dx j = d d + 2 dθ dθ + 2 sin 2 θ dφ dφ. (3.18 Since gîĵ = e î e ĵ, it is obvious that g îĵ = δ îĵ if the vecto e î ae othonomal. We have aleady found in (3.16 that e ī ae othogonal, so we can make them othonomal by scaling them with the invese of thei lengths. Thus we obtain e eˆ = = e e e 1 = sin θ cos φ e x + sin θ sin φ e y + cos θ e z, e θ eˆθ = = e θ eθ e θ = cos θ cos φ e x + cos θ sin φ e y sin θ e z, e φ e ˆφ = = e φ eφ e φ sin θ = sin φ e x + cos φ e y. (3.19 Poblem 4 Pat (a If A µν is antisymmetic and S µν is symmetic, then S µν = g µα g νβ S αβ = g νβ g µα S βα = S νµ (4.1

9 Solutions Ph 236a Week 2 Page 9 of 13 so we know that S µν is also symmetic. So finally A µν S µν = A νµ S µν = A νµ S νµ = A µν S µν. (4.2 Hee the fist step used antisymmety of A µν, the second step used symmety of S µν, and the thid step elabeled dummy indices. Since A µν S µν is equal to minus itself, it must be zeo. Pat (b Let A µν and S µν be as above and let V µν be an abitay tenso. Fo equation (4 in the poblem set, we can see that V µν A µν = 1 2 (V µν A µν + V µν A µν = 1 2 (V µν A µν V µν A νµ = 1 2 (V µν A µν V νµ A µν = 1 2 (V µν V νµ A µν (4.3 Similaly to get equation (5 in the poblem set V µν S µν = 1 2 (V µν S µν + V µν S µν = 1 2 (V µν S µν V µν S νµ = 1 2 (V µν S µν + V νµ S µν = 1 2 (V µν + V νµ S µν (4.4 Pat (c To show that these tansfomation maticies ae inveses of each othe, simply opeate with both of them in succession. e µ = Λ µ µ e µ = Λ µ µ(λ ν µ e ν (4.5 Since we ae tansfoming back into the baed fame, that means e µ = δ ν µ e ν Λ µ µλ ν µ = δ ν µ (4.6

10 Solutions Ph 236a Week 2 Page 10 of 13 The evese can be shown be doing the same thing except with 1-foms instead. Now we know by definition that tensos can be witten as T α γ β = T( w α, e β, w γ (4.7 We also know that it a linea opeato and so plugging in the tansfomation matices we have T ᾱ γ β = T( wᾱ, e β, w γ = T(Λᾱα w α, Λ β β e β, Λ γ γ w γ = ΛᾱαΛ β βλ γ γt( w α, e β, w γ = ΛᾱαΛ β βλ γ γt α γ β (4.8 Pat (d Since g µν is a tenso, its indices can be aised and loweed as any othe tenso. g αβ = g αµ g βν g µν (4.9 g αβ = g αµ g βν g µν (4.10 To show that g α β = δα β, just aise o lowe one of the indices and use equation (8 fom the homewok set. g α β = g αγ g γβ = δ α β (4.11 Pat (a Poblem 5 Case 1: Fist conside the case whee α, β, and γ ae not unique. In that case ɛ αβγρ = 0 fo all ρ, which means the ight-hand side will be 0. If α, β, and γ ae not unique, it makes no sense to talk about even o odd pemutations of them and so this case falls unde othewise. The same agument holds if µ, ν, and λ ae not unique. Case 2: Thus we ae left with the cases whee α, β, and γ ae unique and also µ, ν, and λ ae unique. Howeve, since each index can be 0, 1, 2, o 3, the two sets of indices do not necessaily have to be the same (e.g. we could have α, β, γ = 0, 1, 2 and µ, ν, λ = 1, 2, 3. Conside a case like this (i.e. the set {α, β, γ} is diffeent fom the set {µ, ν, λ}. Since the Levi-Civita tenso is only non-zeo if all fou indices ae distinct, thee is only one value of ρ, say ρ 1, that makes ɛ αβγρ non-zeo. Similaly, thee is only one value of ρ, say ρ 2, that makes

11 Solutions Ph 236a Week 2 Page 11 of 13 ɛ µνλρ non-zeo. Since {µ, ν, λ} is diffeent fom {α, β, γ}, it follows that ρ 1 ρ 2 and thus all fou tems in the sum ove ρ ae 0 and so the total sum is 0. Since the two sets of indices ae not the same, one is not a pemutation of the othe and so this case also falls unde othewise. Thus we ae left with the cases whee the two sets of indices ae the same. In that case (α, β, γ is always eithe an even o odd pemutation of (µ, ν, λ. Recall that the Levi-Civita tenso does not change unde an even pemutation of its indices. Case 3: Conside the case whee (α, β, γ is an even pemutation of (µ, ν, λ. Since we can pefom any even pemutations we like without changing the Levi- Civita tenso, we can conside the specific case (α, β, γ = (µ, ν, λ without loss of geneality. Thee is only one value of ρ, say ρ 0 that is diffeent fom α, β, and γ. This is the only value of ρ fo which ɛ αβγρ is non-zeo. Thus the sum ove ρ educes to one non-zeo tem, namely δ αβγ µνλ = ɛαβγρ ɛ µνλρ = ɛ αβγρ0 ɛ αβγρ0 = (±1( 1 = +1, (5.1 since ɛ αβγδ = ɛ αβγδ. Case 4: The only emaining case now is whee (α, β, γ is an odd pemutation of (µ, ν, λ. In that case we can take (α, β, γ = (ν, µ, λ without loss of geneality, because even pemutations do not change the Levi-Civita tensos. Again, we only have one value of ρ, say ρ 0 fo which ɛ αβγρ is non-zeo. The sum ove ρ again educes to one tem, namely δ αβγ µνλ = ɛαβγρ ɛ µνλρ = ɛ αβγρ0 ɛ βαγρ0 = +ɛ αβγρ0 ɛ αβγρ0 = +(±1( 1 = 1. (5.2 Thus we have shown that +1 if (α, β, γ is an even pemutation of (µ, ν, λ δ αβγ µνλ = 1 if (α, β, γ is an odd pemutation of (µ, ν, λ 0 othewise. (5.3 Note that δ αβ µν = 1 2 δαβγ µνγ (5.4 Fom what we found above, it is obvious that δ αβ µν = 0 if α = β, µ = ν, o {α, β} {µ, ν}, which fall unde othewise. If (α, β is an even pemutation of (µ, ν, then we must have that (α, β = (µ, ν, and thee ae only two values of γ, say γ 1 and γ 2, fo which δ αβγ is non-zeo. Thus the sum ove γ educes to two tems, namely δ αβ µν = 1 2 δαβγ µνγ = 1 2 (δ αβγ1αβγ1 + δ αβγ2αβγ2 = 1 (1 + 1 = +1. (5.5 2

12 Solutions Ph 236a Week 2 Page 12 of 13 Finally, if (α, β is an odd pemutation of (µ, ν, then we must have that (α, β = (ν, µ, and thee ae again only two values of γ, say γ 1 and γ 2, fo which δ αβγ is non-zeo. Thus the sum ove γ again educes to two tems, namely δ αβ µν = 1 2 δαβγ µνγ = 1 (δ αβγ1βαγ1 + δ αβγ2βαγ2 = 1 ( = 1. (5.6 Thus we have shown that +1 if (α, β is an even pemutation of (µ, ν δ αβ µν = 1 if (α, β is an odd pemutation of (µ, ν 0 othewise. (5.7 Pat (b Since J is a 3-index antisymmetic tenso, J is a 1-index tenso. We find J αβγ = J µ ɛ µαβγ = J µ ɛ µ αβγ = 1 3! J νλρ ɛ νλρµ ɛ µ αβγ = 1 3! J νλρɛ νλρµ ɛ µαβγ = 1 3! J νλρɛ νλρµ ɛ αβγµ = 1 3! J νλρ( δ νλρ αβγ = 1 6 J νλρδ νλρ αβγ. (5.8 Now ecall that J is an antisymmetic tenso. Thus J αβγ = 0 if the 3 indices ae not distinct. Similaly, δ νλρ αβγ = 0 if α, β, and γ ae not distinct. Similaly, the tems in the sum above ae 0 if ν, λ, and ρ ae not distinct. Also δ νλρ αβγ = 0 if {ν, λ, ρ} {α, β, γ}. Thus thee ae only 6 tems in the sum fo which δ νλρ αβγ 0, namely the tems whee (ν, λ, ρ is a pemutation of (α, β, γ. We find J αβγ = 1 6 J νλρδ νλρ αβγ = 1 6 (J αβγ J αγβ + J γαβ J γβα + J βγα J βαγ = 1 6 (J αβγ + J αβγ + J αβγ + J αβγ + J αβγ + J αβγ = 1 6 (6J αβγ = J αβγ, (5.9 whee we have used that J is antisymmetic, hence J αγβ J αβγ, fo example. Fo the 2-index antisymmetic tenso F we find F αβ = 1 F µν ɛ µναβ = 1 F µν ɛ µν αβ 2! 2! = 1 2! 1 2! F λρ ɛ λρµν ɛ µν αβ = 1 4 F λρɛ λρµν ɛ µναβ = 1 4 F λρɛ λρµν ɛ αβµν = 1 4 F λρ ( 2δ λρ αβ = 1 2 F λρδ λρ αβ. (5.10

13 Solutions Ph 236a Week 2 Page 13 of 13 Fom pat (a we know that δ λρ αβ Thus we find = 0 unless (λ, ρ = (α, β o (λ, ρ = (β, α. F αβ = 1 2 F λρδ λρ αβ = 1 2 (F αβ F βα = 1 2 (F αβ + F αβ = F αβ, (5.11 whee we have used that F is antisymmetic, hence F αβ = F βα. Finally, fo the 1-index antisymmetic tenso B we find B α = 1 B µνλ ɛ µνλα = 1 B µνλ ɛ µνλ α = 1 3! 3! 3! Bρ ɛ ρµνλ ɛ µνλ α = 1 3! B ρɛ ρµνλ ɛ µνλα = 1 3! B ρɛ ρµνλ ɛ αµνλ = 1 6 B ( ρ 2δ ρµ αµ = 1 3 B ρδ ρµ αµ. (5.12 Note that δ ρµ αµ = 0 if ρ α because in that case the two sets of indices ae always diffeent. Thus δ ρµ αµ is only non-zeo if ρ = α, in that case we find δ ρµ αµ = δ ρ0 ρ0 + δρ1 ρ1 + δρ2 ρ2 + δρ3 ρ3 = 3, (5.13 because ρ will be 0, 1, 2, o 3. Thus we found that δ ρµ αµ = 3δ ρ α, (5.14 whee δ ρ α is the usual Konecke delta. Thus (5.12 becomes B α = 1 3 B ρδ ρµ αµ = 1 3 B ρ3δ ρ α = B α, (5.15 which is what we need to show.