Depatment of Physics, UCSD Physics 5B, Geneal Relativity Winte 05 Homewok, solutions. (a) Fom the Killing equation, ρ K σ ` σ K ρ 0 taking one deivative, µ ρ K σ ` µ σ K ρ 0 σ µ K ρ σ ρ K µ 0 ρ µ K σ ` ρ σ K µ 0. Adding these we have p µ ρ ` ρ µ qk σ σ, µ sk ρ ` σ, ρ sk µ. Add µ, ρ sk σ to both sides to obtain (twice) what we ae looking fo in tems of commutatos of deivatives: µ ρ K σ σ, µ sk ρ ` σ, ρ sk µ ` µ, ρ sk σ. Fo the vey definition of cuvatue we have, fo any vecto field A λ, µ, σ sa ρ R ρλµσ A λ so we have immediately that µ ρ K σ pr ρλσµ ` R µλσρ ` R σλµρ qk λ. Now we just need to shuffle indices aound a bit. The second tem is aleady of the fom we want, R µλσρ R σρµλ. Wite the fist tem as R ρλσµ R σµρλ and combine with the thid tem using anti-symmety in the last thee indices, R σµρλ ` R σλµρ R σρλµ R σρµλ, which doubles the second tem and the esult follows, µ ρ K σ R σρµλ K λ. (0.) (b) We ae going to use the Bianchi identity µr λ R λ µ. Contacting indices in the solution to pat (a), Eq. (0.), we have µ ρ K µ R ρλ K λ, and taking ρ of this, ρ µ ρ K µ ρ pr ρλ K λ q p ρ R ρλ qk λ ` R ρλ ρ K λ. (0.)
Since the Ricci tenso is symmetic in its indices we can eplace pρ K λq fo ρ K λ in the last tem on the ight hand side, buy by Killing s equation pρ K λq 0. So to pove that p ρ R ρλ qk λ 0 we have to show that the left hand side in (0.) vanishes: ρ µ ρ K µ ρ, µ s ρ K µ (again, since pρ K µq 0) ρ, µ s ρ K µ R µ ρµλ ρ K λ ` R λ ρµλ µ K ρ R ρλ ρ K λ R ρµ µ K ρ 0. (once again, since pρ K µq 0) So p ρ R ρλ qk λ 0, o, by the Bianchi identity, p λ RqK λ 0.(a) The pullback is ĝ ij Bxµ By i Bx ν By j g µν, whee x µ ae coodinates on R and y i ae coodinates on P, and the map x µ py i q is Compute the deivatives: x ρ cos φ y ρ sin φ z ρ Bx cos φ Bx By ρ sin φ Bφ Using g µν δ µν we have: By sin φ Bφ ρ cos φ Bz ρ Bz Bφ 0 ĝ ρρ Bxµ ĝ ρφ ĝ φρ Bxµ ĝ φφ Bxµ Bφ Bx ν δ µν cos φ ` sin φ ` 4ρ ` 4ρ Bx ν Bφ δ µν cos φp ρ sin φq ` sin φpρ cos φq ` ρp0q 0 Bx ν Bφ δ µν p ρ sin φq ` pρ cos φq ` p0q ρ
which is summaized as dŝ p ` 4ρ qdρ ` ρ dφ (b)as a matix, the invese of ĝ ij is ĝ ij 0 `4ρ. 0 ρ The push-fowad, which i will denote by g µν, geneally, is given by Computing: g xx ˆBx g xy Bx By g xz Bx Bz g yz By Bz ˆBy g yy g zz ˆBz ` 4ρ ` g µν Bxµ By i Bx ν By j ĝij. ˆBx Bφ ρ cos φ ` 4ρ ` sin φ j x z ` 4z ` y ` 4ρ ` Bx By sin φ cos φ sin φ cos φ 4ρ sin φ cos φ 4xy ρ ` 4ρ ` 4ρ ` 4z ` 4ρ ` Bx Bz ρ cos φ ρ ` 4ρ x ` 4z ` 4ρ ` By Bz ρ sin φ ρ ` 4ρ y ` 4z ˆ By ` 4ρ ` Bφ ρ sin φ ` 4ρ ` cos φ j y z ` 4z ` x ˆ Bz ` 4ρ ` Bφ ρ 4ρ ` 4ρ 4z ` 4z Keep in mind that these ae defined only on the sub-manifold P (so, in a sense, it is bette to keep the expessions fo g as given in tems of ρ and φ. (c) Not much to do hee. g µν δ µν is vey diffeent fom g µν, but thee was no eason to expect them to be the same.. (a) The integal cuve of V µ pxq x µ is the solution to dx µ ptq dt V µ pxptqq x µ ptq. We want a solution that satisfies x µ p0q x µ o. The integal is simple: x µ ptq x µ o e t.
Fo an integal cuve though the oigin, x µ o 0, the solution above is not a cuve but just a map fom the eal line to a single point. The eason is that at that point the tangent field vanishes: the integal cuve needs to know in what diection to move! (b) The map φ t : R n Ñ R n takes x µ o to y µ x µ ptq x µ o e t. (c) The push fowad is given by Explicitly, this is pφ t W q µ po Byµ Bx ν pw ν p pφ t W q µ px o q Bpxµ e t q Bx ν W ν px o e t q e t W µ px o e t q. Notice the minus sign in the exponential. This is because the map φ t takes y µ x µ ptq ÞÑ x µ o which we ae witing as x µ ÞÑ e t x µ. The Lie deivative is, by definition, L V W lim tñ0 t pφ t W q µ px o q W µ px o qs B e t W µ px o e t q W µ px o q ` x ν Bt ob ν W µ px o q (d) V, W s µ x ν B ν W µ W ν B ν x µ x ν B ν W µ W µ. This, of couse, agees with the esult of pat (c). 4. Fist calculate integal cuves of That is, we look fo solutions to A y x B Bx y ` x B By dx dt y x dy dt y ` x Note that the vecto field A has magnitude? eveywhee, is not defined at the oigin and is tangential to a cicle about the oigin, pointing in the clockwise diection. So we expect the integal cuves to gow towads the oigin as they ciculate clockwise. Now, the fact that A µ depends on x{ and y{ cies out fo a desciption in a pola coodinate system, exactly what the whole fomalism is suppose to do fo us automatically. That is, if ξ µ is a new coodinate system, with ξ µ ξ µ px ν q and if we denote the vecto field A components in the new coodinate system by õ, then à µ Bξµ Bx ν Aν 4
So we take fo new coodinates ξ µ p, φq defined so that x cos φ and y sin φ, that is φ actanpy{xq a x ` y If I can still compute deivatives, Bξ µ Bx ν cos φ sin φ sin φ cos φ. I have witten the esult in tems of the coodinates ξ so we can wite õ in tems of those coodinates: à cos φpsin φ cos φq ` sin φp sin φ cos φq à φ sin φpsin φ cos φq ` cos φp sin φ cos φq The equations fo the integal cuve ae now simple, d dt, dφ dt. If the initial point is p 0, φ 0 q at t 0, the solution is ptq 0 t and φptq φ 0 `lnp t{ 0 q. In tems of the oiginal coodinates we have then xptq p 0 tq cospφ 0 ` lnp t{ 0 qq, yptq p 0 tq sinpφ 0 ` lnp t{ 0 qq. Hee is a plot of the cuves. I show thee cuves, going though p0, q, p0, q and p0, q: 5 4 4 5
Things ae much simple fo the B field, because it is easy to integate diectly. We have dx dt xy, dy dt y. The equation fo yptq can be integated immediately, yptq y 0 {p ` y 0 tq. Inseting this in the equation fo xptq we have xptq x 0 p ` y 0 tq. Note that this has xptqyptq x 0 y 0 constant, so the paametic plot is easily dawn:.0.5.0 0.5 0.5.0.5.0 Hee I have taken cuves that go though x 0. at y,,. Of couse, thee ae also analogous integal cuves on the othe thee quadants of the catesian plane. We can also see both sets of integal cuves togethe: 0.4 0. 0. 0.4 0.6 0.8.0 6
Finally, compute C L A B A, Bs, o C µ A ν B ν B µ B ν B ν A µ : C x y x B x pxyq ` y ` x y x y x ı B y pxyq xyb x ` p y qb y ypy xq xpy ` xq ` xy py ` xq y4 ` x y x y x 4 C y y x B x p y q ` ypy ` xq y ` x ` xy py xq ypx ` xy ` y q B y p y q xyb x y ` x ` p y qb y y ` x Below is a plot of the vecto field C supeimposed on the integal cuves of A and B. You can sketch the integal cuves of C in an obvious way (by stinging vectos togethe): ı 0.4 0. 0. 0.4 0.6 0.8.0 7