Solutions Ph 236a Week 5

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1 Solutions Ph 236a Week 5 Page 1 of 10 Solutions Ph 236a Week 5 Kevin Barkett, Jonas Lippuner, and Mark Scheel October 28, 2015 Contents Problem Part (a Part (b Part (c Part (d Part (e Problem Part (a Part (b Part (c Part (d Part (e Problem Part (a Part (b Problem Part (a Part (b Problem

2 Solutions Ph 236a Week 5 Page 2 of 10 Part (a In Homework 2 we found that Problem 1 e r = r = sin θ cos φ e x + sin θ sin φ e y + cos θ e z e θ = θ = r cos θ cos φ e x + r cos θ sin φ e y r sin θ e z, e φ = φ = r sin θ sin φ e x + r sin θ cos φ e y, (1.1 and the metric for these basis vectors is g ij = 0 r 2 0. ( r 2 sin 2 θ by Since ( e r, e θ, e φ is a coordinate basis, the connection coefficients are given Γ i jk = 1 2 gil (g jl,k + g lk,j g jk,l. (1.3 We use Mathematica to do all the tedious calculations. vector of variable x, the metric g, and its inverse ginv. First we define the x = {r, t, p} g = {{1, 0, 0}, {0, r^2, 0}, {0, 0, r^2 Sin[t]^2}} ginv = Inverse[g] Then we define the function ConCoef that calculates Γ i jk according to (1.3. ConCoef[i_, j_, k_] := 1/2 Sum[gInv[[i, l]](d[g[[j, l]], x[[k]]] + D[g[[l, k]], x[[j]]] - D[g[[j, k]], x[[l]]], {l, 1, 3}] We then evaluate Γ i jk with the Table function. Table[ConCoef[1, j, k], {j, 1, 3}, {k, 1, 3}] // MatrixForm Table[ConCoef[2, j, k], {j, 1, 3}, {k, 1, 3}] // MatrixForm Table[ConCoef[3, j, k], {j, 1, 3}, {k, 1, 3}] // MatrixForm

3 Solutions Ph 236a Week 5 Page 3 of 10 We find that the only non-zero connection coefficients are Γ r θθ = r, Γ r φφ = r sin 2 θ, Γ θ rθ = Γ θ θr = 1 r, Γ θ φφ = sin θ cos θ, Γ φ rφ = Γφ φr = 1 r, Γ φ θφ = Γφ φθ = cot θ. (1.4 Part (b Since g ij = δ ij in an orthonormal basis, all the metric derivatives are 0 and so the connection coefficients are given by Γîĵˆk = 1 2 gîˆl(cˆlĵˆk + cˆlˆkĵ c ĵˆkˆl = 1 + c 2 (cîĵˆk cîˆkĵ î ĵˆk = 1 ˆk (c + c ĵ 2 îĵ c î, (1.5 îˆk ĵˆk since gîĵ = δ îĵ and so we can raise and lower indices as we please. In Homework 4 we found that the only non-zero commutation coefficients are c ˆr ˆθ ˆθ = c = 1 ˆθ r, ˆφ ˆθˆr c ˆr ˆφ = c ˆφ = 1 ˆφˆr r, ˆφ cˆθ ˆφ = c ˆφ We enter this information in Mathematica as c. ˆφˆθ = cos θ r sin θ. (1.6 c = {{{0, 0, 0}, {0, -1/r, 0}, {0, 0, -1/r}}, {{0, 1/r, 0}, {0, 0, 0}, {0, 0, -Cos[t]/(r Sin[t]}}, {{0, 0, 1/r}, {0, 0, Cos[t]/(r Sin[t]}, {0, 0, 0}}} And we compute the connection coefficients as follows. ConCoef[i_, j_, k_] := 1/2 (c[[i, j, k]] + c[[i, k, j]] - c[[j, k, i]] Table[ConCoef[1, j, k], {j, 1, 3}, {k, 1, 3}] // MatrixForm Table[ConCoef[2, j, k], {j, 1, 3}, {k, 1, 3}] // MatrixForm Table[ConCoef[3, j, k], {j, 1, 3}, {k, 1, 3}] // MatrixForm

4 Solutions Ph 236a Week 5 Page 4 of 10 We find that the only non-zero connection coefficients are Part (c Recall that we have Γˆr ˆθ ˆθ = Γˆr ˆφ ˆφ = 1 r, Γˆθˆr ˆθ = 1 r, Γˆθ ˆφ = cot θ, ˆφ r Γ ˆφˆr ˆφ = 1 r, Γ ˆφ ˆθ cot θ =. (1.7 ˆφ r A k ;k = A k,k + Γ k ika i. (1.8 In the Cartesian coordinate basis we have Γ i jk = 0 and so Ak ;k = Ak,k = ka k. In the basis ( e r, e θ, e φ we use the connection coefficients we have found in (1.4 to find ( A k ;k = Ar r + Aθ θ + Aφ φ + Ar r A θ ( cot θ r + A φ ( = Ar r + Aθ θ + Aφ φ + 2 r Ar + cot θ A θ = 1 r 2 r ( r 2 A r + 1 sin θ ( sin θ A θ + Aφ θ φ. (1.9 Finally, in the basis ( eˆr, eˆθ, e ˆφ we use the connection coefficients we have found in (1.7 and obtain Aˆk Aˆr = ;ˆk r + 1 Aˆθ r θ + 1 A ˆφ ( r sin θ φ + Aˆr r + 1 ( + Aˆθ cot θ r r + A ˆφ ( Aˆr = r + 1 Aˆθ r θ + 1 A ˆφ r sin θ = 1 ( r 2 Aˆr r r r sin θ φ + 2 Aˆr + cot θ Aˆθ r r ( sin θ Aˆθ + 1 A ˆφ r sin θ φ, (1.10 θ which is the familiar divergence in spherical coordinates from (e.g. Jackson. Note that electromagnetism textbooks use an orthonormal basis, not a coordinate basis, for spherical (and cylindrical coordinates!

5 Solutions Ph 236a Week 5 Page 5 of 10 Part (d Note that det g = r 4 sin 2 θ. If we take ɛ ijk = aˆɛ ijk, (1.11 where +1 if (i, j, k is an even permutation of (1,2,3, ˆɛ ijk = 1 if (i, j, k is an odd permutation of (1,2,3, 0 otherwise (1.12 is the Levi-Civita tensor in the Cartesian coordinate basis ( e x, e y, e z. Note that since we are in 3-dimensional space and there are no minus signs in the metric, we do not pick up a minus sign when changing all covariant indices to contravariant ones. We find We want ɛ 123 = g 1i g 2j g 3k ɛ ijk = r 4 sin 2 θ aˆɛ 123 = r 4 sin 2 θ a. (1.13 ɛ 123 = + det g = r 2 sin θ, (1.14 so it follows that a = 1/(r 2 sin θ and we have + 1 r 2 if (i, j, k is an even permutation of (1,2,3, sin θ ɛ ijk = 1 r 2 if (i, j, k is an odd permutation of (1,2,3, sin θ 0 otherwise. (1.15 Part (e We have (curl A i = ɛ ijk A k;j = ɛ ijk ( A k,j Γ l kja l = ɛ ijk A k,j ɛ ijk Γ l kja l. (1.16 Note that in the basis ( e r, e θ, e φ, the connection coefficients are symmetric in the last two indices and thus the second term above vanishes, because it is a contraction of an anti-symmetric with a symmetric object. Thus for the basis ( e r, e θ, e φ we find (curl A r = (curl A θ = (curl A φ = ( 1 Aφ r 2 sin θ θ ( Ar 1 r 2 sin θ 1 r 2 sin θ A θ, φ φ A φ, r ( Aθ r A r θ. (1.17

6 Solutions Ph 236a Week 5 Page 6 of 10 In the basis ( eˆr, eˆθ, e ˆφ the connection coefficients are not symmetric in the last two indices and so we the second term in (1.16 does not vanish. Also note that the Levi-Civita tensor in the basis ( eˆr, eˆθ, e ˆφ is the same as in 3- dimensional Cartesian coordinates, because the metric is the identity. Thus we find (curl Aˆr = 1 r A ˆφ = 1 r sin θ (curl Aˆθ = 1 r sin θ = 1 r θ 1 r sin θ Aˆθ φ + cot θ r ( θ (sin θ A ˆφ Aˆθ φ Aˆr φ A ˆφ r 1 r A ˆφ, ( 1 Aˆr sin θ φ r (ra ˆφ Aˆr (curl A ˆφ = Aˆθ r 1 r θ + 1 r Aˆθ = 1 ( r r (raˆθ Aˆr, (1.18 θ which is the usual curl in spherical coordinates found in electromagnetism textbooks (these textbooks, as we see, use an orthonormal basis and not a coordinate basis. A ˆφ, Problem 2 Part (a First note that from the notes that g λµ,γ = Λ λµγ + Γ µλγ. So then we can write g α,γ = g λµ,γ g µ g λα = (Γ λµγ + Γ µλγ g µ g λα = Γ α µγg µ Γ λγ gλα (2.1 Part (b In coordinate frame, Γ µ α = 1 2 gµν (g να, + g ν,α g α,ν. in the case where α = µ as in this problem, then the last two terms will cancel, leaving Γ α α = 1 2 gαν g να,. From the previous set, we have that g, g = gαν g να, so then Γ α α = 1 g, 2 g = 1 ( 2 (log( g, = log( g 1 2 (2.2,

7 Solutions Ph 236a Week 5 Page 7 of 10 Part (c We will use the results from parts (a and (b above. g µν Γ α µν = g α, Γ λ gλα = g α, (log( g 1 2 ( = g αν,ν log( g 1 2 g λα,λ g να,ν by part (a by part (b relabeling indices = g αν,ν ( g 1 2,ν g αν ( g 1 2 = 1 (g αν ( g 1 ( g 1 2,ν (2.3 2 Part (d Again, we will use the result from part (b to find A α ;α =A α,α + Γ α αa =A α,α + 1 (( g 1 ( g 1 2, A 2 = 1 (( g 1 ( g 1 2 A α,α (2.4 2 Part (e Start with the definition of the Levi-Civita tensor as ɛ µνρσ = gˆɛ µνρσ where we know that ˆɛ 0123 = 1 and each permutation results in a sign change. Now let us expand out the covariant derivative and use the fact that for coordinate basis that Γ α γ = Γα γ so that ɛ µνρσ;α = ( gˆɛ µνρσ,α Γ µα gˆɛνρσ Γ να gˆɛµρσ Γ ρα gˆɛµνσ Γ σα gˆɛµνρ (2.5 First treat the case where µ, ν, ρ, σ are not all unique. Then the first term in Eq. (2.5 vanishes. But what about the other terms? Suppose µ = ν. Then the last two terms of Eq. (2.5 vanish individually because of the antisymmetry of Levi-Civita, and the remaining terms can be written Γ µα gˆɛνρσ + Γ να gˆɛµρσ, (2.6 which is antisymmetric in µ and ν, and thus vanishes when µ = ν. Therefore ɛ µνρσ;α = 0 if µ = ν. The same argument can be used if any of the components are equal, and shows that ɛ µνρσ;α = 0 if µ, ν, ρ, σ are not all unique.

8 Solutions Ph 236a Week 5 Page 8 of 10 So now consider the case where µ, ν, ρ, σ are all unique. Examine the first connection term in Eq. (2.5: Γ µα gˆɛνρσ. The only value for that can make a nonzero contribution to this term is = µ, since any other value would result in a duplicated index in the Levi-Civita tensor. So then this term becomes Γ µ µα gˆɛµνρσ (where there is no sum on the µ index. Similarly, the second connection term is Γ ν να gˆɛµνρσ (where there is no sum on the ν index, and so on. Summing the four connection terms therefore yields Γ α gˆɛµνρσ (where there is a sum on. Therefore, for µ, ν, ρ, σ all unique, ɛ µνρσ;α = ( ( g,α Γ α From above in part (b, we can use Γ α = ( log( g ( 1 2 and then the equation reduces to ɛ µνρσ;α = g ˆɛ µνρσ (2.7,α ( ( ( g,α log( g ( 1 2 g ˆɛ µνρσ = 0 (2.8,α Part (a Problem 3 To write out the components of u ũ = 0, use the property of taking the covariant derivative on a lower index, u ũ = 0 =u α u ;α =u α (u,α Γ γ α u γ = du dλ Γγ α u γu α = 0 (3.1 which is the parallel transport equation for ũ. Part (b To show this is equivalent to the geodesic equation for vector components, start with the 1-form geodesic equation and manipulate it into the form of the vector geodesic equation we are given. Specfically, du µ g µ dλ Γ γ α u γu α = duµ dλ g µ + u µ u ν g µ,ν Γ γ α u γu α = = duµ dλ g µ + u µ u ν (Γ µν + Γ µν Γ γ α u γu α = = duµ dλ g µ + u µ u ν (g µδ Γ δ ν + g δ Γ δ µν Γ γ α u γu α = duµ dλ g µ + u δ u ν Γ δ ν + u µ u ν g δ Γ δ µν Γ γ α u γu α (3.2

9 Solutions Ph 236a Week 5 Page 9 of 10 where in going from line 1 to line 2 the relation g µ,ν = Γ µν + Γ µν was used and then the definition of Γ µν = g µδ Γ δ ν was used to get to line 3. On the last line of the equation above, note that the second and last terms cancel after a relabeling of indicies. After that, what s left is du µ dλ g µ + u µ u ν g δ Γ δ µν = duµ dλ g µ + g µ Γ µ γαu γ u α =g µ ( du µ dλ + Γµ γαu γ u α =0 (3.3 as the equation in the parenthesis is simply the the geodesic equation for vectors. Note that the geodesic equation for vectors is not just the naively raised geodesic equation for 1-forms. Part (a Problem 4 A geodesic with tangent vector u is spacelike, timelike or null according to whether u u is > 0, < 0, = 0. But u u is conserved along the geodesic since u ( u u = 2 u u u = 0 since the geodesic equation is u u = 0. Part (b The general case gives Euler-Lagrange equation, using the definitions of F, y and ẋ in the problem statement 0 = F x d F ds ẋ = F y y x d ( F y ds y ẋ = F y y x y ( d F ẋ ds y F y d ds ( y ẋ Now the partial can be reexpressed in the following manner ( ( d F = ds y s + y s y + x s x + ẋ ( F = 2 F y s ẋ y y 2 s (4.1 (4.2 where the fact that F (y only explicitly depends on y to produce the last equality. Then the Euler-Lagrange equation can be simplified as y 2 F y ẋ y 2 s + F [ y y x d ( ] y = 0 (4.3 ds ẋ

10 Solutions Ph 236a Week 5 Page 10 of 10 Now, because s is an affine parameter y s = 0 and since F is a monotonic function of y, its derivative is nonzero, or F y 0. The equation then becomes 0 = y 2 [ F F y (0 + ẋ y2 y x d ( ] y ds ẋ = F y 0 = y x d ds [ y x d ( ] y ds ẋ ( y ẋ which gives the Euler-Lagrange equation for geodesics we are looking for. (4.4 Problem 5 Let v α = S α γ M γ, then the left-hand side is (S α γ M γ ;α = v α ;α = v α,α + Γ α µαv µ = (S α γ M γ,α + Γ α µαs µ γ M γ = S α γ,αm γ Expanding the right-hand side yields S α γ;αm γ + Sα γ M γ ;α + Sα γ M γ,α + Γα µαs µ γ M γ. (5.1 = M γ (Sα γ,α + Γ α µαs µ γ + Γ ναs αν γ Γ λ γαs α λ + S α γ (M γ,α Γδ αm γ δ + Γ γ ɛαm ɛ = S α γ,αm γ + Sα γ M γ,α + Γα µαs µ γ M γ + (Γ ναs αν γ M γ + ( Γ λ γαs α λ M γ = S α γ,αm γ Γδ αs α γ M γ δ + Γγ ɛαs α γ M ɛ + Sα γ M γ,α + Γα µαs µ γ M γ + (0 + (0, (5.2 since the terms inside the parenthesis exactly cancel because all indices are dummy indices. Thus we have shown that (S α γ M γ ;α = S α γ;αm γ + Sα γ M γ ;α. (5.3