Solutions Ph 236a Week 5

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 1................................... 2 Part (a................................... 2 Part (b................................... 3 Part (c................................... 4 Part (d................................... 5 Part (e................................... 5 Problem 2................................... 6 Part (a................................... 6 Part (b................................... 6 Part (c................................... 7 Part (d................................... 7 Part (e................................... 7 Problem 3................................... 8 Part (a................................... 8 Part (b................................... 8 Problem 4................................... 9 Part (a................................... 9 Part (b................................... 9 Problem 5................................... 10

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 1 0 0 g ij = 0 r 2 0. (1.2 0 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

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

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 0 + 1 r + 1 + A θ (0 + 0 + cot θ r + A φ (0 + 0 + 0 = 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 0 + 1 r + 1 ( + Aˆθ 0 + 0 + cot θ r r + A ˆφ (0 + 0 + 0 Aˆr = r + 1 Aˆθ r θ + 1 A ˆφ r sin θ = 1 ( r 2 Aˆr r 2 + 1 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!

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

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,

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.

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

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

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