[abc] (2) b 2 =a 2 (a 2 b 1 ) b 1 b 1. b 1 (4) b 1 a 3 b 2 b 2 b 2

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1 Problem 1. (a) Show that if Volumes and dual bases ɛ abc = g[abc] (1) Then show ɛ abc (defined from ɛ abc by raising indices, e.g. v a = g ab v b ) is 1 g [abc] (2) (b) Consider three vectors a 1, a 2, a 3, which span a parallel piped of volume a 1 (a 2 a 3 ) > 0. The Gram-Schmidt decompostion constructs a set of orthogonal vectors from a 1, a 2, a 3 b 1 =a 1 (3) b 2 =a 2 (a 2 b 1 ) b 1 b 1 b 1 (4) b 3 =a 3 (a 3 b 1 ) b 1 b 1 b 1 a 3 b 2 b 2 b 2 b 2 (5) (i) Briefly interpret the decompisition graphically, and show that b 1, b 2, b 3 are orthogonal. (ii) Show using the properties of determinants that a 1 (a 2 a 3 ) = det(b 1, b 2, b 3 ) and that a 1 (a 2 a 3 ) = b 1 b 2 b 3. No long proofs please just a fiew lines. Briefly interpret graphically. (c) Consider given three basis vectors g 1, g 2, g 3. (i) Show that the dual basis is g 1 = g 2 g 3 where = g 1 (g 2 g 3 ). g 2 = g 3 g 1 g 3 = g 1 g 2 (ii) Using the properties of the dual basis and determinants, show (in no more than three lines!) that Ω = g 1 (g 2 g 3 ) = 1/. (6) 1

2 Problem 2. Consider a 2d-coordinate system A dot product in non-orthogonal coordinates x =u 1 + 2u 2 (7) y =u 2 + u 1 (8) (a) Given the components of two vectors v a = (v 1, v 2 ) and w a = (w 1, w 2 ) so that v = v a g a etc, explictly determine the dot product v w in terms of these (lower) components. Problem 3. Spherical coordinates Spherical coordinates are defined by x =r sin θ cos φ (9) y =r sin θ sin φ (10) z =r cos θ (11) (a) Determine the basis vectors g r, g θ, g φ as an expansion in cartesian basis vectors ˆx, ŷ, ẑ, and illustrate them graphically. (b) Determine the metric tensor g ab and ds 2, and g r, g θ, g φ. (c) Determine the volume measure dv using g ab. (d) Compute all Christoffel symbols by computing derivatives, e.g. compute θ g r (12) and reexpand the result in g r, g φ, g θ. Give a graphical explanation for the ratio of Γ θ φφ to Γ r φφ. (e) Compute Γ r φφ and Γθ φφ using the famous formula Γ c ab = 1 2 gcd ( a g db + b g ad d g ab ). (13) and verify that this agrees with the results in the previous item (f) Consider cylindrical coordinates (look at lecture notes). some tragicomical 1 student writes Every year on the comps, 2π Show that the correct result is 1 2 ˆx. 0 dφ cos φ ˆρ = 0 crazily wrong! (14) 2π 1 Definition of tragicomic. 1 : of, relating to, or resembling tragicomedy. 2 : manifesting both tragic and comic aspects. 2

3 (g) The curl of vector field is A = e i ɛ ijk j A k (15) Given this definition in cartesian coordinates, show by coordinate transformation that in a general coordinate system A = g a ɛ abc b A c (16) Argue that for the curl (and only the curl!) we may use a partial instead of covariant derivative A = g a ɛ abc b A c (17) and use this result to show that for general orthogonal coordiantes [ ] [ ] ( V ) = eˆ1 (h 3 V ˆ3 ) (h 2V ˆ2 ) + eˆ2 (h 1 V ˆ1 ) (h 3V ˆ3 ) h 2 h 3 u 2 u 3 h 1 h 3 u 3 u 2 and here eâ = g a /h a. + eˆ3 h 1 h 2 [ (h 2 V ˆ2 ) (h 1V ˆ1 ) u 1 u 2 (h) For the specific surface shown below (i.e. the surface bounded by the red contour at φ = 0 which forms a square in r, θ space), use Eq. (18) to prove the Stokes theorem for this specific surface z ] (18) y x (i) Recall that for a viscous fluid the force per area F/ A of two streams flowing past each other with different velocities v x is F x = η vx A y y (19) Here F x is the force on the upper (faster) stream by the lower (and slower) stream 3

4 A y F x v x (y) I put a y-index on A y to indicate that area vector we are considering, A = A n, is pointing in the y direction. The force per area defines the stress tensor in the system. Thus, the stress tensor for the viscous fluid we have described above has the nonvanishing component T xy = η vx y Indeed, the stress tensor T ij is very generally interpretted as T ij = (20) force in the i-th direction area in the j-th direction = F i (21) A j The stress tensor in cartesian coordinates of a viscous fluid is T ij visc = η( i v j + j v i 2 3 δij l v l ) (22) In the simple case where the x-velocity is a function of y (and all other velocity components vanish), T xy = η y v x is the only non-vanishing component. According to Landau and Lifshitz Fluid Mechanics (a standard text not on general relativity, which therefore uses normalized coordinate vectors), the divergence of the the velocity is (but they leave off the hats!) v = 1 r r(r 2 vˆr ) r sin θ θ(sin θv ˆθ) + 1 r sin θ φv ˆφ (23) Derive this results using the formula involving covariant derivatives, a v a. Also compute it using the general expression v = 1 g a ( gv a ) (24) According to Landau and Lifshitz one of the stress tensor components of a viscous fluid are T ˆrˆθ is (but they leave off the hats!) ( ) T ˆrˆθ = η r v ˆθ + 1 r θvˆr v ˆθ (25) r Derive this result, given its cartesian counter part, Eq. (22) 4

5 (j) Using the setup of the previous problem, suppose that at an angle θ, but φ = 0, T ˆrˆθ = T ˆθˆr is the only non-vanishing component. What are the only non-vanishing cartesian components T ij and how are they related to T ˆrˆθ? Draw a picture to explain your result. 5