Appendix Exercises: Answers and Solutions Exercise Chapter 1 1.1 Complex Numbers as 2D Vectors (p. 6) Convince yourself that the complex numbers z = x + iy are elements of a vector space, i.e. that they obey the rules (1.1)–(1.6). Make a sketch to demonstrate that z1 + z2 = z2 + z1, with z1 = 3 + 4i and z2 = 4 + 3i, in accord with the vector addition in 2D. Exercise Chapter 2 2.1 Exercise: Compute Scalar Product for given Vectors (p. 14) Compute the length, the scalar products and the angles betwen the three vectors a, b, c which have the components {1, 0, 0}, {1, 1, 0}, and {1, 1, 1}. Hint: to visualize the directions of the vectors, make a sketch of a cube and draw them there! The scalar products of the vectors with themselves are: a · a = 1, b · b = 2, c · c = 3, and consequently √ √ a =|a|=1, b =|b|= 2, c =|c|= 3. The mutual scalar products are a · b = 1, a · c = 1, b · c = 2. The cosine of the angle ϕ between these vectors are √1 , √1 , √2 , 2 3 6 © Springer International Publishing Switzerland 2015 389 S. Hess, Tensors for Physics, Undergraduate Lecture Notes in Physics, DOI 10.1007/978-3-319-12787-3 390 Appendix: Exercises… respectively. The corresponding angles are exactly 45◦ for the angle between a and b, and ≈70.5◦ and ≈35.3◦, for the other two angles. Exercises Chapter 3 3.1 Symmetric and Antisymmetric Parts of a Dyadic in Matrix Notation (p. 38) Write the symmetric traceless and the antisymmetric parts of the dyadic tensor Aμν = aμbν in matrix form for the vectors a :{1, 0, 0} and b :{0, 1, 0}. Compute the norm squared of the symmetric and the antisymmetric parts and compare with Aμν Aμν and Aμν Aνμ. In matrix notation, the tensor Aμν is equal to ⎛ ⎞ 010 A = ⎝ 000⎠. (A.1) 000 The trace of this matrix is zero. So its symmetric part coincides with its symmetric traceless part ⎛ ⎞ 010 1 A = ⎝ 100⎠. (A.2) 2 000 The antisymmetric part of this tensor is ⎛ ⎞ 010 1 Aasy = ⎝ −100⎠. (A.3) 2 000 The tensor product A · A yields ⎛ ⎞ 100 1 ⎝ ⎠ Aμλ Aλν = 010 , (A.4) 4 000 and consequently 1 Aμλ Aλμ = . 2 Similarly, the product of the antisymmetric part with its transposed, viz. asy asy Aμλ Aνλ Appendix: Exercises… 391 yields the same matrix as in (A.4). Thus one has also 1 Aasy Aasy = . μλ μλ 2 Due to Aμν Aμν = 1, Aμν Aνμ = 0, this is in accord with (3.10), viz. 1 asy asy Aμν Bνμ = Aλλ Bκκ + Aμν Bνμ + Aμν Bνμ , 3 with Bνμ = Aμν. 3.2 Symmetric Traceless Dyadics in Matrix Notation (p. 39) (i) Write the symmetric traceless parts of the dyadic tensor Cμν = Cμν(α) = 2aμbν in matrix form for the vectors a = a(α) :{c, −s, 0} and b = b(α) :{s, c, 0}, where c and s are the abbreviations c = cos α and s = sin α. Discuss the special cases α = 0 and α = π/4. The desired tensor is ⎛ ⎞ 2cs c2 0 ⎝ ⎠ Cμν = −s2 −2cs 0 , (A.5) 000 and consequently, due to 2cs = sin 2α, c2 − s2 = cos 2α, one obtains ⎛ ⎞ sin 2α cos 2α 0 ⎝ ⎠ Cμν (α) = cos 2α − sin 2α 0 . (A.6) 000 For α = 0 and α = π/4, this tensor reduces to ⎛ ⎞ ⎛ ⎞ 010 100 ⎝ 100⎠, ⎝ 0 −10⎠, (A.7) 000 000 respectively. The diagonal expression follows from the first of these tensors when the Cartesian components of the vectors and tensors are with respect to a coordinate system rotated by 45◦. (ii) Compute the product Bμν(α) = Cμλ (0) Cλν (α), determine the trace and the symmetric traceless part of this product. Determine the angle α, for wich one has Bμμ = 0. 392 Appendix: Exercises… The result is ⎛ ⎞ cos 2α − sin 2α 0 ⎝ ⎠ Bμν(α) = sin 2α cos 2α 0 . (A.8) 000 Consequently. one has ⎛ ⎞ 10 0 1 ⎝ ⎠ Bμν (α) = cos 2α 01 0 , (A.9) 3 00−2 ◦ and Bμμ = 2 cos 2α. Thus one has Bμμ = 0forα = π/4, or 45 . For this angle, the two tensors (A.7) are ‘orthogonal’ in the sense that the trace of their product vanishes. 3.3 Angular Momentum in Terms of Spherical Components (p. 43) Compute the z-component of the angular momentum in terms of the spherical components. For a particle with mass m,thez-component of the angular momentum is Lz = m(x y˙ − yx˙), in cartesian coordinates. In polar coordinates, cf. Sect.2.1.4, one has x = r sin ϑ cos ϕ, y = r sin ϑ sin ϕ, z = r cos ϑ. The time change of x and y is − x˙ =˙rr 1x + ϑ˙ r cos ϑ cos ϕ −˙ϕr sin ϑ sin ϕ, − y˙ =˙rr 1 y + ϑ˙ r cos ϑ sin ϕ +˙ϕr sin ϑ cos ϕ. ˙ In the calculation of Lz, the terms involving r˙ and ϑ cancel, the remaining terms add up to 2 Lz = mr ϕ.˙ 3.4 Torque Acting on an Anisotropic Harmonic Oscillator (p. 44) Determine the torque for the force F =−kr · ee − (r − r · ee), where the parameter k and unit vector e are constant. Which component of the angular momentum is constant, even for k = 1? The torque is T = r×F =−(k−1)(r·e)r×e. Clearly, the torque vanishes for k = 1. For k = 1, the torque still is zero, when r × e = 0or(r · e) = 0 hold true. The first case corresponds the one-dimensional motion along a line parallel to e which passes through the origin r = 0. This is a one-dimensional harmonic oscillator. The second case is a motion in the plane perpendicular to e. This corresponds to an isotropic two-dimensional harmonic oscillator. Appendix: Exercises… 393 3.5 Velocity of a Particle Moving on a Screw Curve (p. 46) Hint: Use α = ωt for the parameter occurring in the screw curve (3.48), ω is a frequency. Differentiation with respect to the time t yields the velocity ω v = ρω[−e sin(ωt) + u cos(ωt)] + χ e × u, 2π where it is assumed that not only the orthogonal unit vectors e and u, but also the radius ρ and the pitch parameter χ are constant. Exercise Chapter 4 4.1 2D Dual Relation in Complex Notation (p. 54) Let the two 2D vectors (x1, y1) and (x2, y2) be expressed in terms of the complex numbers z1 = x1 + iy1 and z2 = x2 + iy2. Write the dual relation corresponding to (4.25) in terms of the complex numbers z1 and z2. How about the scalar product of these 2D vectors? Hint: the complex conjugate of z = x + iy is z∗ = x − iy. ∗ + + ( − ) The product z1z2 is x1x2 y1 y2 i x1 y2 x2 y1 , thus the dual scalar is 1 x y − x y = (z∗z − z z∗). 1 2 2 1 2i 1 2 1 2 Similarly, the scalar product of the two vectors is 1 x y + x y = (z∗ z + z z∗). 1 1 2 2 2 1 2 1 2 In other words, the scalar product is the real part and the dual scalar is the imaginary ∗ part of z1z2. Exercises Chapter 5 5.1 Show that the Moment of Inertia Tensors for Regular Tetrahedra and Octahedra are Isotropic (p. 62) Hint: Use the coordinates (1, 1, 1), (−1, −1, 1), (1, −1, −1), (−1, 1, −1) for the four corners of the tetrahedron and (1, 0, 0), (−1, 0, 0), (0, 1, 0), (0, −1, 0), (0, 0, 1), (0, 0, −1), for the six of the octahedron. 394 Appendix: Exercises… (i) Tetrahedron: the position vectors of the corners are u1 = ex + ey + ez, u2 =−ex − ey + ez, u3 = ex − ey − ez, u4 =−ex + ey − ez. i i x y x z y z In the products uμuν the mixed terms involving eμeν, eμeν, eμeν have the signs (+, +, +), (+, −, −), (−, −, +), (−, +, −) for i = 1, 2, 3, 4, respectively. The sum of these mixed terms vanishes and one finds 4 i i x x y y z z uμuν = 4(eμeν + eμeν + eμeν) = 4 δμν, i=1 thus the moment of inertia tensor Θμν = 8 m δμν is isotropic. 6 i i ( x x + y y + z z) (ii) Octahedron: here the sum i=1 uμuν yields 2 eμeν eμeν eμeν and conse- quently Θμν = 4 m δμν. 5.2 Verify the Relation (5.51) for the Triple Product of a Symmetric Traceless Tensor (p. 72) Hint: use the matrix notation ⎛ ⎞ a 00 ⎝ 0 b 0 ⎠ , 00c with c =−(a + b), for the symmetric traceless tensor in its principal axis system. Compute the expressions on both sides of (5.51) and compare. In matrix notation, the left hand side of 1 a · a · a = a(a : a) 2 is ⎛ ⎞ 2a3/3 − b3/3 − c3/30 0 ⎝ 02b3/3 − c3/3 − c3/30⎠ . (A.10) 002c3/3 − a3/3 − b3/3 Appendix: Exercises… 395 Due to c3 =−(a3 +3a2b+3ab2 +b3), the diagonal elements are equal to a3 +a2b+ ab2 = a(a2 + ab+ b2), b3 + a2b + ab2 = b(a2 + ab+ b2), and −a3 − b3 − 2a2b − 2ab2. On the other hand, the 11-element of a(a : a) is equal to (a2 + b2 + c2)a = 2a3 + 2b3 + 2a2b = 2a(a2 + ab + b2). Similarly, one finds for the 22-element 2b(a2 + ab+ b2). The 33-element is (a2 + b2 + c2)c =−2(a + b)(a2 + ab+ b2) = −2(a3 + b3 + 2a2b + 2ab2). Comparison of the diagonal matrix elements shows the validity of the relation (5.51).