Classical Mechanics Solutions 1.

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1 Classical Mechanics Solutions 1. HS 2015 Prof. C. Anastasiou Prologue Given an orthonormal basis in a vector space with n dimensions, any vector can be represented by its components1 ~v = n X vi e i. 1) i=1 In order to make formulae involving vectors less cumbersome, it is very useful to adopt the Einstein summation convention: repeated indices are implicitly summed over and the sign that indicates the sum omitted. For instance, we shall write ~v = vi e i, instead of the above formula. We will also be using extensively the Kronecker delta 1 if i = j, δij = 0 otherwise. 2) 3) As an example, the orthonormality condition reads e i e j = δij, 4) ~v w ~ = vi e i wj e j = vi wj δij = vi wi. 5) and a scalar product Two indices that are paired and summed over as in the last step on the right are sometimes said to be contracted. Exercise 1. The Levi-Civita symbol. Given a vector space of dimension n, the Levi-Civita symbol is an object with n indices defined by the property ε...i...j... ε...j...i..., 6) together with ε12...n = +1. We say that ε is totally antisymmetric under the exchange of any two indices. i) ii) iii) iv) What is εi1...in equal to when two indices take the same value? Assuming sij = sji, what can you say about ε...i...j... sij? For n = 2, enumerate all values of the Levi-Civita symbol εij and put them in a matrix. For n = 3, list all non-zero values of the Levi-Civita symbol εijk. 1 In differential geometry, it is important to distinguish between upper and lower indices. For this course such distinction is not required if you are wondering why, the reason is that we will only deal with euclidian spaces). 1

2 When two indices are equal, from the definition we get ε...i...i... = ε...i...i... = 0. S.1) Again using the definition we get ε...i...j...s ij = ε...j...i...)s ij = ε...j...i...)s ji = ε...l...k... s lk = 0. i,j i,j i,j k,l In these steps sums were written out explicitly to emphasize that, because i and j are dummy indices, their name is not really important and may be changed at one s own leisure. The Levi-Civita symbol for a 2-dimensional vector space carries two indices, and is therefore written as ε ij with i, j {1, 2}. From the first point we immediately get ε 11 = ε 22 = 0, by definition ε 12 = +1 and from the fundamental property 6) we see that ε 21 = 1. Thus, as a matrix, ε = ) S.2). S.3) For a tridimensional vector space, the entries ε ijk vanish unless all indices take different values. Therefore, by exchanging indices repeatedly, we get ε 123 = +1, ε 213 = 1, ε 231 = +1, ε 321 = 1, ε 312 = +1, ε 132 = 1. S.4) The practical examples of this course will mostly be set in euclidean space in three dimensions. Therefore we are going to work almost exclusively with ε ijk, which will enable us to handle vector calculus in a very convenient way see the next exercises). Given the following identity for the product of two Levi-Civita symbols δ in δ il δ im ε ijk ε nlm = det δ jn δ jl δ jm ; 7) δ kn δ kl δ km ) δjl δ v) Show that ε ijk ε ilm = det jm = δ jl δ km δ jm δ kl. vi) Show that ε ijk ε ijm = 2δ km. vii) Show that ε ijk ε ijk = 6. δ kl δ km These identities all follow from one another inserting a δ and summing. We start from ε ijk ε nlm = δ inδ jl δ km δ inδ jmδ kl + δ il δ jmδ kn δ il δ jnδ km + δ imδ jnδ kl δ imδ jl δ kn ; S.5) using δ ab δ bc = δ ac and δ aa = 3 repeatedly we find ε ijk ε ilm = δ inε ijk ε nlm = 3δ jl δ km 3δ jmδ kl + δ nl δ jmδ kn δ nl δ jnδ km + δ nmδ jnδ kl δ nmδ jl δ kn = 3δ jl δ km 3δ jmδ kl + δ jmδ kl δ jl δ km + δ jmδ kl δ jl δ km = δ jl δ km δ jmδ kl, S.6) ε ijk ε ijm = δ jl ε ijk ε ilm = 3δ km δ km = 2δ km, S.7) ε ijk ε ijk = δ km ε ijk ε ijm = 3δ km. S.8) One of the possible definitions of the vector product reads v w ε ijk v j w k ê i. 8) viii) Show that, also according to this definition, v w is orthogonal to both v and w, and that its length is equal to the area spanned by a parallelogram with sides v and w. Can you say to which property of ε the right-hand rule is related? 2

3 Using point ii) we find immediately v v w) = ε ijk v iv jw k = 0, S.9) which means that v v w), and the same for w. Applying v) yields v w) 2 = ε ijk v jw k )ε ilm v l w m) = v 2 w 2 v w) 2 = v 2 w 2 1 cos 2 ϑ) = v w sin ϑ) 2, S.10) which is the area of the described parallelogram. The sign of v w) would always be flipped if ε had the opposite sign, as can be seen from the definition. Thus, assuming the coordinate system is right-handed, if we had taken ε 123 = 1 we would ended up with the vector product being given by a left-hand rule. Exercise 2. Vector Identities Prove the following identities: 1. a b c) = a c) b a b) c 2. a b c) 2 = a c) 2 b 2 + a b) 2 c 2 2 a c) a b) b c) 3. a b) c = a c) b b c) a 4. a b) c d) = a c) b d) a d) b c) 5. Ra Rb = R a b) 6. ψ = 0 7. A) = 0 8. A) = A) A 9. A B) = B A) A B) 10. A B) = A ) B + B ) A + A B) + B A) 11. A B) = A B) B A) + B ) A A ) B where a, b, c and d are vectors, A, B are vector fields, ψ is a function and R SO3). Moreover assume that all components A i, B j and also ψ are in C2), i.e. two times continuously differentiable. Don t write out cross products explicitly, but use the index notation involving the Levi-Civita symbol ε ijk. 1. a b c)) i = ε ijk a jε klm b l c m = ε kij ε klm a jb l c m = a jb ic j a jb jc i = a c)b i a b)c i where we used ε ijk = ε kij and ε kij ε klm = δ il δ jm δ imδ jl 3

4 2. Here one can either square the previous result which is trivial or write down the cross products explicitly and contract the indices from the getgo. 3. a b) c) i = ε ijk ε jlm a l b m)c k = ε jki ε jlm a l b mc k = a k b ic k a ib k c k = a c)b i b c)a i 4. where we used ε ijk = ε kij and ε jki ε jlm = δ kl δ im δ km δ il a b) c d) = ε ijk a jb k ε iml c md l = a jb k c jd k a jb k c k d j = a c) b d) a d) b c) where we used ε ijk ε ilm = δ jl δ km δ jmδ il 5. First, in the spirit of the expansion of the determinant of a matrix M we observe ε jml M jim mnm ls = ε insdetm). S.11) Hence, we find with R 1 = R T and detr) = 1 R 1 Ra Rb) ) i = R ji Ra Rb) j = R jiε jml R mna nr ls b s = ε insa nb s = a b) i 6. ψ) i = ε ijk j k ψ = 0 since the partial derivatives commute and ε ijk is antisymmetric. 7. A) = ε ijk i ja k = 0 since the partial derivatives commute and ε ijk is antisymmetric. 8. A)) i = ε ijk ε klm j l A m = ε kij ε klm j l A m = i ma m j ja i = A) A) i where we used ε ijk = ε kij and ε kij ε klm = δ il δ jm δ imδ jl 9. A B) = ia B) i = iε ijk A jb k = ε ijk i[a jb k ] = ε ijk ia j)b k + ε ijk A j ib k ) = B k ε kij ia j) A jε jik ib k ) = B A) A B) where we used ε ijk = ε kij, ε ijk = ε jik and the product rule for derivatives. 4

5 10. A ) B + B ) A + A B) + B A)) i = A j jb i + B j ja i + ε ijk ε klm A j l B m + ε ijk ε klm B j l A m = A j jb i + B j ja i + ε kij ε klm A j l B m + ε kij ε klm B j l A m = A j jb i + B j ja i + A j ib j A j jb i + B j ia j B j ja i = A j ib j + B j ia j = i A jb j) = A B)) i 11. where we used ε ijk = ε kij and ε kij ε klm = δ il δ jm δ imδ jl A B)) i = ε ijk ja B) k ) = ε ijk jε klm A l B m) = ε ijk ε klm [ ja l )B m + A l jb m)] = ε kij ε klm [ ja l )B m + A l jb m)] = δ il δ jm δ imδ jl ) ja l )B m) + δ il δ jm δ imδ jl )A l jb m) = ja ib j ja jb i + A i jb j A j jb i = B ) A B A) + A B) A ) B) i where again we used ε ijk = ε kij and ε kij ε klm = δ il δ jm δ imδ jl, as well as the product rule for differentiation. Exercise 3. Jacobian In this exercise, we will have a closer look at the Jacobian for the example of the polar coordinate transformation. a) Calculate the Jacobian matrix Jr, Θ) for the polar coordinate transformation x = r cos Θ, y = r sin Θ. b) Show that dx = cos Θdr r sin ΘdΘ, dy = sin Θdr + r cos ΘdΘ and calculate the area element dx dy in terms of dr and dθ. c) In the expression for dx dy, where does the Jacobian come in? d) When does the determinant of Jr, Θ) vanish? The inverse function theorem states that if a continuously differentiable function has a non-zero Jacobian determinant at some point, the function is invertible in an open region containing that point. What does this theorem tell you about the polar coordinate transformation we looked at in this exercise? 1. The Jacobian matrix is given by Jr, Θ) = x r y r x Θ y Θ ) = cosθ) sinθ) r sinθ) r cosθ) ). 5

6 2. Using d x i = x i x j dx j, we immediately find that dx = cos Θdr r sin ΘdΘ and dy = sin Θdr + r cos ΘdΘ. Now, the area calculates as dx dy = cos Θdr r sin ΘdΘ) sin Θdr + r cos ΘdΘ) = r cos 2 Θdr dθ r sin 2 ΘdΘ dr = rcos 2 Θ + sin 2 Θ)dr dθ = rdr dθ. 3. What we have really calculated above is dx dy = x y x y )dr dθ. Here, the expression in the r Θ Θ r brackets is none other than the determinant of the matrix J: in this case, det Jr, Θ) = r. 4. In the example of the polar coordinates, the determinant only) vanishes at r = 0. The inverse function theorem tells us now that for each point except r = 0, an inverse transformation exists for the polar coordinate transform for an open region around that point. At r = 0, however, we cannot find such an inverse, as any arbitrary Θ could be assigned to that point x = 0, y = 0). Exercise 4. Vector Calculus This exercise requires using various techniques that you have learnt in this sheet. Calculate r/ x i where r = x = x 2 + y 2 + z 2. A vector field u is given by u = a r + a x)x r 3. Find the Jacobian J ij = u i / x j and deduce that u = 0. Using the suffix notation one may write r = x k x k. Then r = 2x kδ ik x i 2 = xi x k x k r. This implies J ij = ui x j = aixj r 3 3a kx k x ix j + a kx k δ ij + δ kj x i) r 5 r 3 = 1 a x aixj + ajxi) 3 r3 r 5 xixj + a x r δij. 3 Note that this splits into a symmetric and antisymmetric part. We have u = u i x i = J ii. If i = j, the antisymmetric part vanishes and so does the symmetric part once we use x ix i = r 2 and δ ii = 3. 6