Index notation in 3D. 1 Why index notation?


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1 Index notation in 3D 1 Why index notation? Vectors are objects that have properties that are independent of the coordinate system that they are written in. Vector notation is advantageous since it is elegant and deals with vectors as single objects rather than collections of components. This helps to make the physical and geometric meaning of equations manifest. However vector notation has some difficulties, a major one being that there is a whole heap of vector algebraic and differential identities that are hard to remember and hard to derive using vector notation. For example: a ( b c) = b ( c a) = c ( a b) (1.1) a ( b c) = b( c a) c( a b) (1.2) ( a b) ( c d) = ( a c)( b d) ( a d)( b c) (1.3) ( a b) ( c d) = ( a c d) b ( b c d) a (1.4) f = 0 (1.5) ( a) = 0 (1.6) ( a) = ( a) 2 a (1.7) (α b) = α b + b α (1.8) (α b) = α b b α (1.9) ( a b) = ( a ) b + ( b ) a + a ( b) + b ( a) (1.10) ( a b) = b ( a) a ( b) (1.11) ( a b) = a( b) b( a) + ( b ) a ( a ) b. (1.12) Most of these identities you should have seen before and all of these identities are used throughout physics, e.g. in classical mechanics, quantum mechanics, electrodynamics etc. (A similar notation is also used in general relativity and other more advanced topics). When using index notation to represent the above identities the proofs become very simple. Thus you don t even need to remember the identities, you can just rederive whichever result you happen to need as you go. To move from vector notation to an indexed notation we just need to choose an orthonormal basis in which to expand the vectors. Then any vector can be written as a triple in that basis, for example in Cartesian coordinates a vector can be represented as v = (v x, v y, v z ) = v xˆx + v y ŷ + v z ẑ. In general we choose the basis { e i, i = 1, 2, 3} and write v = (v 1, v 2, v 3 ) = v 1 e 1 + v 2 e 2 + v 3 e 3. Orthonormal means that e 1 e 1 = e 2 e 2 = e 3 e 3 = 1 and e 1 e 2 = e 2 e 3 = e 3 e 1 = 0. 1
2 The trick to index notation is to write any vector as an indexed symbol: so the vector v R 3 is represented by v i R where i takes the values 1, 2 or 3. Then all of the operations of vector calculus can be written using indexed objects. 2 Algebraic relations 2.1 Free and dummy indices Any index that occurs precisely once on each side of an equation is a free index. An equation with a free index represents a vector equation, as it has to be satisfied for each value the index can take. So for example if A and B are vectors then A i = B i A = B. For any index that is repeated on one side of the equation, we normally use the Einstein summation convention. That is we sum over all possible values of that index (in 3D that is 1,2 and 3). For example A i B ij = 3 A i B ij. i=1 These indices are then called dummy indices because they just stand in for the summation index and can be replaced with any other index without changing the meaning (just like integration variables are dummy varibles). 2.2 Dot product To write the dot product in index notation we need the Kronecker δ symbol. It is just the identity matrix, 1, written in an indexed form: 1 0 δ ij = (2.13)..... Then we see that the orthonormality condition on the basis vectors can be written as e i e j = δ ij. (2.14) Using the linearity of the dot product we can now see how it works in index notation: a b = (a i e i ) (b j e j ) = a i b j ( e i e j ) = a i δ ij b j. (2.15) Since δ ij β j = β i (which can be seen from the definition above or from the vector equivalent 1 b = b) we see the dot product take the normal form of the sum of the components: a b = a i δ ij b j = a i b i. (2.16) 2
3 The symmetry of the dot product can be seen either from the symmetry of δ ij combined with the fact that the components a i and b j are just numbers and therefore commute. The magnitude of a vector is just 2.3 Cross product a 2 = a a = a i a i. Once again we expand two vectors wrt the basis and use linearity to get a b = a i b j ( e i e j ). we know that the cross product of a vector times itself is zero, and using the standard definition of the cross product (right hand rule etc) we see that e 1 e 2 = e 3, e 2 e 3 = e 1, e 3 e 1 = e 2, the others being obtained via antisymmetry of the cross product. This can be summarised by defining a new symbol ε ijk such that e i e j = ε ijk e k. This symbol is known as the LeviCivita symbol, it is also called the totally antisymmetric tensor or the permutation symbol. It is defined by: 0 if any of i, j or k are the same ( ) ε ijk = 1 if ijk is an even permutation of 123 = sgn. (2.17) i j k 1 if ijk is an odd permutation of 123. An equivalent definition is ε ijk = ε jki = ε jik, ε 123 = 1. (2.18) We can illustrate this by looking at, for example, the first component. So setting i = 1 and expanding the implied summation ( a b) 1 = ε 1jk a j b k = ε 123 a 2 b 3 + ε 132 a 3 b = a 2 b 3 a 3 b 2, where the dots are terms with repeated indices on the LeviCivita symbol and thus zero. The other two components can be similarly found. The basic antisymmetry of the cross product can be thought of in terms of the antisymmetry of ε ijk relabelling the dummy indices to get ( a b) i = ε ijk a j b k = ε ikj a j b k = ε ikj b k a j ( a b) i = ε ijk b j a k = ( b a) i. 3
4 2.4 Relating δ and ε The basic identity that relates these two symbols is δ il δ jl δ kl ε ijk ε lmn = δ im δ jm δ km, (2.19) δ in δ jn δ kn but this is overkill. The only identity that is normally needed is obtained by setting k = l above to get ε ijk ε kmn = δ im δ jn δ in δ jm. (2.20) This identity can be derived/checked via brute force (try a few cases) or by arguments based on invariant tensors under spatial rotations. It is also good to note that it respects all of the symmetries that it should, i.e. both sides are antisymmetric under exchange of i j or equivalently m n. 2.5 Scalar triple product and determinants The scalar triple product is easily seen to be a ( b c) = a i (ε ijk b j c k ) = ε ijk a i b j c k (2.21) and by cycling the indices ε ijk a i b j c k = ε kij c k a i b j = ε jki b j c k a i, we get the identities a ( b c) = c ( a b) = b ( c a). (2.22) The determinant of a 3 3 matrix can also be written in indicial form and can be shown to be a 1 a 2 a 3 b 1 b 2 b 3 = a ( b c) = ε ijk a i b j c k. (2.23) c 1 c 2 c Vector triple product The identity we want to prove is a ( b c) = b( a c) c( a b). Diving in ( a ( b c)) i = ε ijk a j ε kmn b m c n = (δ im δ jn δ in δ jm )a j b m c n = b i a j c j c i a j b j = ( b( a c) c( a b)) i, (2.24) we see that it s just a simple consequence of the identity (2.20). 2.7 More vector products Products of more vectors can always be reduced down to repeated applications of the two triple products above, but they can still be computed directly (and efficiently) using index notation. 4
5 3 Vector Calculus The basic element of vector calculus is the operator. It is a vector operator and is sometimes written as. Its defining property is that for any unit vector ˆ s and function f(x, y, z) = f( r), ˆ s f is the partial derivative of f in the direction of ˆ s. Its components, with r = (x, y, z), i = r i = i take partial derivatives in the r i direction. Since each component of i is just a first order partial derivative, then each component must obey the Leibniz (product) rule. This implies that itself obeys the Leibniz rule. For any scalar functions f = f( r) and g = g( r), The familiar gradient, divergence and curl are written as i (fg) = ( i f) g + f i g. (3.25) grad(f) = f = i f, (3.26) div( v) = v = i v i and (3.27) curl( v) = v = ε ijk j v k, (3.28) where we introduced the notational convenience v( r) = v = v i. An important (and obvious) result is For any constant vector b this immediately leads to it also implies that curl( r) = 0 (prove me!). 3.1 Five Leibniz type identities i r j = r j r i = δ ij. (3.29) ( r b) = i r j b j = δ ij b j = b i = b, (3.30) All of these proofs can be done using only the definitions and properties of δ ij and ε ijk and the identities (3.25) and (2.20). The first three are trivial: (a b) = a b + b a (3.31) ( a b) = b a a b (3.32) (a b) = a b b a. (3.33) To prove we write ( a b) = ( b ) a + a( b) ( a ) b b( a), (3.34) LHS = ε ijk j (ε klm a l b m ) = (δ il δ jm δ im δ jl )(b m j a l + a l j b m ) = b j j a i + a i j b j a j j b i b i j a j = RHS. 5
6 The proof of the last identity is left as an exercise: ( a b) = a ( b) + ( a ) b + b ( a) + ( b ) a (3.35) Note that this identity has (3.30) as a special case. 3.2 Repeated derivatives Assuming we are working in the space of smooth functions, all partial derivatives commute: i j f = j i f. Using this, the standard identities are easy to prove: curl gradf = ( f) = 0, (3.36) div curl v = ( v) = 0, (3.37) curl curl v = v = ( v) 2 v = grad div v v. (3.38) where the Laplacian is defined as = 2 =. 4 Exercises Anything that is stated but not derived in these notes! Also you could try: div( r) = r =... (4.39) grad( r ) = r =... (4.40) f( r ) =... (4.41) a( r ) =... etc. (4.42) In the Kepler problem of classical mechanics there is one conserved scalar, the energy, and two conserved vectors, the angular momentum ( L = r p) and the LaplaceRungeLenz vector. An important component of the LRL vector is p L. Show that it is equal to p 2 r ( p r) p. In quantum mechanics the momentum operator is ˆ p = i so the angular momentum is ˆ L = r ˆ p = i r. Prove that ( r a) = a + r( a) + i(ˆ L a). 6
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