Vector Spaces in Physics 8/6/2015
Chapter 2. The Special Symbols ij and ijk , the Einstein Summation Convention, and some Group Theory
Working with vector components and other numbered objects can be made easier (and more fun) through the use of some special symbols and techniques. We will discuss two symbols with indices, the Kronecker delta symbol and the Levi-Civita totally antisymmetric tensor. We will also introduce the use of the Einstein summation convention.
References. Scalars, vectors, the Kronecker delta and the Levi-Civita symbol and the Einstein summation convention are discussed by Lea [2004], pp. 5-17. Or, search the web. One nice discussion of the Einstein convention can be found at http://www2.ph.ed.ac.uk/~mevans/mp2h/VTF/lecture05.pdf . You may find other of the lectures at this site helpful, too.
A. The Kronecker delta symbol, .
This symbol has two indices, and is defined as follows: 0, i j ij , i, j 1,2,3 Kronecker delta symbol (2-1) 1, i j Here the indices i and j take on the values 1, 2, and 3, appropriate to a space of three- component vectors. A similar definition could in fact be used in a space of any dimensionality.
We will now introduce new notation for vector components, numbering them rather than naming them. [This emphasizes the equivalence of the three dimensions.] We will write vector components as
Ax A A, i 1,3 (2-2) yi Az We also write the unit vectors along the three axes as ˆˆˆ ˆ i, j , k ei , i 1,3 (2-3) The definition of vector components in terms of the unit direction vectors is Ai A eˆi , i 1,3 (2-4) The condition that the unit vectors be orthonormal is
eˆi eˆ j ij (2-5) This one equation is equivalent to nine separate equations: iˆ iˆ 1, ˆj ˆj 1, kˆ kˆ 1, iˆ ˆj 0 , ˆj iˆ 0 , iˆ kˆ 0 , kˆ iˆ 0 , ˆj kˆ 0, kˆ ˆj 0!!! [We have now stopped writing "i,j=1,3;" it will be understood from now on that, in a 3-dimensional space, the "free indices" (like i and j above) can take on any value from 1 to 3.]
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Example: Find the value of ˆj kˆ obtained by using equation (2-5). ˆ ˆ Solution: We substitute eˆ2 for j and eˆ3 for k , giving ˆ ˆ j k eˆ2 eˆ3 23 0 , correct since and are orthogonal.
B. The Einstein summation convention.
The dot product of two vectors A and B now takes on the form 3 A B Ai Bi . (2-6) i1 This is the same dot product as previously defined in equation (1-40), except that AxBx has been replaced by A1B1 and so on for the other components.
Now, when you do a lot of calculations with vector components, you find that the sum of an index from 1 to 3 occurs over and over again. In fact, occasions where the sum would not be carried out over all three of the directions are hard to imagine. Furthermore, when a sum is carried out, there are almost always two indices which have the same value - the index i in equation (2-6) above, for example. So, the following practice makes the equations much simpler:
The Einstein Summation Convention. In expressions involving vector or tensor indices, whenever two indices are the same (the same symbol), it will be assumed that a sum over that index from 1 to 3 is to be carried out. This index is referred to as a paired index; paired indices are summed. An index which only occurs once in a term of an expression is referred to as a free index, and is not summed.
This sounds a bit risky, doesn't it? Will you always know when to sum and when not to? It does simplify things, though. The reference to tensor indices means indices on elements of matrices. We will see that this convention is especially well adapted to matrix multiplication.
So, the definition of the dot product is now A B Ai Bi , (2-7) the same as equation (2-6) except that the summation sign is omitted. The sum is still carried out because the index i appears twice, and we have adopted the Einstein summation convention.
To see how this looks in practice, let's look at the calculation of the x-component of a vector, in our new notation. We will write the vector A , referring to the diagram of figure 1-11, as
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A iˆA ˆjA kˆA x y z , (2-8) eˆi Ai where in the second line the summation over i = {1,2,3} is implicit. Now use the definition of the x-component, Ax A1 A eˆ1 . (2-9) Combining (2-8) and (2-9), we have
A11 A eˆ
()eˆˆii A e1
Aii() eˆˆ e1 (2-10)
Aii 1
A1 In the next to last step we used (2-5), the orthogonality condition for the unit direction vectors.
Next we carried out one of the most important operations using the Kronecker delta symbol, summing over one of its indices. This is also very confusing to someone seeing it for the first time. In the last line of equation (2-10) there is an implied summation over the index i. We will write out that summation term by term, just this once:
Ai i1 A111 A2 21 A3 31 (2-11) Now refer to (2-1), the definition of the Kronecker delta symbol. What are the values of the three delta symbols on the right-hand side of the equation above? Answer: 11 = 1, 21 = 0, 31 = 0. Substituting these values in gives
Ai i1 A111 A2 21 A3 31
A1 1 A2 0 A3 0 (2-12)
A1 What has happened? The index "1" has been transferred from the delta symbol to A.
C. The Levi-Civita totally antisymmetric tensor.
The Levi-Civita symbol is an object with three vector indices,
ijk ,i 1,2,3; j 1,2,3; k 1,2,3 Levi-Civita Symbol 1,i , j , k an even permutation of 1,2,3 (2-13) ijk 1,i , j , k an odd permutation of 1,2,3 0 otherwise All of its components (all 27 of them) are either equal to 0, -1, or +1. Determining which is which involves the idea of permutations. The subscripts (i,j,k) represent three numbers, each of which can be equal to 1, 2, or 3. A permutation of these numbers scrambles them up, and it is a good idea to approach this process systematically. So, we are going to discuss the permutation group.
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Groups. A group is a mathematical concept, a special kind of set. It is defined as follows: Definition: A group G is a set of objects ABC, , ,...with multiplication of one member by another defined, closed under multiplication, and with the additional properties: (i) The group contains an element I called the identity, such that, for every element A of the group, AI IA A (2-14) (ii) For every element A of the group there is another element B, such that AB BA I . (2-15) B is said to be the inverse of A: AB 1 . (2-16) (iii) Multiplication must be associative: A BC AB C . (2-17) There is an additional property which only some groups have. If multiplication is independent of the order in the product, the group is said to be Abelian. Otherwise, the group is non-Abelian. AB BA Abelian group. (2-18)
This may seem fairly abstract. But the members of groups used in physics are usually operators, operating on interesting things, such as vectors or members of some other vector space. Right now we are going to consider permutation operators, operating on sets of indices.
The Permutation Group. We will start by defining the objects operated on, then the operators themselves. Consider the numbers 1, 2, and 3, in some order, just like the indices on the Levi-Civita symbol: (,,)abc . (2-19) Here each letter represents one of the numbers, and they all three have to be represented. It is pretty easy to convince yourself that the full set of possibilities is (abc , , ) 1,2,3 , 1,3,2 , 2,1,3 , 2,3,1, 3,1,2 , 3,2,1. (2-20) Now the permutation group of the third degree consists of operators which re-arrange the three numbers as follows:
P a,,,, b c a b c P123 I P a,,,, b c a c b P132 P a,,,, b c b a c P213 . (2-21) P a,,,, b c b c a P231 P a,,,, b c c a b P 312 P a,,,, b c c b a P 321
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The second form of the notation shows where the numbers (1,2,3) would end up under that permutation. The first entry is the permutation which doesn't change the order, which is evidently the identity for the group. The group consists of just these six members
Examples of permutations operating on triplets of indices:
P123 2,3,1 2,3,1 P 2,3,1 2,1,3 132 . (2-22) P321 2,3,1 1,3,2
PP132 321 2,3,1 1,2,3 Do you follow the fourth line? First the permutation P321 is carried out, giving (1,3,2); and then the permutation P132 operates on this result, giving (1,2,3). This brings us to the subject of multiplication of group elements. This fourth line shows us that the product of the given two permutation-group elements is itself a permutation, namely
PPP132 321 312 . (2-23) Try this yourself, and verify that
P312 2,3,1 1,2,3 . (2-24) From this example, it is pretty clear that the group of six elements given above is closed under multiplication. There is an identity, the permutation which doesn't change the order. And it is pretty easy to identify the inverses within the group. Example: Show that 1 PP312 231 . Proof: Try it out on the triplet (a,b,c): p a,,,, b c c a b 312 . P231 c,,,, a b a b c The inverse permutation P231 just reverses the effect of P312.
There are some simpler permutation operators related to the Pijk, the binary permutation operators which just interchange a pair of indices, while leaving the third one unchanged.
P12 a,,,, b c b a c P13 a,,,, b c c b a . (2-25) P23 a,,,, b c a c b It is easy to see that the six group members given in equation (2-21) can be written in terms of the binary permutation operators:
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2 2 2 PIPPP123 12 13 23 PP132 23 PP 213 12 . (2-26) PPP231 12 13 PPP 312 13 12 PP321 13 (Remember, the right-hand operator in a product operates first.)
There is a special subset of permutations of a series of objects called the circular permutations, where the last index comes to the front and the others all move over one (see Figure 2-1).
(a,b,c,d,e,f,g) ---> (g,a,b,c,d,e,f)
Figure 2-1. A circular permutation.
For the six objects listed in eq. (2-21), three of them are circular permutations of (1,2,3), namely
(,abc ,)circular 1,2,3,3,1,2,2,3,1 . (2-27) Each of these is produced by an even number of binary permutations, and the other three are produced by an odd number of binary permutations. So, the group divides up into three "even permutations" and three "odd permutations:"
P123 P even permutations 312 P231 . (2-28) P213 P132 odd permutations P321
The Levi-Civita symbol. Now we can finally use the idea of even and odd permutations to define the Levi-Civita symbol: 1, (ijk) an even permutation of (123) ijk -1, (ijk) an odd permutation of (123) Levi - Civita Symbol (2-29) 0 otherwise Notice that there are only six non-zero symbols, three equal to +1 and three equal to -1. And any binary permutation of the indices (interchanging two indices) changes the sign. This is the key property in many calculations using the Levi-Civita symbol.
Example: Give the values of 312, 213, and 322,
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Answer: (312) is an even permutation of (123), so 312 = +1. (213) is obtained from (312) by permuting the first and last numbers, so it must be an odd permutation, and 213 = -1. (322) is not a permutation of (123) at all, so 322 = 0.
Question: Is the permutation group Abelian? What about the subgroup consisting of the three circular permutations? Answering these questions will be left to the problems.
D. The cross product.
In the last chapter we found the following result for the cross product of two vectors A and B in terms of their components: ˆ A B i(Ay Bz Az By ) ˆ j(Az Bx AxBz ) (1-48) ˆ k(AxBy Ay Bx ) Notice that there are a lot of permutations built into this definition. In particular, each term involves a permutation of (x,y,z), with the first letter indicating the unit vector, the second, the component of A, and the third, the component of B. Here is an elegant way of re-writing this expression using the Levi-Civita symbol: A Bi ijk A j Bk (2-30)
It may be less than obvious at first glance that (2-30) is the equivalent of (1-48). First let's just examine the index structure of the expression. The left-hand side has a single unpaired, or free, index, i. This means that it represents any single one of the components of the vector A B - we would say that it gives the i-th component of A B . Now look at the right-hand side. There is only one free index, and it is i, the same as on the left- hand side. This is the way it has to be. In addition, there are two paired indices, j and k. These have to be summed. If we were not using the Einstein summation convention, this 3 3 expression would read A Bi ijk Aj Bk . We have decided to follow the j1 k 1 Einstein convention and so we will not write the summation signs. However, for any given value of i, there are nine terms to evaluate.
To see exactly how this works out, let's evaluate the result of (2-30) for i=2. This should give the y-component of . Here it is: A B2 2 jk Aj Bk A B A B A B 211 1 1 212 1 2 213 1 3 (2-31) 221 A2 B1 222 A2 B2 223 A2 B3
231 A3 B1 232 A3 B2 233 A3 B3 But, most of these terms are equal to zero, because two of the indices on the Levi-Civita symbol are the same. There are only two non-zero L.-C. symbols: 213 = -1, and 231 = +1. Using these facts, we arrive at the answer
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A B2 A1B3 A3 B1 (2-32) This is the same as the y-component of equation (1-48), if the correct substitutions are made for numbered instead of lettered components. So, the two versions of the cross product agree.
Example: Use the tensor form of the cross product, equation (2-30), to prove that A B B A . Proof: There was a similar relation for the dot product - but with a plus sign! Let's see how this works in tensor notation: A Bi ijk Aj Bk definition of the cross product A B permuting two indices of ε gives a minus sign ikj j k ikj Bk Aj the A j and the Bk can be written in any order B Ai definition of the cross product
In regard to the last step of this example, it is worth remarking that particular name given to a summed index doesn't matter - it is sort of like the dummy variable inside a definite integral. What matters in the definition of the cross product A B is that the index of the components of A match with the second index of , and the index of the components of B, with the third index of .
E. The triple scalar product. BxC height of A rectangular There is a famous way of solid making a scalar out of three area of parallelogram = magnitude of BxC. vectors. It is illustrated in figure 2-2, where the vectors C A , B and C form the three independent sides of a B parallelopiped. The cross Figure 2-2. A parallelepiped, with its sides defined by product of and gives vectors A, B and C. The area of the parallelogram forming the area of the base of the the base of this solid is equal to BC sin , where is the parallelopiped (a angle between B and C. This is just the magnitude of the parallelogram), and dotting cross product BxC. When BxC is dotted into A, the area of with gives the volume: the base is multiplied by the height of the solid, giving its volume. Volume A BC . (2-31) Putting in the forms of the dot and cross product using , we have
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Volume A BC A BC Ai BCi . (2-33)
Aiijk B jCk
ijk Ai B jCk
There is an identity involving the triple scalar product which is easy to demonstrate from this form: A BC BC A C A B. (2-34) In the geometrical interpretation of the triple scalar product, these three forms correspond to the three possible choices of which side of the parallelepiped to call the base (see figure 2-2).
F. The triple vector product.
There is another way to combine three vectors, this time giving another vector: D ABC (triple vector product) (2-35) In tensor notation this becomes D ABC i i ijk Aj BCk (2-36)
ijk Aj klm BlCm This is not very encouraging. It is not simple, and furthermore it conjures up the prospect of more cross products. Do we have to live in dread of DABC and all of her big sisters?
The Epsilon Killer. Happily there is a solution. There is an identity which guarantees that there will never be more than one ijk in an expression, by reducing a product of two epsilons to Kronecker deltas. Here it is:
ijk ilm jl km jm kl (2-37) This is the epsilon killer! Here are the important structural features. There are two epsilons, with the first index of each one the same, so there is a sum over that index. The other four indices (two from each epsilon) are all different, and so are not summed. We will not prove eq. (2-37), but it is not too difficult, if you just consider all the possibilities for the indices.
We will now use this identity to simplify the expression for the vector triple product: A BCi ijk Aj klm BlCm definition of cross product A B C cyclic permutation of indices kij klm j l m (2-38) ( il jm im jl )Aj BlCm use epsilon killer
Am BiCm Al BlCi sum over paired indices of deltas
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The last step has used the "index transferring" property of a sum over one index of a delta symbol illustrated in equation (2-12). In the last line of (2-38) we can see two sums over paired indices, A C AC and A B A B. This gives m m l l A B Ci Bi A C Ci A B (2-39) or, in pure vector form, A BC BA C CA B (2-40) This is sometimes referred to as the "BAC - CAB" identity. It occurs regularly in advanced mechanics and electromagnetic theory.
PROBLEMS
In the problems below, repeated indices imply summation, according to the Einstein summation convention. Sum from 1 to 3 unless otherwise stated.
Problem 2-1. Consider ij and ijk as defined in the text, for a three-dimensional space.
(a) How many elements does ij have? How many of them are non-zero? (b) Give the following values:
11
23
31
(c) How many elements does have? How many are equal to zero? Which elements are equal to -1? (d) Give the following values:
111 321 123
132 Problem 2-2. Evaluate the following sums, implied according to the Einstein Summation Convention.
ii
12 j j3
12k1k
1 jj
Problem 2-3. Consider a possible group of permutations operating on three indices, but consisting of only the two members
IP, 12 (3-25) (a) Is this set of operators closed under multiplication? Justify your answer.
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(b) Is this set of operators Abelian? Justify your answer.
Problem 2.4. Consider a possible group of permutations operating on three indices, but consisting of only the four members
IPPP,,,12 13 23 (3-25) (a) Is this set of operators closed under multiplication? Explain your answer. (b) Is this set of operators Abelian? Explain your answer.
Problem 2.5. Consider the full permutation group, operating on three indices. (a) Is the group Abelian? Explain your answer. (b) What about the subgroup consisting of just the two circular permutations (and the identity)? Explain your answer. [You might approach these questions by simply trying two successive permutations, and then reversing the order.]
Problem 2-6. Assume that the cross product DAB is defined by the relation
DABABi ijk j k . i
Show using tensor notation (rather than writing out all the terms) that the magnitude of this vector agrees with the geometrical definition of the cross product. That is, show that D has a magnitude equal to |ABsin|. [Hint: Evaluate DD using the ''epsilon-killer'' identity.]
Problem 2-7. Use tensor notation (rather than writing out all the terms) to prove the following identity for three arbitrary vectors A , B , and C . A BC B C A C A B
Problem 2-8. (a) Use tensor notation (rather than writing out all the terms) to prove the following identity for two arbitrary vectors and . A A B 0
[Hint: Use the symmetries of the Levi-Civita symbol to prove that A A B A A B . This implies that both sides of the equation are equal to zero.] (b) Make a geometrical argument, based on the direction of AB , to show that this identity has to be satisfied.
Problem 2-9. Let eˆi ,i 1,2,3 be the usual three directional unit vectors of a 3- dimensional Cartesian coordinate system, satisfying the orthonormality relation eˆi eˆ j ij . In terms of components, and can be written as
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A A eˆ , mm B Bjj eˆ . Using these definitions for A and B and using tensor notation, show that A B Ai Bi .
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