1 Index Gymnastics and Einstein Convention

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1 Index Gymnastics and Einstein Convention Part (i) Let g δ µνe µ e ν be the metric. Then: Part (ii) x x g (x, x) δ µνe µ (x σ e σ) e ν (x ρ e ρ) δ µνx σ x ρ e µ (e σ) e ν (e ρ) δ µνx σ x ρ δ µ σ δ ν ρ δ µνx µ x ν x µx µ The same analysis proceeds: Part (iii) x y g (x, y) δ µνe µ (x σ e σ) e ν (y ρ e ρ) δ µνx σ y ρ e µ (e σ) e ν (e ρ) δ µνx σ y ρ δ µ σ δ ν ρ δ µνx µ y ν x µy µ We wish to show δ µνδ νρ δ µρ. Explicitly, this is: δ µνδ νρ ν δ µνδ νρ δ µδ ρ + δ µδ ρ + δ µδ ρ Now we can proceed on a case by case basis:. µ : only δ µδ ρ δ δ ρ δ ρ survives, so δ µνδ νρ i ρ µ.. µ : only δ µδ ρ δ δ ρ δ ρ survives, so δ µνδ νρ i ρ µ.. µ : only δ µδ ρ δ δ ρ δ ρ survives, so δ µνδ νρ i ρ µ. In all possible cases, we have that δ µνδ νρ i ρ µ; so this is identically δ µρ. Part (iv) We wish to show that a νδ νµ a µ. Explicitly, we have: a νδ νµ a δ µ + a δ µ + a δ µ Again, we can proceed on a cases by cases basis, and we will see that only the ν µ term for the summation survives; so a νδ νµ a µ. Part (v) We wish to show that δ µµ (in -dimensions). Explicitly, this is the statement that: δ µµ µ That this is true is directly veried: In general, we have that: δ + δ + δ + + δ µµ N for N is the dimension of the vector space. Part (a) We wish to show that: This can be veried directly: µ Part (b) (A µ B µ) (C ν D ν) (A µ C ν ) (B µd ν) (A µ B µ) (C ν D ν) ν µ ν µ ν A µ B µc ν D ν A µ C ν B µd ν (A µ C ν ) (B µd ν) If A µν A νµ and B µν B νµ, we have that: A µνb µν µ ν A µνb µν A B + A B + A B +A B + A B + A B +A B + A B + A B Noting now that the term A ijb ij (no summation) is equal to A jib ji, then: A B + A B 0 A B + A B 0 A B + A B 0 and by skew-symmetry, we must have A A A 0. Therefore, A µνb µν 0 Alternatively, we can simply write that: A µνb µν A νµb νµ (by symmetry / antisymmetry) So we end up with: A µνb µν (by changing dummy indices µ ν) A µνb µν A µνb µν This can be only true if A µνb µν 0.

2 Antisymmetry Part (a) Consider that in dimensions, the even permutations of {,, } are: (,, ), (,, ), (,, ) while the odd permutations are: Therefore we have that: (,, ), (,, ), (,, ) ε ε ε ε ε ε so the cyclic permutations have parity of +. Consider that in 4 dimensions, the even permutations of {,,, 4} are: (,,, 4), (,, 4, ), (, 4,, ), (,, 4, ), (,,, 4), (, 4,, ), (,,, 4), (,, 4, ), (, 4,, ), (4,,, ), (4,,, ), (4,,, ). while the odd permutations are: (,, 4, ), (,,, 4), (, 4,, ), (,,, 4), (,, 4, ), (, 4,, ), (,, 4, ), (,,, 4), (, 4,, ), (4,,, ), (4,,, ), (4,,, ). In this case then, it is instead that ε 4 but ε 4. Part (b) We wish to derive the general expression: δ il δ im δ in ε ijk ε lmn det 4 δ jl δ jm δ jn δ kl δ km δ kn Consider that ε ijk is eectively dened as: ε ijk sgn (σ) where σ S is the permutation that takes (,, ) to (i, j, k). Since ε ijk is the sign of the permutation from (,, ) to (i, j, k) and ε lmn is the sign of the permutation from (,, ) to (l, m, n), then eectively ε ijk ε lmn is the sign of the permutation τ 0 that takes (i, j, k) to (l, m, n). As such, we can dene: ε ijk ε lmn sgn (τ 0) Now, we can articially extend the denition as: ε ijk ε lmn σ S sgn (σ) δ σ(i)l δ σ(j)m δ σ(k)n since if σ τ 0, then δ σ(i)l δ σ(j)m δ σ(k)m 0. Note however that this is the very denition of the determinant; if we make the identication p a δ lσ(i), p b δ mσ(j), p c δ nσ(k), then we have: ε ijk ε lmn σ S sgn (σ) δ σ(i)l δ σ(j)m δ σ(k)n σ S sgn (σ) δ lσ(i) δ mσ(j) δ nσ(k) a,b,c ε abc p ap b p c det (p) δ li δ lj δ lk det 4 δ mi δ mj δ mk δ ni δ nj δ nk but since the determinant of a matrix and its transpose are the same, we have that: δ il δ im δ in ε ijk ε lmn det 4 δ jl δ jm δ jn δ kl δ km δ kn Expanding this gives us: ε ijk ε lmn δ il (δ jmδ kn δ jnδ km ) In homework notation, this is: Part (c) +δ im (δ jnδ kl δ jl δ kn ) +δ in (δ jl δ km δ jmδ kl ) ε ijk ε i j k δ ii (δ jj δ kk δ jk δ kj ) +δ ij (δ jk δ ki δ ji δ kk ) +δ ik (δ ji δ kj δ jj δ ki ) We wish to show the contraction formula: ε ajk ε amn δ jmδ kn δ jnδ km Consider from part (b), replacing i a, l a, and summing over a: Part (d) ε ajk ε amn δ aa (δ jmδ kn δ jnδ km ) +δ am (δ jnδ ka δ jaδ kn ) +δ an (δ jaδ km δ jmδ ka ) (δ jmδ kn δ jnδ km ) + (δ jnδ km δ jmδ kn ) + (δ jnδ km δ jmδ kn ) δ jmδ kn δ jnδ km Proceeding in a manner similar to part (b), we will have that: ε ijkl ε abcd det 6 4 δ ia det 4 δ ia δ ib δ ic δ id δ ja δ jb δ jc δ jd δ ka δ kb δ kc δ kd δ la δ lb δ lc δ ld +δ ic det 4 δ jb δ jc δ jd δ kb δ kc δ kd δ lb δ lc δ ld δ ja δ jb δ jd δ ka δ kb δ kd δ la δ lb δ ld 7 δ ib 4 δ ja δ jc δ jd δ ka δ kc δ kd δ la δ lc δ ld δ id det 4 δ ja δ jb δ jc δ ka δ kb δ kc δ la δ lb δ lc δ iaε jkl ε bcd δ ib ε jkl ε acd + δ icε jkl ε abd δ id ε jkl ε abc

3 So we have that: ε ijkl ε abcd δ iaε jkl ε bcd δ ib ε jkl ε acd + δ icε jkl ε abd δ id ε jkl ε abc Taking i s, a s, and contracting: ε sjkl ε sbcd δ ssε jkl ε bcd δ sb ε jkl ε scd + δ scε jkl ε sbd δ sd ε jkl ε sbc So we have that: 4ε jkl ε bcd ε jkl ε bcd + ε jkl ε cbd ε jkl ε dbc 4ε jkl ε bcd ε jkl ε bcd ε jkl ε bcd ε jkl ε bcd ε jkl ε bcd ε sjkl ε sj k l ε jklε j k l Vector Products Part (i) We wish to show that for vectors a, b, c, we have that: a (b c) b (c a) c (a b) Note that this is really a consequence of the preservation of ε ijk under cyclic permutations. Consider that: and likewise: Part (ii) a (b c) ε ijk a i b j c k ε jki a i b j c k ε ijk a k b i c j ε ijk b i c j a k b (c a) b (c a) ε ijk b i c j a k ε jki b i c j a k ε ijk b k c i a j ε ijk c i a j b k c (a b) We wish to show that a (b c) (a c) b (a b) c. This is: a (b c) ε ijk a j ε kmn b m c n ε ijk ε kmn a j b m c n Part (iii) ε kij ε kmn a j b m c n (δ imδ jn δ inδ jm) a j b m c n δ imδ jna j b m c n δ inδ jma j b m c n a n b i c n a m b m c i (a c) b (a b) c We wish to show that (a b) (c d) (a c) (b d) (a d) (b c). This can be obtained from using parts (i) and (ii). From part (i), we have that: (a b) (c d) c (d (a b)) and from part (ii), we have that: c (d (a b)) c (d b) a c (d a) b (a c) (b d) (a d) (b c) Part (iv) We wish to show that (i) is the spherical sine law, and that (iii) is the spherical cosine law. First we show (i). The spherical sine law is the statement that: sin A sin a sin B sin b sin C sin c where a, b, c are the lengths of the sides opposite to angles A, B, C, respectively. Let a, b, c be unit vectors at the origin, such that A is the angle at the endpoint of a, B is the angle at the endpoint of b, and C is the angle at the endpoint of c. Then a is the line segment traced between b and c, b is the line segment traced between a and c, and c is the line segment traced between a and b. Now, suppose we can show that: a ((a b) (a c)) a (b c) Then we will have solved our problem; because by (i), we have that a (b c) is invariant under cyclic permutations of the arguments a, b, c; so likewise we will have: a ((a b) (a c)) b ((b c) (b a)) c ((c a) (c b)) Now, the quantity a ((a b) (a c)) is given as: a ((a b) (a c)) a (a b) (a c) cos (θ) where θ is the angle between the vectors a and (a b) (a c); however, by their construction, these two vectors are necessarily colinear, so that cos (θ). Additionally, since a is a unit vector, then a. Therefore, we have that: a ((a b) (a c)) (a b) (a c) Using the fact that u v u v sin (φ uv), we obtain: a ((a b) (a c)) a b a c sin (φ a b,a c ) Note again, however, that by its construction, we have that φ a b,a c A (the angle between a b,a c is exactly the angle between the tangent vectors at a in the direction along b and c. Therefore, we have: a ((a b) (a c)) a b a c sin (A) sin (c) sin (b) sin (A) Finally, cyclically permutating the variables a, b, c (and repeating the same procedure) gives us: sin (c) sin (b) sin (A) sin (a) sin (c) sin (B) sin (b) sin (a) sin (C) Now, dividing everything by sin (a) sin (b) sin (c) gives us: sin (A) sin (a) This is the spherical sine law. To actually show that: sin (B) sin (b) sin (C) sin (c) a ((a b) (a c)) a (b c)

4 we have that: a ((a b) (a c)) a (((a b) c) a ((a b) a) c) ((a b) c) a ((a b) a) (a c) Now, a b is perpendicular to a so ((a b) a) 0. Likewise, for a is a unit vector, a. Therefore we obtain: a ((a b) (a c)) (a b) c c (a b) Using (i), we have: a ((a b) (a c)) a (b c) Now the cosine law. Using (iii), we have that: (a b) (c d) (a c) (b d) (a d) (b c) Taking b d, we have: Explicitly, this is: (a b) (c b) (a c) (b b) (a b) (b c) a b c b cos (θ a b,c b ) cos (b) cos (c) cos (a) Again, by our construction we have θ a b,c b a b sin (c), etc., we have: sin (c) sin (a) cos (B) cos (b) cos (c) cos (a) Rearranging terms give us: cos (b) cos (c) cos (a) + sin (c) sin (a) cos (B) This is the spherical cosine law. B; and for 4 Bernoulli and Vector Products Given Euler's equation for uid motion: In Cartesian coordinates, this is: We wish to show that: v s x s dv + (v ) v h dv i + vs x s v i x sh v i ε ijk v j ε kab x av b + x i v On the RHS, we have that: ε ijk v j ε kab x av b ε ijk ε kab v j x a v b δ ia δ jb v j x a v b + δ ib δ ja v j x a v b v j x i v j + v j x j v i Therefore we have: Note however that: Therefore: RHS v s x i (v s ) + v s x s x i v x i vs v s v i + x i v vs x i (v s ) + x i (vs ) v s v s x i (v s ) RHS v s x s v i so we have proven our claim. As such, we have shown that: (v ) v v ( v) + v Therefore Euler's equation becomes: from which we obtain: «dv + v ( v) + v h «dv + v ω v + h for which we have identied ω v as the vorticity. As such, if dv 0, then we are left with: v ω v + h Along a streamline r (t), we have (by denition) that: As such, we have: v dr ω v dr d ( r) d (0) 0 As such, v ω 0. This gives us that: v + h 0 along a streamline. Therefore, we must have: v + h const 4

5 Antisymmetry and Determinants Part (a) We begin with the classical denition of the determinant of A, an n n matrix: We wish to show that: det (A) ε i i...i n A i A i... A nin det (A) ε j j n...j n A i j A i j... A inj n We begin by using the denition of det (A) to obtain that: det (A) ε j...j n A j... A njn Now, since the product A j... A njn is commutative, we can imagine rearranging the order of the product, at no cost, so that the rst index k of A kl goes in the order of i... i n: det (A) ε j...j n A j... A njn ε j...j n Now, we can rearrange the indices of ε j...j n into the order of j i... j in, at the cost of a permutation sign: det (A) ε j...j n sgn (σ) ε ji...j in where σ is the permutation... n i... i n. However, this is exactly the denition of. That is, sgn (σ). So this give us: det (A) sgn (σ) ε ji...j in (sgn (σ)) ε ji...j in ε ji...j in Finally, the indicies j ik are just dummy indicies, so we may as well relabel them j ik j k : det (A) ε ji...j in This gives us our desired equality. As an example, take n 4. We have: ε j...j n A i j... A inj n ε ijkl det (A) ε ijkl ε abcd A aa b A ca 4d Let us take, say, ijkl 4. Then: ε 4 det (A) ε 4ε abcd A aa b A ca 4d ε 4ε abcd A 4d A aa ca b sgn (σ) ε 4ε dacb A 4d A aa ca b where σ : abcd dacb. Equivalently, this is σ : 4 4. So sgn (σ) ε 4, so at the end of the day we have: ε 4 det (A) ε dacb A 4d A aa ca b And then we can relabel the dummy indices as we wish. From this now, we wish to show that det (A) det (B) det (AB). We have: det (A) det (B) det (B) det (A) Part (b) (i) B i... B nin ε j...j n A j... A njn ε j...j n B i... B nin A j... A njn B j i... B jni n A j... A njn (AB) i... (AB) nin det (AB) Given V is an n-dimensional vector space. Let ω : V n C be a completely anti-symmetric multi-linear form. First, we wish to show that, up to a multiplicative constant, only one such map ω exists. That is, we wish to show that the space of all ω is one-dimensional. Let {e i} n be a basis for i V ; then for x V, we have x xi e i. Consider then (), x (),..., x (n). We can rewrite this as: ω x i () ei,..., x in (n) ein Explicitly, this is: x i ()... xin (n) ω (ei,..., e in ) x i ()... xin (n) ω (ei,..., e in ) i...i n However, due to the complete antisymmetry of ω, we need not sum over all indicies i... i n, and instead only over those that are permutations of,,..., n. Let S n be the set of permutations of,..., n. Then we can equally write: i...i n S n x i ()... xin (n) ω (ei,..., e in ) Now, we can re-order the arguments e i,... e in of ω, and pick up a permutation sign in the process: x i ()... xin (n) ω (ei,..., e in ) i...i n S n x i ()... xin (n) sgn (σ) ω (e,..., en) i...i n S n where σ is the permutation σ : i,..., i n,..., n. That means σ :,..., n i,..., i n. But for sgn (σ) sgn `σ, we can equally write sgn (σ). Therefore, we can more compactly write: x i ()... xin (n) εi...i n ω (e,..., e n) D ω (e,..., e n) where D is the proportionally factor D x i ()... xin (n) εi...i n, (the letter D suggestively chosen to stand for determinant). As such, we see that any map ω is completely determined by its action on ω (e,..., e n). If ω (e,..., e n) 0, then ω is identically zero. We can therefore consider, w/o l.o.g., only functions with ω (e,..., e n) 0 (if ω were identically zero, then it's trivially proportional to any other completely antisymmetric multilinear function).

6 Suppose now we have another such function σ. By the same procedure, we will also arrive at: σ `x D σ (e,..., e n) where the D is the same D as in ω. In fact, D makes no reference to the map ω or σ, and depends only on the arguments x (k). As such: D σ `x ω (e,..., e n) σ (e,..., e n) so: ω (e,..., e n) σ (e,..., e σ `x n) Therefore, we deduce that x ()... x (k) V n, we will have that σ `x. So every such function is proportional to another such function by a multiplicative constant. (ii) Assuming ω is not identically zero. We wish to show that, given a set of vectors x (),..., x (n), we will have that x (),..., x (n) is linearly independent i 0. First, we show that if x (),..., x (n) is linearly dependent, then 0. For x(),..., x (n) is linearly dependent, then we can w/o l.o.g. write x (n) P n i aix (i). As such, we will have that:! n ω x (),..., a ix (i) i n a `x iω (),..., x (i) i However, throughout this summation, every index i will coincide once with another argument ω. Therefore, (),..., x (i) always has a repeated argument, and so is zero. Therefore, 0. Next, we show that if x (),..., x (n) is linearly independent, then 0. The key point now is that since x (),..., x (n) is a set of n linearly independent vectors in an n- dimensional vector space, then this set x (),..., x (n) forms a basis for V. Hence, any vectors y V can be expressed as y y i x (i). Consider now: ω `y (),..., y (n) y i ()... yin (n) εi...i n where y (k) are arbitrary vectors in V. As such, we cannot have that 0; otherwise ω `y(),..., y (n) 0 for arbitrary y's, meaning that ω would be identically zero, contradicting our original assumption. Therefore, 0 i x(),..., x (n) are linearly independent. We now dene the determinant of a linear map A : V V as: det (A) ω `Ax(),..., Ax (n) to ω. Let this constant of proportionality be D. We will write ω A D ω. This means that x (),..., x (n) V n, we have that ω A `x D ω `x. If we can compute the value of D and show that D det (A), then we are done. Since ω A D ω, then in particular this relationship must hold when evaluated on a basis {e i}. Our choice of basis will be s.t. A e i A i, where A i is the i th column of the matrix form of A. More explicitly, this is A e i A ik e k. Using this basis, we have: and: ω A (e,..., e n) D ω (e,..., e n) ω A (e,..., e n) ω (A e,..., A e n) Consequently, we have that: ω (A i e i,..., A nin e in ) A i... A nin ω (e,..., e n) det (A) ω (e,..., e n) D ω (e,..., e n) det (A) ω (e,..., e n) and since e i are a basis, they are linearly independent, so ω (e,..., e n) 0. As such, we must have: D det (A) Hence our new denition coincides with our old one. Finally, we wish to show that det (AB) det (A) det (B). Consider: det (AB) ω `ABx (),..., ABx (n) ω A `Bx(),..., Bx (n) det (A) ω `Bx (),..., Bx (n) det (A) ω B `x det (A) det (B) since this is true for arbitrary x's, then we must have: det (AB) det (A) det (B) We must show that this coincides with the original denition. Dene: ω A `x ω `Ax(),..., Ax (n) It is easily seen that ω A is also a complete anti-symmetric, multilinear map. Therefore, we must have that ω A is proportional 6