INTEGRALSATSER VEKTORANALYS. (indexräkning) Kursvecka 4. and CARTESIAN TENSORS. Kapitel 8 9. Sidor NABLA OPERATOR,
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1 VEKTORANALYS Kursvecka 4 NABLA OPERATOR, INTEGRALSATSER and CARTESIAN TENSORS (indexräkning) Kapitel 8 9 Sidor 83 98
2 TARGET PROBLEM In the plasma there are many particles (10 19, per m 3 ), strong magnetic fields and electric currents. How can we describe the behaviour of the plasma? Magnetohydrodynamics (MHD) Simple example: THE THETA PINCH z r θ plasma J B When the plasma is in equilibrium, MHD equations are: grad p = j B rotb = μ 0 j 1 grad p = rotb B μ ( ) 0 We need to introduce: p is the pressure j is the current density Operators Nabla And then? How to continue? 1
3 What is a function? OPERATOR A function is a law defined in a domain X that to each element x in X associates one and only one element y in Y. y Example: X=[0,2] f(x)=x x The slope of f(x) is its derivative: y gx ( ) df ( x ) = dx g(x) is still a function. 0 2 x So the derivative is a rule that associates a function to another function. The derivative is an example of operator 2
4 OPERATOR DEFINITION An operator T is a law that to each function f in the function class D t associates a function T(f). DEFINITION An operator T is linear if T(af+bg)= )=at(f)+bt(g at(f)+bt(g), where f and g are functions belonging to D t. EXAMPLE: =? d T is it linear dx YES daf ( + bg) df dg Taf ( + bg ) = = a + b = at( f ) + bt ( g ) dx dx dx SUM AND PRODUCT OF OPERATORS Sum of two operators Product of two operators ( T + U)( f) = T( f) + U( f) ( TU )( f ) = T ( U ( f )) 3
5 We studied gradient, divergence and curl:,, x y z is common to all three definitions grad diva φ φ, φ, φ x y z A A x y A z + + x y z rot A eˆ ˆ ˆ x ey ez x y z A A A x y z NABLA,, x y z gradφ = φ diva = A rot A = A This operator is called NABLA 4
6 LAPLACE OPERATOR and something more The divergence of the gradient is called laplacian or Laplace operator 2 φ = φ is the laplacian of the scalar field φ. Sometimes written as: Δφ In cartesian coordinates: = =,,,, = x y z x y z x y z φ φ φ φ = x y z The nabla can be used to define new operators like: A or A Example: = = + + ( x, y, z),, x y z A A A A A A A x y z x y z B B B A B = A + A + A x y z so: ( ) x y z ( a ) r EXERCISE: calculate where a is constant Note that: ( A ) B A( B) EXERCISE: calculate a( r) 5
7 VECTOR IDENTITIES φ and ψ : scalar fields A and B : vector fields ( φψ ) ( φ ) ψ φ ( ψ ) = + ( ) ( ) ( ) ( ) φa = φ A+ φ A φa = φ A+ φ A ( A B) B ( A) A ( B) = ( A B) ( B ) A B( A) ( A ) B A( B) ( A B) ( B ) A ( A ) B B ( A) A ( B) = + = See next slides for the proof of some of these identities ( φ ) = 0 ( A ) 0 2 ( ) ( ) = A = A A ID1 ID2 ID3 ID4 ID5 ID6 ID7 ID8 ID9 6
8 Let s consider ID2: NABLARÄKNING x z z = ( φ A ),, ( φ A ) Nabla contains derivatives and we know that: This seems almost like a vector! Can we simply use the vector algebra rules? NO! So we have to remember that nabla must be applied to all the fields in the bracket. How? By adding dots to each field and rewriting the expression as a sum: ( φ A ) = ( φa) + ( φa) & & Then we must remember that nabla will be applied only to the field with the dot. Now we can rewrite the expression using vector algebra rules: ( φ A ) = ( φa) + ( φa) & & ( ) ( ) & & ( ) ( ) n ca + n ca = = A φ+ φ A n a c + n a c = & & a n c + n a c= a n c + cn a & & & & d( fg) df dg = g+ f ID1 dx dx dx ID2 rewriting the expression using vector algebra EXERCISE: prove that ( φa ) = ( φ) A+ φ A ID3 7
9 NABLARÄKNING To correctly perform the nabla calculation you need three steps. ( ) We want to calculate the following expression: φ, A, ψ, B, K Where can be: (gradient) or (divergence) or (curl) STEP 1 Rewrite the expression as a sum with N terms, where N is the number of (scalar or vector) fields in the expression. Every term in the sum must be identical to the original expression, but the i-th field in the i-th term must have a dot. Then, the nabla operator will be applied only to the field with the dot dot. ( φ, A, ψ, B, K) ( φ, A, ψ, B, K) ( φ, A, ψ, B, K) & & ( φ, A, ψ, B, ) ( φ, A, ψ, B, ) = + + K + K + K & & STEP 2 The nabla can now be formally considered as a vector. Each term must be rewritten using vector algebra rules in order that only the field with the dot dot will appear after the nabla. STEP 3 Finally, you can remove the dot. (But remember that the nabla is NOT a vector) 8
10 NABLARÄKNING: EXAMPLES Prove ID4: ( A B) = B ( A) A ( B) ( A B) = ( A B) + ( A B) = & & Now nabla can be treated as vector. since: We obtain: = B ( A) A ( B) = B rota A rotb & & n ( A B) = B ( n A) = A ( n B) ID4 Prove ID7: ( φ ) = 0 ID7 ( φ ) = ( φ ) = & = ( φ ) = 0 & ( A ) = ( A ) = & since: We obtain: ( ) ( ) 0 n nλ = λ n n = 2 Prove ID9: ( A ) = ( A) A ID9 since: n ( n c) = n( n c) c( n n) We obtain: ( ) ( ) ( ) 2 = A A= A A & & 9
11 TARGET PROBLEM r J θ z plasma B 1 0 ( ) grad p = rotb B μ 1 p B B 0 ( ) = μ 2 B 1 p + = ( B ) B 2 μ0 μ0 but a ( n b) = n( a b) b( a n) ( ) ( ) ( ) ( ) B B = B B = B B + B B = & & & & 1 2 ( ) = + 2 B B B ( ) ( ) ( ) 2 B = B B = B B + B B = 2 ( B B ) & & & particle pressure magnetic pressure Forces due to bending and parallel compression of the field In our case field lines are straight and parallel 2 B 2 p + = μ B p + constant 2 μ = 0 p B r 10
12 A BIT OF HISTORY Why the word nabla nabla? The theory of nabla operator was developed by Tait (a co-worker of Maxwell ). It was one of his most important achievements. But he was also a good musician in playing an old assyrian instrument similar to an harp. The name of this instrument in greek is nabla. The name nabla for the operator was suggested by James Clerk Maxwell to make a joke on Tait s hobby 11
13 WHICH STATEMENT IS WRONG? WHICH STATEMENT IS WRONG? 1- grad, div and rot can be expressed with the help of (yellow) 2- is a vector (red) 3- x( x( φ)=0 (green) 4- ( xα)=0 (blue) 12
14 INTEGRALSATSER 13
15 But how does the Arkimedes principle work? We can use the Arkimedes principle. What is the force that makes it floating? A body is floating in the water TARGET PROBLEM 14
16 TARGET PROBLEM df V ˆn S How to continue? Apply Guass theorem? A ds = divadv S V df = pnds ˆ where p [N/m 2 ] is the pressure F = df = ( pnds ˆ ) = pds S S But A is vector, while p is a scalar! z 15
17 In previous lessons we saw that: Q P φ dr = φ( Q ) φ( P ) A ds = A dr S L AdV = A ds V S (Stokes) (Gauss) (1) (2) (3) What do they have in common? They all express the integral of a derivative of a function in terms of the values of the function at the integration domain boundaries. In this sense, theorems (1), (2) and (3) are a generalization of: b a df dx = f ( b ) f ( a ) dx We can further generalize the Gauss theorem : ( ) ( ) ds K = dv K S V Generalized Gauss theorem where ( ) can be substituted with everything that gives a well defined meaning to both sides of the expression. EXERCISE: give three examples for the term ( ) 16
18 ( ) ( ) ds K = dv K S V ( K ) = A (A) If, we obtain the Gauss theorem (already proved) ( K ) = φ (B) If, we obtain: PROOF ds φ = dv φ S V (Gauss) ID2 eˆ φds = φeˆ ds = ( φeˆ) dv = i i i S S V (( ) ˆ ˆ ) ˆ ˆ = φ e + φ e dv = φ edv = e φdv i i i i V V V ( K ) = A (C) If, we obtain: PROOF ˆi e S V ( ) ds A = A dv Multiply by, use the Gauss theorem and then ID4 17
19 We can further generalize also the Stokes theorem : ( ) ( )( ) dr = ds K K Generalized Stokes theorem L S ( K ) = A (A) If, we obtain the Stokes theorem where ( ) can be substituted with everything that gives a well defined meaning to both sides of the expression. (already proved) ( K ) = φ (B) If, we obtain: PROOF ˆi e φ dr = ds grad φ L S Multiply by, use the Stokes theorem and then ID3 ( K ) = A (C) If, we obtain: dr A = ( ds ) A L S PROOF Multiply by e ˆi and use the Stokes theorem. 18
20 z TARGET PROBLEM df = pnds ˆ S V df ˆn where p [N/m 2 ] is the pressure F = df = ( pnds ˆ ) = pds S S A ds = divadv How to continue? Apply Gauss theorem? S V But A is vector, while p is a scalar! We apply the generalized Gauss theorem, with ( )=φ. F = pds = p dv 0 S V p = p ρ gz ( 0,0, ρ ) p= g φds = φdv S V Arkimedes principle F = ρge dv = ρgve V ˆ ˆ z z where ρ is the water density and g the gravitational acceleration 19
21 WHICH STATEMENT IS WRONG? WHICH STATEMENT IS WRONG? 1- Gauss and Stokes theorems imply that the integral of the derivative of a function can be expressed as the value of the function at the integration domain boundaries. (yellow) low) φdr L 2- is a vector (red) φds S 3- is a vector (green) 4- is a scalar (blue) S ds A 20
22 INDEXRÄKNING (suffix notation) AND CARTESIAN TENSORS 21
23 INDEXRÄKNING ( A B ) we used the nablaräkning = ( A B) + ( A B) = B rota A rotb & & To simplify this expression Can we use smarter methods? YES! (sometimes)! These are called suffix notation methods ( indexr indexräkning kning ) and come from the study of tensors. To understand this method, we start with a (brief) look at Cartesian tensors 22
24 Ohm s s law: j = σ E PHYSICAL EXAMPLE ELECTRICAL CONDUCTIVITY Current density If E = (0, E,0) Electrical conductivity then j = (0, σ E,0) y y Electric field But for many materials this is not true!! j = ( jx, jy, jz) Is the Ohm s law wrong? NO! σ is not a scalar σ is a cartesian tensor of rank 2 j x σxx σxy σ xz E x j = σe jy = σ yx σ yy σ yz Ey j σ σ σ E z zx zy zz z x If E = (0, E,0) y then j = ( σ E, σ E, σ E ) xy y yy y zy y z E j j j i = σ ike k y 23
25 1- Indices x, y, z can be substituted with 1, 2, 3 2- Coordinates x, y, z with x 1, x 2, x 3. Examples: A = A x 1 ( Ax, Ay, Az) = ( A1, A2, A3) eˆ ˆ x = e eˆ = eˆ y 1 2 eˆ ˆ z = e3 φ A = φ = φ = y y A x 2,2 1,2 c = a + b ci = ai + bi SUFFIX NOTATION x 1 x z 3 A 2 y A A 3 z A x 1 y x 2 in suffix notation this corresponds to the 3 equations obtained using i=1,2,3 The suffix i is called free suffix The choice of the free suffix is arbitrary: c = a + b j j j c = a + b m m m represent the same equation! But the same free suffix must be used for each term of the equation 24
26 SUFFIX NOTATION 3- Summation convention: a b ab a b a b ab = = i = 1,3 i i whenever a suffix is repeated in a single term in an equation, summation from 1 to 3 is implied. The repeated suffix is called dummy suffix. The choice of the dummy suffix is arbitrary: we can write also a b = akb k No suffix appears more than twice in any term of the expression: ( a b )( c d) = abc i i jd j i i a b ab = we cannot use i also here! But the ordering of terms is arbitrary: Example: abc k h k dummy suffix free suffix ( )( ) abc i i jd j= c jad i jb i= cka md kb m= a b c d acb ac = k k h = b k k h = ( a c ) b k free suffix EXERCISE. Write this expression using vectors: ab i ka nc ka i ( ) 2 = a b c a dummy suffixes 25
27 TENSORS The Ohm s law is: j = σ E But σ is not a scalar : j x σxx σxy σ xz E x j y = σ yx σ yy σ yz E y j σ σ σ E z zx zy zz z In suffix notation this can be written very concisely: j i = σ ike k σ is a cartesian tensor of rank 2 in the 3-D space. And it has 3 2 components the rank is the number of suffixes A tensor of rank M in the n-d space has n M components t ij is a tensor of rank 2 and can be regarded as a matrix if it is defined in the 2D space, then i,j=1,2 and it has 2 2 components in the 3D space, then i,j=1,2,3 and it has 3 2 components in the 4D space, then i,j=1,2,3,4 and it has 4 2 components t m is a tensor of rank 1 and can be regarded as a vector A tensor is Cartesian if the coordinate system is Cartesian 26
28 The Kronecker delta The Kronecker delta is a tensor of rank 2 defined as: δ ij It can be visualized 1 i = j as a nxn identity matrix 1 0 = (where n is the dimension 0 otherwise of the space) 0 1 Some properties of the Kronecker delta: δ ii = 3 3 δ = δ = δ + δ + δ = ii ii i = (in a 3D space) summation convention δ a = δ a = aδ + a δ + + a δ + = a δ km am = ak km m km m k k m km k m K K δ km ljm = ljk l δ = l δ = l δ + l δ + + l δ + = l jm km jm km j k j k jm km jk m summation convention all zeros, unless k=m 27
29 The alternating tensor (Levi Civita tensor or permutationssymbolen) ε The alternating tensor ε ijk (a tensor of rank 3) is defined as: 0 if any of i, j, k are equal = ˆ e ( ˆ e ˆ e ) = + 1 if ( i, j, k) = (1, 2, 3) or (2, 3,1) or (3,1, 2) 1 if ( i, j, k ) = (1,3,2) or (2,1,3) or (3,2,1) ijk i j k ( even permutation of 1,2,3) ( odd permutation of 1,2,3) a b =ε a b i The alternating tensor can be used to express the cross product: ( ) ijk j k PROOF: ( ) ˆ ( ) ˆ ( ˆ ) ( ˆ ) ˆ ( ˆ ˆ ) a b = e a b = e a e b e = e e e a b = ε a b i i i j j k k i j k j k ijk j k EXAMPLE FOR THE x COMPONENT (i=1): ( a b) = a2b3 a3 b = = + = ε ab ε ab ε ab ε ab ab ab 1 jk j k 1 jk j k j= 1 k= 1 Some properties: ijk jki kij ε = ε = ε εijk = ε jik (odd permutations change the sign) Very useful to simplify expressions ε ijk εklm = δilδ jm δim δ jl involving two cross products (even permutations does NOT change the sign) 28
30 GRADIENT, DIVERGENCE AND CURL IN SUFFIX NOTATION GRADIENT φ φ φ φ φ φ φ =,, =,, = φ, φ, φ x y z x x x (,1,2,3 ) φ = φ i So, the component i of the gradient is: ( ),i DIVERGENCE A A A A A A = + + = + + = = y A x z Aii, Aii, x y z x1 x2 x3 i So, the divergence is: A = A ii, CURL A A = = = x y z x x Az y 3 A2 ( A ) 2 3 A A = ε A + ε A = ε A 3,2 2, , ,3 1 jk k, j So, the component i of the curl is: ( ), A =ε A i ijk k j 29
31 CARTESIAN TENSORS (the definition) Assume that the matrix R defines a rotation in a Cartesian coordinate system x 3 x 3 In the new coordinate system the vector is: A ' = RA A A and in suffix notation is: A i = RikA k A Cartesian tensor T of order M (or rank M) is: a quantity in the 3D Euclidean space that has M indices and 3 M components Ti, j, K, o i, j,... o= 1,2,3 M indexes x 1 x 1 and which under a rotation of coordinates R ij transforms as: T = R R... R T i, j, K, o i, p j, q o, w p, q, K, w P x 2 x 2 30
32 Nablaräkning and Indexräkning use of tensors in the calculation of nabla expressions Calculate: ( a r) where r = ( x, y, z ) and a is constant 1- Nablaräkning n ( a b) = a ( b n) ( a r) = ( a r) + ( a r) = 0+ a ( r ) = a ( r) = 0 & & & a is a constant =0 2- Indexräkning ( ) ( ) ( ) a r = ε ar = ε a r + ar = ε ar = i ikl k l, ikl k, i l k l, i ikl k l, i 0 r only if l = i li, 0 If l = i then ε = ijk 0 31
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