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The Maple computer algebra environment In the presentation that - PDF document

Dec 09, 2023 •156 likes •1.72k views

The Maple computer algebra environment In the presentation that follows we use the Maple worksheet mode, where input lines are identified by a prompt: > We communicate with the computer entering our computation typing in

  1. x , y , z R x , y , z d x d y d z (38) (38) > > Bra psi . P R . Ket psi 2 d x d y d z x , y , z (39) (39) This is a bracket rule for the scalar product of a state-vector of the discrete basis u and . > > %Bracket Bra u , n , Ket psi = psi n u n = n (40) (40) > > Setup % bracketrules = u v = f u , v , R x , y , z = x , y , z , u n = n (41) (41) This rule for u n permits projecting onto the u basis, which is equivalent to inserting a projector between and . Note the use of delay evaluation quotes surrounding the Bracket; the operation is performed in the next line by using %. > > ' Bracket psi, P u , psi ' P u (42) (42) > > % N 1 2 n (43) (43) n = 0 A Ket can have different types of spaces associated with its quantum numbers. In the following example, Kets from a basis B have four quantum numbers, two of which, B 3 and B 4 , are associated with continuous spaces, and the dimension of the space associated with each quantum number is different. > > Setup quantumcontinuousbasis = B 3 , B 4 , quantumbasisdimension = B 1 = 1 /2 ..1 /2, B 2 = 0 .. N , B 3 = a .. a , B 4 = infinity ..infinity 1 2 .. 1 quantumbasisdimension = R = , u = N , B 1 = 2 , B 2 = 0 .. N , B 3 = a .. a , B 4 (44) (44) = .. , quantumcontinuousbasis = R , B 3 , B 4 This is the projector onto the basis B : > > Projector Ket B , n , m , x , y (45) (45)

  2. a N 1 B B d x d y (45) (45) 1 1 m = 0 n = 0 n 2 , m , x , y n 2 , m , x , y a Quantum operators, eigenvectors, eigenvalues and commutators To indicate to the system that a letter represents a quantum operator, use the Setup command; this sets B as a quantum operator. > > Setup op = B * Partial match of 'op' against keyword 'quantumoperators' quantumoperators = B (46) (46) Note that after having entered Setup mathematicalnotation = true the extended typesetting displays noncommutative objects in different colors; to change this color, see ?Setup. Because B is now a quantum operator, B m , n , x , y is an eigenvector of the four operators B 1 , B 2 , B 3 , and B 4 , with eigenvalues m , n , x , and y , respectively. For example, > > B 2 . Ket B , m , n , x , y n B m , n , x , y (47) (47) Quantum operators can also be 3-D Euclidean vectors; for that purpose, you must load the Physics[Vectors] subpackage. > > with Vectors &x , ` ` , `.` , ChangeBasis , ChangeCoordinates , Component , Curl , DirectionalDiff , (48) (48) Divergence , Gradient , Identify , Laplacian , , Norm , Setup , diff Set the vectors L , r , and p as quantum operators (note the use of the option redo to erase previous definitions of quantum operators). > > Setup op = L_ , r_ , p_ , x , y , z , p x , p y , p z , L x , L y , L z , redo * Partial match of 'op' against keyword 'quantumoperators' quantumoperators = L , L x , L y , L z , p , p x , p y , p z , r , x , y , z (49) (49) Define L as the angular momentum operator L = r p , and set commutation rules for the components of r and p . > > L_ r_ &x p_ L r p (50) (50)

  3. > > r_ x _i y _j z _k r i x j y k z (51) (51) > > p_ p x _i p y _j p z _k ; p i p x j p y k p z (52) (52) Vector calculus with noncommutative components preserves A B = B A and A B = B A . This is achieved by symmetrizing and antisymmetrizing, respectively, each of these products. For example, below is the Component of L along the x- axis. > > _i . L_ y p z z p y (53) (53) To enter the commutation rules between each component of r and p with each other, you can write these commutators and pass the whole set to Setup . When there are many, as in this case, it is more convenient to use a Matrix and an indexing function. Enter the core information as a procedure: C represents the Commutator of the Components of the vectors a and b . > > C a_ , i , b_ , j %Commutator Component a_ , i , Component b_ , j C a , i , b , j a i , b j (54) (54) So, given i and j from 1 to 3 identifying the components of r and p , an algebra can be set as set as follows. > > algebra i , j C r_ , i , p_ , j = I * KroneckerDelta i , j , C r_ , i , r_ , j = 0, C p_ , i , p_ , j = 0 algebra i , j C r , i , p , j = I i , j , C r , i , r , j = 0, C p , i , p , j = 0 (55) (55) Now all of the commutators between each component of r and p can be constructed with one call to Matrix. > > Matrix 3, 3, algebra x , p x = I, x , x = 0, p x , p x = 0, x , p y = 0, x , y = 0, p x , p y = 0, (56) (56) x , p z = 0, x , z = 0, p x , p z = 0 , , y , p x = 0, y , x = 0, p y , p x = 0, y , p y = I, y , y = 0, p y , p y = 0, y , p z = 0, y , z = 0, p y , p z = 0 , , z , p x = 0, z , x = 0, p z , p x = 0, z , p y = 0, z , y = 0, p z , p y = 0, z , p z = I, z , z = 0, p z , p z = 0 ,

  4. Pass this Matrix to Setup to set the algebra rules. > > Setup % algebrarules = p x , p y = 0, p x , p z = 0, p y , p z = 0, x , p x = I, x , p y (57) (57) = 0, x , p z = 0, x , y = 0, x , z = 0, y , p x = 0, y , p y = I, y , p z = 0, y , z = 0, z , p x = 0, z , p y = 0, z , p z = I , Set, for instance, the values of Lx , Ly , and Lz , the components of L . > > L x _i . L_ L x y p z z p y (58) (58) > > L y _j . L_ L y z p x x p z (59) (59) > > L z _k . L_ L z x p y y p x (60) (60) Verify the commutator algebra for these components of L . > > Commutator L x , L y = I L z I y p x I x p y = I x p y y p x (61) (61) > > Commutator L z , L x = I L y I x p z I z p x = I z p x x p z (62) (62) > > Commutator L y , L z = I L x I z p y I y p z = I y p z z p y (63) (63) The three equations above are identically true. Other operators frequently used in different contexts are the Annihilation and Creation operators: they augment or diminish the value of a quantum number by one. These operators are suitable, for instance, for working with multi-particle vector states; in that context the quantum numbers are called occupation numbers. This constructs a pair of annihilation/creation operators acting on the basis A involving only one quantum number. > > am Annihilation A am a (64) (64)

  5. > > ap Creation A ap a (65) (65) Annihilation and Creation operators act on Kets belonging to discrete bases and assume that the "lower" state happens when the quantum number is equal to zero (frequently called "vacuum": a ket with occupation number equal to zero represents a state with "no particles"). > > am . Ket A , n n A n (66) (66) 1 > > am . % n n 1 A n (67) (67) 2 > > ap . Ket A , n n 1 A n (68) (68) 1 > > ap . % n 1 n 2 A n (69) (69) 2 The Commutator of the operators a- and a are automatically set when these operators are constructed, and satisfy (note the use of the inert form %Commutator ): > > %Commutator am , ap a , a (70) (70) > > value % 1 (71) (71) To indicate that the Kets of a basis are fermionic, use an anticommutative variable to label the basis. To set the prefix identifier of anticommutative variables use the Setup command. > > Setup anticommutativeprefix = Theta anticommutativeprefix = , _ (72) (72) > > type Theta, anticommutative true (73) (73) > > Ket Theta (74) (74) Construct Annihilation and Creation operators acting on this basis; use the option notation = explicit so that the basis and the quantum numbers onto which these operators act are explicit. > > Am Annihilation Theta, notation = explicit Am a (75) (75) 1

  6. > > Ap Creation Theta, notation = explicit Ap a (76) (76) 1 The AntiCommutator of these operators satisfy > > %AntiCommutator Am , Ap a , a (77) (77) 1 1 > > value % 1 (78) (78) According to Pauli's exclusion principle, only one fermionic particle can be in a given state, so starting from the vacuum, > > Ket Theta, 0 (79) (79) 0 > > Ap . % (80) (80) 1 > > Ap . % 0 (81) (81) > > Am . %% (82) (82) 0 And as is always the case, the annihilation operator acting on the vacuum returns zero > > Am . % 0 (83) (83) * Unitary Operators in Quantum Mechanics * Eigenvalues of an unitary operator and exponential of Hermitian operators • • Show that the eigenvalues of an unitary operator are all on the unit circle, their modulus is 1. Show that an operator e i H is unitary provided that H is Hermitian ( H = H † ) • • and is any real parameter. Solution > > restart ; with Physics : interface imaginaryunit = i :

  7. > > Setup unitaryoperators = U unitaryoperators = U (84) (84) If U is a normalized eigenvector of U with eigenvalue > > U Ket U , = U Ket U , U U = U (85) (85) > > Dagger (85) U † = U U (86) (86) So, to show that the eigenvalues have modulus equal to 1, multiplying sides by sides > > (86) . (85) 2 1 = (87) (87) To show that, when H is Hermitian, then V = e i H is unitary, > > Setup quantumoperators = V , hermitianoperators = H , realobjects = hermitianoperators = H , quantumoperators = H , U , V , realobjects = (88) (88) > > V = exp i H V = e i H (89) (89) > > Dagger (89) V † = e i H (90) (90) Again multiply sides by sides > > (89) . (90) V V † = 1 (91) (91) > > (90) . (89) V † V = 1 (92) (92) Therefore, V is unitary > > * Properties of unitary operators Consider two set of kets a n and b n , each of them constituting a complete orthonormal basis of the same space .

  8. *Verify that U = b k a k , maps one basis into the other, i.e.: k = 0 b n = U a n > > restart ; with Physics : Tell the system that a n and b n , are complete orthonormal basis > > Setup quantumoperators = U , bracketrules = %Bracket Bra a , m , Ket a , n = KroneckerDelta m , n , %Bracket Bra b , m , Ket b , n = KroneckerDelta m , n bracketrules = a m a n = m , n , b m b n = m , n , quantumoperators = U (93) (93) > > U = Ket b , k Bra a , k k = 0 U = b k a k (94) (94) k = 0 Apply this operatorial equation to a m > > ' % . Ket a , m ' U = b k a k a m (95) (95) k = 0 > > % U a m = b m (96) (96) > > *Show that U = b k a k is unitary k = 0 Recalling the expansion of the operator U > > (94) U = b k a k (97) (97) k = 0 > > Dagger (94) U † = a k b k (98) (98) k = 0

  9. Again multiply sides by sides > > ' (98) . (94) ' U † = a k b k U = b k a k (99) (99) k = 0 k = 0 > > % U † U = a k1 a k1 (100) (100) k1 = 0 > > (94) . (98) U U † = b k1 b k1 (101) (101) k1 = 0 and since a n and b n form two complete basis of the same space , the right-hand sides are equal to the identity operator , and so U is unitary. > > *Show that the matrix elements of U in the a n and b n basis are equal Recalling the expansion of the operator U > > (94) U = b k a k (102) (102) k = 0 Compute now the matrix elements of U in the a n and b n basis > > ' Bra a , n . (94) . Ket a , m ' a n U = b k a k a m (103) (103) k = 0 > > % a n U a m = a n b m (104) (104) Likewise > > ' Bra b , n . (94) . Ket b , m ' b n U = b k a k b m (105) (105) k = 0 > > % (106) (106)

  10. b n U b m = a n b m (106) (106) > > Schrödinger equation and unitary transform Consider a ket t that solves the time-dependant Schrödinger equation: i t = H t t t and consider t = U t t , where U t is a unitary operator. Does t evolves according a Schrödinger equation i t t = t t and if yes, which is the expression of t ? Solution > > restart ; with Physics : interface imaginaryunit = i : > > Setup automaticsimplification = true , mathematicalnotation = true , quantumoperators = , hermitianoperators = H , unitaryoperators = U , realobjects = t, automaticsimplification = true , hermitianoperators = H , mathematicalnotation (107) (107) = true , quantumoperators = , H , U , realobjects = , t , unitaryoperators = U > > CompactDisplay U , H , t U t will now be displayed as U H t will now be displayed as H t will now be displayed as (108) (108) > > Ket , t = U t Ket , t (109) (109)

  11. = U (109) (109) t t Compute now the evolution of t > > i diff (109) , t i t = i U t U (110) (110) t t t t Simplify this equation taking into account Schrödinger's equation for : > > i t Ket , t = H t Ket , t i t = H (111) (111) t t > > simplify (110) , (111) , t Ket , t i t = i U t U H (112) (112) t t t Now, from > > (109) = U (113) (113) t t U t * > > rhs = lhs (109) U † U = U † (114) (114) t t > > simplify (114) = U † (115) (115) t t Inserting this result in (112) > > subs (115) , (112) t = i U t U † U H U † i (116) (116) t t t the amiltonian for t is given by the coefficient of on the right-hand side t > > t = Coefficients rhs (116) , Ket , t = i U t U † U H U † (117) (117) So satisfies a Schrodinger equation and as one can expect, is Hermitian t > > Dagger (117) (117) (118) (118)

  12. i U U † U † U H i U t U † U H U † † = (118) (118) t Recalling that U t satisfies U t * = U t U t * > > U t U U † = 1 (119) (119) > > diff (119) , t U t U † U U † t = 0 (120) (120) > > subs (120) , (118) i U U † U † U H i U t U † U H U † † = (121) (121) t In the time independent case, i.e. U t = U , reduced to: U = U (122) (122) > > subs U t = U , (117) = i U t U † U H U † (123) (123) > > % = U H U † (124) (124) > > Translation operators using Dirac notation i a P In this section, we focus on the operator T a = e Settings > > restart ; with Physics : interface imaginaryunit = i : > > Setup realobjects = a , x , , m , x 1 , x 2 , unitaryoperators = T , hermitianoperators = , X , P , quantumcontinuousbasis = X , P hermitianoperators = , P , X , quantumcontinuousbasis = P , X , realobjects (125) (125) = , a , m , x , x 1 , x 2 , unitaryoperators = T > > Setup bracketrules = Bracket Bra P , p , Ket = p , Bracket Bra X , x ,

  13. 1 2 exp i Ket = x , Bracket Bra X , x , Ket P , p = 2 x p i x p 2 e bracketrules = P p = p , X x = x , X x P p = (126) (126) 2 > > Assume 0 :: 0, (127) (127) Useful closure relations > > 1 = Projector Ket X , x 1 = X x X x d x (128) (128) To have equivalent projectors with different integration variables, we use as the identity 1 = , 1 = operator: 2 : > > 1 = Projector Ket P , p , 2 = Projector Ket P , q 1 = P p P p d p , 2 = P q P q d q (129) (129) > > i a P The Action (translation) of the operator T a = e on a ket Considering a general ket , introduce a closure relation > > Ket = 1 Ket = 1 (130) (130) > > subs (129) , % = P p P p d p (131) (131) > > Bra X , x . % (132) (132)

  14. i x p 1 2 e p x = d p (132) (132) 2 Which gives after a variable change x = y a > > PDEtools :- dchange x = y a , % , y , known = : subs y = x , % i x a p 1 2 e p x a = d p (133) (133) 2 i a P Let's now evaluate the action of e on in the X , x basis > > (131) = P p P p d p (134) (134) i a P > > Bra X , x . e . % i p a x 1 2 e p i a P X x e = d p (135) (135) 2 Comparing the above with (133) > > % (133) i p a x 1 2 e p i a P X x e x a = d p (136) (136) 2 i x a p 1 2 e p d p 2 > > simplify (136) i a P X x e x a = 0 (137) (137) > > isolate % , x a

  15. i a P x a = X x e (138) (138) > > Action of T a on an operator V X Let's consider an operator V X , that can be written as a formal power series: v n x n . V x = n = 0 Its matrix elements are: > > %Bracket = Bracket Bra X , x 1 , V X , Ket X , x 2 X x1 V X X x2 = V x 2 x 2 x 1 (139) (139) Using the closure relation > > (128) 1 = X x X x d x (140) (140) V X can also be represented in the X , x basis as > > V X (128) V X = V x X x X x d x (141) (141) Let's now introduce two closure relations to evaluate V X in the momentum basis P > > Ket X , x = 1 Ket X , x , Bra X , x = Bra X , x 2 X x = 1 X x , X x = X x 2 (142) (142) > > subs % , (141) V X = V x 1 X x X x 2 d x (143) (143) Recalling > > (129) 1 = P p P p d p , 2 = P q P q d q (144) (144) > > subs % , %% V X = V x P p P p d p X x X x P q P q d q d x (145) (145)

  16. > combine (145) > V X = V x P p P p X x X x P q P q d q d p d x (146) (146) > > eval % , `*` = `.` i x p q V x e P p P q V X = d q d p d x (147) (147) 2 Apply now the translation operator T a i > > T a = exp a P i a P T a = e (148) (148) % . (147) . % * > > i p q a x V x P p P q e † = T a V X T a d q d p d x (149) (149) 2 Making a variable change x = y a > > PDEtools :- dchange x = y a , % , y , known = V : subs y = x , % i x p q V x a P p P q e † = T a V X T a d q d p d x (150) (150) 2 Evaluate the matrix element of this result and compute the integral > > Bra X , x 1 % Ket X , x 2 i x x1 p i x x2 q V x a e † X x2 X x1 T a V X T a = (151) (151) 2 2 4 d q d p d x > > value % † X x2 X x1 T a V X T a = V x 1 a x 1 x 2 (152) (152) > >

  17. * The quantum operator components of L satisfy L j , L k = i j, k , m L m > > restart ; with Physics : interface imaginaryunit = i : > > Setup spaceindices = lowercaselatin , automaticsimplification = true ; automaticsimplification = true , spaceindices = lowercaselatin (153) (153) Define L , r and p as tensors of the 3-D Euclidean space embedded in > > Define L , r , p Defined objects with tensor properties L , p , r , , , , g , , a , b , , , (154) (154) , , , Now set L, p, r as quantum operators and the related Commutator rules for the algebra in tensor notation > > Setup quantumoperators = L , p , r , %Commutator p j , p k = 0, %Commutator r j , p k = i KroneckerDelta j , k , %Commutator r j , r k = 0 algebrarules = p j , p k = 0, r j , p k = i j , k , r j , r k = 0 , quantumoperators (155) (155) = L , p , r , The definition of L j > > L j = LeviCivita j , k , m r k p m k , m L j = r k p m j (156) (156) The rule to be verified: > > %Commutator L j , L k = i LeviCivita j , k , m L m = i j , k , m L m L j , L k (157) (157) Substitute now the operator L i by its tensor form in terms r k and p m in the commutator above > > Library :- SubstituteTensor (156) , (157) a , m , r b p c k b , c a , b , m r a p m j = i j , k , m r a p b (158) (158) Simplify, all in one go, we expect an identity > > Simplify (158)

  18. i r j p k r k p j = i r j p k r k p j (159) (159) The same one step at a time, first expand the commutator on the left-hand side > > expand (158) b , c r a p m r b p c a , m k a , b , m r b p c r a p m = i j , k , m r a p b (160) (160) j > > Simplify (160) i r j p k r k p j = i r j p k r k p j (161) (161) > > * Quantum Commutation Rules Basics, F X , p = i F X Derive the commutation rules, in the coordinates representation, between an arbitrary function of the coordinates and the related momentum, departing from the differential representation p n = i n This exercise illustrate how to have full control of the computational process by using different elements of the Maple language, including inert representations of abstract vectorial differential operators, Hermitian operators, algebra rules, etc. It also illustrate the ability to perform computations algebraically, using the product operator, but with differential operators , and transform the products into the application of the operators only when we want that, as we do with paper and pencil. Solution > > restart : with Physics : with Physics Vectors : interface imaginaryunit = i : Set the problem: all of x , y , z , p x , p y , p z are Hermitian operators all of x , y , z commute between each other tell the system only that the operators x , y , z are the differentiation variables of the corresponding (differential) operators p x , p y , p z but do not tell what is the form of the operators > > Setup differentialoperators = p_ , x , y , z , hermitianoperators = p , x , y , z , algebrarules = %Commutator x , y = 0, %Commutator x , z = 0, %Commutator y , z

  19. = 0 , quiet algebrarules = x , y = 0, x , z = 0, y , z = 0 , differentialoperators = p , (162) (162) x , y , z , hermitianoperators = p , x , y , z , Assuming F X is a smooth function, the idea is to apply the commutator F X , p to an arbitrary ket of the Hilbert space , perform the operation explicitly after setting a x , y , z differential operator representation for p , and from there get the commutation rule between F X and p . Start introducing the commutator, to proceed with full control of the operations we use the inert form %Commutator > > alias X = x , y , z : > > CompactDisplay F X F X will now be displayed as F (163) (163) > > %Commutator F X , p_ Ket , X F , p (164) (164) x , y , z This product of a commutator, involving a differential operator and and a function of the coordinates, and a Ket that depends on the coodinates, can be transformed into the (traditional in computer algebra) application of the differential operator when desired, as follows: > > (164) = Library :- ApplyProductsOfDifferentialOperators (164) F , p = F p p F (165) (165) x , y , z x , y , z x , y , z Note that, in p F X , the application of p is not expanded: at this point nothing is x , y , z known about p , it is not necessarily a linear operator. In the Quantum Mechanics problem at hands, however, it is. So give now the operator p an explicit representation as a linear vectorial differential operator (we use the inert form %Nabla, , to be able to proceed with full control one step at a time) > > p_ f i %Nabla f p f i f (166) (166) The expression (165) becomes > > (165) F , p = i F i F (167) (167) x , y , z x , y , z x , y , z Activate now the inert operator and simplify taking into account the algebra rules for the

  20. coordinate operators x , y = 0, x , z = 0, y , z = 0 > > value (167) F , p = i F x i y j z k i F x (168) (168) x , y , z x , y , z x , y , z x , y , z , F x i F y F y j F z x , y , z x , y , z x , y , z x , y , z x , y , z F z k x , y , z > > Simplify (168) F , p = i i F x i j F y i k F z (169) (169) x , y , z x , y , z x , y , z x , y , z To make explicit the gradient in disguise on the right-hand side, factor out the arbitrary ket x , y , z > > Factor (169) F , p = i F x i F y j F z k (170) (170) x , y , z x , y , z Combine now the expanded gradient into its inert (not-expanded) form > > Gradient = %Gradient F X F x i F y j F z k = F (171) (171) > > subs (171) , (170) F , p = i F (172) (172) x , y , z x , y , z Since (172) is true for all , this ket can be removed from both sides of the equation. x , y , z One can do that either taking coefficients (see Coefficients ) or multiplying by the "formal inverse" of this ket, arriving at the (expected) form of the commutation rule between F X and p > > (172) Inverse Ket , x , y , z F , p = i F (173) (173) > > > > Schrödinger vs Heisenberg picture Within the Schrödinger picture of Quantum Mechanics, the time evolution of the state of a system, represented by a Ket t , is determined by Schrödinger's equation:

  21. i d = H t t d t where H , the Hamiltonian, as well as the quantum operators O S representing observable quantities, are all time-independent. Within the Heisenberg picture, a Ket representing the state of the system does not evolve with time , but the operators O H t representing observable quantities, and through them the Hamiltonian H , do. Problem: Departing from Schrödinger's equation, a) Show that the expected value of a physical observable in Schrödinger's and Heisenberg's representations is the same, i.e. that t O S = O H t t b) Show that the evolution equation of an observable O H in Heisenberg's picture, equivalent to Schrödinger's equation, is given by: i O H t , H . O H t = where in the right-hand-side we see the commutator of O H with the Hamiltonian of the system. > > Solution Let O S and O H respectively be operators representing one and the same observable quantity in Schrödinger's and Heisenberg's pictures, and H be the operator representing the Hamiltonian of a physical system. All of these operators are Hermitian. So we start by setting up the framework for this problem accordingly, including that the time t and Planck's constant are real. To automatically combine powers of the same base (happening frequently in what follows) we also set combinepowersofsamebase = true. > > with Physics : interface imaginaryunit = i : > > Setup hermitianoperators = H , O H , O S , realobjects = t , , combinepowersofsamebase = true , mathematicalnotation = true combinepowersofsamebase = true , hermitianoperators = H , O H , O S , p , x , y , z , (174) (174) mathematicalnotation = true , realobjects = , i , j , k , , r , , , , r , , t , , x , y ,

  22. z Let's consider Schrödinger's equation > > i diff Ket , t , t = H Ket , t i t = H (175) (175) t t Now, H is time-independent, so (175) can be formally solved: t is obtained from the solution 0 at time t = 0, as follows: i H t > > T exp i t H T e (176) (176) > > Ket , t = T Ket , 0 i t H = e (177) (177) t 0 To check that (177) is a solution of (175) , substitute it in (175) : > > eval (175) , (177) i t H i t H H e = H e (178) (178) 0 0 Next, to relate the Schrödinger and Heisenberg representations of an Hermitian operator O representing an observable physical quantity, recall that the value expected for this quantity at time t during a measurement is given by the mean value of the corresponding operator (i.e., bracketing it with the state of the system ). t So let O S be an observable in the Schrödinger picture: its mean value is obtained by bracketing the operator with equation (177) : > > Dagger (177) O S (177) i t H i t H t O S = 0 e O S e (179) (179) t 0 The composed operator within the bracket on the right-hand-side is the operator O in Heisenberg's picture, O H t : > > Dagger T O S T = O H t i t H i t H e O S e = O H t (180) (180) Analogously, inverting this equation,

  23. > T (180) Dagger T > i t H i t H O S = e O H t e (181) (181) As an aside to the problem, we note from these two equations, and since the operator i t H T = e is unitary (because H is Hermitian), that the switch between Schrödinger's and Heisenberg's pictures is accomplished through a unitary transformation. Inserting now this value of O S from (181) in the right-hand-side of (179) , we get the answer to item a) > > lhs (179) = eval rhs (179) , (181) t O S = 0 O H t (182) (182) t 0 where, on the left-hand-side, the Ket representing the state of the system is evolving with time (Schrödinger's picture), while on the the right-hand-side the Ket 0 is constant and it is O H t , the operator representing an observable physical quantity, that evolves with time (Heisenberg picture). As expected, both pictures result in the same expected value for the physical quantity represented by O. To complete item b), the derivation of the evolution equation for O H t , we take the time derivative of the equation (180) : > > diff rhs = lhs (180) , t i t H i t H i t H i t H i H e O S e i e O S H e O Ht = (183) (183) To rewrite this equation in terms of the commutator O S , H , it suffices to re-order the i t H product H e in the first term, placing the exponential first: i t H > > Library :- SortProducts (183) , e , H , usecommutator i t H i t H i t H i t H i e H O S O S , H e i e H O S e O Ht = (184) (184) > > Normal (184) (185) (185)

  24. i t H i t H i e O S , H e O Ht = (185) (185) Finally, to express the right-hand-side in terms of O H t , H instead of O S , H , we take the commutator of the equation (181) with the Hamiltonian > > (181) i t H i t H O S = e O H t e (186) (186) > > Commutator (181) , H i t H i t H O S , H = e O H t , H e (187) (187) Combining these two expressions, we arrive at the expected result for b) , the evolution equation of a given observable O H in Heisenberg's picture > > eval (185) , (187) i O H t , H O Ht = (188) (188) > > Quantization of the energy of a particle in a magnetic field Show that the energy of a particle of charge q and mass m in a constant magnetic field B oriented along the z axis can be written as 1 c a † a H = 2 c = q B where a † and a are creation and annihilation operators and m > > Solution The classical Hamiltonian is given by

  25. 2 q A p c H = 2 m The underlying quantum mechanics algebra rules are r i , p j = i , j , r i , r = 0, p i , p j = 0 j > > restart ; with Physics : with Vectors : interface imaginaryunit = i : > > Setup hermitianoperators = A , H , , , p , p , x , y , z , quantumoperators = a , realobjects = , B , c , m , q , : c > > Setup %Commutator x , p x = i , %Commutator x , p y = 0, %Commutator y , x = 0, %Commutator y , p x = 0, %Commutator y , p y = i , %Commutator p y , p x = 0 algebrarules = x , p x = i , x , p y = 0, y , x = 0, y , p x = 0, y , p y (189) (189) = i , p y , p x = 0 , Using q > > = p_ A_ x , y c q A x , y = p (190) (190) c The Hamiltonian can be written as 2 > > H = 2 m 2 H = (191) (191) 2 m In Coulomb's gauge, the following vector potential gives the magnetic field of the problem, B = B k B y B x > > A_ x , y = _i _j ; 2 2 1 1 A x , y = 2 B i y 2 B j x (192) (192)

  26. > > CompactDisplay A_ x , y A x , y will now be displayed as A (193) (193) Indeed we have > > Divergence (192) A = 0 (194) (194) > > Curl (192) A = B k (195) (195) Derive now the commutation rule for x , y > > = x _i y _j ; = i j (196) (196) x y > > p = p x _i p y _j p = i p x j p y (197) (197) > > (190) q A = p (198) (198) c > > subs (192) , (196) , (197) , (190) 1 1 q 2 B i y 2 B j x i j y = i p x j p y (199) (199) x c > > Component (199) , 1 q B y x = p x (200) (200) 2 c > > Component (199) , 2 q B x y = p y (201) (201) 2 c > > Commutator (200) , (201) = i q B x , (202) (202) y c > > Setup (202) algebrarules = x , p x = i , x , p y = 0, y , x = 0, y , p x = 0, y , p y (203) (203)

  27. = i q B = i , x , , p y , p x = 0 , y c Time to bring in annihilation and creation operators c > > a = i x y 2 q B 2 c i x y a = (204) (204) 2 q B (204) * > > 2 c i a † = x y (205) (205) 2 q B Verify the normalization of this definition > > Commutator (204) , (205) a , a † = 1 (206) (206) > > Setup (206) a , a † algebrarules = = 1, x , p x = i , x , p y = 0, y , x = 0, y , p x = 0, (207) (207) = i q B y , p y = i , x , , p y , p x = 0 , y c To express the Hamiltonian in terms of a , a † > > (191) 2 H = (208) (208) 2 m > > subs (196) , % 2 i j x y H = (209) (209) 2 m > > (204) , (205) 2 c i 2 c i , a † = x y x y a = (210) (210) 2 q B 2 q B > > solve % , x , y (211) (211)

  28. i q B a † 2 a 2 q B a † a 2 x = , y = (211) (211) 2 c c > > subs (211) , (209) 2 q B a † q B a † i a 2 i j a 2 2 c 2 c H = (212) (212) 2 m > > simplify expand (212) 2 a a † q B 1 H = (213) (213) 2 m c > > Library :- SortProducts (213) , Dagger a , a , usecommutator 2 a † a q B 1 H = (214) (214) 2 m c c = q B This is the Hamiltonian of an harmonic oscillator with frequency m . The possible 1 values for the energy are known: E = c n , where n is a positive integer. 2 > > Quantization of the Lorentz Force We consider the case of a quantum, non-relativistic, particle with mass m and charge q evolving under the action of an arbitrary time-independent magnetic field B = A , where A is the vector potential. The Hamiltonian for this system is 2 p q A H = 2 m where p is the momentum of the particle, and the force acting in this particle, also called the Lorentz force, is given by . F = m v where v is the quantized velocity of the particle, and all of H , p , v , B , A and F are Hermitian quantum operators representing observable quantities.

  29. In the classic (non-quantum) case , the Lorentz force F for such a particle in the absence of electrical field is given by F = q v B , Problem: Departing from the Hamiltonian, show that in the quantum case the Lorentz force is given by F = q v B B v 2 [1] Photons et atomes, Introduction à l'électrodynamique quantique, p. 179, Claude Cohen- Tannoudji, Jacques Dupont-Roc et Gilbert Grynberg - EDP Sciences janvier 1987. > > Solution We choose to tackle the problem in Heisenberg's picture of quantum mechanices , where the state of a system is static and only the quantum operators evolve in time according to . t = i O H , O t Also, the algebraic manipulations are simpler using tensor abstract notation instead of the standard 3D vector notation. We then start setting the framework for the problem, a system of coordinates X, indicating the dimension of the tensor space to be 3 and the metric Euclidean, and that we will use lowercaselatin letters to represent tensor indices. In addition, not necessary but for convenience, we set the lowercase latin i to represent the imaginary unit and we request automaticsimplification so that the output of everything comes automatically simplified in size. > > restart ; with Physics : interface imaginaryunit = i : > > Setup mathematicalnotation = true , automaticsimplification = true , coordinates = X , dimension = 3, metric = Euclidean , spacetimeindices = lowercaselatin , quiet automaticsimplification = true , coordinatesystems = X , dimension = 3, (215) (215) mathematicalnotation = true , metric = 1, 1 = 1, 2, 2 = 1, 3, 3 = 1 , spacetimeindices = lowercaselatin

  30. Next we indicate the letters we will use to represent the quantum operators with which we will work, and also the standard commutation rules between position and momentum, always the starting point when dealing with quantum mechanics problems > > Setup quantumoperators = F , hermitianoperators = r , x , p , v , A , B , H , realobjects = , m, q , algebrarules = %Commutator x k , x l = 0, %Commutator p k , p n = 0, %Commutator x k , p l = i KroneckerDelta k , l algebrarules = p k , p n = 0, x k , p l = i k , l , x k , x l = 0 , (216) (216) hermitianoperators = A , B , H , p , r , v , x , quantumoperators = A , B , F , H , p , r , v , x , realobjects = , m , q , x1 , x2 , x3 , Note that we start not indicating F as Hermitian, in order to arrive at that result. The quantum operators A , B , and F are explicit functions of X, so to avoid redundant display of this functionality on the screen we use > > CompactDisplay A , B , F X A X will now be displayed as A B X will now be displayed as B F X will now be displayed as F (217) (217) Define now as tensors the quantum operators that we will use with tensorial notation (recalling: for these, Einstein's sum rule for repeated indices will be automatically applied when simplifying) > > Define x , p , v , A , B , F , quiet A , B , F , p , v , x , a , a , X a , a , g a , b , a , b , (218) (218) a , b , c The Hamiltonian, 2 p q A H = 2 m in tensorial notation, is given by 1 2 > > H = p n q A n X 2 m 2 p n q A n H = (219) (219) 2 m

  31. Generally speaking to arrive at F = q v B B v what we now need to do is 2 1) Express this Hamiltonian (219) in terms of the velocity And, recalling that, in Heisenberg's picture, quantum operators evolve in time according to . t = i O H , O t 2) Take the commutator of H with the velocity itself to obtain its time derivative and, from . , that commutator is already the force up to some constant factors. F = m v To get in contact with the basic commutation rules between position and momentum behind quantum phenomena, the quantized velocity itself can be computed as the time derivative of the position operator, i.e as the commutator of x k with H i > > Commutator (219) , x k i q 2 A n , A n , x k i q p n , A n , x k 2 q A n p n k , n i H , x k = (220) (220) 2 m This expression for the velocity, that involves commutators between the potential A n , the position x k and the momentum p n , can be simplified taking into account the basic quantum algebra rules between position and momentum. We assume that A n (X) can be decomposed into a formal power series (possibly infinite) of the x k , hence all the A n commute between themselves as well as with all the x k : > > %Commutator A k X , x l = 0, %Commutator A k X , A l X = 0 A k , x l = 0, A k , A l = 0 (221) (221) (Note: in some cases, this is not true, but those cases are beyond the scope of this worksheet.) Add these rules to the algebra rules already set so that they are all taken into account when simplifying things > > Setup algebrarules = (221) (222) (222)

  32. algebrarules = p k , p n = 0, x k , p l = i k , l , x k , x l = 0, A k , x l = 0, A k , (222) (222) A l = 0 , > > Simplify (220) i H , x k q A k p k = (223) (223) m The right-hand side of (223) is then the k th component of the velocity tensor quantum operator, the relationship is the same as in the classical case > > v k = rhs (223) q A k p k v k = (224) (224) m and with this the Hamiltonian (219) can now be rewritten in term of the velocity completing step 1) > > simplify (219) , SubstituteTensorIndices k = n , rhs = lhs (224) 2 m v n H = (225) (225) 2 For step 2), to compute i m H , v k . = F = m v we need the commutator between the different components of the quantized velocity which, contrary to what happens in the classical case, do not commute. For this purpose, take the commutator between (224) with itself after replacing the free index > > Commutator (224) , SubstituteTensorIndices k = n , (224) q A k , p n p k , A n v k , v n = (226) (226) m 2 To simplify (226) , we use the fact that if f is a commutative mapping that can be decomposed into a formal power series in all the complex plan (which is assumed to be the case for all A k (X) ), then p k , f x , y , z = i k f x , y , z where p k = i k is the momentum operator along the x k axis. This relation reads in tensor notation: > > Commutator p k , A n X = i d_ k A n X

  33. p k , A n = i k A n (227) (227) Add this rule to the rules previously set in order to automatically take it into account in (226) > > Setup (227) algebrarules = p k , p n = 0, p k , A n = i k A n , x k , p l = i k , l , x k , (228) (228) x l = 0, A k , x l = 0, A k , A l = 0 , > > (226) i q n A k k A n v k , v n = (229) (229) m 2 Also add this other rule so that it is taken into account automatically > > Setup (229) algebrarules = p k , p n = 0, p k , A n = i k A n , v k , v n (230) (230) i q n A k k A n = , x k , p l = i k , l , x k , x l = 0, A k , x l = 0, m 2 A k , A l = 0 , Recalling now the expression of the Hamiltonian (225) as a function of the velocity, one can i m H , v k . = compute the components of the force operator F k = m v k i > > F k X = m %Commutator rhs (225) , v k 2 m v n i m , v k 2 F k = (231) (231) Simplify this expression for the quantized force taking the quantum algebra rules (230) into account > > Simplify (231) q n A k v n k A n v n v n n A k v n k A n F k = (232) (232) 2 It is not difficult to verify that this is the antisymmetrized vector product v B . Departing from B = A expressed using tensor notation,

  34. > > B c X = LeviCivita c , n , m d_ n A m X B c = c , m , n n A m (233) (233) and taking into account that v B k = b , c , k v b B c multiply both sides of (233) by b , c , k v b , getting > > LeviCivita k , b , c v b (233) b , c , k v b B c = b , c , k c , m , n v b n A m (234) (234) > > Simplify (234) b , c , k v b B c = v m k A m v n n A k (235) (235) Finally, replacing the repeated index m by n > > SubstituteTensorIndices m = n , (235) b , c , k v b B c = v n k A n v n n A k (236) (236) Likewise, for B v k = b , c , k B b v c multiplying (233) , this time from the right instead of from the left, we get > > Simplify (233) LeviCivita k , b , c v b b , c , k B c v b = k A m v m n A k v n (237) (237) > > SubstituteTensorIndices m = n , (237) b , c , k B c v b = k A n v n n A k v n (238) (238) Simplifying now the expression (232) for the quantized force taking into account (236) and (238) we get > > simplify (232) , rhs = lhs (236) , rhs = lhs (238) q b , c , k v b B c B c v b F k = (239) (239) 2 i.e. F = q v B B v 2 in tensor notation. Finally, we note that this operator is Hermitian as expected > > (239) Dagger (239) (240) (240)

  35. † = 0 F k F k (240) (240) > > * The Hidden SO4 symmetry of the Hydrogen Atom Let's consider the Hydrogen atom and its Hamiltonian 2 p H = r , 2 m e where p is the electron momentum, m e its mass, a real positive constant, and r the distance of the electron from the proton located at the origin. We assume that the proton's mass is infinite. Introducing the angular momentum and Runge-Lenz vector Z : L = r p , 1 r Z = L p r . m e where Z is a constant of the motion, i.e. d d t Z t = 0. Switching to Quantum Mechanics, this condition reads H , Z = 0. The expression of Z must be symmetrized in order to be Hermitian 1 r Z = L p p L r . 2 m e Now: departing from the basic commutation rules between position r , momentum p in tensor notation, show that

  36. H , L n = 0 and H , Z n = 0, L m , Z n = i m , n , o Z o , = 2 i Z m , Z n H m , n , o L o . m e Finally, since H commutes with both L and Z , defining m e M n = 2 H Z n , the set of relations above can be rewritten as an SO(4) Lie algebra L m , L n = i m , n , o L o , L m , M n = i m , n , o M o , M m , M n = i m , n , o L o . I Commutation rules and useful identities Quantum commutation rules basics and the Hamiltonian of the hydrogen atom Set macros for M = m e > > restart ; with Physics : with Library : interface imaginaryunit = i : macro M = m e : Set the context: Cartesian coordinates, 3D Euclidean space, lowercase letters representing tensor indices, use automatic simplification (automatically apply simplify/size on everything, before returning it on the screen). Here V = 1 r will represent the potential of the central force. > > Setup coordinates = cartesian , hermitianoperators = X , realobjects = , , m e , automaticsimplification = true , dimension = 3, metric = Euclidean ,

  37. spacetimeindices = lowercaselatin , mathematicalnotation = true , quiet automaticsimplification = true , coordinatesystems = X , dimension = 3, (241) (241) hermitianoperators = X , mathematicalnotation = true , metric = 1, 1 = 1, 2, 2 = 1, 3, 3 = 1 , realobjects = , , m e , x , y , z , spacetimeindices = lowercaselatin Setting quantum (Hermitian) operators and related commutators: - Z is Hermitian, but we derive that property further below. - The potential V(X) of the hydrogen atom is assumed to commute with position, not with momentum - the commutation rule with p is derived further ahead - The commutator rules for angular momentum are an easy problem, we take them as the departure point - The last two commutators involving G(X) are for the differential operators approach only, not really part of the problem > > Setup quantumoperators = Z , hermitianoperators = V , G , H , L , X , p , algebrarules = %Commutator X k , X l = 0, %Commutator p k , p l = 0, %Commutator X k , p l = i KroneckerDelta k , l , %Commutator L j , L k = i LeviCivita j , k , n L n , %Commutator p j , L k = i LeviCivita j , k , n p n , %Commutator X j , L k = i LeviCivita j , k , n X n , %Commutator X k , V X = 0, %Commutator V X , G X = 0, %Commutator X k , G X = 0 algebrarules = V X , G X = 0, L j , L k = i j , k , n L n , p j , L k (242) (242) = i j , k , n p n , p k , p l = 0, X j , L k = i j , k , n X n , X k , G X = 0, X k , V X = 0, X k , p l = i k , l , X k , X l = 0 , hermitianoperators = G , H , L , V , p , x , y , z , quantumoperators = G , H , L , V , Z , p , x , y , z , Define the tensors > > Define p k = p x , p y , p z , L k = L x , L y , L z , Z k = Z x , Z y , Z z , quiet (243) (243)

  38. a , L k , a , X a , Z k , a , g a , b , p k , a , b , (243) (243) a , b , c > > CompactDisplay V X , G X V X will now be displayed as V G X will now be displayed as G (244) (244) The Hamiltonian for the hydrogen atom p l 2 > > H = V X 2 M 2 p l H = V (245) (245) 2 m e > > 2 = V and V 3 X n , V 3 X l Identities (I): n V = V = 0 For more compact calculus, we use the dimensionless potential V X 1 > > V X = sqrt X o 2 1 V = (246) (246) 1 2 2 X o The gradient of V X is > > d_ n (246) 1 n V = X n (247) (247) 3 2 2 X o So that (246) 3 , (247) > > subs rhs = lhs V 3 X n n V = (248) (248) Equivalently, V X can be written > > SumOverRepeatedIndices (246) 1 V = (249) (249) 1 2 x 2 y 2 z 2

  39. 2 = V X , that will often be used afterwards 3 X l from which one can deduce V X (249) 3 x 2 y 2 z 2 > > 1 V 3 x 2 y 2 z 2 = (250) (250) 1 2 x 2 y 2 z 2 = X l 2 , (250) x 2 y 2 z 2 > > subs rhs = lhs (249) , 2 = V V 3 X l (251) (251) And finally V X = 0 > > SumOverRepeatedIndices dAlembertian (249) 1 4 x 2 4 y 2 4 z 2 3 5 2 x 2 y 2 z 2 1 V = 3 (252) (252) 3 4 2 x 2 y 2 z 2 > > Factor (252) V = 0 (253) (253) > > Identities (II): the commutation rules between L, p and the potential V X One has > > L q = LeviCivita q , m , n X m p n L q = m , n , q X m p n (254) (254) > > Commutator (254) , V X L q , V = m , n , q X m p n , V (255) (255) > > %Commutator = Commutator p q , V X p q , V = p q , V (256) (256) 3 > > %Commutator = Commutator p q , V X p q , V 3 p q , V 3 = (257) (257) At this point, setup differentialoperators and set some commutators with an arbitrary function G(X), to be used in the alternative demonstrations based on using differentialoperators and to derive the commutation rules between L, p and V(X) > > Setup differentialoperators = p k , x , y , z

  40. differentialoperators = p k , X (258) (258) Now, apply the differential operators found in the commutators above to a generic function G X that will be removed afterwards > > lhs = ApplyProductsOfDifferentialOperators @ rhs (255) G X L q , V G = m , n , q X m p n V G V p n G (259) (259) > > lhs = ApplyProductsOfDifferentialOperators @ rhs (256) G X p q , V G = p q V G V p q G (260) (260) > > lhs = ApplyProductsOfDifferentialOperators @ rhs (257) G X p q , V 3 G = p q V 3 G V 3 p q G (261) (261) > > ApplyProductsOfDifferentialOperators p l G X = p l G X p l G = p l G (262) (262) Define now the momentum operator as an indexed procedure > > p proc local Ind op procname ; return i Physics :- d_ Ind args ; end : > > (259) L q , V G = i m , n , q X m n V G (263) (263) So that > > SubstituteTensor (248) , (263) Inverse G X = i m , n , q X m V 3 X n L q , V (264) (264) and finally > > Simplify (264) L q , V = 0 (265) (265) > > (260) Inverse G X p q , V = i q V (266) (266) > > SubstituteTensor (248) , (266) = i V 3 X q p q , V (267) (267) > > (261) Inverse G X

  41. p q , V 3 q V V 2 V 2 q V = i V q V V (268) (268) > > lhs (268) = Simplify SubstituteTensor (248) , rhs (268) p q , V 3 = 3 i V 5 X q (269) (269) To remove l G X from the equation above, use (262) , which, after defining p as a procedure in (263) , becomes > > (262) i l G = p l G (270) (270) > > isolate (262) , l G X i p l G l G = (271) (271) Add these new rules to Setup > > (265) , (267) , (269) = i V 3 X q , p q , V 3 = 3 i V 5 X q L q , V = 0, p q , V (272) (272) > > Setup (272) ; algebrarules = L j , L k = i j , k , n L n , L q , V = 0, X j , L k = i j , k , n X n , (273) (273) X k , X l = 0, X k , p l = i k , l , X k , G = 0, X k , V = 0, p j , L k p q , V 3 = 3 i V 5 X q , = i V 3 X q , = i j , k , n p n , p k , p l = 0, p q , V V , G = 0 , Now undo differentialoperators to work using two different approaches, with and without differentialoperators > > Setup differentialoperators = none differentialoperators = none (274) (274) > > II H , L n = 0 and H , Z n = 0 H , L n = 0 Recalling the Hamiltonian of the hydrogen atom and the definition of angular momentum (254) > > (245) ; (254) ;

  42. 2 p l H = V 2 m e L q = m , n , q X m p n (275) (275) We get > > Commutator (245) , (254) X m V 3 X n m e i m , n , q p l p n l , m H , L q = (276) (276) m e > > Simplify (276) H , L q = 0 (277) (277) > > H , Z n = 0 Setting up the problem and Z n is hermitian More difficult. Start from the definition of the quantum Runge-Lenz vector 1 > > Z k = 2 M LeviCivita a , b , k L a p b p a L b V X X k a , b , k L a p b p a L b Z k = V X k (278) (278) 2 m e From that, one has the hermicity of Z k > > Dagger (278) (278) † Z k Z k (279) (279) 2 V X k m e 2 X k V m e a , b , k L a p b L b p a p a L b p b L a = 2 m e > > Simplify (279) † Z k Z k = 0 (280) (280) Since the system now knows about the commutation rule between linear and angular momentum, > > %Commutator = Commutator L a , p b L a , p b = i a , b , n p n (281) (281) the expression for Z k can be simplified

  43. > > Simplify (278) i p k V X k m e a , b , k p a L b Z k = (282) (282) m e and the angular momentum removed from the defining expression for Z k > > (254) L q = m , n , q X m p n (283) (283) > > Simplify SubstituteTensor (254) , (282) 2 i p k V X k m e X k p n X m p k p m Z k = (284) (284) m e Here we set up the problem, to show that H , Z k = 0 > > Simplify Commutator (245) , (284) 1 V 3 X k V 5 X k X l 2 2 p k V 3 2 i X k X l p l V 3 H , Z k = 2 i X m (285) (285) 2 m e 2 i V X k X m p m V 2 2 i p k V > > Algebraic approach For the term with V 5 we use the derived identity (251) > > (251) 2 = V V 3 X l (286) (286) 2 (251) X k > > V X V 5 X l 2 X k = V 3 X k (287) (287) > > Normal (287) 2 = V 3 X k V 5 X k X l (288) (288) > > subs (288) , (285) H , Z k (289) (289) V 3 X k 2 p k V 3 i X k X l p l V 3 i V X k X m p m V 2 i X m i p k V = m e

  44. > > Simplify (289) V 5 X k X m 2 V 3 X k 2 p k V 3 i X m i p k V H , Z k = (290) (290) m e Another term with V 5 appeared > > SubstituteTensorIndices l = m , (288) 2 = V 3 X k V 5 X k X m (291) (291) > > subs (291) , (290) 2 V 3 X k 2 p k V 3 i X m i p k V H , Z k = (292) (292) m e Make X m and V be contiguous to further apply (251) > > SortProducts (292) , p k , X m , usecommutator 2 V 3 X k 2 2 i k , m X m V 3 i p k X m i p k V H , Z k = (293) (293) m e > > Simplify (293) p k V 3 X m 2 i p k V H , Z k = (294) (294) m e > > p k SubstituteTensorIndices l = m , (286) 2 = p k V p k V 3 X m (295) (295) > > subs (295) , (294) H , Z k = 0 (296) (296) And this is the result we wanted to prove. In the next section there is an alternative derivation that could be seen as more abstract or more direct > > Alternative approach using differential operators As done in the previous section when deriving the commutators between linear and angular momentum, on the one hand, and the central potential V on the other hand, the idea here is again to use differential operators taking advantage of the ability to compute with them as operands of a product, that get applied only when it appears convenient for us

  45. > > Setup differentialoperators = p k , x , y , z differentialoperators = p k , X (297) (297) So take the starting point (285) > > (285) 1 V 3 X k V 5 X k X l 2 2 p k V 3 2 i X k X l p l V 3 H , Z k = 2 i X m (298) (298) 2 m e 2 i V X k X m p m V 2 2 i p k V and to show that the left-hand side is equal to 0, multiply by a generic function G X followed by transforming the products involving p k by the application of this differential operator > > (285) G X 1 V 3 X k V 5 X k X l 2 2 p k V 3 H , Z k G = 2 i X m (299) (299) 2 m e 2 i V X k X m p m V 2 G 2 i X k X l p l V 3 2 i p k V > > ApplyProductsOfDifferentialOperators (299) 1 H , Z k G = V k G k V G V X k X m m V V (300) (300) m e V 2 m G 2 k V V 2 V m V G X m V k V V V 2 k V V 3 k G l V V 2 G X k X l V l V V V 3 X k G V 5 X k X l 2 G V 2 l V V 3 l G G 2 2 > > Simplify (300) 1 2 k V G V 2 H , Z k G = V k G k V G X m (301) (301) m e 2 k V G V V 2 X m 2 k V G V 3 X m 2 k G V X m G V 3 X k G V 5 X k X l 2 X k X l l V G V 2 2 2 2 Use now the derived identities for the gradients of V and G and then remove the generic function G from the equation by multiplying by the inverse of G

  46. > > (248) ; (271) ; V 3 X n n V = i p l G l G = (302) (302) > > Simplify SubstituteTensor (248) , (271) , (301) Inverse G X 3 V 3 X k 3 V 5 X k X l 2 i V 3 X m 2 p k i V p k 2 2 H , Z k = (303) (303) m e To show that the right-hand side is actually 0, recalling (251) > > (251) 2 = V V 3 X l (304) (304) > > rhs = lhs (251) p k V p k = V 3 X l 2 p k (305) (305) This and using (288) again > > Simplify SubstituteTensor (305) , (288) , (303) H , Z k = 0 (306) (306) Reset differentialoperators in order to proceed to the next section working without differential operators > > Setup differentialoperators = none differentialoperators = none (307) (307) > > III L m , Z n = i m , n , k Z k Strategy: 1. 1. Express L m and Z k in terms of X a and p b from previous sections 2. 2. Construct the left-hand and right-hand sides of L q , Z k = i k , q , u Z u , the formula we want to prove 3. 3. Simplify the result Step 1 is given by

  47. > > (254) L q = m , n , q X m p n (308) (308) > > (284) 2 i p k V X k m e X k p n X m p k p m Z k = (309) (309) m e Step 2. The left-hand side of the identity to be proved is the left-hand side of this commutator > > Commutator (254) , (284) 1 V 3 X n X k m e L q , Z k = m , n , q i X m V k , n k , m p n (310) (310) m e 2 k , n i X m p k p a a , n i X m p b 2 i X k p b p n b , m i X a p a k , m p k a , m p n The right-hand side of the identity to be proved is the left-hand side of this equation > > i LeviCivita q , k , u SubstituteTensorIndices k = u , (284) 2 i k , q , u i p u V X u m e X u p n X m p u p m i k , q , u Z u = (311) (311) m e Step 3. Take one minus the other one and the right-hand side must be equal to 0 > > (310) (311) 1 V 3 X n X k m e m , n , q L q , Z k i k , q , u Z u = i X m V k , n (312) (312) m e 2 k , n m , n , q i X a p a k , m p k a , m p n m , n , q i V X u m e k , q , u i X m p b 2 k , q , u i X u p n 2 i X k p b p n b , m m , n , q i X m p k p a a , n m , n , q i X m p u p m k , q , u p n k , m m , n , q p u k , q , u > > Simplify (312) L q , Z k i k , q , u Z u = 0 (313) (313) > >

  48. 2 i IV Z m , Z n = H m , n , o L o m e Algebraic approach Here again the starting point is ?? , the definition of the quantum Runge-Lenz vector > > SubstituteTensorIndices k = q , (284) 2 i p q V X q m e X q p n X m p q p m Z q = (314) (314) m e In this section the strategy is the same as in the previous section: construct the left-hand and right-hand sides of the identity to be proved (in the title of this section), take one minus the other one, and show that the right-hand side is equal to 0. Start with the left-hand side of the identity to be proved > > Commutator (284) , (314) 1 Z k , Z q = i X m a , k p q p a p m a , m p k p q p a m e X m i p k (315) (315) 2 m e V 3 X m X q V 3 X k X q p m i X q 2 p b V 3 X b V m , q i V k , q V 3 3 i 3 i V 5 X b 2 V 3 p b X b X k m e 2 p b X b 2 V 3 X k X q m e 2 V 3 X q X k m e 2 i a , n X k p n p q p a 2 2 i b , m X q p b p k p m 2 i V X q p b m e b , k 2 i n , q X k p n p b 2 2 2 k , q p n 2 i b , k X q p b p n i X a p a m , q p q a , m p k p m 2 k , q 2 m , q p k p m 2 p q p a a , k 2 p b 2 i X k V p n n , q 2 p n V 3 X n V 3 3 i 3 i V 5 X n 2 V 3 p n X n X q 2 p n X n 2 i X a p q V 3 X a V 3 X q p a X k m e i V X a p a k , q p q a , k

  49. 2 2 p m 2 i X a p a k , q p q a , k p n i X m k , q p b m , q p k p b Now the right-hand side, constructed from the definition of the angular momentum > > (254) L q = m , n , q X m p n (316) (316) 2 i > > H LeviCivita q , k , u SubstituteTensorIndices q = u , (254) M 2 i k , q , u H L u 2 i k , q , u m , n , u H X m p n = (317) (317) m e m e Replace the Hamiltonian H by its expression quadratic in the momentum > > (245) 2 p l H = V (318) (318) 2 m e > > lhs (317) = SubstituteTensor (245) , rhs (317) 2 p l 2 i V X m p n m , n , u k , q , u 2 i k , q , u H L u 2 m e = (319) (319) m e m e Now set up the problem, taking (315) minus (319) , working towards having it in final form, ready for solving, in (321) > > (315) (319) 2 i k , q , u H L u 1 Z k , Z q = 2 i b , m X q p b p k p m i X a p a k , q (320) (320) 2 m e m e 2 2 p m p q a , k p n i X m a , k p q p a p m a , m p k p q p a i X m k , q p b 2 m , q p k p b i X a p a m , q p q a , m p k p m 2 i a , n X k p n p q p a 2 p l 2 V 3 X k X q m e V 3 X q X k m e 2 i n , q X k p n p b 2 i 2 m e 2 V X m p n m , n , u k , q , u m e 2 i V X q p b m e b , k 2 i b , k X q p b p n

  50. i X a p q V 3 X a V 3 X q p a X k m e i V X a p a k , q p q a , k m e 2 2 m , q p k p m a , k p q p a k , q p b k , q p n i X m p k V 3 X m X q V 3 X k X q p m V m , q V k , q 2 i X k V p n n , q 2 p n V 3 X n V 3 3 i 3 i V 5 X n 2 V 3 p n X n X q 2 p n X n 2 i X q 2 p b V 3 X b V 3 3 i 3 i V 5 X b 2 p b X b 2 V 3 p b X b X k m e 2 i k , q , u H L u So the starting point to prove that Z k , Z q = is m e > > Simplify (320) 2 i k , q , u H L u Z k , Z q m e 1 V 5 X a 2 X k X q m e = (321) (321) 2 m e m e i X a X k X q p a V 3 m e i V X k X m X q p m V 2 m e i X q p k V m e 2 X k p q V 3 m e 2 X q p k V 3 m e i X a i X m i X k p q V m e 2 2 p m 2 2 p k i g k , q X a p a p n i g k , q X m p b 2 i X q p k p n 2 i X q p a were the proof is achieved showing that the right-hand side of this equation is indeed equal to 0. Start checking the repeated indices, as we would do by hand > > Check (321) , all The products in the given expression check ok. The repeated indices per term are: ... , ... , ... ; the free indices are: ... u , k , q = a , b , m , n , k , q (322) (322) Check in which terms - that involve V - are these repeated indices appearing > > for term in select has , map op , indets (321) , ` ` , V do term = Check term , repeated , quiet od i X k p q V m e =

  51. 2 X q p k V 3 m e = i X m m i V X k X m X q p m V 2 m e = m i X q p k V m e = 2 X k p q V 3 m e = i X a a i X a X k X q p a V 3 m e = a V 5 X a 2 X k X q m e = a (323) (323) By eye, m = a results in a simplification in that two terms become equal > > SubstituteTensorIndices m = a , (321) 2 i k , q , u H L u Z k , Z q m e 1 V 5 X a 2 X k X q m e = (324) (324) 2 m e m e i X a X k X q p a V 3 m e i V X k X a X q p a V 2 m e i X q p k V m e 2 X k p q V 3 m e 2 X q p k V 3 m e i X a i X a i X k p q V m e 2 2 p a 2 2 p k i g k , q X a p a p n i g k , q X a p b 2 i X q p k p n 2 i X q p a > > Simplify (324) 2 i k , q , u H L u Z k , Z q m e 1 = i X k p q V m e X q p k V m e (325) (325) 2 m e m e 2 X k p q V 3 m e 2 X q p k V 3 m e 2 2 p k X a X a 2 X q p k p n 2 X q p a One more round of the same: > > Check (325) , all The products in the given expression check ok. The repeated indices per term are: ... , ... , ... ; the free indices are: ... u , k , q = a , n , k , q (326) (326) > > SubstituteTensorIndices a = n , (325) 2 i k , q , u H L u Z k , Z q m e 1 = i X k p q V m e X q p k V m e (327) (327) 2 m e m e 2 X k p q V 3 m e 2 X q p k V 3 m e 2 2 p k X n X n 2 X q p k p n 2 X q p n > > Simplify (327) 2 i k , q , u H L u Z k , Z q m e (328) (328) m e

  52. 2 p q V 3 2 X q p k V 3 i X q p k V X k X n X n X k p q V = m e Run now a loop re-sorting products in the key terms of the right-hand side above, approximating X and V (move p to the left). On the way, create a table of helpful substitution equations > > eq table : for term in select has , map op , indets (328) , ` ` , V do eq term term = SortProducts term , p k , p q , X k , X q , X n , usecommutator , totheleft od eq Xq pk V X q p k V = p k X q i k , q V 2 p q V 3 = 2 eq X k X n p q X k X n 2 i n , q X k X n 2 pq V 3 Xk Xn 2 V 3 i k , q X n eq Xk pq V X k p q V = p q X k i k , q V 2 X q p k V 3 = p k X q X n 2 V 3 2 V 3 eq X n i k , q X n (329) (329) 2 Xq pk V 3 Xn 2 i k , n X q X n V 3 Substitute now > > subs convert eq , list , (328) 2 i k , q , u H L u Z k , Z q m e 1 = i p k X q i k , q V (330) (330) m e m e 2 V 3 2 2 V 3 p q X k X n 2 i n , q X k X n i k , q X n p k X q X n 2 V 3 2 i k , n X q X n V 3 i k , q X n p q X k i k , q V There are still terms containing V 3 X 2 to be simplified. So again from > > (251) 2 = V V 3 X l (331) (331) and noting that these two commute 3 , X l 2 > > %Commutator = Commutator V X V 3 , X l 2 = 0 (332) (332)

  53. rewrite (251) switching the order X n 2 V X 3 = V X > > 2 V 3 = V X n (333) (333) > > p k X q (333) 2 V 3 = p k X q V p k X q X n (334) (334) Substitute now > > subs (334) , (330) 2 i k , q , u H L u Z k , Z q m e 1 = i p k X q i k , q V (335) (335) m e m e 2 V 3 2 p q X k X n 2 i n , q X k X n i k , q X n p k X q V 2 V 3 2 i k , n X q X n V 3 i k , q X n p q X k i k , q V > > Simplify (335) p q V 3 X k X n 2 2 i k , q , u H L u Z k , Z q m e i p q V X k = (336) (336) m e m e Finally, removing the last term that involves V 3 > > p q SubstituteTensorIndices l = n , (251) X k p q V 3 X n 2 X k = p q V X k (337) (337) > > Normal (337) 2 = p q V X k p q V 3 X k X n (338) (338) > > subs (338) , (336) 2 i k , q , u H L u Z k , Z q m e = 0 (339) (339) m e which is the identity we wanted to prove. In the next section the same result is obtained using differential operators > > Alternative approach using differential operators The main idea: make p k be a differential operator, then (321) by a generic function G(X), apply the products of differential operators, then use tensorial simplifications: > > (248) ; (271) ; (251)

  54. V 3 X n n V = i p l G l G = 2 = V V 3 X l (340) (340) The goal is again to show that the right-hand side of (321) is equal to 0, so set p k to be a differential operator, multiply (321) by a generic function G X followed by applying p k where it corresponds > > Setup differentialoperators = p k , x , y , z differentialoperators = p k , X (341) (341) > > (321) G X 2 i k , q , u H L u Z k , Z q m e G 1 V 5 X a 2 X k X q m e = (342) (342) 2 m e m e i X a X k X q p a V 3 m e i V X k X m X q p m V 2 m e i X q p k V m e 2 X k p q V 3 m e 2 X q p k V 3 m e i X a i X m i X k p q V m e 2 2 p m 2 2 p k G i g k , q X a p a p n i g k , q X m p b 2 i X q p k p n 2 i X q p a Apply now p k > > ApplyProductsOfDifferentialOperators (342) 2 i k , q , u H L u Z k , Z q m e G 1 g k , q 3 X a a = G (343) (343) 2 m e m e g k , q 3 X m m G m e X k q V G V q G m e X q k V G V k G m e V X k X m X q m V V V 2 m G 2 X k q V V 2 V m V G m e X a V q V V V 2 q V V 3 q G 2 X q k V V 2 G m e X m V k V V V 2 k V V 3 k G a V V 2 G m e X a X k X q V a V V V 2 a V V 3 a G V 5 X a 2 X k X q m e G G We want to show that the right-hand side is equal to 0; start simplifying with respect to algebra rules and using Einstein's sum rule for repeated indices > > Simplify (343) (344) (344)

  55. 2 i k , q , u H L u G m e Z k , Z q G 1 G V 5 X a 2 X k X q 2 = (344) (344) m e m e 2 X q k V G V 2 2 X q k V G V X k q V G X m V X m V 2 X m 2 X q k V G V 3 X m 2 X q k G V X q k G X q k V G 2 X k q V G V 2 2 X k q V G V V X k q G X a V X a V 2 X a 2 X k q V G V 3 X a 2 X k q G X a X k X q a V G V 2 Next, using (248) , the gradient of V can be removed > > (248) V 3 X n n V = (345) (345) > > SubstituteTensor (345) , (344) 2 i k , q , u H L u G m e Z k , Z q G 1 G V 5 X a 2 X k X q 2 = (346) (346) m e m e X k V 3 X q G 2 X q V 3 X k G V 2 2 X q V 3 X k G V V 2 X m 2 X q V 3 X k G X m V X m V 3 X m 2 X q k G X q V 3 X k G V X q k G V X k q G 2 X k V 3 X q G V 2 2 X k V 3 X q G V V 2 X a 2 X k V 3 X q G X a V X a V 3 X a 2 X k q G X a X k X q V 3 X a G V 2 > > Simplify (346) 2 i k , q , u H L u G m e Z k , Z q G (347) (347) m e 2 V 3 X m 2 X q k G V 3 X a 2 X k q G V X q k G V X k q G = m e From (271) the gradient of G can be removed > > (271) i p l G l G = (348) (348) > > SubstituteTensor (271) , (347) i V 3 X m 2 X q p k G 2 i k , q , u H L u G m e Z k , Z q G 1 2 = (349) (349) m e m e

  56. i V 3 X a 2 X k p q G V X q i p k G 1 V X k i p q G 1 By eye there are terms that are similar; check the repeated indices > > Check (349) , all The products in the given expression check ok. The repeated indices per term are: ... , ... , ... ; the free indices are: ... u , k , q = a , m , k , q (350) (350) > > SubstituteTensorIndices a = m , (349) 2 i k , q , u H L u G m e Z k , Z q G (351) (351) m e i V 3 X m 2 X q p k G V 3 X k X m 2 p q G V X q p k G V X k p q G = m e And there are still two terms of the form X 2 V 3 that can be removed using (251) > > SubstituteTensorIndices l = m , (251) 2 = V V 3 X m (352) (352) > > (352) X q p k G X ; X q V 3 X m 2 p k G = X q V p k G (353) (353) > > rhs = lhs (352) X k p q G X ; X k V p q G = X k V 3 X m 2 p q G (354) (354) > > Simplify subs (353) , (354) , (351) Inverse G X 2 i k , q , u H L u Z k , Z q m e = 0 (355) (355) m e Which is already the result we wanted to obtain. > > Classical Field Theory (functional differentiation) 4 model The field equations for the 4 model, the corresponding Action, and the field equations: The Lagrangian of the

  57. > > restart ; with Physics : Setup automaticsimplification = true , coordinatesystems = cartesian Default differentiation variables for d_, D_ and dAlembertian are: X = x , y , z , t Systems of spacetime Coordinates are: X = x , y , z , t automaticsimplification = true , coordinatesystems = X (356) (356) > > g_ 1 0 0 0 0 1 0 0 g , = (357) (357) 0 0 1 0 0 0 0 1 > > CompactDisplay X Phi X will now be displayed as (358) (358) 2 m 2 4 d_ mu Phi X lambda Phi X 2 > > L 2 Phi X 2 4 2 4 m 2 L (359) (359) 2 2 4 > > S Intc L , X 2 4 m 2 S d x d y d z d t (360) (360) 2 2 4 > > ' Fundiff ' S , = 0 2 4 m 2 d x d y d z d t = 0 (361) (361) 2 2 4 > > % 3 m 2 = 0 (362) (362) > > show 3 m 2 X X X = 0 (363) (363) * Maxwell equations departing from the 4-dimensional Action for Electrodynamics Maxwell equations result from equation to zero the functional derivative o the Action with respect

  58. to the 4-D potential A > > restart ; with Physics : Setup automaticsimplification = true , coordinatesystems = cartesian Default differentiation variables for d_, D_ and dAlembertian are: X = x , y , z , t Systems of spacetime Coordinates are: X = x , y , z , t automaticsimplification = true , coordinatesystems = X (364) (364) The 4-D electromagnetic potential > > Define A mu X Defined objects with tensor properties A , , , X , , g , , , , (365) (365) , , , > > CompactDisplay A X A X will now be displayed as A (366) (366) The electromagnetic field tensor F , > > F mu, nu d_ mu A nu X d_ nu A mu X ; F , A A (367) (367) Equate to 0 the functional derivative of the corresponding Action > > ' Fundiff ' Intc F mu, nu ^2, X , A rho = 0 2 d x d y d z d t = 0 A A (368) (368) A > > (368) , , 2 A A g 2 A A g = 0 (369) (369) Simplify the contracted spacetime indices Simplify (369) > > 4 A A = 0 (370) (370) > > * The field equations for a quantum system of identical particles Problem: derive the field equation describing the ground state of a quantum system of identical

  59. particles (bosons), that is, the Gross-Pitaevskii equation (GPE). This equation is particularly relevant to describe Bose-Einstein condensates (BEC). Solution Two steps: • • Construct the Lagrangian for the system, and with it write the action functional • • Minimize the action by equating to zero its functional derivative with respect to the boson field. > > restart ; with Physics : with Physics Vectors : > > interface imaginaryunit = i : > > macro Psi = psi x , y , z , t : > > CompactDisplay , V x , y , z , t psi x , y , z , t will now be displayed as V x , y , z , t will now be displayed as V (371) (371) The energy density E for a quantum system of identical boson particles is (see [3]) 2 G 2 V x , y , z , t abs Psi 2 2 abs Psi 4 ; > > E 2 m Norm %Gradient Psi 2 4 2 G 2 E V (372) (372) 2 m 2 4 x , y , z , t is a complex field, V x , y , z , t an external potential, the term G is the atom- 2 atom interaction. > > Setup realobjects = t , m , , G , V x , y , z , t : The Lagrangian density L in terms of the Energy E i > > L conjugate Psi diff Psi, t Psi diff conjugate Psi , t E 2 2 4 i 2 G 2 t t L V (373) (373) 2 2 m 2 Construct the action and equate to zero the functional derivative

  60. > > ' Fundiff ' Intc L , x , y , z , t , psi = 0 2 4 i 2 G 2 t t V (374) (374) 2 2 m 2 d x d y d z d t = 0 > > (374) 2 2 y , y 2 2 2 G i t V m x , x z , z = 0 (375) (375) 2 m Make the Laplacian explicit > > Laplacian = %Laplacian Psi 2 z , z = (376) (376) x , x y , y > > simplify conjugate (375) , (376) 2 2 2 i t m 2 m G V = 0 (377) (377) 2 m The standard form of the Gross–Pitaevskii equation: > > i isolate (377) , diff Psi, t 2 2 2 m G V i t = (378) (378) 2 m > > collect convert expand (378) , abs , psi 2 2 2 i t = G V (379) (379) 2 m > > Phys. Rev. A 74, 043405, (2006) [1] Gerritsma and R. J. C. Spreeuw, Topological constraints on magnetostatic traps , Phys. Rev. A 74, 043405 (2006) We consider a device constructed with a set of electrical wires fed with constant electrical currents. Those wires can have an arbitrary complex shape. The device is operated in a regime such that, in some region of interest, the moving particles experience a magnetic field that varies slowly compared to the Larmor spin precession frequency. In this region, the effective potential is proportional to the modulus of the field: B x , y , z , this potential has a minimum and, close to

  61. this minimum, the device behaves as a magnetic trap. Figure 1: Schematic representation of a Ioffe-Pritchard magnetic trap. It is made of four infinite rods and two coils. _________________________________________ Following [1], we show that:

  62. a) For a time-independent magnetic field B x , y , z in vacuum, up to order two in the relative coordinates X i = x , y , z around some point of interest, the coefficients of orders 1 and 2 in this expansion, v i , j and c i , j , k , respectively the gradient and curvature, contain only 5 and 7 independent components. 2 b) All stationary points of B x , y , z (nonzero minima and saddle points) are confined to a curved surface defined by det j B i = 0. c) The effective potential, proportional to B x , y , z , has no maximum, only a minimum. Finally, we draw the stationary condition surface for the case of the widely used Ioffe-Pritchard magnetic trap. The independent components of v i , j and c i , j , k entering 1 B i = u i v i , j X j 2 c i , j , k X j X k > > restart > > with Physics : > > Setup coordinates = cartesian , dimension = 3, metric = Euclidean , spacetimeindices = lowercaselatin , quiet , minimizetensorcomponents = true coordinatesystems = X , dimension = 3, metric = 1, 1 = 1, 2, 2 = 1, 3, 3 (380) (380) = 1 , minimizetensorcomponents = true , spacetimeindices = lowercaselatin > > g_ 1 0 0 0 1 0 g a , b = (381) (381) 0 0 1 2 , around We are interested in determining the location of the stationary points of B x , y , z which the device behaves as a magnetic trap. Up to order two in the relative coordinates X i = x , y , z around a point of interest x i within this region (that we take as origin of the system of references), 1 > > i = u i v i , j X j 2 c i , j , k X j X k 1 i = u i v i , j X j 2 c i , j , k X j X k (382) (382)

  63. where i is the truncated expansion of B i (computers are picky, don't like recursive definitions) and, following [1], we introduce the notation u i = B i X X j = 0 v i , j = j B i X X j = 0 c i , j , k = k B i X j X j = 0 Here, v i , j denotes the gradient tensor and c i , j , k the curvature of B at the point x i = 0. > > CompactDisplay i X , B i X X will now be displayed as B X will now be displayed as B (383) (383) > > Define B i , (382) Defined objects with tensor properties i , B i , a , a , X a , c i , j , k , a , g a , b , u i , v i , j , a , b , (384) (384) a , b , c At this stage, the gradient tensor v i , j has no known symmetry; it thus has up to 9 independent components and, in general, a rank 3 tensor like c i , j , k has 3 3 = 27 independent components, > > Library :- NumberOfIndependentTensorComponents v ; 9 (385) (385) > > Library :- NumberOfIndependentTensorComponents c 27 (386) (386) but c i , j , k = k B i X is invariant by a permutation of its second and third j X j = 0 indices; indicate this symmetry and this last number is reduced to 18: > > Define redo , c i , j , k , symmetric = 2, 3 , quiet i , B i , a , a , X a , c a , b , d , a , g a , b , u i , v i , j , a , b , (387) (387) a , b , c > > Library :- NumberOfIndependentTensorComponents c 18 (388) (388)

  64. > > c 1, j , k , matrix c 1, 1, 1 c 1, 1, 2 c 1, 1, 3 c 1, 1, 2 c 1, 2, 2 c 1, 2, 3 c 1, j , k = (389) (389) c 1, 1, 3 c 1, 2, 3 c 1, 3, 3 For a stationary magnetic field B in vacuum, we also have B = 0 B = 0 so that the number of independent components of c i , j , k can be further reduced. > > B = 0 Starting with B = 0, from (382) > > (382) 1 i = u i v i , j X j 2 c i , j , k X j X k (390) (390) > > LeviCivita a , b , i d_ b (382) c i , j , k X j g b , k g b , j X k a , b , i b = a , b , i v i , j g b , j (391) (391) 2 i > > Simplify (391) a , b , i b = X j c i , j , k v i , k a , i , k (392) (392) i The right-hand-side of (392) must be zero no matter what the X j are. This means that the following two terms, the coefficients of X j , are equal to 0. > > Coefficients (392) , X j a , b , i b = a , i , k v i , k , 0 = c i , j , k a , i , k (393) (393) i From the first of these equations, one can then see that v i , j is actually symmetric:

  65. > > 0 = rhs (393) 1 0 = a , i , k v i , k (394) (394) > > TensorArray % 0 = v 2, 3 v 3, 2 0 = v 1, 3 v 3, 1 0 = v 1, 2 v 2, 1 (395) (395) Add this symmetry to the definition of v i , j > > Define redo , v i , j , symmetric , quiet i , B i , a , a , X a , c a , b , d , a , g a , b , u i , v a , b , a , b , (396) (396) a , b , c Check that the symmetry is explicitly there > > v v 1, 1 v 1, 2 v 1, 3 v 1, 2 v 2, 2 v 2, 3 v a , b = (397) (397) v 1, 3 v 2, 3 v 3, 3 > > Library :- NumberOfIndependentTensorComponents v 6 (398) (398) From the second equation in (393) , > > (393) 2 0 = c i , j , k a , i , k (399) (399) c i , j , k is also symmetric under permutation of its 1st and 3rd indices, and because it is already symmetric under permutation of its 2nd and 3rd indices, c i , j , k is actually fully symmetric. Although this is sort of obvious, it can be verified as follows: redefine c i , j , k indicating the symmetries 1, 3 , 2, 3 and check the symmetries of the resulting tensor: > > Define redo , c i , j , k , symmetric = 1, 3 , 2, 3 Defined objects with tensor properties i , B i , a , a , X a , c a , b , d , a , g a , b , u i , v a , b , a , b , (400) (400) a , b , c > > Library :- GetTensorSymmetryProperties c 1, 2, 3 , (401) (401) Or, directly count the number of independent components: > > Library :- NumberOfIndependentTensorComponents c (402) (402)

  66. 10 (402) (402) Indeed, a fully symmetric tensor constitutes a vector space with a dimension given by the binomial of the dimension + rank - 1 and the rank; that is dimension rank 1 > > VectorSpaceDim dimension , rank : rank > > VectorSpaceDim 3, 3 10 (403) (403) > > B = 0 Again, starting from (382) > > (382) 1 i = u i v i , j X j 2 c i , j , k X j X k (404) (404) > > d_ i (382) c i , j , k X j g i , k g i , j X k = v i , j g i , j (405) (405) i 2 i > > Simplify (405) = X k c j , j , k v j , j (406) (406) i i The right-hand-side of (406) must be zero no matter what the X j are. This means that the following two terms, the coefficients of X k , are equal to 0. > > Coefficients (406) , X k = v j , j , 0 = c j , j , k (407) (407) i i This time the resulting equations don't increase the symmetry of the tensors as in B = 0, but permit reducing the number of independent components. From the first of these equations, v i , j is traceless > > v trace = 0 v 1, 1 v 2, 2 v 3, 3 = 0 (408) (408) > > isolate (408) , v 3, 3

  67. v 3, 3 = v 1, 1 v 2, 2 (409) (409) > > Library :- RedefineTensorComponent (409) v 1, 1 v 1, 2 v 1, 3 v 1, 2 v 2, 2 v 2, 3 v a , b = (410) (410) v 1, 3 v 2, 3 v 1, 1 v 2, 2 > > Library :- NumberOfIndependentTensorComponents v 5 (411) (411) From the second equation in (407) , > > TensorArray (407) 2 0 = c 1, 1, 1 c 1, 2, 2 c 1, 3, 3 , 0 = c 1, 1, 2 c 2, 2, 2 c 2, 3, 3 , 0 = c 1, 1, 3 c 2, 2, 3 (412) (412) c 3, 3, 3 , Redefine one component using each of these equations: > > map u isolate u , op 1, rhs u , (412) c 1, 3, 3 = c 1, 1, 1 c 1, 2, 2 , c 2, 3, 3 = c 1, 1, 2 c 2, 2, 2 , c 3, 3, 3 = c 1, 1, 3 (413) (413) c 2, 2, 3 , > > Library :- RedefineTensorComponent (413) 1..3 x 1..3 x 1..3 Array Data Type: anything c a , b , d = (414) (414) Storage: rectangular Order: Fortran_order > > c i , j , 3, matrix c 1, 1, 3 c 1, 2, 3 c 1, 1, 1 c 1, 2, 2 c 1, 2, 3 c 2, 2, 3 c 1, 1, 2 c 2, 2, 2 c i , j , 3 = (415) (415) c 1, 1, 1 c 1, 2, 2 c 1, 1, 2 c 2, 2, 2 c 1, 1, 3 c 2, 2, 3 Now count the number of independent components of the curvature: > > Library :- NumberOfIndependentTensorComponents c 7 (416) (416) By the way, due to the symmetries of the curvature, B is not just Curl and Divergence free, but also Laplacian free. From (382) ,

  68. > > SubstituteTensor i = B i X , (382) 1 B i = u i v i , j X j 2 c i , j , k X j X k (417) (417) > > dAlembertian (417) B i = c i , k , k (418) (418) > > TensorArray (418) B 1 = 0 B 2 = 0 B 3 = 0 (419) (419) > > The stationary points are within the surface det j B i = 0 In order to determine the location of the stationary points of the square of the potential, 2 , we need a copy of U ~ B i with different repeated dummy indices. > > SubstituteTensorIndices j = m , k = n , (382) 1 i = u i v i , m X m 2 c i , m , n X m X n (420) (420) > > (382) (420) 1 1 2 = u i v i , j X j 2 c i , j , k X j X k u i v i , m X m 2 c i , m , n X m X n (421) (421) i > > Simplify (421) 1 2 = 4 X k X a X m X n c i , m , n X k X a u i X j X k X a v i , j c a , i , k v i , j X j v i , m X m (422) (422) i 2 2 v i , j X j u i u i Removing higher order terms (> 2) with respect to the coordinates, > > STV op indets (422) , specfunc SpaceTimeVector STV X a , X j , X k , X m , X n (423) (423) > > select u degree u , STV identical 3, 4 , expand rhs (422) 1 4 X k X a X m X n c a , i , k c i , m , n X j X k X a c a , i , k v i , j (424) (424) > > U X = Simplify rhs (422) (424) : SubstituteTensorIndices a = i , m = k , expand % 2 U X = X j X k c i , j , k u i v i , j v i , k X j X k 2 v i , j X j u i u i (425) (425) > > CompactDisplay U X (426) (426)

  69. U X will now be displayed as U (426) (426) > > collect (425) , STV , distributed 2 U = c i , j , k u i v i , j v i , k X j X k 2 v i , j X j u i u i (427) (427) For U to be stationary, all of its first derivatives p U must cancel at X = 0. > > d_ n (427) n U = c i , j , k u i v i , j v i , k X j g k , n g j , n X k 2 v i , j u i g j , n (428) (428) > > Simplify (428) n U = 2 X j c i , j , n 2 v i , n u i 2 X j v i , j v i , n (429) (429) Evaluation at X = 0 : > > SubstituteTensor X j = 0, (429) n U = 2 v i , n u i (430) (430) Except for the trivial solution u i = 0, for U to be stationary, u i must be an eigenvector of v i , j with eigenvalue 0. Now, the determinant of v i , j is the product of its eigenvalues, hence the stationary points occur where det v i , j = 0. In turn, v i , j = j B i x k where x k is some point within the magnetic trap, hence the stationary points are the x k of the 2D surface det j B i = 0 > > 2 U = B has only minima, no maxima To see that U has no maxima, only minima, we need to insert u i v i , n = 0 in the definition (425) of U and consider the second derivative with respect to the coordinates: > > (425) 2 U = X j X k c i , j , k u i v i , j v i , k X j X k 2 v i , j X j u i u i (431) (431) > > 0 = v i , j u i 0 = v i , j u i (432) (432)

  70. > > 2 (432) X j 0 = 2 v i , j X j u i (433) (433) > > (425) (433) 2 U = X j X k c i , j , k u i v i , j v i , k X j X k u i (434) (434) The second derivative is given by the coefficient with respect to X k X j > > t j , k = Coefficients rhs (434) , X j X k , 1 t j , k = c i , j , k u i v i , j v i , k (435) (435) Now, U has only minima, no maxima, if this second derivative is always positive (its trace is positive definite). So take the trace of this expression: > > Define (435) Defined objects with tensor properties i , B i , a , a , X a , c a , b , d , a , g a , b , t j , k , u i , v a , b , a , b , (436) (436) a , b , c > > t trace 2 2 2 2 2 2 v 1, 1 2 v 1, 1 v 2, 2 2 v 1, 2 2 v 1, 3 2 v 2, 2 2 v 2, 3 (437) (437) Add and subtract: 2 > > v 1, 1 v 2, 2 2 v 1, 1 v 2, 2 (438) (438) > > t trace (438) expand (438) 2 2 2 2 2 2 v 1, 1 2 v 1, 2 2 v 1, 3 v 2, 2 2 v 2, 3 v 1, 1 v 2, 2 (439) (439) This trace is a sum of the squares of real quantities. It is therefore always positive. This recovers a well known result: there is no local maximum of a static magnetic field in free space. Indeed, such a maximum would require the three eigenvalues of v i , j to be negative, which would also imply a negative trace. Therefore, the stationary condition can only be a local minimum or a saddle point. See [1] for a more thorough discussion. > > Drawing the Ioffe-Pritchard Magnetic Trap The magnetic field of the Ioffe-Pritchard trap, quadratic in the relative coordinates [x,y,z], is approximated as

  71. x z 0 x y z C > > 0 y B j = A 2 1 z 2 x 2 y 2 U 0 2 1 A x 2 C x z 1 A y 2 C y z B j = (440) (440) x 2 y 2 C z 2 2 2 U 2 > > Define (440) Defined objects with tensor properties i , B i , a , a , X a , c a , b , d , a , g a , b , t j , k , u i , v a , b , a , b , (441) (441) a , b , c The surface of stationary points is defined by > > d_ j B i j B i (442) (442) The matrix behind: > > TensorArray (442) C z C x A 0 2 2 C z C y 0 A (443) (443) 2 2 C x C y C z 2 2 The stationary condition det j B i = 0 : > > factor LinearAlgebra :- Determinant (443) = 0 C 2 x 2 z C 2 z y 2 2 C 2 z 3 2 C x 2 A 2 A C y 2 8 z A 2 C = 0 (444) (444) 8 is scaled by a single parameter = 2 A . C 8 2 A > > simplify C 3 (444) , = C

  72. y 2 z 2 z 3 x 2 x 2 y 2 2 2 = 0 (445) (445) from which we can deduce the 2D stationary manifold. For = 1: > > plots :- implicitplot3d subs epsilon = 1, (445) , x = 4 ..4, y = 4 ..4, z = 1.5 ..1.5, style = surface , scaling = constrained , grid = 50, 50, 50 , caption = The Ioffe Pritchard magnetic trap for epsilon = 1 ; The Ioffe-Pritchard magnetic trap for = 1 > > General Relativity * Database of solutions to Einstein's Equations G g = 8 T , , ,

  73. Main reference: - Stephani, H., Kramer, D., MacCallum, M., Hoenselaers, C. Herlt, E. Exact Solutions of Einstein's Field Equations , Cambridge Monographs on Mathematical Physics, second edition. Cambridge University Press, 2003. The authors reviewed more than 4,000 papers containing solutions to Einstein’s equations in the literature and organized the material into chapters according to the physical properties of these solutions. These solutions are digitized within Maple since 2016, so that it is now possible to actually compute with them . Examples Load Physics, consider the case of one metric. Set it, for instance, to be Schwarzschild ' s solution > > restart ; with Physics : > > g_ sc _______________________________________________________ Systems of spacetime Coordinates are: X = r , , , t Default differentiation variables for d_, D_ and dAlembertian are: X = r , , , t The Schwarzschild metric in coordinates r , , , t Parameters: m _______________________________________________________ r 0 0 0 r 2 m r 2 0 0 0 g , = (446) (446) 2 r 2 sin 0 0 0 r 2 m 0 0 0 r And that is all we do. Everything else gets automatically computed on background (the only information saved in the database is the metric) The tensor components of the general relativity tensors related to this solution get derived automatically from their definition > > Christoffel definition

  74. g , g , g , = (447) (447) , , 2 2 2 These are the 2x2 matrix components of the Christoffel symbols of the second kind with the first index contravariant with value 1 > > Christoffel ~1, alpha, beta, matrix m 0 0 0 r r 2 m 0 r 2 m 0 0 1 = (448) (448) 2 , 0 0 r 2 m sin 0 2 m 2 m r 0 0 0 r 3 > > Riemann definition R , = g , (449) (449) , , , , , , , , For example, the Riemann invariants using the standard formulas by Carminati and McLenaghan > > Riemann invariants r 0 = 0, r 1 = 0, r 2 = 0, r 3 = 0, w 1 = 6 m 2 , w 2 = 6 m 3 , m 1 = 0, m 2 = 0, m 3 = 0, m 4 (450) (450) r 6 r 9 = 0, m 5 = 0 The related Weyl scalars in the context of the Newman-Penrose formalism > > Weyl scalarsdefinition , , , , , , 0 = C l m l m , 1 = C l n l m , 2 = (451) (451) , , , , , , C l m m n , 3 = C l n m n , 4 = , , , C n m n m > > Weyl scalars m 0 = 0, 1 = 0, 2 = r 3 , 3 = 0, 4 = 0 (452) (452) The Killing vectors > > Define K , quiet :

  75. > > KillingVectors K , cos sin K = 0, 0, 0, 1 , K = 0, sin , 0 , K = 0, cos , , 0 , (453) (453) tan tan K = 0, 0, 1, 0 > > Define (453) 2 Defined objects with tensor properties , , K , , R , , R , , , C , , , X , , g , , , , G , , , , (454) (454) , , , , , , > > LieDerivative K g_ mu, nu g , K g , K (455) (455) > > TensorArray (455) 0, 0, 0, 0 , (456) (456) 2 cos 1 tan cos cos 2 0, 0, r 2 sin 2 sin tan tan sin cos cos r 2 cos , 0 , , tan 2 cos 1 tan cos cos 2 0, r 2 sin r 2 cos 2 sin tan tan sin cos cos sin cos sin 2 , 2 r 2 sin , 0 tan tan sin , , 0, 0, 0, 0 > > simplify (456) 0 0 0 0 0 0 0 0 (457) (457) 0 0 0 0 0 0 0 0

  76. > > Geodesics tensornotation d 2 d d 2 X X X (458) (458) , d d d > > CompactDisplay r , theta, phi, t tau r will now be displayed as r theta will now be displayed as phi will now be displayed as t will now be displayed as t (459) (459) > > Geodesics 2 r 2 cos r sin r sin cos 2 r = , = , (460) (460) , , r r sin 2 1 r 2 1 r 3 r , = 4 cos 1 cos m 2 m r 3 2 r 2 2 r r 2 m r 2 r 2 , t , 4 r 3 t 2 m 4 m m = 2 2 2 m r t , r r 2 m This system of ODEs, as is, it is out of reach of the DE solvers of the system mainly due to the presence of non-rational objects like sin and cos having for arguments one of the unknowns of the system, . On the other hand, we know the geodesics for the Schwarzschild metric describe the motion of particles in the gravitational field of a central fixed large mass. So to investigate the solvability of these equations one can assume is constant an due to the rotational symmetry choose a value for it that simplifies the equations, for example, = 2 . > > (460) = 2 2 r = , 0 = 0, r , (461) (461) , r 2 2 r r 2 4 r 3 t 2 m r 2 r 2 m 4 m m 2 2 = , t , = 2 m r 3 r

  77. 2 m r t , r r 2 m > > dsolve (461) 1 ..2 # 10 seconds ... r = 6 m , = _C2 _C3 , t = 6 6 m _C1 , t = 6 6 m _C1 , r (462) (462) r m r r m r = _C4 , = _C2 _C3 , t = d _C1 , t = d m m _C1 > > • • One can query the database, directly from the spacetime metrics command (g_). For example, these are the solutions (metrics) to Einstein's equations that appear in the book and related to Levi-Civita, the Italian mathematician > > g_ civi _______________________________________________________ ____________________________________________________________ 12, 16, 1 = "Authors" = "Bertotti (1959)", "Kramer (1978)", "Levi-Civita (1917)", "Robinson (1959)" , "PrimaryDescription" = "EinsteinMaxwell", "SecondaryDescription" = "Homogeneous" ____________________________________________________________ 12, 18, 1 = "Authors" = "Bertotti (1959)", "Kramer (1978)", "Levi-Civita (1917)", "Robinson (1959)" , "PrimaryDescription" = "EinsteinMaxwell", "SecondaryDescription" = "Homogeneous" ____________________________________________________________ 12, 19, 1 = "Authors" = "Bertotti (1959)", "Kramer (1978)", "Levi-Civita (1917)", "Robinson (1959)" , "PrimaryDescription" = "EinsteinMaxwell", "SecondaryDescription" = "Homogeneous" , "Comments" = "_lambda=_zeta" ____________________________________________________________ 22, 7, 1 = "Authors" = "Levi-Civita (1917), Frehland (1971)" , "PrimaryDescription" = "Vacuum", "SecondaryDescription"

  78. = "Cylindrically-Symmetric" , "Comments" = "Locally static, Weyl class_m=0,1 - flat, _m=1/2, 2, -1 - PetrovType D" Warning, found more than one match for the keyword 'civi', as seen above. Please (463) (463) refine your 'keyword' or re-enter the metric 'g_[...]' with the list of three numbers identifying the metric, for example as in g_ 12 , 16 , 1 or Setup(metric = 12 , 16 , 1 ) > > g_ 12, 16, 1 _______________________________________________________ Systems of spacetime Coordinates are: X = t , x , , Default differentiation variables for d_, D_ and dAlembertian are: X = t , x , , The Bertotti (1959) metric in coordinates Kramer (1978) Parameters: k , 0 , Resetting the signature of spacetime from - - - to - in order to match the signature in the database of metrics _______________________________________________________ k 2 sinh x 2 0 0 0 k 2 0 0 0 g , = (464) (464) k 2 0 0 0 2 k 2 sin 0 0 0 The search can also be done visually, by properties; this is the only solution in the database that is: • • A Pure Radiation solution, • • of Petrov Type "D", • • Plebanski-Petrov Type "O" • • and that has Isometry Dimension equal to 1: > > DifferentialGeometry :- Library :- MetricSearch Set the solution, and everything related to work with it, in one go > > g_ 28, 74, 1 _______________________________________________________ Systems of spacetime Coordinates are: X = u , , r , y Default differentiation variables for d_, D_ and dAlembertian are: X = u , , r , y

  79. The Frolov and Khlebnikov (1975) metric in coordinates u , , r , y Parameters: 0 , m u , b , d Comments: With m(u) = constant, the metric is Ricci flat and becomes 28.24 in Stephani. _______________________________________________________ g , = (465) (465) 2 3 2 m u 3 6 m u 2 r r 2 r 3 6 b m u 2 b d , r m u 2 r 2 , 1, 0 , m u r 2 r 2 , , 0, 0 , 3 m u 2 b d 1, 0, 0, 0 , 3 0, 0, 0, r 2 2 b d Check the types > > Tetrads :- PetrovType "D" (466) (466) > > Tetrads :- SegreType "O", "[(1,11)1]" (467) (467) To conclude, how many solutions are digitized? > > DifferentialGeometry :- Library :- Retrieve "Stephani", 1 8, 33, 1 , 8, 34, 1 , 12, 6, 1 , 12, 7, 1 , 12, 8, 1 , 12, 8, 2 , 12, 8, 3 , 12, 8, (468) (468) 4 , 12, 8, 5 , 12, 8, 6 , 12, 8, 7 , 12, 8, 8 , 12, 9, 1 , 12, 9, 2 , 12, 9, 3 , 12, 9, 4 , 12, 9, 5 , 12, 9, 6 , 12, 9, 7 , 12, 12, 1 , 12, 12, 2 , 12, 12, 3 , 12, 12, 4 , 12, 13, 1 , 12, 14, 1 , 12, 16, 1 , 12, 18, 1 , 12, 19, 1 , 12, 21, 1 , 12, 23, 1 , 12, 23, 2 , 12, 23, 3 , 12, 24.1, 1 , 12, 24.2, 1 , 12, 24.3, 1 , 12, 26, 1 , 12, 27, 1 , 12, 28, 1 , 12, 29, 1 , 12, 30, 1 , 12, 31, 1 , 12, 32, 1 , 12, 34, 1 , 12, 35, 1 , 12, 36, 1 , 12, 37, 1 , 12, 37, 2 , 12, 37, 3 , 12, 37, 4 , 12, 37, 5 , 12, 37, 6 , 12, 37, 7 , 12, 38, 1 , 12, 38, 2 , 12, 38, 3 , 12, 38, 4 , 12, 38, 5 , 13, 1, 1 , 13, 1, 2 , 13, 1, 3 , 13, 2, 1 , 13, 2, 2 ,

  80. 13, 2, 3 , 13, 3, 1 , 13, 7, 1 , 13, 7, 2 , 13, 7, 3 , 13, 7, 4 , 13, 7, 5 , 13, 7, 6 , 13, 7, 7 , 13, 7, 8 , 13, 9, 1 , 13, 14, 1 , 13, 14, 2 , 13, 14, 3 , 13, 15, 1 , 13, 15, 2 , 13, 15, 3 , 13, 15, 4 , 13, 17, 1 , 13, 19, 1 , 13, 22, 1 , 13, 31, 1 , 13, 32, 1 , 13, 35, 1 , 13, 46, 1 , 13, 48, 1 , 13, 49, 1 , 13, 49, 2 , 13, 51, 1 , 13, 53, 1 , 13, 55, 1 , 13, 56, 1 , 13, 57, 1 , 13, 58, 1 , 13, 59, 1 , 13, 59, 2 , 13, 60, 1 , 13, 60, 2 , 13, 60, 3 , 13, 60, 4 , 13, 60, 5 , 13, 60, 6 , 13, 60, 7 , 13, 60, 8 , 13, 61, 1 , 13, 61, 2 , 13, 62, 1 , 13, 62, 2 , 13, 62, 4 , 13, 62, 6 , 13, 63, 1 , 13, 63, 2 , 13, 63, 3 , 13, 63, 4 , 13, 64, 1 , 13, 64, 2 , 13, 64, 3 , 13, 64, 4 , 13, 65, 1 , 13, 67, 1 , 13, 67, 2 , 13, 69, 1 , 13, 71, 1 , 13, 72, 1 , 13, 73, 1 , 13, 74, 1 , 13, 74, 2 , 13, 74, 3 , 13, 76, 1 , 13, 77, 1 , 13, 77, 2 , 13, 79, 1 , 13, 79, 2 , 13, 80, 1 , 13, 81, 1 , 13, 83, 1 , 13, 84, 1 , 13, 84, 2 , 13, 84, 3 , 13, 85, 1 , 13, 85, 2 , 13, 86, 1 , 13, 87, 1 , 14, 6.1, 1 , 14, 6.2, 1 , 14, 6.3, 1 , 14, 7, 1 , 14, 8.1, 1 , 14, 8.2, 1 , 14, 8.3, 1 , 14, 9.1, 1 , 14, 9.2, 1 , 14, 10, 1 , 14, 10, 2 , 14, 12, 1 , 14, 12, 2 , 14, 12, 3 , 14, 14, 1 , 14, 14, 2 , 14, 15, 1 , 14, 15.1, 2 , 14, 15.2, 2 , 14, 15.3, 2 , 14, 16, 1 , 14, 16, 2 , 14, 17, 1 , 14, 18, 1 , 14, 18, 2 , 14, 19, 1 , 14, 20, 1 , 14, 21, 1 , 14, 21, 2 , 14, 21, 3 , 14, 22, 1 , 14, 23, 1 , 14, 24, 1 , 14, 25, 1 , 14, 26, 1 , 14, 26, 2 , 14, 26, 3 , 14, 26, 4 , 14, 27, 1 , 14, 28, 1 , 14, 28, 2 , 14, 28, 3 , 14, 29, 1 , 14, 30, 1 , 14, 31, 1 , 14, 32, 1 , 14, 33, 1 , 14, 35, 1 , 14, 37, 1 , 14, 38, 1 , 14, 38, 2 , 14, 38, 3 , 14, 39, 1 , 14, 39, 2 , 14, 39, 3 , 14, 39, 4 , 14, 39, 5 , 14, 39, 6 , 14, 40, 1 , 14, 41, 1 , 14, 42, 1 , 14, 46, 1 , 15, 3, 1 , 15, 3, 2 , 15, 4, 1 , 15, 4, 2 , 15, 4, 3 , 15, 9, 1 , 15, 10, 1 , 15, 12, 1 , 15, 12, 2 , 15, 12, 3 , 15, 12, 4 , 15, 12, 5 , 15, 12, 6 , 15, 17, 1 , 15, 17, 2 , 15, 17, 3 , 15, 17, 4 , 15, 18, 1 , 15, 19, 1 , 15, 19, 2 , 15, 20, 1 , 15, 21, 1 , 15, 21, 2 , 15, 22, 1 , 15, 23, 1 , 15, 23, 2 , 15, 24, 1 , 15, 24, 2 , 15, 25, 1 , 15, 25, 2 , 15, 26, 1 , 15, 26, 2 , 15, 27, 1 , 15, 27, 2 , 15, 27, 3 , 15, 27, 4 , 15, 27, 5 , 15, 27, 6 , 15, 27, 7 , 15, 27, 8 , 15, 28, 1 , 15, 29, 1 , 15, 30, 1 , 15, 31, 1 , 15, 32, 1 , 15, 34, 1 , 15, 34, 2 , 15, 34, 3 , 15, 43, 1 , 15, 43, 2 , 15, 43, 3 , 15, 50, 1 , 15, 50, 2 , 15, 50, 3 , 15, 50, 4 , 15, 50, 5 , 15, 50, 6 , 15, 62, 1 , 15, 62, 2 , 15, 62, 3 , 15, 63, 1 , 15, 63, 2 , 15, 63, 3 , 15, 65, 1 , 15, 65, 2 , 15, 66, 1 , 15, 66, 2 , 15, 66, 3 , 15, 75, 1 , 15, 75, 2 , 15, 75, 3 , 15, 77, 1 , 15, 77, 2 , 15, 77, 3 , 15, 78, 1 , 15, 79, 1 , 15, 81, 1 , 15, 81, 2 , 15, 81, 3 , 15, 82, 1 , 15, 82, 2 , 15, 82, 3 , 15, 83, 1.1 , 15, 83, 1.2 , 15, 83, 2 , 15, 83, 3.1 , 15, 83, 3.2 , 15, 83, 4 , 15, 84, 1 , 15, 85, 1 , 15, 85, 2 , 15, 85, 3 , 15, 86, 1 , 15, 86, 2 , 15, 86, 3 , 15, 87, 1 , 15, 87, 2 , 15, 87, 3 , 15, 87, 4 , 15, 87, 5 , 15, 88, 1 , 15, 89, 1 , 15, 90, 1 , 16, 1, 1 , 16, 1, 2 , 16, 1, 3 , 16, 1, 4 , 16, 1, 5 , 16, 1, 6 ,

  81. 16, 1, 7 , 16, 1, 8 , 16, 1, 9 , 16, 1, 10 , 16, 1, 11 , 16, 1, 12 , 16, 1, 13 , 16, 1, 14 , 16, 1, 15 , 16, 1, 16 , 16, 1, 17 , 16, 1, 18 , 16, 1, 19 , 16, 1, 20 , 16, 1, 21 , 16, 1, 22 , 16, 1, 23 , 16, 1, 24 , 16, 1, 25 , 16, 1, 26 , 16, 1, 27 , 16, 14, 1 , 16, 14, 2 , 16, 14, 3 , 16, 14, 4 , 16, 14, 5 , 16, 14, 6 , 16, 14, 7 , 16, 14, 8 , 16, 14, 9 , 16, 14, 10 , 16, 14, 11 , 16, 14, 12 , 16, 14, 13 , 16, 14, 14 , 16, 14, 15 , 16, 14, 16 , 16, 14, 17 , 16, 14, 18 , 16, 14, 19 , 16, 14, 20 , 16, 18, 1 , 16, 19, 1 , 16, 20, 1 , 16, 22, 1 , 16, 24, 1 , 16, 24, 2 , 16, 43, 1 , 16, 45, 1 , 16, 45, 2 , 16, 46, 1 , 16, 47, 1 , 16, 50, 1 , 16, 51, 1 , 16, 54, 1 , 16, 61, 1 , 16, 63, 1 , 16, 66, 1 , 16, 66, 2 , 16, 66, 3 , 16, 67, 1 , 16, 71, 1 , 16, 72, 1 , 16, 73, 1 , 16, 74, 1 , 16, 75, 1 , 16, 76, 1 , 16, 77, 1 , 16, 77, 2 , 16, 77, 3 , 16, 78, 1 , 17, 4, 1 , 17, 4, 2 , 17, 5, 1 , 17, 9, 1 , 17, 14, 1 , 17, 15, 1 , 17, 15, 2 , 17, 16, 1 , 17, 20, 1 , 17, 22, 1 , 17, 23, 1 , 17, 24, 1 , 17, 24, 2 , 17, 26, 1 , 17, 27, 1 , 17, 27, 2 , 17, 28, 1 , 17, 28, 2 , 17, 29, 1 , 17, 29, 2 , 17, 30, 1 , 17, 31, 1 , 18, 2, 1 , 18, 2, 2 , 18, 2, 3 , 18, 2, 4 , 18, 2, 5 , 18, 2, 6 , 18, 2, 7 , 18, 2, 8 , 18, 48, 1 , 18, 48, 2 , 18, 49, 1 , 18, 50, 1 , 18, 64, 1 , 18, 64, 2 , 18, 64, 3 , 18, 65, 1 , 18, 66, 1 , 18, 67, 1 , 18, 71, 1 , 18, 75, 1 , 19, 17, 1 , 19, 17, 2 , 19, 21, 1 , 20, 3, 1 , 20, 4, 1 , 20, 5, 1 , 20, 7, 1 , 20, 8, 1 , 20, 9, 1 , 20, 10, 1 , 20, 11, 1 , 20, 12, 1 , 20, 13, 1 , 20, 15, 1 , 20, 16, 1 , 20, 17, 1 , 20, 20, 1 , 20, 21, 1 , 20, 23, 1 , 20, 27, 1 , 20, 28, 1 , 20, 29, 1 , 20, 32, 1 , 20, 34, 1 , 20, 36, 1 , 20, 38, 1 , 20, 38, 2 , 20, 38, 3 , 20, 44, 1 , 20, 46, 1 , 20, 54, 1 , 20, 57, 1 , 20, 57, 2 , 21, 1, 1 , 21, 1, 2 , 21, 1, 3 , 21, 4, 1 , 21, 5, 1 , 21, 6, 1 , 21, 7, 1 , 21, 10, 1 , 21, 10, 2 , 21, 11, 1 , 21, 16, 1 , 21, 17, 1 , 21, 17, 2 , 21, 20, 1 , 21, 22, 1 , 21, 22, 2 , 21, 24, 1 , 21, 28, 1 , 21, 30, 1 , 21, 30, 2 , 21, 30, 3 , 21, 31, 1 , 21, 35, 1 , 21, 41, 1 , 21, 52, 1 , 21, 57, 1 , 21, 58, 1 , 21, 59, 1 , 21, 60, 1 , 21, 61, 1 , 21, 61, 2 , 21, 61, 3 , 21, 61, 4 , 21, 61, 5 , 21, 70, 1 , 21, 71, 1 , 21, 72, 1 , 21, 73, 1 , 21, 74, 1 , 21, 74, 2 , 21, 74, 3 , 21, 74, 4 , 22, 3, 1 , 22, 4, 1 , 22, 4, 2 , 22, 5, 1 , 22, 6, 1 , 22, 6, 2 , 22, 7, 1 , 22, 8, 1 , 22, 8, 2 , 22, 8, 3 , 22, 8, 4 , 22, 8, 5 , 22, 11, 1 , 22, 12, 1 , 22, 13, 1 , 22, 14, 1 , 22, 15, 1 , 22, 16, 1 , 22, 17, 1 , 22, 18, 1 , 22, 18, 2 , 22, 19, 1 , 22, 22, 1 , 22, 23, 1 , 22, 24, 1 , 22, 27, 1 , 22, 28, 1 , 22, 29, 1 , 22, 34, 1 , 22, 34, 2 , 22, 34, 3 , 22, 34, 4 , 22, 34, 5 , 22, 47, 1 , 22, 48, 1 , 22, 49, 1 , 22, 50, 1 , 22, 51, 1 , 22, 52, 1 , 22, 53, 1 , 22, 59, 1 , 22, 63, 1 , 22, 64, 1 , 22, 67, 1 , 22, 67, 2 , 22, 70, 1 , 23, 1, 1 , 23, 2, 1 , 23, 2, 2 , 23, 2, 3 , 23, 2, 4 , 23, 3, 1 , 23, 4, 1 , 23, 6, 1 , 23, 6, 2 , 23, 7, 1 , 23, 8, 1 , 23, 8, 2 , 23, 8, 3 , 23, 11, 1 , 23, 12, 1 , 23, 13, 1 , 23, 13, 2 , 23, 13, 3 , 23, 13, 4 , 23, 14, 1 , 23, 14, 2 , 23, 14, 3 , 23, 14, 4 , 23, 14, 5 ,

  82. 23, 14, 6 , 23, 14, 7 , 23, 15, 1 , 23, 15, 2 , 23, 15, 3 , 23, 15, 4 , 23, 15, 5 , 23, 16, 1 , 23, 16, 2 , 23, 16, 3 , 23, 16, 4 , 23, 16, 5 , 23, 17, 1 , 23, 17, 2 , 23, 18, 1 , 23, 18, 2 , 23, 19, 1 , 23, 19, 2 , 23, 20, 1 , 23, 20, 2 , 23, 20, 3 , 23, 21, 1 , 23, 21, 2 , 23, 22, 1 , 23, 23, 1 , 23, 24, 1 , 23, 25, 1 , 23, 26, 1 , 23, 26, 2 , 23, 26, 3 , 23, 26, 4 , 23, 26, 5 , 23, 26, 6 , 23, 26, 7 , 23, 26, 8 , 23, 30, 1 , 23, 31, 1 , 23, 32, 1 , 23, 32, 2 , 23, 32, 3 , 23, 33, 1 , 23, 33, 2 , 23, 33, 3 , 23, 34, 1 , 23, 35, 1 , 23, 36, 1 , 23, 36, 2 , 23, 36, 3 , 23, 37, 1 , 23, 38, 1 , 23, 38, 2 , 23, 38, 3 , 23, 39, 1 , 23, 39, 2 , 23, 40, 1 , 23, 40, 2 , 23, 40, 3 , 23, 40, 4 , 23, 41, 1 , 23, 41, 2 , 23, 41, 3 , 23, 42, 1 , 23, 43, 1 , 23, 44, 1 , 23, 44, 2 , 23, 45, 1 , 23, 46, 1 , 23, 47, 1 , 23, 48, 1 , 23, 48, 2 , 23, 49, 1 , 23, 50, 1 , 23, 51, 1 , 23, 52, 1 , 23, 53, 1 , 24, 2, 1 , 24, 2, 2 , 24, 2, 3 , 24, 2, 4 , 24, 2, 5 , 24, 2, 6 , 24, 2, 7 , 24, 2, 8 , 24, 2, 9 , 24, 2, 10 , 24, 2, 11 , 24, 21, 1 , 24, 22, 1 , 24, 28, 1 , 24, 35, 1 , 24, 37, 1 , 24, 37, 2 , 24, 37, 3 , 24, 37, 4 , 24, 37, 5 , 24, 37, 6 , 24, 37, 7 , 24, 37, 8 , 24, 37, 9 , 24, 38, 1 , 24, 38, 2 , 24, 40, 1 , 24, 46, 1 , 24, 46, 2 , 24, 46, 3 , 24, 47, 1 , 24, 51, 1 , 25, 2, 1 , 25, 5, 1 , 25, 5, 2 , 25, 6, 1 , 25, 9, 1 , 25, 12, 1 , 25, 16, 1 , 25, 16, 2 , 25, 16, 3 , 25, 16, 4 , 25, 22, 1 , 25, 24, 1 , 25, 26, 1 , 25, 26, 2 , 25, 26, 3 , 25, 28, 1 , 25, 30, 1 , 25, 31, 1 , 25, 35, 1 , 25, 36, 1 , 25, 36, 2 , 25, 39, 1 , 25, 43, 1 , 25, 45, 1 , 25, 55, 1 , 25, 56, 1 , 25, 56, 2 , 25, 57, 1 , 25, 57, 2 , 25, 59, 1 , 25, 61, 1 , 25, 62, 1 , 25, 65, 1 , 25, 68, 1 , 25, 68, 2 , 25, 68, 3 , 25, 74, 1 , 26, 5, 1 , 26, 5, 2 , 26, 5, 3 , 26, 5, 4 , 26, 6, 1 , 26, 11, 1 , 26, 13, 1 , 26, 14, 1 , 26, 14, 2 , 26, 15, 1 , 26, 16, 1 , 26, 16, 2 , 26, 16, 3 , 26, 16, 4 , 26, 16, 5 , 26, 16, 6 , 26, 19, 1 , 26, 21, 1 , 26, 22, 1 , 26, 23, 1 , 26, 25, 1 , 26, 25, 2 , 26, 25, 3 , 27, 27, 1 , 27, 37, 1 , 28, 8, 1 , 28, 12, 1 , 28, 16, 1 , 28, 17, 1 , 28, 21, 1 , 28, 21, 2 , 28, 21, 3 , 28, 21, 4 , 28, 21, 5 , 28, 21, 6 , 28, 21, 7 , 28, 24, 1 , 28, 25, 1 , 28, 26, 1 , 28, 26, 2 , 28, 26, 3 , 28, 37, 1 , 28, 41, 1 , 28, 42, 1 , 28, 43, 1 , 28, 44, 1 , 28, 44, 2 , 28, 44, 3 , 28, 44, 4 , 28, 44, 5 , 28, 44, 6 , 28, 45, 1 , 28, 45, 2 , 28, 46, 1 , 28, 46, 2 , 28, 47, 1 , 28, 50, 1 , 28, 53, 1 , 28, 53, 2 , 28, 55, 1 , 28, 55, 2 , 28, 56.1, 1 , 28, 56.2, 2 , 28, 56.2, 3 , 28, 56.3, 1 , 28, 56.4, 1 , 28, 56.5, 1 , 28, 56.6, 1 , 28, 58.2, 1 , 28, 58.3, 1 , 28, 58.3, 2 , 28, 58.4, 1 , 28, 59, 1 , 28, 60, 1 , 28, 61, 1 , 28, 64, 1 , 28, 66, 1 , 28, 67, 1 , 28, 68, 1 , 28, 71, 1 , 28, 72, 1 , 28, 73, 1 , 28, 74, 1 , 28, 78, 1 , 29, 13, 1 , 29, 34, 1 , 29, 38, 1 , 29, 46, 1 , 29, 53, 1 , 29, 60, 1 , 29, 62, 1 , 29, 64, 1 , 29, 71, 1 , 29, 74, 1 , 29, 75, 1 , 30, 14, 1 , 30, 20, 1 , 30, 22, 1 , 30, 26, 1 , 30, 27, 1 , 30, 27, 2 , 30, 28, 1 , 30, 28, 2 , 30, 28, 3 , 30, 30, 1 , 30, 33, 1 , 30, 34, 1 , 30, 36, 1 , 30, 37, 1 , 30,

  83. 43, 1 , 30, 45, 1 , 30, 46, 1 , 30, 47, 1 , 30, 50, 1 , 30, 51, 1 , 30, 51, 2 , 30, 51, 3 , 30, 52, 1 , 30, 58, 1 , 30, 59, 1 , 30, 62, 1 , 30, 64, 1 , 30, 65, 1 , 30, 69, 1 , 30, 71, 1 , 30, 73, 1 , 30, 76, 1 , 31, 7, 1 , 31, 8, 1 , 31, 26, 1 , 31, 34, 1 , 31, 34, 2 , 31, 38, 1 , 31, 40, 1 , 31, 41, 1 , 31, 43, 1 , 31, 43, 2 , 31, 43, 3 , 31, 49, 1 , 31, 50, 1 , 31, 56, 1 , 31, 57, 1 , 31, 58, 1 , 31, 60, 1 , 31, 61, 1 , 32, 31, 1 , 32, 43, 1 , 32, 47, 1 , 32, 59, 1 , 32, 60, 1 , 32, 62, 1 , 32, 71, 1 , 32, 78, 1 , 32, 80, 1 , 32, 80, 2 , 32, 80, 3 , 32, 80, 4 , 32, 94, 1 , 32, 96, 1 , 32, 96, 2 , 32, 99, 1 , 32, 102, 1 , 32, 102, 2 , 32, 104, 1 , 33, 5, 1 , 33, 7, 1 , 33, 8, 1 , 33, 8, 2 , 33, 8, 3 , 33, 9, 1 , 33, 10, 1 , 33, 10, 2 , 33, 11, 1 , 33, 12, 1 , 33, 12, 2 , 33, 13, 1 , 33, 14, 1 , 33, 15, 1 , 33, 16, 1 , 33, 17, 1 , 33, 18, 1 , 33, 19, 1 , 33, 20, 1 , 33, 22, 1 , 33, 23, 1 , 33, 25, 1 , 33, 25, 2 , 33, 28, 1 , 33, 30, 1 , 33, 31, 1 , 33, 34, 1 , 33, 35, 1 , 33, 38, 1 , 33, 40, 1 , 33, 43, 1 , 33, 44, 1 , 33, 45, 1 , 33, 48, 1 , 33, 49, 1 , 34, 23, 1 , 34, 25, 1 , 34, 128, 1 , 35, 6, 1 , 35, 7, 1 , 35, 8, 1 , 35, 9, 1 , 35, 19, 1 , 35, 29, 1 , 35, 30, 1 , 35, 33, 1 , 35, 34, 1 , 35, 35, 1 , 35, 73, 1 , 35, 74, 1 , 35, 75, 1 , 35, 76, 1 , 35, 76, 2 , 35, 77, 1 , 35, 78, 1 , 35, 79, 1 , 35, 80, 1 , 36, 11, 1 , 36, 12, 1 , 36, 13, 1 , 36, 14, 1 , 36, 15, 1 , 36, 18, 1 , 36, 18, 2 , 36, 19, 1 , 36, 20, 1 , 36, 22, 1 , 36, 23, 1 , 36, 24, 1 , 36, 25, 1 , 36, 26, 1 , 36, 28, 1 , 36, 30, 1 , 36, 31, 1 , 36, 32, 1 , 36, 33, 1 , 36, 34, 1 , 36, 35, 1 , 36, 36, 1 , 36, 37, 1 , 37, 13, 1 , 37, 21, 1 , 37, 39, 1 , 37, 40, 1 , 37, 45, 1 , 37, 49, 1 , 37, 50, 1 , 37, 51, 1 , 37, 53, 1 , 37, 55, 1 , 37, 57, 1 , 37, 58, 1 , 37, 58, 2 , 37, 58, 3 , 37, 64, 1 , 37, 65, 1 , 37, 66, 1 , 37, 68, 1 , 37, 83, 1 , 37, 84, 1 , 37, 98, 1 , 37, 104, 1 , 37, 106, 1 , 38, 1, 1 , 38, 2, 1 , 38, 3, 1 , 38, 4, 1 , 38, 5, 1 , 38, 6, 1 > > nops (468) 991 (469) (469) > > Tetrads and Weyl scalars in canonical form Generally speaking a canonical form is obtained using transformations that leave invariant the tetrad metric in a tetrad system of references, so that theWeyl scalars are fixed as much as possible (conventionally, either equal to 0 or to 1). Bringing a tetrad in canonical form is a relevant step in the tackling of the equivalence problem between two spacetime metrics.

  84. The implementation is as in " General Relativity, an Einstein century survey ", edited by S. W. Hawking (Cambridge) and W. Israel (U. Alberta, Canada), specifically Chapter 7 written by S. Chandrasekhar, page 388: Residual 0 1 2 3 4 invariance Petrov type 0 0 0 1 0 none I Petrov type 0 0 0 1 0 none II Petrov type 0 0 0 1 0 none III Petrov type 0 0 0 0 0 2 D remains invariant under rotations of Class III Petrov type 0 0 0 0 1 4 N remains invariant under rotations of Class II The transformations (rotations of the tetrad system of references) used are of Class I, II and III as defined in Chandrasekar's chapter - equations (7.79) in page 384, (7.83) and (7.84) in page 385. Transformations of Class I can be performed with the command Physics:-Tetrads:- TransformTetrad using the optional argument nullrotationwithfixedl_ , of Class II using nullrotationwithfixedn_ and of Class III by calling TransformTetrad ( spatialrotationsm_mb_plan, boostsn_l_plane ), so with the two optional arguments simultaneously. The determination of appropriate transformation parameters to be used in these rotations, as well as

  85. the sequence of transformations happens all automatically by using the optional argument, canonicalform of TransformTetrad . > > restart ; with Physics : with Tetrads ; _______________________________________________________ Setting lowercaselatin_ah letters to represent tetrad indices Defined as tetrad tensors see ?Physics,tetrads , a , , a , b , a , b , c , a , b , c Defined as spacetime tensors representing the NP null vectors of the tetrad formalism see ?Physics,tetrads : see ?Physics,tetrads , l , n , m , m _______________________________________________________ IsTetrad , NullTetrad , OrthonormalTetrad , PetrovType , SegreType , TransformTetrad , e_ , (470) (470) eta_ , gamma_ , l_ , lambda_ , m_ , mb_ , n_ > > Petrov type I The numbers below used to enter the metric always refer to the equation number in the " Exact solutions to Einstein's field equations " textbook > > g_ 12, 21, 1 _______________________________________________________ Systems of spacetime Coordinates are: X = t , x , y , Default differentiation variables for d_, D_ and dAlembertian are: X = t , x , y , The McLenaghan, Tariq (1975) metric in coordinates Tupper (1976) Parameters: a , k , 0 Comments: k parametrizes the most general electromagnetic invariant with respect to the last 3 Killing vectors Resetting the signature of spacetime from - - - to - in order to match the signature in the database of metrics _______________________________________________________ (471) (471)

  86. 1 0 0 2 y a 2 0 0 0 x 2 g , = (471) (471) a 2 0 0 0 x 2 x 2 4 y 2 2 y 0 0 > > PetrovType "I" (472) (472) > > SegreType "D", "[(1,1)(11)]" (473) (473) The default tetrad computed by the Physics package routines > > e_ 1 0 0 2 y a 0 0 0 x = (474) (474) a , a 0 0 0 x 0 0 0 x The corresponding Weyl scalars > > Weyl scalars 0 = I x 1 4 = I x , 1 = 0, 2 = a 2 , 3 = 0, (475) (475) a 2 x a 2 x ... there is abs around. Let's assume everything is positive to simplify the presentation of formulas > > Assume x 0, y 0, a 0 a :: 0, , x :: 0, , y :: 0, (476) (476) The scalars are now simpler, although still not in "canonical form" because 0 and 4 1. 3 > > Weyl scalars I 1 I 0 = a 2 , 1 = 0, 2 = a 2 , 3 = 0, 4 = (477) (477) a 2 For Petrov type I, the Weyl scalars are in canonical form when 0 = 0, 4 = 0 and 3 = 1.

  87. > > TransformTetrad canonicalform 2 5 3 2 2 5 2 5 3 2 5 1 , , 2 a x , (478) (478) 2 a 2 2 a x 1 2 5 3 2 2 5 x 2 5 3 2 6 a 2 2 5 3 5 x 6 y , , a 2 2 5 3 4 5 5 2 , 10 3 a 3 2 5 a 3 2 5 5 5 2 2 , , 10 x 10 x 1 3 a 2 4 2 5 5 5 2 x 2 5 3 2 30 2 5 3 5 x 6 y , , 2 5 3 5 2 5 3 2 I 2 5 3 5 10 4 5 I 2 5 3 2 , 4 2 5 3 2 I 5 5 I 2 4 5 5 2 a , 20 x 10 I I 2 5 3 2 5 a , 1 10 I I 2 5 20 x 60 3 2 5 2 y 2 5 3 2 2 2 5 3 5 y , 3 x , I 2 5 3 2 2 5 3 2 I 2 5 3 5 4 4 5 2 5 3 5 , 10 2 5 3 2 I 5 5 I 2 4 5 5 2 a , 20 x 10 I I 2 5 3 2 5 a 1 , 10 I I 2 5 20 x 60

  88. 3 2 5 2 y 2 5 3 2 2 2 5 3 5 y , 3 x Despite the fact that the result is a much more complicated tetrad, this is an amazing result: the Weyl scalars are all fixed (see below). Let's first verify that this is indeed a tetrad, and that now the Weyl scalars are in canonical form > > IsTetrad (478) Type of tetrad: null true (479) (479) Set (478) to be the tetrad in use and recompute the Weyl scalars > > Setup tetrad = (478) : Indeed we now have 0 = 0, 3 = 1 and 4 = 0 > > simplify Weyl scalars 1 3 I 2 2 1 I 0 = 0, 1 = , 2 = , 3 = 1, 4 = 0 (480) (480) a 4 a 2 So Weyl scalars computed after setting the canonical tetrad (478) to be the tetrad in use are in canonical form. Great! NOTE: computing the canonical Weyl scalars is not really the difficult part, and within the code, these scalars (480) are computed before arriving at the tetrad (478) . What is really difficult (from the point of view of computational complexity and simplifications) is to compute the actual canonical form of the tetrad (478) . > > Petrov type II Consider this other solution to Einstein's equation (again, the numbers in g_[[24,37,7]] always refer to the equation number in the " Exact solutions to Einstein's field equations " textbook) > > g_ 24, 37, 7 _______________________________________________________ Systems of spacetime Coordinates are: X = u , v , x , y Default differentiation variables for d_, D_ and dAlembertian are: X = u , v , x , y The Stephani metric in coordinates u , v , x , y Parameters: f x , a , 1 u , x , y Comments: Case 6 from Table 24.1:Psi1(u, x, y): diff(Psi1(u, x, y), x, x) diff(Psi1(u, x, y), y, y) = 0, diff(x*diff(_M(u, x, y), x), x) x*diff(_M(u, x, y), y, y) = _kappa0* (diff(_Psi(u, x, y), x)^2 diff(_Psi(u, x, y), y)^2)

  89. _______________________________________________________ 2 x f x y a x 0 0 x 0 0 0 1 0 0 0 g , = (481) (481) x 1 0 0 0 x Check the Petrov type > > PetrovType "II" (482) (482) The starting tetrad > > e_ x f x y a 0 0 0 x x f x y a 0 0 f x y a I 2 2 = (483) (483) 2 a , 0 0 2 x 1 4 x 1 4 I 2 2 2 0 0 2 x 1 4 x 1 4 results in Weyl scalars not in canonical form: > > Weyl scalars 1 3 I a 2 x f x 3 f x 0 = 0, 1 = 0, 2 = 2 , 3 = 0, 4 = (484) (484) 8 x 3 x 4 y a 4 f x For Petrov type "II", the canonical form is as for type "I" but in addition 1 = 0. Again let's assume positive, not necessary, but to get simpler formulas around > > Assume f x 0, x 0, y 0, a 0 a :: 0, , x :: 0, , f x :: , 0 , f x :: 0, , y :: 0, (485) (485) Compute now a canonical form for the tetrad, to be used instead of (483) > > TransformTetrad canonicalform (486) (486)

  90. 3 3 I a 2 x f x 3 f x , 0, 0, 0 , (486) (486) 8 x 3 x 3 2 x 3 I a 8 2 x f x 3 f x 3 y a 3 f x , 9 3 I a 2 x f x 3 f x 3 x 3 2 8 , 3 3 I a 2 x f x 3 f x 4 x 5 4 2 3 I a 2 x f x 3 f x 3 I a 2 x f x 3 f x , 3 3 I a 2 x f x 3 f x 3 I a 2 x f x 3 f x 2 x 5 4 2 I f 4 x x 3 I f x I 3 I a 2 x f x 3 f x 3 a , , 3 3 I a 2 x f x 3 f x 3 I a 2 x f x 3 f x 3 I a 2 x f x 3 f x 3 x , 0, 3 3 I a 2 x f x 3 f x 2 4 , x 1 2 3 I a 2 x f x 3 f x I 2 3 I a 2 x f x 3 f x 2 , , x 1 4 3 I a 2 x f x 3 f x 3 I a 2 x f x 3 f x 3 x , 0, 3 3 I a 2 x f x 3 f x 2 4 , x x 1 2 3 I a 2 x f x 3 f I 2 3 I a 2 x f x 3 f x 2 , x x 1 4 3 I a 2 x f x 3 f Set this tetrad and check the Weyl scalars again > > Setup tetrad = (486) : > > Weyl scalars 1 0 = 0, 1 = 0, 2 = 2 , 3 = 1, 4 = 0 (487) (487) 8 x 3 This result (487) is fantastic. Compare these Weyl scalars with the ones (484) before

  91. transforming the tetrad. > > * Petrov type III > > g_ 12, 35, 1 _______________________________________________________ Systems of spacetime Coordinates are: X = u , x , y , z Default differentiation variables for d_, D_ and dAlembertian are: X = u , x , y , z The Kaigorodov (1962) metric in coordinates Cahen (1964) Parameters: _______________________________________________________ Warning, for the signature - , that is with the timelike component in position 1 , the spacetime metric indicated has g 0, 0 = g 1, 1 = 0 , and so the corresponding system of reference cannot be realized with real bodies (e.g. you cannot define proper time nor synchronize clocks in any infinitesimal region of space). Note as well that the corresponding 3 -dimensional space metric is singular. e 2 z 0 0 0 e 2 z e 4 z 2 e z 0 g , = (488) (488) 2 e z 2 e 2 z 0 0 3 0 0 0 > > Assume z 0, Lambda 0 :: 0, , z :: 0, (489) (489) The Petrov type and the original tetrad > > PetrovType "III" (490) (490) > > SegreType "O", "[(1,111)]" (491) (491) > > e_

  92. I I 2 e 4 z 2 e 2 z I e z 2 2 2 2 1 0 I I 2 e 4 z 2 2 e 2 z I e z 1 2 2 2 0 = (492) (492) I 2 3 2 e 4 z a , 2 0 0 2 I 2 3 2 e 4 z 2 0 0 2 This tetrad results in the following scalars > > Weyl scalars 0 = 11 2 3 3 2 2 2 , 1 = , 2 = 4 , 3 = , (493) (493) 4 4 2 4 4 2 = 11 2 2 4 that are not in canonical form, which for Petrov type III is as in Petrov type II but in addition we should have 2 = 0. Compute now a canonical form for the tetrad > > TransformTetrad canonicalform I I 3 2 e 2 z 2 e z 0 2 2 2 I 13 I 9 I 2 e 4 z 2 e 2 z 2 e z 2 8 8 7 3 3 2 8 (494) (494) 3 I I 3 2 e 2 z 2 e z 0 4 4 4 3 I I 3 2 e 4 z 2 e 2 z 2 e z I 4 4 4 Set this one to be the tetrad in use and recompute the Weyl scalars > > Setup tetrad = (494) : > > Weyl scalars 0 = 0, 1 = 0, 2 = 0, 3 = 1, 4 = 0 (495) (495)

  93. > > Petrov type N > > g_ 12, 6, 1 _______________________________________________________ Systems of spacetime Coordinates are: X = u , v , y , z Default differentiation variables for d_, D_ and dAlembertian are: X = u , v , y , z The Defrise (1969) metric in coordinates u , v , y , z Parameters: , 0 Comments: Lambda 0 required for a pure radiation solution _______________________________________________________ Warning, for the signature - , that is with the timelike component in position 1 , the spacetime metric indicated has g 0, 0 = g 1, 1 = 0 , and so the corresponding system of reference cannot be realized with real bodies (e.g. you cannot define proper time nor synchronize clocks in any infinitesimal region of space). Note as well that the corresponding 3 -dimensional space metric is singular. 3 0 0 0 2 y 2 3 3 0 0 2 y 2 y 4 g , = (496) (496) 3 0 0 0 y 2 3 0 0 0 y 2 > > Assume y 0, Lambda 0 :: 0, , y :: 0, (497) (497) > > PetrovType "N" (498) (498) > > SegreType "O", "[(2,11)]" (499) (499) The original tetrad and related Weyl scalars are not in canonical form:

  94. > > e_ I I 4 2 3 2 2 3 2 3 0 y 2 2 y I I 4 2 3 2 2 3 2 3 0 y 2 = (500) (500) 2 y a , 2 3 2 3 0 0 4 2 y 2 3 2 3 0 0 4 2 y > > Weyl scalars 1 = I I 0 = 4 , 4 , 2 = 4 , 3 = 4 , 4 = (501) (501) 4 For Petrov type "N", the canonical form has 0 and all the other n = 0. 4 Compute a canonical form, set it to be the tetrad in use and recompute the Weyl scalars > > TransformTetrad canonicalform I 2 2 3 0 0 0 y 2 I 2 2 3 I 2 3 2 3 0 y 2 y (502) (502) I I 2 2 3 2 2 3 2 3 0 y 2 2 y y I I 2 2 3 2 2 3 2 3 0 y 2 2 y y > > Setup tetrad = (502) : > > Weyl scalars 0 = 0, 1 = 0, 2 = 0, 3 = 0, 4 = 1 (503) (503)

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