From (2) follows, if z0 = 0, then z = vt, thus a2 =?va (2.3) Then 2:3 beomes z0 = z (z? vt) (2.4) t0 = bt + b2z Consider the onsequenes of (3). A ligh


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1 Chapter 2 Lorentz Transformations 2. Elementary Considerations We assume we have two oordinate systems S and S0 with oordinates x; y; z; t and x0; y0; z0; t0, respetively. Physial events an be measured in either system, and the Lorentz transformation give the relation between the oordinates x0; y0; z0; t0 and x; y; z; t. The transformation has to fulll the following requirements:. The transformation should be linear. Otherwise a spei system S, or a point in spae or time would be distint. (For a linear transformation, the inverse has the same form.) 2. Eah point in R 3 given by x0; y0; z0 in S0 moves with onstant veloity ~v with respet to a point x; y; z in S. 3. A measurement of the speed of light should give C in both systems. We assume the uniform motion is in z?diretion. Then x0 = x ; y0 = y : (2.) From () follows z0 = az + a2t (2.2) t0 = bt + b2z 26
2 From (2) follows, if z0 = 0, then z = vt, thus a2 =?va (2.3) Then 2:3 beomes z0 = z (z? vt) (2.4) t0 = bt + b2z Consider the onsequenes of (3). A light pulse is sent at and x = y = z = t = 0 in systems S (2.5) x0 = y0 = z0 = t0 = 0 in systems S0 ; (2.6) where S is the rest frame and S0 some moving frame. In both systems, the speed of light is given by, i.e., From this follows rest frame S : moving frame S0 : q x 2 + y 2 + z 2 = t (2.7) q x0 2 + y0 2 + z0 2 = t0 (2.8) x0 2 + y0 2 + z0 2 = x 2 + y 2 + a 2 (z? vt) 2 (2.9) = x 2 + y 2 + a 2 z? v q x 2 + y 2 + z 2 2 = 2 t0 2 = 2 (bt + b2z) 2 = 2 b q x 2 + y 2 + z 2 + b2z! 2 This must be idential for x; y; z and x 2 + y 2 ; z p x 2 + y 2 + z 2 ; z 2. Terms proportional to z x 2 + y 2 :? a 2 q x 2 + y 2 + z 2 :? a2 v 2 2 = 2 b2 2 (2.0) = 2 b b2! 2 z 2 : a 2 + v2 = 2 b b2 2! 27
3 These are three equations for three unknowns, whih give for a; b; b2: a = b = q? v2 2 (2.) b2 =? v 2 a Thus the Lorentz transformations beome z0 = z? vt q? v2 2 (2.2) t0 = t? v 2 z q? v2 2 The inverse transformations are obtained by replaing v with?v : z0 + vt0 z = (2.3) q? v2 2 t = t0 + v z0 q 2? v Lorentz Transformations as Orthogonal Transformations in Four Dimensions A Lorentz transformation leaves the expression invariant. Proof by inserting x 2 + x x 2 3? 2 t 2 (2.4) x0 2 + x0 2 + x0 2 3? 2 t0 2 = x 2 + x 2 + (x 3? vt) 2? 2 (t? v x2) 2 2 (2.5)? v2? v2 2 2 = x 2 + x x 2 3? 2 t 2 28
4 Dening x4 : it, the invariant quantity an be written as 4X = x 2 = x x = x 0 x0 (2.6) This means that the length of a vetor x in R 4 remains unhanged under Lorentz transformations. Thus, a Lorentz transformation an be interpreted as rotation in a fourdimensional spae, x 0 = a x (2.7) with a a =. For the speial ase that S0 moves with v in x3?diretion, one has a () = a (?) = 0 B i 0 0?i C A (2.8) where v and = p? 2. Let us onsider vetors and tensors in a fourdimensional spae. A is a fourdimensional vetor, if its omponents transform as the oordinates of a fourdimensional vetor spae, i.e., A 0 = a A : (2.9) A tensor of rank 2 transforms as A 0 = a a A : (2.20) Those vetors have a speial meaning. Sine aording to Einstein's priniple of relativity all systems moving with a uniform veloity are equivalent, the laws of physis must obey the same equations in e.g., system S and system S0. Compare rotations in three dimensions. The laws of physis are independent from the position of one oordinate system with respet to the other. Physial quantities are desribed by salars, vetors and tensors, and the physial laws are given by ombination of those quantities. Sine the physis is independent from the position of the oordinate system, the form of the equation is the same in eah oordinate system. Sine the laws of physis an be desribed in eah moving frame, and aording to speial relativity all moving frame are equivalent, the laws of physis must be desribed by 29
5 equations whih do not hange when going from one frame to another. Thus, they must be expressed as salars, vetors and tensors in a fourdimensional spae. (The Maxwell equations are Lorentz invariant. Newtonian mehanis not.) The quantity A B is a foursalar. If A A = 0 in a Minkovski metri (imaginary omponents), then this does not imply that all omponents are zero. A vetor is spaelike if A A > 0, and timelike if A A < 0. For the vetor produt in three dimension, the totally antisymmetri " tensor " ijk = i i2 i3 j j2 j3 k k2 k3 (2.2) with " ijk " lmk = il im = il jm? jl im (2.22) jl jm played a speial role. Analogously we introdue a fourdimensional totally antisymmetri tensor " = (2.23) with " " = (2.24) = + +??? " " = 2? 2 Now we an dene a `vetor produt' C = " A B (2.25) 30
6 whih is an antisymmetri tensor of rank 2 with six independent omponents. antisymmetri tensor of rank 2 an be assoiated with a dual tenor Every ~C = 2 " C (2.26) with ~ C = C. If C stands for a vetor produt, then ~C = A B? A B : (2.27) Most operations an be arried through from three to four dimensions x gradient vetor (2.28) A x gradient = salar R = " A x = antisymmetri tensor of rank 2 ~R = A x? A x 3
7 2.3 Eletrodynamis in Four Vetors From the vetor potential A ~ and the salar potential, one an onstrut a four vetor A = (A; A2; A3; i) with A x = ~ r ~ A + t = 0 ; (2.29) whih states that the divergene of the four potential to zero (Lorentz ondition). Similarly one introdues a four urrent J = (J; J2; J3; i ), and the ontinuity equation beomes J x = r ~ J ~ + t = 0 : (2.30) The wave equation an be written as 2A =? 4 J ; (2.3) where 2 := 2 x x =? 2 2 t 2 : (2.32) This equation is idential with the two equations? 2 2 t 2!? 2 t 2! ~A =? 4 =?4 ~J (2.33) We now dene a totally antisymmetri tensor of rank 2 F := A x? A x (2.34) and the orresponding dual tensor 32
8 ~F = 2 " F = " A x (2.35) Expliitly F is given by F = ~F = 0 B 0 B 0 B3?B2?iE?B3 0 B?iE2 B2?B 0?iE3 ie ie2 ie3 0 0?iE3 ie2 B ie3 0?iE B2?iE2 ie 0 B3?B?B2?B 3 0 C A = " ijk B k ie i ie i 0 C A =?i" ijk E k B i?b i 0!! (2.36) These tensors allow to write the Maxwell equations in a very ompat fashion F ~ = 0, r E ~ + x F x = 0, r ~ B? B t E t = 0 ; r ~ B = 0 (2.37) = 0 ; r ~ E = 4 33
9 2.4 Lorentz Transformations in a Four Vetor Notation Aside of setting x4 = it, it is also ustomary to use ~x = (x; x2; x3) and x 0 = x = t. If one onsiders x i =? x i ; i = ; 2; 3, then the invariant line element 2 t 2? x 2? x 2 2? x 2 3 an be written as x x = 2 t 2? ~x 2. x := (x0; x; x2; x3) = (x 0 ;?x ;?x 2 ;?x 3 ) is the ovariant four vetor, x := (x 0 ; x ; x 2 ; x 3 ) is the ontravariant four vetor. The invariant line elements is written as where g is the ovariant metri tensor with g = d 2 = g dx dx (2.38) 0 B ? ? ? C A : (2.39) In speial relativity the metri tensor is onstant. The ontravariant metri tensor g is given by the relation g g = (2.40) whih is equivalent to g = g. Thus one has x = g x (2.4) x = g x (2.42) (2.43) Salars, vetors and tensors are given by their transformation properties. An arbitrary tensor of higher rank may have ovariant as well as ontravariant omponents, e.g., A = g A : (2.44) 34
10 2.5 Elements of Speial Relativity Expliitly a Lorentz transformation is given as x 0 = x ; (2.45) whih leaves the salar produt x 2 = (x 0 ) 2? ~x 2 (2.46) invariant. From the invariane of the salar produt, i.e., x 02 = x 2 follows T g = g : (2.47) This ondition is fullled by matries of the form = 0 B osh? sinh 0 0? sinh osh C A (2.48) The parameter is alled rapidity and depends on the veloity ~v. Expliitly tanh = v : (2.49) and osh = q =? ( v )2 sinh = v q =? ( v )2 (2.50) If one adds to the Lorentz transformations (2.45) translations in spae and time x 0 = x + a (2.5) where a = a 0 ~a! (2.52) is an arbitrary fourvetor, one obtains the symmetry group whih haraterizes the theory of speial relativity: x 0 = x + a with T g = g : (2.53) 35
11 This group is alled Poinare or inhomogeneous Lorentz group. A general approah to relativisti quantum mehanis an be designed the following way: Aording to the general priniples of quantum mehanis, the Poinare group has to be represented in the Hilbert spae of quantum states as unitary (or antiunitary) operators: (; a)?! U(; a) : (2.54) The problem is then to determine the irreduible (unitary) representations of the Poinare group. This problem was solved in 939 by Wigner. He obtained the result: For every real number m 0 and eah j = 0; 2 ; ; 2 3 ; exists an irreduible representation of the Poinare group haraterized by (m; j). The "quantum numbers" m and j an here be assoiated with the mass and spin of a orresponding partile (partile physis). A more elementary (and historial) approah is to look for a Shrodinger equation, whih fullls Lorentz ovariane. Obviously, one annot start from the wellknown expressions ih d dt j s(t)i = H( ~ P ; ~ Q) j s (t)i (2.55) or d dt A H(t) = i h [H( ~ P ; ~ Q); A h (t)] ; (2.56) sine both equations ontain time t and spae Q ~ in an asymmetrial fashion, namely t as number and Q ~ as hermitian operator. A symmetri form is only ahieved if either t beomes an operator or Q a number vetor. In the rst ase, one would have to introdue in analogy to [P j ; Q k ] = h i jk (2.57) the ommutator [H; t] = h i : (2.58) However, the latter ontradits the physial requirement that the spetrum of H has to be bounded from below. Thus, we hoose the seond possibility and work exlusively in the oordinate representation, where the operator ~ Q appears only as number vetor ~r, and the wave funtion is represented as Here spae and time enter on the same footing. (x) (x 0 ; x ; x 2 ; x 3 ) = (t; ~x) : (2.59) 36
12 The dierent equation for (x), whih has to orrespond to (2.55), has to ontain the relativisti relation between energy and momentum, i.e., for a partile moving without the inuene of a fore, the relation q E = ~p 2 + (m) 2 (2.60) has to be valid. Energy and momentum are in quantum mehanis replaed as E?! ih t ~p?! h i ~r ~r : (2.6) The above relations orrespond to E = h! ~p = h ~ k (2.62) whih reet the translational invariane with respet to spae and time, and whih shall stay valid in a relativisti quantum mehanis. In order to bring the relations (2.6) into a formal ovariant form, dimensions and signs have to be onsidered: E := p 0?! ih (t) = ih x 0?(~p) k := p k?!? h i (~ r ~r ) k = ih x k (k = ; 2; 3) (2.63) As a Reminder: The derivative with respet to the ontravariant omponents form the ovariant omponents of a fourvetor. Consider x applied to a salar funtion of x 2 = x x x f(x2 ) = f 0 (x 2 ) x (x x ) = f 0 (x 2 ) 2x : (2.64) Thus, x := ; = 0; ; 2; 3 and p?! ih ; (2.65) whih shows that p transforms as a ovariant vetor. Inserting (2.63) into (2.60) leads to q E? ~p 2 + (m) 2?! ih 0? 2 k + (m) 2 = ih 0? 37 s?h 2 X k q?h 2 + (m) 2 : (2.66)
13 The last expression is unfortunately again asymmetri with respet to the dierentiation with respet to time and spatial oordinates. This asymmetry an be avoided when starting from the squared energy momentum relation E 2 = ~p 2 + (m) 2 (2.67) in the form (m) 2 = E 2? ~p 2 = p 2 0? ~p 2 = p p : (2.68) One obtains: p p? (m) 2?! (ih) 2? (m) 2 =?h " 2 m + h 2 # (2.69) Here the derivatives with respet to time and spae oordinates are symmetri, and one denes 2 := = 2 t 2? : (2.70) Aording to the previous onsiderations, the relativisti relation between energy and momentum is fullled, if the wave funtion (x) obeys the dierential equation 2 + m h 2! whih is the soalled KleinGordon equation. Remark on Natural Units: (x) = 0 ; (2.7) In priniple, one would like to have fundamental units for spae, time and mass, whih are derived from the fundamental laws of physis. elementary length `0 elementary time t0 elementary mass m0 A modern andidate for `0 is the Plank length s gh `P lank = 2 = :6 0?33 m (2.72) (g: gravitational onstant) 38
14 Even if `0 is not expliitly xed, one an use and h to x units for t0 and m0: t0 = `0 := elementary time = time, whih light needs to pass `0. m0 = h `0 := elementary mass from the Compton relation. If one hooses `0; t0 and m0 as elementary units, then: The speed of light has the numerial value : =. The Plank onstant h has the numerial value : h =. Starting from these natural units, all other units an be related to the unit of length, e.g., m. Some important relations aquire the form q E = p 2 + m 2 E =! ~p = ~ k = e 2 ~ = e 2m g ~ s S (magneti moment) (2 + m 2 ) (x) = 0 (Klein? Gordon equation) (2.73) 39
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