(a) We desribe physics as a sequence of events labelled by their space time coordinates: x µ = (x 0, x 1, x 2 x 3 ) = (c t, x) (12.

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1 2 Relativity Postulates (a) All inertial observers have the same equations of motion and the same physial laws. Relativity explains how to translate the measurements and events aording to one inertial observer to another. (b) The speed of light is onstant for all inertial frames 2. Elementary Relativity Mehanis of indies, four-vetors, Lorentz transformations (a) We desribe physis as a sequene of events labelled by their spae time oordinates: x µ = (x 0, x, x 2, x 3 ) = ( t, x) (2.) The spae time oordinates of another inertial observer moving with veloity v relative to the first measures the oordinates of an event to be x µ = (x 0, x, x 2 x 3 ) = ( t, x) (2.2) (b) The oordinates of an event aording to the first observer x µ determine the oordinates of an event aording to another observer x µ through a linear hange of oordinates known as a Lorentz transformation: x µ x µ = L µ ν(v)x ν (2.3) I usually think of x µ as a olumn vetor so that without indies the transform x 0 x x 2 (2.4) x 3 ( x ) (x) = ( L ) (x) (2.5) where L is the a matrix and (x) signifies olumn vetors like Eq. (2.4) Then to hange frames from K to an observer K moving to the right with speed v relative to K the transformation matrix is γ γβ (L) = (L µ ν ) = γβ γ (L) µ ν = L µ ν (2.6) with β = v/ and γ = / β 2. Here L 0 = γβ is the entry in the 0 -th row and -st olumn 5

2 52 CHAPTER 2. RELATIVITY A short exerise done in lass shows that a this boost ontrats the x + x 0 +x diretion (i.e. t+x) and expands the x x 0 x diretion (i.e. t x). Thus, x + and x are eigenvetors of Lorentz boosts in the x diretion x + β = + β x+ (2.7) x + β = β x (2.8) () Instead of using v we sometimes use the rapidity y and note that y β for small β tanh y = v or y = 2 ln + β β (2.9) With this parametrization we find that the Lorentz boost appears as a hyperboli rotation matrix osh y sinh y (L) = (L µ ν) = sinh y osh y (2.0) Then x + = e y x + x = e y x (2.) (d) Sine the spead of light is onstant for all observers we demand that (t) 2 + x 2 = (t) 2 + x 2 (2.2) under Lorentz transformation. follow (group) requirements: We also require that the set of Lorentz transformations satisfy the L( v)l(v) =I (2.3) L(v 2 )L(v ) =L(v 3 ) (2.4) here I is the identity matrix. These properties seem reasonable to me, sine if I transform to frame moving with veloity v and then transform bak to a frame moving with veloity v, I shuld get bak the same result. Similarly two Lorentz transformations produe another Lorentz transformation. (e) Sine the ombination (t) 2 + x 2 (2.5) is invariant under lorentz transformation, we introdued an index notation to make suh invariant forms manifest. We formalized the lowering of indies with a metri tensor: In this way we define a dot produt x µ = g µν x ν x µ = ( t, x) (2.6) g 00 = g = g 22 = g 33 = (2.7) x x = x µ x µ = (t) 2 + x 2 (2.8) is manifestly invariant. Similarly we raise indies x µ = g µν x ν (2.9)

3 2.. ELEMENTARY RELATIVITY 53 with g µν = (2.20) Of ourse the proess of lowering and index and then raising it agiain does nothing: g µ ν = g µσ g σν = δ µ ν = identity matrix = (2.2) (f) Generally the upper indies are the normal thing. We will try to leave the dimensions and name of the four vetor, orresponding to that of the spatial omponents. Examples: x µ = (t, x), A µ = (Φ, A), J µ = (ρ, j), and P µ = (E/, p). (g) Four vetors are anything that transforms aording to the lorentz transformation A µ = (A 0, A) like oordinates A µ = L µ νa ν (2.22) Given two four vetors, A µ and B µ one an always onstrut a Lorentz invariant quantity. A B = A µ B µ = A µ g µν B ν = A 0 B 0 + A B = A 0 B 0 + A B = A µ g µν B µ = A µ B µ = A B (2.23) (h) Notation. We denote the transformation matrix (L) (2.24) A matrix just has rows and olumns and has no idea what is a row with an upper index µ versus a lower index Then entries (L) µν of the matrix are labelled by rows (µ) and olumns (ν). You are free to move this row and olumn index up and down at will the first index labels the row, the seond the olumn. In this way (L) µν = ( L ) νµ = (L)µ ν = ( L ) µ ν = L µ ν (2.25) is all the same numerial number L µ ν for speified µ and ν. However, the muh preferred plaemnt of the indies surrounding the matrix is just a visual reminder of the individual entries L µ ν whih together form the matrix, (L) and (L ), and that is all, e.g. x µ = L µ νx ν = (L) µ νx ν = x ν (L ) µ ν (2.26) The indies labelling L µ ν an not be raised and lowered randomly, but are raised and lowered with the metri tensor, i.e. multiplying the matrix (L) with the matrix (g). Thus and (gl) µν = g µρ L ρ ν L µν (2.27) (glg) ν µ = g µρl ρ σg σν L ν µ (2.28) (i) From the invariane of the inner produt we see that the lower (ovariant) omponents of four vetors transform with the inverse transformation and as a row, I usually think of x µ (with a lower index) as a row x µ x ν = x µ (L ) µ ν. (2.29) (x 0 x x 2 x 3 ) (2.30)

4 54 CHAPTER 2. RELATIVITY So the transformation rule in terms of matries is (x 0 x x 2 x 3 ) = (x 0 x x 2 x 3 ) (L ) (2.3) In this way the inner produt A µ B µ = (A 0 A A 2 A 3 ) (L )( L) B 0 B B 2 B 3 = A µb µ (2.32) is invariant. If you wish to think of x µ as a olumn, then it transforms under lorentz transformation with the inverse transpose matrix x 0 ) x 0 x x 2 (L = x x 2 (2.33) x 3 x 3 (j) As is lear from Eq. (2.23), the metri tensor is an invariant tensor, i.e. g µν = L µ ρl ν σg ρσ = (L) µ ρ(l) ν σg ρσ (2.34) is the same tensor diag(,,, ) in all frames (so I dont need to put an underline g µν on the LHS). From Eq. (2.34) it follows that the inverse (transpose) Lorentz transform an be found by raising and lowering the indies of the transform matrix, i.e. L σ ρ g ρµ L µ νg νσ = (L ) σ ρ = (L ) σ ρ (2.35) where we have defined Lρ σ. Thus if one wishes to think of a lowered four vetor A µ as a olumn, one has A ν = L ν µ A µ (2.36) Thus, a short exerise (done) in lass shows that if then there is a onsisteny hek T µν =L ν ρl µ σt σρ (2.37) =(L) µ σ T σρ (L ) ν ρ (2.38) T µ ν =Lµ σ L ρ ν T σ ρ (2.39) =(L) µ σ T σ ρ (L ) ρ ν (2.40) i.e. that lower indies transform like rows with the inverse matrix (L ) upstairs indies transform like olumns with the regular matrix (L). Doppler shift, four veloity, and proper time. (a) The frequeny and wave number form a four vetor K µ = ( ω, k), with k = ω/. This an be used to determine a relativisti dopler shift. (b) For a partile in motion with veloity v p and gamma fator γ p, the spae-time interval is ds 2 dx µ dx µ = (dt) 2 + dx 2 = (dτ) 2. (2.4)

5 2.. ELEMENTARY RELATIVITY 55 ds 2 is assoiated with the liks of the lok in the partiles instantaneous rest frame, ds 2 = (dτ) 2, so we have in any other frame dτ ( ) 2 dx ds 2 / = dt / dt 2 (2.42) = dt γ p (2.43) () The four veloity of a partile is the distane the partile travels per proper time so Note U µ U µ = 2. U µ dxµ dτ = (u0, u) = (γ p, γ p v p ) (2.44) U µ = L µ νu ν (2.45) (d) The transformation of the four veloity under Lorentz transformation should be ompared to the transformation of veloities. For a partile moving with veloity v p in frame K, then in another frame K moving to the right with speed v the partile moves with veloity v p = v p v (2.46) v pv/ 2 v p = v p γ p ( v pv/ 2 ) (2.47) where v p and vp are the omponents of v p parallel and perpendiular to v. These are easily derived from the transformation rules of U µ and the fat that v p = u/u 0. Energy and Momentum Conservation (a) Finally the energy and momentum form a four vetor ( ) E P µ =, p (2.48) The invariant produt of P µ with itsself the rest energy P µ P µ = (m) 2 (2.49) This an be inverted giving the energy in terms of the momentum, i.e. the dispersion urve E(p) = p 2 + (m) 2 (2.50) (b) The relation between energy and momentum determines the veloity. At rest E = m 2. Then a boost in the negative v p diretion shows that a partile with veloity v p has energy and momentum ( ) E P µ =, p = m (γ p, γ p β p ) = mu µ (2.5) i.e. v p = p (E/) = E(p) p Thus as usual the derivative of the dispersion urve is the veloity. (2.52)

6 56 CHAPTER 2. RELATIVITY () Energy and Momentum are onserved in ollisions, e.g. for a reation w have P µ + P µ 2 = P µ 3 + P µ 4 (2.53) Usually when working with ollisions it makes sense to suppress or just make the assoiation: E E p is short for p (2.54) m m 2 A starting point for analyzing the kinematis of a proess is to square both sides with the invariant dot produt P 2 P P. For example if P + P 2 = P 3 + P 4 then: (P + P 2 ) 2 =(P 3 + P 4 ) 2 (2.55) P 2 + P P P 2 =P P P 3 P 4 (2.56) m 2 m 2 2 2E E 2 + 2p p 2 = m 2 3 m 2 4 2E 3 E 4 + 2p 3 p 4 (2.57)

7 2.2. COVARIANT FORM OF ELECTRODYNAMICS Covariant form of eletrodynamis (a) The players are: i) The derivatives µ ( x µ = µ x µ = ( t, ) t, ) (2.58) (2.59) ii) The wave operator iii) The four veloity U µ = (u 0, u) = (γ p, γ p v p ) iv) The urrent four vetor v) The vetor potential vi) The field strength is a tensor whih ultimately omes from the relations In indies we have = µ µ = 2 t (2.60) J µ = (ρ, J) (2.6) A µ = (Φ, A) (2.62) F αβ = α A β β A α (2.63) E = ta Φ (2.64) B = A (2.65) F 0i =E i E i =F 0i (2.66) F ij =ɛ ijk B k B i = 2 ɛ ijkf jk (2.67) In matrix form this anti-symmetri tensor reads 0 E x E y E z F αβ = E x 0 B z B y E y B z 0 B x (2.68) E z B y B x 0 Raising and lowering indies of F µν an hange the sign of the zero omponents, but does not hange the ij omponents, e.g. E i = F 0i = F i0 = F i 0 = F i 0 = F 0i = F 0 i = F 0i (2.69) vii) The dual field tensor implements the replaement As motivated by the maxwell equations in free spae E B B E (2.70) E =0 (2.7) te + B =0 (2.72) B =0 (2.73) tb E =0 (2.74)

8 58 CHAPTER 2. RELATIVITY whih are the same before and after this duality transformation. The dual field stength tensor is 0 B x B y B z F αβ = B x 0 E z E y B y E z 0 E x (2.75) B z E y E x 0 The dual field strength tensor F µν = 2 ɛµνρσ F ρσ (2.76) where the totally anti-symmetri tensor ɛ µνρσ is + even perms 0,,2,3 ɛ µνρσ = odd perms 0,,2,3 (2.77) 0 0 otherwise viii) The stress tensor is Or in terms of matries Θ µν em = F µλ F ν λ + g µν ( 4 F αβf αβ) (2.78) Θ µν em = u em S em / S em / T ij (2.79) Note that Θ 0i = S i em/ = g i em, and T ij = ( E i E j + 2 δij E 2 ) + ( B i B j + 2 δij B 2 ). You an remember the stress tensor Θ µν by realling that it is quadrati in F, symmetri under interhange of µ and ν, and traeless Θ µ µ = 0. These properties fix the stress tensor up to a onstant. (b) The equations are i) The ontinuity equation: µ J µ =0 (2.80) t ρ + J =0 (2.8) ii) The wave equation in the ovariant gauge A µ =J µ / (2.82) Φ =ρ (2.83) A =J/ (2.84) This is true in the ovariant gauge µ A µ =0 (2.85) tφ + A = 0 (2.86) iii) The fore law is: dp µ dτ = ef µ U ν ν (2.87) de dt =ee v (2.88) dp dt =ee + ev B (2.89) If these equations are multiplied by γ they equal the relativisti equations to the left.

9 2.2. COVARIANT FORM OF ELECTRODYNAMICS 59 iv) The soured field equations are : µ F µν = J ν (2.90) te + B = J E =ρ (2.9) (2.92) v) The dual field equations are : µ F µν =0 (2.93) B =0 (2.94) tb E =0 (2.95) as might have been inferred by the replaements E B and B E. The dual field equations an also be written in terms F µν, and this is known as the Bianhi identity: where ρ, µ, ν are yli. Or (for the mathematially inlined) the Bianhi identity reads ρ F µν + µ F νρ + ν F ρµ = 0, (2.96) [µ F µ2µ 3] = 0, (2.97) where the square brakets denote the fully antisymmetri ombination of µ, µ 2, µ 2, i.e. the order is like a determinant [µ F µ2µ 3] 3![ ( µ F µ2µ 3 µ2 F µµ 3 + µ3 F µµ 2 ) + ( µ F µ3µ 2 + µ2 F µ3µ µ3 F µ2µ ) ] (2.98) The seond line is the same as the first sine F µν is antisymmetri. Eq. (2.97) is the statement that F µν is an exat differential form. vi) The dual field equations are equivalent to the statement that that F µν (or E, B) an be written in terms of the gauge potential A µ (or Φ, A) F µν = µ A ν ν A µ (2.99) B = A (2.00) E = ta Φ (2.0) The potentials are not unique as we an always make a gauge transform: A µ A µ + µ Λ (2.02) A A + Λ (2.03) Φ Φ + tλ (2.04) vii) The onservation of energy and momentum an be written in terms of the stress tensor: ( ) u em + (S µ Θ µν em = F µ J ν em /) = E J/ (2.06) t ν (2.05) ( S j ) em/ + i T ij = ρe j + (J/ B) j (2.07) t The energy and momentum transferred from the fields F µν to the partiles is µ Θ µν meh =F µ J ν ν (2.08)

10 60 CHAPTER 2. RELATIVITY Or µ Θ µν meh + µθ µν em =0 (2.09)