Relativisti Kinematis -a projet in Analytial mehanis Karlstad University Carl Stigner 1th January 6 Abstrat The following text is a desription of some of the ontent in hapter 7 in the textbook Classial Mehanis, by H Goldstein, C Poole and J Safko. The purpose is to desribe som aspets of relativisti kinematis. To reah this goal, the onept of four-vetors will be introdued. While doing this it will also be a short disussion on tensors. Two sorts of ollisions will be disussed.- Elasti and inelasti ollisions. The first in more detail than the latter. We will see that there is a onsiderable differene in how muh energy we need to put into a system in order to produe new matter in different situations. There will also be a desription on how to extend the onept of angular momentum to the speial theory of relativity. An analouge to the enter of mass will also be disussed. 1
CONTENTS CONTENTS Contents I Basi onepts 3 1 The Einstein summation onvention 3 Four-vetors 3.1 Four-vetors.......................................... 3.1.1 The position four-vetor............................... 3.1. The veloity four-vetor............................... 4.1.3 The momentum four-vetor............................. 4.1.4 The Minskowski Fore................................ 4. The metri tensor, salar produt and invariant quantities.................. 4.3 Why four-vetors?...................................... 5 3 Tensors and one-forms 6 3.1 Definition of a tensor..................................... 6 3. 1-forms............................................ 6 3.3 Tensorprodut and wedge produt.............................. 6 II Relativisti kinematis and systems of partiles 7 4 Sattering proesses 7 4.1 Inelasti sattering...................................... 7 4.1.1 Prodution of an anti-proton............................. 9 4.1. Veloity of C.M- system and kineti energy energy in the laboratory- frame.... 9 4. Elasti sattering....................................... 9 5 Angular momentum 11 5.1 Total angular momentum, onservation............................ 1 5.1.1 Translation of angular momentum.......................... 1 5.1. Center of energy................................... 13
FOUR-VECTORS Part I Basi onepts In this first part the reader is reminded of the basi onepts of the speial theory of relativity, like the Minskowski spae, the Lorentz-transformation and the metri tensor. We also introdue the onvenient four-vetors. There will also be a short disussion on tensors, tensor produts and what s alled 1-forms. 1 The Einstein summation onvention In this text the Einstein summation onvention is used. That is whenever an index is repeated in an expression, one subsript and one supersript, in one term, a summation is implied. Four-vetors Remember that a point in the 4-dimensional spae-time (The Minskowski spae) is desribed by the four oordinates, t, x, y, z, where t is the time in the speifi inertial frame S, in whih we are desribing the system, and x, y, z are the usual spae oordinates. If these oordinates are put together in a vetor (t, x, y, z) t, the oordinates in another inertial frame, S travelling with a veloity v with respet to S, the oordinates in S are given by: t x y z = L t x y z (1) where L is a 4 4-matrix, desribing the transformation. In it s most general form L is given by: γ γβ x γβ y γβ z γβ x 1 + (γ 1) β x β (γ 1) βxβy L = β (γ 1) βxβz β γβ y (γ 1) βxβy β 1 + (γ 1) β y β (γ 1) βyβz β γβ z (γ 1) βxβz β (γ 1) βyβz β 1 + (γ 1) β z β Where β = v/ with v the relative veloity of the to systems and γ = (1 (β/) ) 1/. In what follows the spae oordinates will be labeled by x 1, x, x 3. The four dimensional vetor is what s alled a four-vetor The energy, E of a partile of mass m travelling with a veloity v is: By subtrating the rest energy m the kineti energy T is:.1 Four-vetors E = γm (3) T = m (γ 1) (4) Consider a partile with mass m moving with a veloity v in some inertial frame S. The motion of this partile is desribed by the following four-vetors..1.1 The position four-vetor As desribed above, the position of an event in spae-time is desribed by the postion four vetor, x, with omponents, x µ, µ =, 1,, 3, where x = t and x i, i = 1,, 3 are the usual spae oordinates. In what follows, we use the onvention that indies labeled by latin letters, are running from 1 to 3, and indies labeled by greek letters are running from to 3. () 3
. The metri tensor, salar produt and invariant quantities FOUR-VECTORS The worldline of a partile is desribed by expressing x µ as funtions of some parameter. Ususally this parameter is hoosen to be τ-the proper time. Hene the position four-vetor x(τ) is a parametrisation of the worldline of some partile..1. The veloity four-vetor The tangent vetor to the postion four-vetor, x, is u-the veloity four vetor: u µ = dxµ dτ (5) or by using d dτ = γ d dt, the omponents of u are: u = γ u i = γv i = γ dxi dt (6) where v i are the omponents of the veloity v of the partile, as measured by an observer at rest in S..1.3 The momentum four-vetor The momentum four-vetor is simply p = mu where u is the four-veloity. Hene, using eq.(3): p = γm = E/ p i = mu i i = 1,, 3 (7).1.4 The Minskowski Fore An extension of Newtons seond law to speial relativity is with omponents: where F i = dvi dt. K = γ de dt dp µ dτ = Kµ (8) K i = γf i, i = 1,, 3 (9). The metri tensor, salar produt and invariant quantities We would like to alulate the length squared, of the four vetors. Sine (length) is a salar, it should be invariant under a Lorentz transformation. Sine the spae is not Euledian, we annot alulate the salar produt by simply adding produts of omponents in an orthogonal basis, we have to find the metri of the spae. The salar produt of two vetors,,u and v, is a funtional, that onverts the two vetors into a salar field. Denote this funtional by g: u v = g(u, v) The salar produt of two basis vetors are denoted aording to: g(e α, e β ) = g αβ (1) If we now onsider a vetor between two events at a small distane from eah other in spae-time: ζ = x α e α, the squared distane between the two events is ζ ζ. But a distane between two losely loated events is ( s) = ( x ) ( x 1 ) ( x ) ( x 3 ). Using ( s) = ζ ζ: ( s) = ( x ) ( x 1 ) ( x ) ( x 3 ) = ( x α e α ) ( x β e β ) = x α x β e α e β = g αβ x α x β (11) 4
.3 Why four-vetors? FOUR-VECTORS Or if we onsider infinitesimal displaements: ds = g αβ dx α dx β (1) Hene the metri tensor for the Minskowski spae-time has the following omponents: g = 1 g ii = 1, i = 1,, 3 g αβ = forα β (13) and the salar produt of two vetors u and v is alulated as: u v = g αβ u α v β (14) The length squared of a four vetor is invariant under any Lorentz transformation. As an exemple, take the (length) of u: u u = γ γ v = γ ( v ) (15) where v is the speed of the partile relative the frame we are looking in. Sine γ 1 = 1 v / = v we have: u u = (16) and it immediately follows that: But we an also express p p as: p p = m (17) p p = g µν p µ p ν = E p (18) where p is the three-momentum as measured by an observer at rest. Therefore m = E.3 Why four-vetors? p or: E = m 4 + p (19) In what follows, two properties of the four-vetors will be of great importane. The first is that they transform orretly under a Lorentz transformation. u, p, and K are nothing but derivatives or a onstant resaling (m) plus a derivative of the position four-vetor, x. Sine the Lorentz transformation of x is a linear transformation, the derivatives transforms in a orret way. Take for example the veloity four-vetor, u: u = d dτ x = d dτ Lx = L d dτ x = Lu () Hene u = Lu. The seond important property is that, as we have seen, the length squared of a four vetor is invariant under a Lorentz transformation. Both of these properties will be used in what follows. 5
3 TENSORS AND ONE-FORMS 3 Tensors and one-forms 3.1 Definition of a tensor Definition 1: A tensor of rank arguments. ( p From this def. we an see that the metri tensor is a tensor of rank R and it s linear in both of it s arguments beause: If u,v,w are vetors and α, β R we have from eq. (14): 1. (αu) v = α(u v) AND (u + v) w = u w + v w. u (βv) = β(u v) AND u (v + w) = u v + u w where 1 shows linearity in the first argument an in the seond. ) is a funtion of p vetors into R, whih is linear in eah of it s ( ). It is a funtion of vetors into 3. 1-forms ( ) Consider the objet g αβ u α in the salar produt. This is from the above definition a tensor of rank 1 Sine it is a linear funtion of 1 vetor into R. This objet is alled a 1-form. and it s a funtion of a vetor, η(v) having values in R. Consider a fourvetor a with omponents a µ and it s 1-form a µ = g µν a ν. Components of 1-forms are written with subsript indies and vetors with supersript. The operation a µ a µ is alled ontration. Now we an see that the length squared of a four vetor is nothing but the ontration of a vetor and it s 1-form. A ontration an be formed between any vetor and 1-form and is denoted η, v = η v + η 1 v 1 + η v + η 3 v 3 (1) ( ) n Definition : A tensor of rank is a funtion of p vetors and n 1-forms into R, whih is linear p in eah of the arguents. ( ) 1 Hene the four vetors above are tensors of rank sine they are linear funtions of one 1-form into R. ( ) 1 We an also see that the Lorentz transformation is a tensor of rank sine ontrating it with one 1 1-form and one vetor produes a salar. 3.3 Tensorprodut and wedge produt ( ) A tensor produt between two vetors produes a tensor of rank. If T is suh a tensor, this operation is denoted T = u v. Sine it s a seond rank tensor it s a funtion of two 1-forms. If η and λ are 1-forms the value of T (η, λ) is T (η, λ) = u v(η, λ) = η, u λ, v () The omponents of T are T αβ = u α v β. Hene T (η, λ) an be written as: T (η, λ) = u α η α v β λ β (3) The wedge produts between two vetors, denoted u v is also a seond rank tensor defined as: or in omponents: u v := u v v u (4) (u v) αβ = u α v β u β v α (5) 6
4 SCATTERING PROCESSES Part II Relativisti kinematis and systems of partiles Generally it s diffiult to make generalisations to speial theory of relativity of the equations of motions for systems of partiles. This is due to the fat that paths in spae-time generally are parametrized by the proper time τ. In a general many partile system the different partiles will have different proper times. Below follows a disussion on ollision proesses between two partiles. Here the momentum four-vetor together Lorentz transformation will be the main tool. There will also be a disussion on a generalisation of angular momentum. 4 Sattering proesses Consider two partiles shot at eah other with high energies. The main appliation of suh a model is in ollision experiments of elementary partiles. There is still a lot to learn about the fores and interation between elementary partiles. Therefore suh ollision experiments are of great interest. What we an do with the help of speial theory of relativity, is that we an study the situation before and after the ollision. When the partiles are separated far enough, they an be desribed by lassial mehanins (i.e the partiles are onsidered to be "lassial". That means no quantum effets are, onsidered). The interation itself is far more ompliated. We will study the inelasti sattering proess and determine some onditions for when some speifi reation an take plae. We will also study the elasti sattering proess and see that at high energy some results impossible in lassial physis will arise. 4.1 Inelasti sattering Consider two partiles, with masses m 1 and m moving with three-momenta p 1 and p in the laboratory frame. The partiles are olliding and produing a new set of partiles with masses m r, r = 3, 4, 5,... Consider the enter of momentum frame, abbreviated C.M-frame. This is the frame in whih the spatial part of the four momentum of a system of partiles is zero. The C.M-frame for a two partile system is the referene frame of minimal energy 1. The total four-momentum-vetor in any frame an be written as P = [ E1+E p 1 + p ] = [ E P ] (6) where E = E 1 + E and p = p 1 + p. In the C.M-frame p = : P = [ E ] (7) Where E = E 1 + E is the total energy in the C.M-frame. Sine P µ P µ = P µ P µ we have: E = E P (8) Or, sine P, (P = only when the two frames oinide) we have: E E (9) where E is the energy in an arbitrary referene frame. Hene the energy in the C.M-frame onstitutes a lower limit for the energy in any frame Note that E = (m 1 + m ) only when the two partiles are at rest in C.M-frame. 1 In the textbook by Goldstein, Poole and Safko they make the strange statement that the energy is larger in the C.M-frame, than in the laboratory frame. This is as far as I an see inoret 7
4.1 Inelasti sattering 4 SCATTERING PROCESSES Let the oordinates of the C.M-system be primed, and the lab system oordinates unprimed. Sine the total three-momentum is zero in the C.M-system, one an model this system as the rest system of a partile of rest mass M = E /. Note that E = 4 ( i m i) only if all partiles are at rest in the C.Mframe, otherwise it will be E = 4 ( i 4 m i + p i ). The length square of the total four momentum is invariant: P µ P µ = P µ P µ (3) Sine by def. the three-momentum is zero in C.M-frame we have: But in the lab-frame we have: P = (E /,,, ) P µ P µ = E (31) P = ((E 1 + E )/, p 1 + p ) P µ P µ = (m 1 + m ) + E 1E p 1 p (3) Where eq.(19) has been used. Setting P µ P µ = P µ P µ yields an expression for the energy E in the C.M-frame: E = 4 (m 1 + m ) + E 1 E p 1 p (33) If now one of the partile is at rest in the lab-frame, say partile, we have p = and E = m. Hene: E = 4 (m 1 + m ) + m E 1 (34) Or by denoting the exess energy E 1 m 1 by T 1 (i.e T 1 is the kineti energy of the inident partile) we get: E = 4 (m 1 + m ) + m T 1 (35) Hene for high energies, the available energy E in the C.M-frame is proportional to T 1. For pratile purposes one often wants to minimize T 1. From eq.(35) one an see that T 1 exhibits a minimum when E does. The energy, in C.M-frame, after the reation will be: E = r m r 4 + p r (36) where p r is the tree-momentum of the r:th partile in the C.M-frame. From eq.(36) it s lear that the minimum of E ours when all reation produts are at rest in the C.M- frame. Equating (35) and (36) with all p r = yields Define: T 1 m 1 = ( r m r) (m 1 + m ) (37) m 1 m ( ) Q := m r (m 1 + m ) (38) r Hene Q is the differene in rest energy due to the reation. Using eq.(38) eq.(37) beomes T 1 m 1 = Q + Q(m 1 + m ) m 1 m 4 (39) We an now see that, when an experiment is performed by shooting partiles at a stationary target, there is a threshold kineti energy below whih a reation with a ertain Q-value an not take plae. 8
4. Elasti sattering 4 SCATTERING PROCESSES 4.1.1 Prodution of an anti-proton As an exemple, onsider the prodution of an anti-proton p by the following proess: p + n p + n + p + p (4) where p is a proton and n is a neutron or a proton. The masses of all partiles involved in the proess are approximately the same, m 938MeV/. We then have from eq.(38): Inserting this value of Q into eq.(37) gives the threshold energy: Q = m (41) T 1 = 6m (4) Or T 1 5.63 MeV. I.e the energy required for the proess is 3 times larger than the rest mass of the produed partiles. If on the other hand the two initial partiles would have been travelling with equal but opposite momenta, p 1 = p and the lab.-system and the C.M-system oinides. Therefore E = (T 1 + m ). In the anti-proton reation we need, if all reation produts are at rest, E = 4m, and therefore: T 1 = m (43) In other words we need to aelerate partiles to six times more kineti energy with a stationary target than in an experiment with olliding partiles. This is why one often prefers to perform suh experiments by olliding two beams of partiles, instead of shooting at some stationary target. 4.1. Veloity of C.M- system and kineti energy energy in the laboratory- frame The zeroth omponent of the momentum four vetor is, in the C.M- system P = M. In another frame P = γm. In the lab frame we have, if we are onsidering the ase of stationary target: P = E 1 + E = E 1 + m Hene the relative speed v of the C.M- system and the lab system is suh that: (44) γ = E 1 + m M (45) with γ = (1 v / ) 1/. If we onsider the threshold initial energy, where all partiles are at rest in the C.M- system after reation γ is given by: γ threshold = T 1 + (m 1 + m ) r m r (46) where again E 1 = m 1 + T 1 has been used. Now that we know the value of γ, the kineti energy of the i:th reation produt in the lab. system is given by: Hene, for the antiproton disussed above we will have : 4. Elasti sattering T i = m i (γ threshold 1) (47) γ threshold = 6m + m 4m = T p = m (48) Let s now turn our attention to the proess of elasti sattering. Consider two partiles again with masses m 1 and m and inital momenta p 1 and p. Let again m be initially at rest in the lab. system. Hene p =. In the language of the previous setion, this is a reation produing two partiles with masses m 3 = m 1 and m 4 = m and momenta p 3 and p 4, in the lab. system. Let p 1 be parallel to the z-axis, suh that the relative motion of C.M-system and and lab-system is along the z-axis. The inoming and outgoing The hoie of axis is made in order to math the usual non- relativisti sattering angle 9
4. Elasti sattering 4 SCATTERING PROCESSES three-momenta are defining a plane. We take this plane to be the xz-plane. In the C.M-system p 3 makes an angle θ with respet to the z-axis and p 4 respet to the same axis. Inserting E = M in eq. (45) with E given by eq. (35), γ is found to be: makes an angle π θ with γ = T 1 + (m 1 + m ) m T 1 + (m 1 + m ) 4 (49) In the C.M- system the spatial part of the momentum four- vetor is zero. In the laboratory system the spatial part is P = γmβe z where βe z is the veloity of C.M- system relative the lab. system (β x = β y = ). On the other hand, P = p 1 = p 1 e z β = p 1 /γm. Using γ from eq.(45) we get: β = β z = p 1 E 1 + m (5) Hene, using eq. () the transformation from the laboratory system to the C.M system is desribed by the Lorentz transformation: γ γβ L = L lab. C.M = 1 1 (51) γβ γ And the inverse transformation is simply a simlar transformation but with relative veloity β instead of β: γ γβ L C.M lab = L 1 = 1 1 (5) γβ γ The two vetors to be transformed are: and: p 1 = p 3 = E 1 p 3 1 E 1 p 1 3 p 3 3 (53) (54) Where the fat that there is no motion in y-diretion has been used. Hene: γ E1 γβp3 1 p 1 = Lp 1 = (55) γβ E1 + γp3 1 and: p 3 = L 1 p 3 = γ(p 3 + βp 3 3 ) p 1 3 γ(βp 3 + p 3 3 ) Due to elasti sattering: p 1 = p 3 whih implies: p 1 3 = p 1 sin θ p 3 3 = p 1 os θ p 3 = p 1 = E 1 (56) (57)
5 ANGULAR MOMENTUM Where p 1 = p 1. Inserting these expressions in eq.(56): p 3 = γ( E 1 + βp 1 os θ) p 1 sin θ γ(β E 1 + p 1 os θ) (58) We now have all neessary relations between the inital and final quantites. Some (rather long not very interesting) algebrai manipulations 3 takes us to the expression: with E 1 = T1 m 1 and ρ = m1 m. Hene the minimum energy of the sattered partile is: In the lassial limit with E 1 << 1, the lassial result: T 3 T 1 = 1 ρ(1 + E 1/) (1 + ρ) + ρe 1 (1 os θ) (59) E 3,min = E 1 (1 ρ) (1 + ρ) + ρe 1 (6) T 3,min (1 ρ) = T 1 (1 + ρ) (61) is obtained. In the ultrarelativisti region ρe 1 >> 1. Assume also ρ << 1 whih means that we are onsidering sattering of a partile from a muh heavier partile. In this situation T 3,min = m 1 E 3 (1 ρ) m 1 ρ = (m m 1 ) m Hene, in the ultrarelativisti region, the miniumum T 3 is independent of T 1, the partile an loose a lot of it s energy, by sattering of a muh heavier partile. This is, as an be seen from eq. (61) not possible in the lassial ase. It s also possible to study the angular dependene in the laboratory frame. No details are disussed here. The result though, is that the relativisti treatment shows that the sattering angle, measured in the laboratory frame, ϑ will be smaller than what would have been expeted from a lassial treatment. 5 Angular momentum In this setion we will develop a relativisti extension of the angular momentum, and derive an equation of motion that redues to the lassial expression L = N in the lassial limit. We start by defining a seond rank tensor: m := x p (6) with omponents: 3 see Goldstein page 37-38 for some details. Misprint: Eq. 7.15 shall be as eq. (59) m µν = x µ p ν x µ p µ (63) 11
5.1 Total angular momentum, onservation 5 ANGULAR MOMENTUM Beause of the anti-symmetry, all diagonal elements are zero. The non-zero elements of the sub-tensor m ij in the lassial limit, are nothing but the omponents of L and L Taking the τ-derivative of eq.(6) yields: dm dτ = dx dτ p + p dp dτ = u p + x K = x K (64) u p = sine u p = µ(u u) where µ is the mass of the partile. u u = beause of the anti-symmetry of the wedge produt. Comparing with the equation of motion for the lassial angular momentum, we an now define a generalisation of torque as: N := x K (65) Therefore we an now write the equation of motion for rotation in Minskowski spae-time: Or in omponents: dm dτ = N (66) µν dm = N µν (67) dτ Taking the non-relativisti limit, the lassial expression L = N follows for the subtensors with indies i, j = 1,, 3. Hene it makes sene to define the relativisti momentum and torque aording to eq.(6) and eq.(65) 5.1 Total angular momentum, onservation The total angular momentum of n partiles is: M = n m k (68) k=1 As already mentioned it s diffiult to reate equations of motion for a system of partiles. Though it s possible to argue for onservation of angular momentum in some situations. When there is no interation between the partiles in an isolated system, no fores or torques are ating on the partiles and the total angular momentum is of ourse onserved. Consider two partiles interating through point interations. Then it s possible to write an equation of motion of the form (67) for a omposite partile when the partiles are "stuk" together. If the fores of interation are equal in magnitude but of opposite diretion, the total torque will be zero, and therefore angular momentum is onserved. It s important to notie that the intertion has to be a point interation. That is whenever the partiles are separated, the fore between them has to be zero. Therefore angular momentum is onserved even for an isolated system of partiles interating only through point interations. 5.1.1 Translation of angular momentum Eq. (68) gives M with respet to the origin. How an we alulate M with respet to some other point in spae-time? Consider an arbitrary point in spae-time a λ. The total angular momentum an be alulated with respet to this referene event: M(a λ ) = k (x k a λ ) p k = k x k p k a λ k p k = M() a λ P (69) where M() is the angular momentum with respet to the origin. Hene M with respet to the origin is the sum of M with respet to the point a λ and a term looking as if the whole system would have been loated at a λ, just as in the non-relativisti ase. 1
REFERENCES REFERENCES 5.1. Center of energy In the non relativisti ase, there is a system in whih one point is always stationary. That is the enter of mass in a system travelling with the same veloity as the enter of mass. Is there something similar in relativisti mehanis? Yes there is! Consider the omponents M j : M j = k (x kp j k xj k p k) = k (tp j k E k xj k ) (7) (71) sine x k = t for all partiles. Looking in the C.M frame, k pj k = for j = 1,, 3. If M is onserved, all omponents of M are onstant. Hene: x j k E k = M j = onst. (7) k If the system is isolated the total energy k E k = onst. Therefore the spatial point with omponents defined as: R j k := xj k E k k E j = 1,, 3 (73) k is a stationary point in the enter of momentum frame. This point is sometimes refered to as the enter of energy. Note that in the lassial limit where T k << m k, R j approahes the lassial enter of mass. It s worth to emphasize that a the enter of energy an only be defined in the C.M-frame, and for onserved M. Referenes [1] Herbert Goldstein. Classial mehanis 3rd edition. Pearson Eduation,. [] Bernard F. Shutz. A first ourse in general relativity. Cambridge University Press, 1999. 13