George Mason University. Physics 540 Spring Notes on Relativistic Kinematics. 1 Introduction 2

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1 George Mason University Physics 540 Spring 2011 Contents Notes on Relativistic Kinematics 1 Introduction 2 2 Lorentz Transformations Position-time 4-vector Velocity 4-Vector Momentum-Energy 4-Vector Momentum Redefined Momentum and Energy of a Free Particle Momentum-energy 4-vector of a particle defined Units Example: decay of a π meson A digression on the metric tensor Relativistic Kinematics of Collisions Transformations to and from the CM frame Examples Reaction Thresholds Invariant Quantities Invariant Mass Momentum Transfer Invariant Volume Element Rapidity 12 5 Problems 12

2 1 Introduction This review discusses and summarizes the topics in relativistic kinematics needed for this course. 2 Lorentz Transformations If the invariance of c, the speed of light in vacuum, is taken as an experimental fact, the correct transformation of the coordinates of a particle between two inertial reference frames, the Lorentz transformation, can be derived by assuming that it is the simplest linear transformation which makes c independent of the frame of reference. Suppose a particle has position vector r in frame S and r in frame S. That is, for relative motion along the common x, x axes, this transformation can be written as x = ax + bt (1) y = y z = z t = ex + ft in which a, b, e and f are constants for a particular pair of inertial frames, but may depend on the relative velocity, v, of the frames, and on universal constants. The above equations should describe the transformation of any point, between the two frames. We may use the following set of specific points to evaluate the constants, and hence, to derive the transformation. Assume the origin of S advances along the +x direction. The specific points and the equations which constrain the constants are: 1. The origin of frame S. Then 2. The origin of frame S. Then x = 0 x = vt x = vt x = 0 3. A point on a wavefront of a light wave pulse emitted, along the +x direction at time t = 0. Then the universality of c requires: x = ct x = ct 4. A point on a wavefront of a light wave pulse emitted, along the +y direction at time t = 0. Then the universality of c again requires: x = 0 y = ct x 2 + y 2 = c 2 t 2 2

3 The result is the Lorentz transformation: where x = γ(x βct) (2) y = y z = z t = γ( (β/c)x + t) γ = 1 1 β 2 β = v c Again v is relative velocity of S with respect to S. The inverse transformation may be obtained by exchanging primed with unprimed quantities, and β with β. 2.1 Position-time 4-vector We may write equations (2) in matrix form: x y z ict = γ 0 0 iβγ iβγ 0 0 γ x y z ict where i = 1. It is necessary to put the i s in to get the correct sign for each term. Writing the transformation in this manner displays an important feature: it is, mathematically, a rotation. The object being operated on by the 4 4 matrix is a 4-component vector, the position-time 4-vector, r = (r, ict). To see this more clearly, consider the submatrix: ( γ iβγ ) iβγ γ and also consider the rotation of a vector A in the x y plane, as shown in Figure 1. Y Y Y θ A θ A X A θ X X Figure 1: Rotation of (left) a vector, and (right), the coordinate system. 3

4 We can see from this figure that rotation of a vector clockwise by an angle θ will yield the same new components as a rotation of the coodinate system counter-clockwise by the same angle. In the latter case, we have A xx + A yy = A x x + A y y in which x, y, x, and y are unit vectors along the two sets of coordinate axes. By dotting both sides of this equation with the appropriate unit vector, we obtain A x = A x cos θ + A y sin θ A y = A x sin θ + A y cos θ In matrix form, the rotation looks like: ( ) ( A x cos θ sin θ A = y sin θ cos θ ) ( Ax A y ) Comparing this result with the Lorentz transformation, it is evident that the latter can be regarded as a rotation provided cos θ = γ Since the Lorentz factor, γ is always 1, it means that θ is imaginary. This is not troublesome, since the parameter θ is not a physical quantity. If the Lorentz transformation (LT) corresponds to rotation of a position-time 4- vector, then the magnitude of the 4-vector must be invariant under the LT, just as the length of a 3-vector is invariant under an ordinary rotation. Moreover, the scalar product of any two 4-vectors must also be invariant under a LT. As an illustration of this, consider the square of the position-time 4-vector, r 2 : r 2 = x 2 + y 2 + z 2 + (ict) 2 Suppose the reference frames S and S overlap at the instant when a light source at the origin of one emits an isotopic pulse of light. Then if (x, y, z) is a point on the wavefront, then special relativity requires that In other words, x 2 + y 2 + z 2 = c 2 t 2 x 2 + y 2 + z 2 = c 2 t 2 x 2 + y 2 + z 2 c 2 t 2 = x 2 + y 2 + z 2 c 2 t 2 which says that r 2 in Lorentz invariant, i.e. a Lorentz scalar. We may generalize these results to say that a 4-vector is a object which transforms under a LT in the same way as the position-time 4-vector, i.e. the transformation matrix is the same. 4

5 2.2 Velocity 4-Vector It can be shown from the transformations of dx, dy, dz, and dt, that the LT of the velocity of a particle, v, is v x = v x v r (3) 1 (v x v r /c 2 ) v y = (4) γ r (1 (v x v r /c 2 )) v z = v z (5) γ r (1 (v x v r /c 2 )) In these equations, γ r contains the relative velocity between reference frames. v x, v y and v z do not transform in the same manner as x,y and z, as can be seen by comparison with Equations 2. However, a velocity 4-vector can be constructed. It is: v x γ/c v ũ = y γ/c v z γ/c iγ In the above expression the v in γ is the velocity of the particle in a given reference frame. This object does transform in the same manner as the position-time 4-vector. For example, LT of ũ x, using the LT matrix, is v y γ v x = γ rγ(v x v r ) (6) The same result can be obtained from from Equations 3,4, and 5. The magnitude of ũ is obtained from the dot product of ũ with itself, giving ũ 2 = 1 which is Lorentz invariant, since it does not depend on any kinematic quantities. One may wonder why we bother with an object with a constant magnitude; its use will be made evident shortly. Further discussion of kinematic 4-vectors may be found in Reference [1]. 2.3 Momentum-Energy 4-Vector Momentum Redefined Before discussing the momentum-energy 4-vector, it is necessary to point out that special relativity reveals that the classical definition of the momentum of a particle p = mv is only an approximation. The correct expression is p = mvγ in which γ is now the Lorentz factor of the particle, i.e. the v in γ is the speed of the particle in a particular rest frame. This redefinition is necessary so that if, in a collision, total momentum is conserved in one inertial frame, then it is also conserved in any other interial frame. The argument for this involves the details of a thought experiment [2][3]. 5

6 2.3.2 Momentum and Energy of a Free Particle For a free particle of (rest) mass m, the total energy and the momentum are related by E = This is seen by using the definition of p, and v = βc to obtain p 2 c 2 + m 2 c 4 (7) p = mβcγ and also the expression for E, which is proven in most elementary textbooks: E = mc 2 γ By squaring and subtracting the above two equations, we get Now because we have So inserting 1 into Equation 8 QED. Another way to write this equation is The right-hand side of this is a constant., E 2 p 2 c 2 = m 2 c 4 γ 2 m 2 c 4 β 2 γ 2 (8) γ 2 = 1 1 β 2 γ 2 β 2 γ 2 = 1 1 β 2 β2 1 β 2 = 1 E 2 = p 2 c 2 + m 2 c 4 p 2 c 2 E 2 = m 2 c 4 (9) Momentum-energy 4-vector of a particle defined We can obtain this 4-vector immediately by simply multiplying the velocity 4-vector, ũ, by the constant mc, where, again, m is the rest mass of the particle. p = mcũ Or p = p x p y p z ie/c 6

7 The dot product of p with itself, must be a Lorentz invariant, and we see from Equation 9 that this is indeed the case: p 2 = m 2 c 2 The Lorentz transformation of the momentum-energy 4-vector between reference frames can be written in terms of the longitudinal and transverse components of momentum, and the total energy by applying the LT matrix of Section 2.1 to p, giving p = γ(p βe/c) (10) p = p E = γ(e βcp ) Here β and γ refer to the relative velocity of S with respect to S. Also, if p = p x, then p = p 2 y + p2 z. Remember that p in the above equations is still relativistic momentum. The transformation from S to S is given by Units p = γ(p + βe /c) (11) p = p E = γ(e + βcp ) There is a system of units for m, p, and E in use by many workers. Suppose the unit of energy is taken to be 1 MeV. Then since E = mc 2 γ if the corresponding unit of mass is taken to be 1 MeV/c 2, then the above equation may be written E = mγ For example, the rest mass of the proton is MeV/c 2. The rest energy is then MeV. Similarly, if the corresponding unit of momentum is MeV/c then we have E = p 2 + m 2 So, using this system of units, we may eliminate all c s from kinematic formulae. We may also, of course take the energy unit to be, say, ev or GeV, provided consistency is maintained. In what follows, we have set c = Example: decay of a π meson As an example of these transformations consider the decay of the charged π meson: π µ + ν 7

8 The known masses of these particles permits calculation of the momentum of the µ or ν in the rest frame of the π. Momentum conservation requires that the two momenta in this frame be in opposite directions and of equal magnitude. The magnitude of this momentum turns out to be 30 MeV/c. For a π of momentum 500 Mev/c, let us determine the corresponding momentum and angle of the µ if it is emitted at an angle of θ = 10 in rest frame of the π. The mass of the π(µ) is Mev/c 2 (105.7 Mev/c 2 ). Solution: 1. The Lorentz factor of the π meson is γ = E/m = p 2 π + m 2 π m π = and also, β = p/e = Let the π rest frame be S. (Remember that when the relative velocity is positive, S moves to the right in S.) Then from trigonometry, the longitudinal and transverse components of the µ momentum in the π rest frame are p = p cos(θ ), and p = p sin(θ ). The µ energy in the π rest frame is 3. Then, using equations 11 and E = p 2 + m 2 µ = Mev p = γ(p + βe ) = MeV/c p = p = 5.21 MeV/c This makes p = MeV/c The energy is MeV. The angle in S, found from tan(θ) = p /p is A digression on the metric tensor If you find the minus sign on the right hand side of Equation 9 displeasing, you are not alone. Many authors choose to get rid of it by modifing the rules for calculating the scalar product of two 4-vectors. Formally we may write p 2 = µ g µν p µ p ν ν where g µν is the metric tensor. This quantity is of great importance in general relativity, but not to us here. We may, for Cartesian coordinates, always take it to be diagonal. Until now, we have been implicitly taking it to be a diagonal matrix with elements +1. If you want to change the sign of the right-hand side of equation 9 you may define the first three diagonal elements of g µν to be -1, and then remove the i from the 4th component 8

9 of all 4-vectors. The LT matrix must also be appropriately modified. Then, the scalar products which were formerly negative will be positive. No physics will be changed. The only important thing is consistency. In what follows, we will stick with the original rules, i.e. +1 for all diagonal elements of g µν, and tolerate complaints about a poor choice of metric. 3 Relativistic Kinematics of Collisions Collisions of microscopic particles usually occur over such a short time that the effects of external forces, such as gravity or static electromagnetic fields, are negligible. This means that both total momentum and energy are conserved. A way of expressing this with 4-vectors is: pi = p f Because we are accounting for total energy, including mass, the above equation is correct even when there are a different number of particles in the initial and final states. Consider a collision with two particles in the initial state. There is a special frame of reference, the Center of Momentum or CM frame, in which the total momentum is zero. In this frame, the total momentum is also zero in the final state, even if new particles are produced. It is possible, in the CM frame, to have all the final state particles at rest. A frequent experimental situation, however, is one in which one of the particles, the target particle, is at rest, and the other is moving it its direction with a specified momentum; let us designate this the lab frame. Figure 2 illustrates these frames. P lab 1. P CM m 2 Figure 2: Incident particles in (left) the laboratory frame, and (right) the CM frame Transformations to and from the CM frame To relate lab and CM quantities, we first find the β and γ of the CM frame in the lab frame. Assuming the incident directions are longitudinal, p CM 1 = γ(p lab 1 βe1 lab (12) p CM 2 = γ(0 βm 2 ) Equating the magnitude of the two momenta in the CM frame gives β = p lab 1 /(E lab 1 + m 2 ) and, with a little algebra, γ = E1 lab + m 2 m m m 2 E1 lab 9

10 By substituting these values back into Equation 12, the CM momentum is found to be p CM = p lab 1 m 2 m m m 2 E1 lab One can use this result to find the CM energy of each particle. An important quantity is the total CM energy; this is freqently given a special designation: s = E CM total = E1 CM + E2 CM We can find s from the above equations. It is easier to obtain, however, by considering the sum of the momentum-energy 4-vectors of the incident particles. The sum of two 4-vectors is another 4-vector; and the square of any 4-vector is a Lorentz invariant i.e. the same in any inertial frame. First consider the sum in the CM frame: ( p 1 + p 2 ) 2 = (p cm 1 + p cm 2 ) 2 (E CM 1 + E CM 2 ) 2 = s Now the same 4-vector, evaluated in the lab frame, must have the same magnitude, so which gives so the total CM energy is ( p 1 + p 2 ) 2 = p lab2 1 (E lab 1 + m 2 ) 2 s = m m m 2E 1 s = m m m 2 E 1 (13) This shows that the total CM energy increases only as the square root of the lab energy in a fixed target collision Examples 1. Consider a fixed target collision of two protons, with the moving proton having an energy of 900GeV. (This energy is available at the Fermi National Accelerator Laboratory.) What is the total CM energy? SOLUTION: Using Equation 13, with m =.938 Gev/c 2, gives s = 41GeV. 2. At the Fermilab colliding beam facility, the energy of particles in each beam is 900GeV. What is total CM energy? SOLUTION: Now the the CM frame is the laboratory, so the total energy is s = 1800GeV. 3.1 Reaction Thresholds We may use Equation 13 for finding the energy threshold for production of new particles. If an additional particle or particles is produced in the final state, the lowest lab energy for which this can occur, i.e. the threshold energy, must be that energy for which the final state particles in the CM frame are at rest. In other words smin = 10 j m j

11 For example, consider the production of a proton-anti-proton pair in a proton-proton collision: p + p p + p + p + p If one of the protons is initially at rest, (i.e. we are in the lab frame) what is the reaction threshold energy for the incident proton? Using the above equation, s min = 16m 2 p and Equation 13, we get E p,lab,min = 7m p = 6.57GeV 3.2 Invariant Quantities The previous example make use of a Lorentz invariant, in this case, s, the total CM energy squared, which was evaluated in two different reference frames. This is, in general, a powerful method of solving many problems in relativistic kinematics. Several other useful Lorentz invariants are described next Invariant Mass For any number of particles, the sum of all energy-momentum 4-vectors is another 4- vector, and its square is a Lorentz invariant. It is usually called the invariant mass of the system, and m 2 inv = ( E j ) 2 ( p j ) 2 j j The rightmost term involves, in general, dot products of different momenta Momentum Transfer Consider the 2-body reaction Energy and momentum conservation require that The 4-momentum transfer is p 1 + p 2 = p 3 + p 4 q = p 3 p 1 = p 2 p 4 The square of this must also be an invariant. A commonly used notation is t = q 2 It can be shown that in the CM of the above system, the 4-momentum and the 3- momentum transfers are the same. 11

12 3.2.3 Invariant Volume Element Another application of the Lorentz transformation of momentum and energy gives the useful result that the momentum volume element d 3 p/e is a Lorentz invariant. This is useful for the expression of cross sections. 4 Rapidity We know from Equation 3 that the transformation of a velocity is non-linear in the relativistic case, because of the denominator. An interesting question to ask, then, is this: Is there a kinematic quantity, let us call it y, which transforms as simply as v x does under a Galilean transformation. That is, y = y + f(v r ) where f(v r ) does not depend on y. The answer is yes and the kinematic quantity, called the rapidity of a particle, is y = 1 [ ] E + 2 ln p E p It can be shown that y = y + tanh 1 v r /c This means that a distribution in y will not change shape in a Lorentz transformation, it simply slides along the y direction. It turns out that a very simple quantity is approximately equal to y. For relativistic particles (i.e. β 1), the quantity η = ln tan(θ/2) y The quantity η is called the pseudo-rapidity of a particle. 5 Problems 1. Starting with Equations (1), evaluate the constants using the four specific cases described, to obtain the Lorentz Transformation (Equations (2)). 2. Why must the transformation between reference frames be linear? 3. Show that Equation 6 follows from Equations 3 and What must be the energy of an incident particle in a (proton-proton) fixed target collision in order to have s = 1800GeV? 12

13 References [1] The Classical Theory of Fields, by L. Landau and E. Lifshitz, translated by Morton Hammermesh, Addison Wesley (1951). [2] Berkeley Physics Course, Vol 1, Mechanics, by Charles Kittel, Walter D. Knight, Malvin A. Ruderman, A. Carl Helmholz, and Burton J. Moyer. Second Edition. McGraw-Hill, [3] Special Relativity, by A. P. French, W. W. Norton, (1968). [4] Spacetime Physics, by Edwin F. Taylor and John Archibald Wheeler, W. H. Freeman and Co, 2nd Edition (1993). 13

Lecture 9 - Applications of 4 vectors, and some examples

Lecture 9 - Applications of 4 vectors, and some examples E. Daw April 4, 211 1 Review of invariants and 4 vectors Last time we learned the formulae for the total energy and the momentum of a particle in