MISN-0-75
RELATIVISTIC SPACE-TIME: FOUR-VECTORS
RELATIVISTIC SPACE-TIME: FOUR-VECTORS by ct Frank Zerilli
an event 1. Description (popping of a flashbulb) a. Events in Spacetime ...... 1 b. Four-vectors ...... 1 c. World Lines ...... 1
y 2. The Lorentz Transformation as a Matrix ...... 2
3. Precise Definition of a 4-Vector ...... 4 x 4. Proof of the Momentum-Energy 4-Vector ...... 5
Acknowledgments...... 8
Project PHYSNET·Physics Bldg.·Michigan State University·East Lansing, MI
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THIS IS A DEVELOPMENTAL-STAGE PUBLICATION Title: Relativistic Space-Time: Four-Vectors OF PROJECT PHYSNET Author: Frank Zerilli, Michigan State University The goal of our project is to assist a network of educators and scientists in Version: 2/1/2000 Evaluation: Stage 0 transferring physics from one person to another. We support manuscript processing and distribution, along with communication and information Length: 1 hr; 20 pages systems. We also work with employers to identify basic scientific skills Input Skills: as well as physics topics that are needed in science and technology. A number of our publications are aimed at assisting users in acquiring such 1. Write a given system of linear equations in matrix form (MISN-0- skills. 301). 2. State down the Lorentz transformation and use it to transform Our publications are designed: (i) to be updated quickly in response to given positions and times (MISN-0-12). field tests and new scientific developments; (ii) to be used in both class- 3. Define relativistic energy and momentum in terms of mass and room and professional settings; (iii) to show the prerequisite dependen- velocity (MISN-0-24). cies existing among the various chunks of physics knowledge and skill, as a guide both to mental organization and to use of the materials; and Output Skills (Knowledge): (iv) to be adapted quickly to specific user needs ranging from single-skill K1. Vocabulary: event, four-vector, world line. instruction to complete custom textbooks. K2. Write the Lorentz transformation in matrix form. New authors, reviewers and field testers are welcome. K3. Show that relativistic energy and momentum are the components of a four-vector. PROJECT STAFF Output Skills (Problem Solving): Andrew Schnepp Webmaster S1. Transform the coordinates of a given event from one inertial frame Eugene Kales Graphics to another. Peter Signell Project Director S2. Transform given energy-momentum four-vectors from one given Lorentz (inertial) frame to another. ADVISORY COMMITTEE Post-Options: D. Alan Bromley Yale University 1. “Topics in Relativity: Doppler Effect and Pair Production” E. Leonard Jossem The Ohio State University (MISN-0-308). A. A. Strassenburg S. U. N. Y., Stony Brook
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c 2001, Peter Signell for Project PHYSNET, Physics-Astronomy Bldg., Mic° h. State Univ., E. Lansing, MI 48824; (517) 355-3784. For our liberal use policies see: http://www.physnet.org/home/modules/license.html.
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RELATIVISTIC SPACE-TIME: FOUR-VECTORS ct world line of an by world line of a arbitrarily moving point Frank Zerilli point object stationary at (x0,y0)
1. Description y0 y 1a. Events in Spacetime. The term “event” is used to specify a x0 point in what we will call spacetime which consists of the usual three- x dimensional euclidean space of location plus the one-dimensional axis of time. As an example, suppose we fire a flashbulb (see Fig. 1). The fir- Figure 2. “World lines” in spacetime. ing of the flashbulb is an event and to specify its location in spacetime we must give four quantities - three quantities (for example, rectangular y(t) = y0 coordinates x, y, z in some inertial frame) which specify its location in z(t) = z space, plus one quantity which specifies its “location” in time (in some 0 inertial frame). ct = ct
1b. Four-vectors. The four quantities constitute a vector called a where x0, y0, z0 are the space coordinates of the stationary point object. “four-vector” in this four-dimensional space. We will, for convenience, Or, we can say that the four-vector of the point object at any time t is: choose to use (ct) as the fourth component of the four-vector rather than t, since the equations we will write are then a little simpler. Also, the (x0, y0, z0, ct) four components of the vector then all have the same dimension: length. In general, a world line can be specified by giving a four-vector which is Note: (ct) is the distance light travels in time t. a function of time, or some other parameter, call it τ: 1c. World Lines. The trajectory of a point object in spacetime is called its “world line.” For example, the world line of the point object [ x(τ), y(τ), z(τ), ct(τ) ]. stationary in the frame described by coordinates (x, y, z, ct) is a straight All physical objects have some extension: we speak of the length, width, line parallel to the t-axis and could be described by the set of equations: and height of objects. Upon reflection we see that for a complete specifi- cation of the extension of an object we should specify not only its spatial x(t) = x0 dimensions but also its extension in time. For example, we might specify the dimension of the building as: 100 ft long, 50 ft wide, 30 ft high, and ct 20 years extended in time (it was built in 1910 and demolished in 1930). an event In a spacetime diagram, we might describe the object by giving the world (popping of a flashbulb) lines for each point of the object (see Fig. 3).
y 2. The Lorentz Transformation as a Matrix We know that the space and time coordinates of a point with respect to one Lorentz (inertial) frame are related to the coordinates with respect x to a second Lorentz (inertial) frame by Lorentz transformation. Let L Figure 1. An “event” in spacetime. be a Lorentz frame with coordinates x, y, z, ct and let L0 be a Lorentz
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ct x x'
L L'
y z z' y y' x Figure 4. Two inertial frames and a relative velocity. Figure 3. A collection of world lines for an extended object. 3. Precise Definition of a 4-Vector frame with coordinate axes x0, y0, z0, ct0 moving uniformly with velocity v parallel to the z-axis of L. Also let the x0, y0, z0 axes of L0 be parallel to Any set of four quantities which transforms from one Lorentz frame the x, y, z axes, respectively, of L. Then the Lorentz transformation may to another just like the coordinates (x, y, z, ct) of a point in spacetime is be written: called a four-vector.1 Thus (x, y, z, ct) is a four-vector. x0 = x One example of a very important four-vector is the energy- y0 = y momentum four-vector whose spatial components are the three compo- nents of relativistic momentum and whose fourth component is the rela- z0 = γz βγct − tivistic energy (divided by c): ct0 = βγz + γct − E p , p , p , . where we have utilized the widely used notation: x y z c µ ¶ v β ; γ (1 β2)−1/2. ≡ c ≡ − Knowing that the components of momentum and energy transform like a four-vector, we can find the momentum and energy of an object as This may be written as a matrix equation: viewed in any Lorentz frame if we are given the values in any one Lorentz frame. For example, suppose we have a particle of mass m at rest in one 0 x 1 0 0 0 x Lorentz frame. Consider a Lorentz frame moving with velocity v along y0 0 1 0 0 y = the x-direction. Then, in this new frame, the particle is moving with z0 0 0 γ βγ · z 0 − velocity v along x -axis and: ct0 0 0 βγ γ ct − − 0 px γ βγ 0 βγm0c 0 = − = − Since the x and y coordinates are not affected in this case we may suppress E /c βγ γ · m0c γm0c them and write: µ ¶ µ − ¶ µ ¶ µ ¶ So: 0 2 z γ βγ z 0 m0v 0 m0c 0 = − px = and E = . ct βγ γ · ct −(1 v2/c2)1/2 (1 v2/c2)1/2 µ ¶ µ − ¶ µ ¶ − − 1This definition is correct only for flat spacetime. In non-flat spacetimes, coordi- nates are not the components of four-vectors.
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1 uz 1 uz = v_ u0 = = x b z uxv ux x' c γ 1 γ 1 β − c2 − c u g = 1 m0 Ö1 - b2 The quantity 1 (u0/c)2) occurs several times in the energy and momentum formulas so−let us calculate it in terms of the velocity in the p v first (unprimed) frame:
z z' 0 2 0 2 0 2 0 2 (u ) = (ux) + (uy) + (uz) 2 −2 2 2 y y' (ux βc) + γ (uy + uz) = − 2 Figure 5. A velocity as seen from a stationary frame and ux 1 β from a moving frame. − c ³ ´ So: This gives us just the expected values for momentum and energy of a particle of mass m0 moving with velocity v. 2 u2 + u2 − ux −2 y z 2 β + γ 2 u0 c − Ã c ! 1 = 1 ³ ´ 2 4. Proof of the Momentum-Energy 4-Vector − c − ux µ ¶ 1 β − c The most straightforward way of showing that momentum and energy 2 ³ 2 2 ´ 2 2 are the components of a 4-vector is to write the momentum and energy ux 2 uxβ 2 uy + uz 1 2 β + 2 (1 β ) 2 of a particle in terms of its real mass and velocity in one Lorentz frame − c − c − − Ã c ! = and then use the Lorentz transformation for velocity (that is, the velocity u 2 1 x β addition law) to find the form of the momentum and energy of the particle − c in a second Lorentz frame. ³ ´ 2 2 u2 u + u 1 β2 1 x 1 β2 y z m ~u m ~u0 − − c2 − − c2 p~ = 0 p~0 = 0 µ ¶ Ã ! 2 0 2 = ¡ ¢ ¡ 2 ¢ 1 (u/c) 1 (u /c) ux − 0 − 1 β E m0c E m0c − c =p =p c 1 (u/c)2 c 1 (u0/c)2 u2 + u2 ³+ u2 ´ u2 − − 1 x y z 1 1 − c2 1 − c2 p p = 2 2 = 2 2 Figure 5 shows a particle of mass m0 moving with velocity ~u in the γ ux γ ux 1 β 1 β first frame. The second frame is moving at velocity v with respect to the − c − c first frame, along the x-axis, and the particle has a velocity ~u0 as seen in ³ ´ ³ ´ the second frame. So, finally, we have the relation we want: −1/2 u As usual, let β = v/c and γ = 1 β2 . The velocity addition 1 x β − 1 law gives us: = γ − c ¡ ¢ 0 2 1/2 2 1/2 0 ux v ux βc [1 (u /c) ] [1 (u/c) ] ux = −u v = −u − − 1 x 1 x β − c2 − c Now we can easily express the components of momentum and energy 0 1 uy 1 uy as seen in the second (primed) frame in terms of the values as seen in the uy = u v = u γ 1 x γ 1 x β − c2 − c
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u first (unprimed) frame: x E0 m0cγ 1 β = − c 0 c ³ 2 1/2 ´ m0ux 0 m0ux u px = 1 2 and px = 1 2 1 u2 / u02 / − c2 1 1 µ m¶u m c − c2 − c2 = βγ 0 x + γ 0 µ ¶ µ ¶ 1/2 1/2 − u2 u2 1 2 1 2 u − c − c 1 x β µ ¶ µ ¶ 0 ux βc − c E p = m0γ − = βγpx + γ . x ux ³ 2 1´/2 1 β u − c − c 1 − c2 ³ ´ µ ¶ If we write the above equations in matrix form we obtain: γm0ux γβm0c E 0 = − = γpx βγ px γ 0 0 βγ px 2 1/2 − c 0 − u py 0 1 0 0 py 1 2 0 = = − c p 0 0 1 0 pz µ ¶ z E0/c βγ 0 0 γ E/c 0 − m0uy 0 m0uy py = 1 2 and py = 1 2 This is precisely the equation for a Lorentz transformation along the u2 / u02 / 1 1 x-axis. Thus, the components of momentum and energy transform exactly − c2 − c2 like the coordinates (x, y, z, ct) and so form a four-vector. µ ¶ µ ¶ u γ 1 x β 0 1 uy − c p = m0 = py Acknowledgments y ux ³ 2 1/´2 γ 1 β u − c 1 2 Preparation of this module was supported in part by the National ³ ´ − c µ ¶ Science Foundation, Division of Science Education Development and Re- 0 m0uz 0 m0uz search, through Grant #SED 74-20088 to Michigan State University. pz = 1 2 and pz = 1 2 u2 / u02 / 1 1 − c2 − c2 µ ¶ µ ¶ u γ 1 x β 0 1 uz − c p = m0 = pz z ux ³ 2 1/´2 γ 1 β u − c 1 − c2 ³ ´ µ ¶ 0 E m0c E m0c = 1 2 and = 1 2 c u2 / c u02 / 1 1 − c2 − c2 µ ¶ µ ¶
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c. Write the world line in the moving coordinate frame using the time PROBLEM SUPPLEMENT in the moving frame as a parameter. 8. A particle of mass m moving relativistically with momentum p is Problems 10-11 also occur in this module’s Model Exam. projected at a second stationary particle also of mass m. a. Find the total 4-momentum (i.e., momentum-energy 4-vector) of 1. A star 1000 light years away explodes at t = 0 (as described in our the system. (earth) coordinate system). Write out the event vector (as column matrix). [Note that c = 1 is units of light years/year.] b. Find a coordinate frame in which the total 3-mom (i.e., 3 space-like components which are the momentum) is zero. 2. A space ship is passing earth at t = 0. The space ship is moving (at ◦ 1/2 c) in the direction of the star (from problem 1). If the space ship 9. Two particles leave a collision point at 90 to the initial direction in observers also agree that t = 0 when the space ship passes earth, what the C.M. frame. If the velocity of the C.M. frame is c what are the is the event vector for the star explosion in the space ship’s coordinate angles in the lab frame? What is the magnitude of the momentum in system? the lab frame?
3. An electron passes at 0.98 c through an accelerator tube. The tube 10. An event is seen in S1 to be at the spacetime point (10.0 m, 0, 0, tc), −6 where t = 1.0 10 s. At what spacetime point is this event in S2, has length L2 according to the electron. In the electron frame what × are the event vectors for: which is moving at 0.6 c (in the y-direction) with respect to S1?
a. the event where the electron enters the tube; 11. A particle of mass m0, momentum p = m0c, hits an unknown object b. the event where the electron leaves the tube. Note that x compo- in such a way that it is stopped. Write its initial and final momentum nent is measured from the electron (i.e., with the vectors in the 4-vectors. Transform the difference of these 4-vectors to find a frame electron frame). where the collision appears to be elastic (i.e., the energy is the same before as after so that the energy component of the 4-vector difference 4. The above accelerator tube has length L1 according to a stationary is zero). Give β for the frame. observer. Give the two event vectors corresponding to the two events of problem 3 in the stationary coordinate frame.
5. Referring again to problems 3 and 4, transform according to the Brief Answers: Lorentz transformation events of problem 4 into the events of prob- lem 3. Find a relation between L2 and L1. Note that the electron sees 1000 1. the accelerator tube to be contracted because it is moving relative to 0 him. µ ¶ 1 6. An object follows a circular path in the x y plane. The radius of γ γ 1000γ − −2 1000 the circle is 6r and at t = 0, x = r, y = 0. It completes the circle in 2. 1 = 1 = · 0 Ã 1000γ ! time T . Write the world line for this object. γ γ µ ¶ −2 −2 2 7. a. Transform the world line of problem 6 to a frame moving at v in 1000 √3 × the +z direction. 1 1000 b. In this frame the object moves in a helical path. What is the period −√3 × (the time for one revolution)? which used:
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1 1 2 γ βγ p γ(p βE/c) 0 γ = = = b. − = − = [1 (1/4)]1/2 (3/4)1/2 √3 βγ γ · E/c ( βp + E/c)γ = 0 − µ − ¶ µ ¶ µ − ¶ µ 6 ¶ 2 2 2 1/2 0 where: E = m c + p + mc 0 c 3. a. b. Therefore we obtain β = pc/E, which is different from the single 0 Ã L2 ! ¡ ¢ µ ¶ v particle formula. Note that this β is not the velocity for any particular 2 2 2 1/2 0 L1 particle since E/c = mc + (m c + p ) . 4. a. b. c 0 L1 µ ¶ Ã v ! γ 0 +βγ 0 9. 0 1 0 p = 0 L1 0 · 1/2 γ βγ +βγ 0 γ m2c2 + p2 5. L2 = − L1 = γL1 βγ γ · βγL1 + 1/2 β µ − ¶ β − β ¡ ¢βγ m2c2 + p2 2 2 1/2 = L2 = γL1(1 β ) = (1 β ) L1 p ⇒ − − ¡ ¢1/2 γ m2c2 + p2 t t 6. r cos 2π , r sin 2π , 0, ct 2 T T 2 2 β 2 2 2 ¡ ¢ · µ ¶ µ ¶ ¸ pL = p + m c + p | | 1 β2 7. a. Lorentz Transformation in the z-direction: − ¡ ¢ 1 0 0 0 r cos(2πt/T ) p(1 β2)1/2 θ = tan−1 − 0 1 0 0 r sin(2πt/T ) 1/2 Ãβ (m2c2 + p2) ! 0 0 γ βγ · 0 − 0 0 βγ γ ct 1 1 − 10. β = 0.6; γ = 1/2 = 0 (1 0.36) 0.8 r cos(2πt/T ) x − r sin(2πt/T ) y0 The Lorentz transformation in the y-direction is: = = vγt z0 1 0 0 0 x − 1 cγt ct0 0 γ 0 βγ 0 − . The event is: , b. From answer to part(a): t0 = γt 0 0 1 0 0 2πt0 0 βγ 0 γ ct1 x0 = r cos − −6 γT where: x1 = 10 m, t1 = 10 sec. µ 0 ¶ 0 2πt y = r sin 1 0 0 0 x1 x1 γT µ ¶ 0 γ 0 βγ 0 vγt1 z0 = vt0 − = − − 0 0 1 0 · 0 0 Therefore the period is: γT = T 0. 0 βγ 0 γ ct1 γct1 − c. 0 0 r cos(2πt /T ) 10 m r sin(2πt0/T 0) 0.6 10 m 3 108 10−6 m vt0 −0.8 × × 225 m − 0 = 0 = − ct ¡ ¢ 0 −6 1 −6 c 1.25 10 s 10 s c × p 0 p 0.8 × 8. a. + = 2 2 2 1/2 µ ¶ E1/c mc m c + p + mc µ ¶ µ ¶ Ã ! ¡ ¢ 15 16 MISN-0-75 PS-5 MISN-0-75 ME-1
m c 0 m c 11. 0 = 0 √2m0c − m0c (√2 1)m0c µ ¶ µ ¶ µ − ¶ MODEL EXAM γ βγ m0c γm0c 1 β √2 1 − = − − βγ γ · √2 1 m0c γm0c β + √2 1 µ − ¶ µ − ¶ µ ¡ − ¡ − ¢¢ ¶ 1. See Output Skills K1-K2 in this module’s ID Sheet. = β = √2 1 ¡ ¢ ¡ ¢ ⇒ − 2. Show that energy and momentum of a particle are the components of a four-vector. Let the particle have velocity ux, uy, uz in one Lorentz 0 0 0 frame and ux, uy, uz in a second frame moving with velocity v along the x-axis with respect to the first frame. The velocity addition law is:
0 ux βc ux = −u (1 x β) − c
0 1 uy uy = u γ (1 x β) − c
0 1 uz uz = u γ (1 x β) − c ux 1 1 β = γ − c u0 u 1 ( )2 1 ( )2 − c r − c r where β = v/c and γ = (1 β2)−1/2. − 3. An event is seen in S1 to be at the spacetime point (10.0 m, 0, 0, tc), −6 where t = 1.0 10 s. At what spacetime point is this event in S2, × which is moving at 0.6 c (in the y-direction) with respect to S1?
4. A particle of mass m0, momentum p = m0c, hits an unknown object in such a way that it is stopped. Write its initial and final momentum 4-vectors. Transform the difference of these 4-vectors to find a frame where the collision appears to be elastic (i.e., the energy is the same before as after so that the energy component of the 4-vector difference is zero). Give β for the frame.
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Brief Answers:
1. See this module’s text. 2. See this module’s text. 3. See this module’s Problem Supplement, problem 10. 4. See this module’s Problem Supplement, problem 11.
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