Relativity. Theory of Space & Time. by Dennis Dunn. Version date: Tuesday, 4 November :04. Time & Space... Time & Space... Time & Space...

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1 Relativity Theory of Space & Time by Dennis Dunn Version date: Tuesday, 4 November 008 4:04 Time & Space... Time & Space... Time & Space... Time & Space... Time & Space... Time & Space... Space & Time... Space & Time... Space & Time... Space & Time... Space & Time... Time & Space... Space & Time... Space & Time... Time & Space... Space & Time... Time & Space... Space & Time... Space & Time... Space & Time... Time & Space... Time & Space... Space & Time... Time & Space... Space & Time... Space & Time... Space & Time... Space & Time... Space & Time... Space & Time... Time & Space... Time & Space... Time & Space... Time & Space... Time & Space... Time & Space...

2 Copyright c 00 Dennis Dunn. USEFUL REFERENCES Your notes from Classical Physics PH00 Your notes from Electromagnetism PH003 Spacetime Physics E F Taylor and J A Wheeler (99) W H Freeman ISBN Introduction to the Relativity Principle G Barton (999) Wiley ISBN

3 Contents Contents 3 Introduction 5 Physics: Experiment and Theory 6 Newtonian Space-Time 8. Time 8. Space 8.3 Distance 8.4 Straight Lines 9.5 A diversion: A rather strange geometry 9.6 Galilean Transformations.7 Galilean Relativity.8 Problems 4 3 Michelson-Morley Experiment 6 3. Maxwell s Electromagnetism 6 3. The Experiment Problems 8 4 Constant Speed of Light 0 4. Consequences of constant speed of light 0 4. Problems 3 5 Lorentz Transformations 4 5. Velocity and Acceleration Transformations 6 5. Simultaneous Events Maximum Speed Problems 30

4 4 6 Space-Time Distances 3 6. Time Dilation and Length Contraction 3 6. Straight Lines Space-Time Diagrams 36 7 Space-Time Vectors Transformations Problems 40 8 Doppler Effects 4 8. Longitudinal Doppler Shift 4 8. Transverse Doppler Shift Problems 45 9 Particle Lifetimes Problems 47 0 Cloc (or Twin) Paradox Problems 50 Electromagnetism 5. The Space-Time Vector Differential Operator 5. Maxwell s Equations 5 Equations of Motion 55. Problems 58. Free Particle Motion 58.3 Motion in a constant force field 59.4 Problem 6.5 Particle Collisions 6.6 Problem 66 Index 67

5 Introduction Intended Learning Outcomes On completion of the module you should be able to: discuss the relationship between the expression for distance and geometry use Galilean transformations describe the conflict between Maxwell s wave equation and Galilean relativity analyse the Michelson-Morley experiment discuss the significance of the result of the experiment describe consequences of assuming a constant speed of light (non-invariance of distances and time-intervals) analyse simple experiments demonstrating time-dilation and length contraction use Lorentz transformations derive the velocity transformation from the Lorentz transformation define the three space-time invariant distances: space-lie distance, time-lie distance and null distance describe the essential property of the straight line between two points with a timelie separation (it is the longest time-lie path between the points) define the properties of space-time-vectors in terms of the four basis vectors determine the relation between components of space-time-vectors in different reference frames use space-time-vectors to study Doppler effects; particle decay times; cloc paradox; and particle dynamics state the relativistic form for particle energy and momentum define and use the vector differential operator express the electric and magnetic fields in terms of a space-time vector potential determine the transformation properties of electric and magnetic fields 5

6 Chapter Physics: Experiment and Theory PHYSICS EXPERIMENT OBSERVATION THEORY MODELS PREDICTIONS Physics begins with observations and experiments experiments are just more precise observations or, alternatively, observations are qualitative experiments. However physics is definitely not simply a list or record of the results of experiments. The main aim of physics is the construction of theories or models. A theory proposes entities which are the ingredients of the system being modelled and some set of rules which specify the properties and behaviour of these entities. The rules usually involve mathematics but this is is mathematics with a definite purpose. (Mathematicians often don t appreciate this!) The construction of a theory often involves the creation of new mathematics. The aim of a theory is to mae predictions. A theory which cannot be used to mae predictions is NOT a theory. The predictions need to be compared to experimental results. If there are no existing experimental results then we need to persuade a friendly experimenter to design a suitable experiment and mae some measurements! If the predictions do not agree with the experimental result then the theory fails. (Note that this concept of a theory failing does not exist in mathematics!) 6

7 7 When a theory fails then we need to invent a new one. Unfortunately there is no prescription for doing this. This is a slightly naive view of physics: Problem arise because of the uncertainties involved in experimental results and, sometimes, in the approximate nature of the quantitative predictions. This means that it is not always immediately obvious whether the prediction and the actual result agree or disagree. This explains why there is an emphasis in physics laboratories on the determination of the uncertainties in the experimental results. Uncertainties in the predictions can arise if the equations which result from the theory are too complicated to solve exactly and approximations have to be employed: Unfortunately this is the usual situation.

8 Chapter Newtonian Space-Time The following is a review of the essential features of the Newtonian theory of space-time.. Time Time is one-dimensional that is it is described by one parameter t. Time is measured by a device called a cloc. If we standardize the cloc unit (eg to be the second) then all observers will measure the same time intervals. That is, time-intervals are invariant.. Space Space is three-dimensional that is, it requires 3 parameters to specify the space points. For each pair of space points P and P there is an invariant quantity called the distance between P and P. The distance is measured by a device called a ruler. If we standardize the ruler unit (to be, say, the metre) then even if two observers use two different sets of parameters to describe the space points x, y, z and u, v, w say they will still agree on the distance between points..3 Distance If we choose to describe the space points by using a reference frame with origin O and mutually perpendicular axes OX, OY, OZ then the distance between two points P, represented by (x, y, z ), and P, represented by (x, y, z ), is s(p, P ) = (x x ) + (y y ) + (z z ) (.) This distance defines the geometry of the space the so-called Euclidean geometry. The distance ds between two neighbouring points (x, y, z) and (x + dx, y + dy, z + dz) is given by ds = dx + dy + dz (.) 8

9 .4 Straight Lines 9 The length of a general path L is defined as s(l) = ds (.3) That is we simply add the lengths of each segment of the path. We could invent a model of space with a more complicated geometry. For example we could define L ds = g xx dx + g yy dy + g zz dz + g xy dxdy + g yz dydz + g zx dzdx (.4) The parameters g ij then define a non-euclidean geometry..4 Straight Lines A straight line can be defined in terms of distance. The straight line joining space-points P and P is the path between these points that has the shortest length. In the Newtonian space there is only one straight line between two points. However in more complicated geometries this need not be so. We can show (not particularly easily!) that in Newtonian geometry there is a unique straight line between two points (x 0, y 0, z 0 ) and (x, y, z ) and this can be written as x (s) = s (x x 0 ) + x 0 y (s) = s (y y 0 ) + y 0 (.5) z (s) = s (z z 0 ) + z 0 where s is some parameter which specifies the position on the line and is zero at (x 0, y 0, z 0 ) and is at (x, y, z )..5 A diversion: A rather strange geometry I want to demonstrate the way in which the expression for distance in a particular model of space influences the geometry of that space. In order to do this I am going to invent a two-dimensional space in which the distance between neighbouring points (x, y) and (x + dx, y + dy) is ds = cos ( y a ) dx + dy (.6)

10 .5 A diversion: A rather strange geometry 0 a is some length. I now consider two paths in this space both oriented in the y-direction, that is the x-coordinate on each path is fixed. y P Q P Q P0 Q0 x I mar three points on each of these paths P 0 -P -P and Q 0 -Q -Q. The co-ordinates of these points are: P 0 = (d, 0) Q 0 = (d + D, 0) P = (d, πa 4 ) Q = (d + D, πa 4 ) P = (d, πa ) Q = (d + D, πa ) (.7) I will now wor out the lengths of some lines involving these points. P 0 Q 0 : I consider the line from P 0 to Q 0 which has y = 0. On this line we have dy = 0. The element of distance then becomes ds = cos( y a ) dx = dx (because cos(0)=). The length of this line P 0 Q 0 is, from (.6), the integral over dx from d to d + D and is therefore (d + D) d = D. P Q : I now consider the line between P and Q which again has a constant value of y (= πa/4). Again on the path dy = 0 and the element of distance becomes ds = cos( y a ) dx dx = (because cos(π/4)=/ ). The length of this line P Q is, from (.6), the integral over dx/ from d to d + D and is therefore (d + D)/ d/ = D/. P Q : Finally consider the corresponding line, that is with constant y, between P and Q. The element of distance then becomes ds = cos( y a ) dx = 0 (because cos(π/)=0). The length is therefore 0!!

11 .6 Galilean Transformations We therefore have two lines both in the y-direction which start with a separation D and end with a separation zero. Two lines having the same direction is one of the ways parallel could be introduced. So parallel lines do meet! How can this be? At this point you are probably thining that I have introduced a model that has no relevance to physics (and even less to you!). Not so! You have certainly used such a geometry. I should point out that the line we considered between P and Q is not a straight line, although the line we considered between P 0 and Q 0 is. However this complication does not affect (or explain) the strangeness of the results..6 Galilean Transformations The concept of the reference frame is vital in Physics and it is equally vital to be able to convert results obtained using one reference frame to those obtained using an other. In this module I will, for simplicity, only consider conversions between reference frames which have the same axes but whose origins are moving relative to each other. An inertial reference frame is one which is not rotating (that is the directions of the axes are not changing) and whose origin is not accelerating. Consider two observers who use two different inertial frames F and F that have two different origins O and O. Suppose also that the velocity of O relative to O is constant and is denoted by V. Y Y Z O X Z r O r X If the two observers label the space points by vectors r and r from the two origins then these are related by r = r OO = r Vt (.8) Strictly speaing this holds true only if the origins O and O coincide at time 0. However if we wor with the displacements r and r that are made in a time-interval t then these are related by r = r V t (.9)

12 .7 Galilean Relativity This relation holds whatever the initial positions of the two origins. Notice that if the displacement between two points is observed at the same time then r = r and the two observers in the two reference frames would agree on the distance between the two points. If we divide (.9) by t and tae the limit t 0 then we obtain the relation between the velocities measured in these two reference frames: v = v V (.0) These equations (.8)-(.0) relating the positions and the velocities are nown as Galilean transformations. We should also note that the theory implicitly assumes that time-intervals in the two reference frames are the same t = t (.).7 Galilean Relativity The relativity postulate is that the laws of physics must loo the same in all inertial reference frames. This relativity postulate, in detail, means the following: Suppose that we have equations representing (what we believe to be) a physical law expressed in terms of the co-ordinates and velocities of a particular reference frame F. Now we use the above Galilean transformations and express the equations in terms of the corresponding variables of reference frame F. Then the transformed equations should have exactly the same form in terms of these transformed variables. We should emphasise that this relativity postulate imposes severe restrictions on the other laws of physics..7. Particle Collisions & Conservation of Kinetic Energy As an example of this we consider a particle scattering experiment in which a set of particles collide and emerge from the collision as a different set of particles. In this experiment the initial set of particles have masses m (I), =,..N (I), and in a particular reference frame F the laboratory say have initial velocities v (I). The final set of particles have masses m (F ), =,..N (F ) and have final velocities v (F ). These velocities are measured in a region of space in which the potential energy is zero.

13 .7 Galilean Relativity 3 Conservation of energy then requires that the initial and final inetic energies must be equal: N (I) = ( m(i) v (I) v(i) ) = N (F ) = ( ) m(f ) v (F ) v (F ) (.) Suppose now another observer records the velocities in a different reference frame F, moving with velocity V with respect to the first frame. The second observer s version of conservation of energy equation will be N (I) = ( m(i) v (I) ) N (F ) v (I) ( ) = m(f ) v (F ) v (F ) = (.3) Suppose we use the Galilean transformation (of velocities) to transform (.3) bac into reference frame F. The required transformation is for each of the particle velocities. Applying this to (.3) gives v = v V N (I) = ( ) ( ) N m(i) v (I) V v (I) V (F ) ( ) ( ) = m(f ) v (F ) V v (F ) V = and after multiplying out the scalar products N (I) = m(i) N (F ) = m(f ) [ v (I) v(i) + V V v(i) ] V = [ v (F ) v (F ) + V V v (F ) V (.4) ] (.5) This can be written as three separate terms: [ N (I) = m(i) [V V] V [ N (I) [ N (I) = [ v (I) m(i) v(i) N (F ) ] N (F ) = = m(i) v(i) N (F ) = = m(f ) m(f ) ] + m(f ) v (F ) [ ] ] v (F ) v (F ) + ] = 0 (.6)

14 .8 Problems 4 According to the law of relativity this equation should be exactly the same the expression for conservation of inetic energy (.). This is clearly not so. The only solution to this dilemma is that the two extra terms must also be zero and they must be zero for any value of the velocity V. So we must have N (I) = m (I) v(i) = N (F ) = m (F ) v (F ) (.7) and N (I) = N (F ) m (I) = = m (F ) (.8) The first of these equations expresses conservation of momentum and the second expresses conservation of mass. So the (Galilean) relativity principle says that if we use conservation of (inetic) energy we must also use conservation of momentum and conservation of mass. We are forced to accept the three conservation laws as a pacage : they are not independent..8 Problems (i) Equipped with only a ruler, how could you define and set up 3 mutually perpendicular axes (ii) Above we have stated that distance defines geometry. As examples try to show that two geometric concepts angle parallel lines can be described entirely in terms of distance. Note: Parallel lines can be defined in more than one way. (iii) Discuss how you could define the straight line between two points A and B. Suppose wanted to extend this straight line (beyond B) by a specified length L. Explain how this could be done. That is how you would choose the new end of the line C. (iv) Give physical meaning to the model space, in the A diversion... section and give examples of such paths P 0 -P -P and Q 0 -Q -Q. If you get stuc consult a fellow student from Meteorology! (v) Go through the particle scattering example in the Galilean Relativity section yourself. Start from the conservation of energy (.3); use (.0) to relate each v to the corresponding v then investigate the conditions which mae the equation equivalent to (.).

15 .8 Problems 5 (vi) (a) Consider two reference frames F and F and suppose that F is moving with speed V x along the x-axis relative to F. A particle has a speed v y in the y- direction in F. What is its velocity in F? (b) Consider the same two reference frames. Suppose a particle is moving in the y- direction in F but we don t now its speed and that in F this particle has speed v but that we don t in which direction. Find the two particle velocities.

16 Chapter 3 Michelson-Morley Experiment In this section we investigate an experiment whose result contradicted that predicted by the Newtonian-Galilean space-time theory; and hence disproved that theory. However there is evidence to suggest that Einstein was not aware of this experiment and that it had therefore no bearing on the development of relativity. I therefore start with a something which almost certainly did influence the theory. 3. Maxwell s Electromagnetism For most theoretical physicists, including Einstein, a significant problem was that the electromagnetic equations of Maxwell do not satisfy the Galilean relativity principle. Maxwell s equations are fundamental in physics. There have been innumerable tests of this theory and there has been nothing to suggest that even a minor modification is required. They still form a major part of the Standard Model which is the most recent basic physics theory. In free space Maxwell s equations give rise to a wave equation. If, purely for simplicity, I assume there is no dependence on y and z this wave equation is d S d t = d S (3.) d x where S is any of the components of the electric or magnetic field. Suppose this equation holds in a particular reference frame F. What form does the equation have in F which is moving parallel to the x-axis with speed V relative to F? The new variables, instead of x and t, are x and t where x = x V t t = t The relationships between the derivatives are 6

17 3. The Experiment 7 d dt = d dt V d dx d dx = If I use these derivatives then the wave equation in frame F becomes d S ( d t = V ) d S d x + V d S (3.) dx dt d dx This is clearly different to equation (3.) and so this equation does not satisfy Galilean Relativity. Hence there is either something wrong with Maxwell s equations or with the Galilean transformations. 3. The Experiment In this famous experiment, which has been repeated many times, the assumption is that there is a special inertial reference frame F the aether which is the basic medium of light and in which light has the same speed in all directions. We then consider another inertial frame F moving with velocity V with respect to the F and mae measurements on the light to try to determine the value of V. In practice frame F was attached to earth so the experiment was trying to measure earth s speed. The apparatus used is the Michelson interferometer.

18 3.3 Problems 8 In the diagram M is a partially-reflecting mirror; M and M 3 are normal mirrors. The distances from the centre of M to M and M 3 are equal (to D). In this apparatus the light can tae two paths before emerging: M M M and M M 3 M. The time taen on these two paths should be different (Why?) and the interferometer essentially measures the time difference t. If the speeds of light along the 4 paths are c c 4 as shown then the time difference is: t = D c + D [ D + D ] c c 3 c 4 (3.3) How do we determine the speeds of the four beams? You should have already solved this problem (Problem(.(v))). We have particles (photons) whose directions we now in F but with unnown speeds; whereas in F we now the speeds (= c) but not the directions. The diagram shows these relations in the case where V is parallel to c 3. If V/c is small then resulting expression for the time difference t is approximately t = D V c 3 (3.4) The Michelson interferometer measures t indirectly via interference of the two light beams. The experimental result was that the velocity of the earth is zero. And moreover it is zero at all times during the year. 3.3 Problems (i) Show that the magnitudes of the 4 speeds of light, in the Michelson-Morley experiment, in frame F are, in the case shown in the diagram, c = c = V c 3 = c V ; c 4 = c + V and hence that the time difference t is given by (3.5)

19 3.3 Problems 9 t = D c V V. (3.6) (ii) In the original version of the Michelson-Morley experiment the distance D was m. (This was achieved by using reflections from several mirrors rather than just two but the above analysis is still valid). The expected speed of the earth was approximately ms. Calculate the expected time difference t using (3.4) and relate this to the period T of a light wave of wavelength 600nm. That is wor out t/t. Why is this fraction relevant? (iii) Explore other, more recent, versions of the Michelson-Morley experiment and summarize the accuracies of the results obtained.

20 Chapter 4 Constant Speed of Light The current interpretation of the zero result of the Michelson-Morley experiment is, not that there is no relative motion, but that there is something wrong with the Galilean transformation equation (.0) and hence the whole Newtonian theory of space-time. The simplest resolution of the result of the Michelson-Morley experiment is to assume that the speed of light has the same value in all inertial reference frames. If this is so then the magnitudes of the magnitudes of the velocities c i are all equal.; the denominators in (3.3) are all the same and the time difference is zero in complete agreement with the experiment. We could arrive at the same conclusion if we consider Maxwell s equations to be paramount. Maxwell s equations give rise to waves moving with speed c. If these equations are required to have exactly the same form in every inertial frame (principle of relativity) then the speed of the waves must also be same in every inertial frame. This gives us a clue as to the form of a new theory. We need to find a replacement for the Galilean transformation which is consistent with the speed of light being the same in all inertial reference frames. Note that the Galilean transformation was deduced from the form of the Newtonian model of space and so we also need a new model for space. What we need to do now is to investigate consequences of the assumption of a constant speed of light and then, later, to incorporate this into a proper theory. It is important to note that in this section we are just exploring ideas that need to be incorporated in any new theory. 4. Consequences of constant speed of light 4.. Clocs and Rulers If the speed of light is a universal constant then this can be used to relate the time and distance units. We can therefore measure time in terms of distance units using a ruler or equivalently measure distance in time units using a cloc. 0

21 4. Consequences of constant speed of light The speed of light is reduced to a conversion factor relating the two (equivalent) sets of units the second and the meter. This is exactly equivalent to.54 being the conversion factor between centimetres and inches. We can no longer measure the speed of light. It is simply defined to be the conversion factor m s If we want to exploit the symmetry of space-time we should really use the same units to measure space and time. 4.. Time Dilation Consider a light beam being reflected bac and forth between two parallel mirrors that are separated by a distance a. Suppose there are two observers taing measurements; one uses the reference frame F in which the mirrors are fixed; the other uses a reference frame F moving with a speed V along the plane of the mirrors. In F the time taen for the light to travel from one mirror to the other and bac again is t = a c That is, the distance travelled divided by the speed. This is what the first observer would measure. Now consider the second observer. In reference frame F the path taen by the light beam is as shown because the mirrors are moving (with velocity V) in this frame. If the period is t in this frame the distance travelled by the light beam in one period is So we have ( V t a + ) t = ( ) V t a c +.

22 4. Consequences of constant speed of light Solving for t gives t = a c V. This is the period measured by the second observer. Clearly the two observers measure two different time intervals for the same events. This means that one of the fundamental assumptions of Newtonian space-time that time-intervals are invariant no longer holds. The ratio of the two measurements is t t = ( ). (4.) V c 4..3 Length Contraction Suppose we perform a similar experiment except that the reference frame used by the second observer is moving in the direction of the normal to the mirrors, that is parallel to the light beam. Suppose that in the frame F the distance between the two mirrors is a. The time taen to travel from one mirror to the other and bac again is t = a c

23 4. Problems 3 as before. Now consider the reference frame F moving with velocity V perpendicular to the plane of the mirrors with respect to F. The diagram shows these mirrors (moving with velocity V in F ) at three times: time 0 when the light just leaves the first mirror; time t when the light just reaches the second mirror; and time t = t + t when the light returns to the first mirror. If I assume that the distance between the mirrors is a (not necessarily equal to a) in this frame then these times are given by Solving these two equations gives t = a V t c t = a + V t c t = a c V If we assume that the relationship between t and t is the same as in (4.) then we find that a a =. ( ) V. (4.) c Clearly the two observers measure two different distances between the same objects. This means that another of the fundamental assumptions of Newtonian space-time that distances are invariant also no longer holds. 4. Problems Calculate the ratios of time and space intervals predicted by (4.) and (4.) for the case V/c=3/5.

24 Chapter 5 Lorentz Transformations The Michelson-Morley experiment demonstrates a failure for the Galilean transformations that are a property of the Newtonian space-time model. We now loo for modified transformations called the Lorentz transformations which are consistent with the concept of a constant speed of light. We then need to incorporate this into another model of space-time Einstein space-time. Consider two reference frames F and F and suppose F is moving with a speed V along the x-axis relative to F. In this case the Galilean transformations equation (.9), in terms of displacements, which we want to replace are x = x V t t = t y = y z = z (5.) We loo for transformations that leave the co-ordinates perpendicular to the relative motion unaltered. Let s try x = α x + β t t = γ x + δ t y = y z = z (5.) This is similar to (5.) but includes changes to the time-intervals. We have to see if we can find four suitable parameters α, β, γ and δ. These parameters are independent of the displacements and must depend only on the relative speed V. We do this by considering simple experiments with mirrors as above. We repeat the experiment first considered in section 3.. (time dilation). The space-time displacement is between the two events when the light beam leaves and then returns to the lower mirror. In frame F this space-time displacement is x = 0 y = 0 z = 0 t = a 4

25 5 The spatial displacements are zero because in this reference frame the light beam returns to its starting point. In reference frame F the mirrors are moving to the left with speed V and so the spacetime displacements are: x = V t ; y = 0; z = 0; t = a c V The detail of the result for t is given in section 3... If these results are inserted into the trial form of the Lorentz equations (5.) we obtain: δ = V β = V V We next perform the similar experiment in which the same two mirrors are now fixed in F and hence are moving with speed V to the right in F. Analysing this experiment gives: α = V γ = V V The final result as the replacement for the Galilean transformation the Lorentz transformation is:

26 5. Velocity and Acceleration Transformations 6 x = x t = V V ( x V ) c x t ( x t Vc x ) (5.3) y = y z = z Notice that by using x t = c t rather than t as the time variable the equations assume a more symmetric form. Notice also that this applies to the components ( x t, x, y, z) of the space-time separation of any two space-time points. The inverse transformation is almost identical: It only involves a change in the direction of the velocity V. ( x = V x + V ) c x t x t = V y = y z = z ( x t + Vc x ) (5.4) 5. Velocity and Acceleration Transformations We now consider a different situation in which ( x t, x, y, z) is not the separation of two arbitrary space-time points but is the physical displacement made by some object. That is, ( x, y, z) is the spatial displacement made in time x t /c. We can determine the velocity transformation corresponding to equation (5.3). In F the components of the velocity, in the limit t 0, are In frame F they are given by v x = x t ; v y = y t ; v z = z t.

27 5. Velocity and Acceleration Transformations 7 v x = x t ; v y = y t ; v z = z t The resulting Lorentz velocity transformation is: v y = v z = v x = V V v x V V v x v y V v x v z V v x (5.5) If V/c is small then these results reduce to the Galilean results. Again, the results for the inverse transformation are very similar: v y = v z = v x = V V v x + V + V v x v y + V v x v z + V v x (5.6) Consider the special case v x = c, v y = 0 and v z = 0 that is a photon moving along the x-axis with speed c. The transformed quantities according to (5.5) are v x = c, v y = 0 and v z = 0. That is the photon is still travelling with speed c. This should not be a surprise since the equations were set up in a way that forced this speed to be constant in every inertial frame. However it s always worth checing that we have got it right! We can similarly determine the acceleration transformation corresponding to equation (5.3). In F the components of the acceleration are

28 5. Velocity and Acceleration Transformations 8 a x = dv x dt ; a y = dv y dt ; a z = dv z dt. In frame F they are given by a x = dv x dt ; a y = dv y dt ; a z = dv z dt The resulting Lorentz acceleration transformation is: ( V ) 3 a y = a z = a x = ( V ) ( V v x ( V v ) 3 a x x ) ( V ) ( V v x ) a y + a z + V a ( x V v x V a ( x V v x ) v y ) v z (5.7) Again, the results for the inverse acceleration transformation are very similar:

29 5. Simultaneous Events 9 ( V ) 3 a y = a z = a x = ( V ) ( + V v x ( + V ) 3 a v x x ) ( V ) ( + V v x ) a y a z V a x ( + V v x V a ( x + V v x ) v y ) v z (5.8) 5. Simultaneous Events The concept of events being simultaneous is no longer universal. Suppose in reference frame F two events have a spatial separation with components x, y, z but these events are simultaneous that is, t = 0. Now we can use the above Lorentz transformation to determine the space and time displacements in the moving frame F : x t = x = V V y = y z = z x ( Vc x ) In the frame F the events do not occur at the same time ( x t 0). (5.9) 5.3 Maximum Speed In order for (5.3)-(5.6) to be valid we require, what I have implicitly assumed, that

30 5.4 Problems 30 V < v x < (5.0) Hence c has the significance of being the maximum allowable speed. 5.4 Problems (i) Complete the second part of the derivation of the Lorentz transformation, that is the determination of α and γ (ii) A distant galaxy is observed from earth. Two events, with a spatial separation of light years, are observed to be simultaneous. The separation is perpendicular to the direction from earth to the galaxy. If the galaxy is moving away from earth with speed c/0 determine the space and time separations in the reference frame fixed in the galaxy. (iii) Explain, in equation (5.7), how you would determine whether x t is positive or negative. That is, which of the events occurs first in F

31 Chapter 6 Space-Time Distances We have seen that the Einstein model of space-time removes two of the fundamental elements of Newtonian space-time: Distances are no longer invariant and neither are time-intervals. The form of the distance element in Newtonian space-time d x + d y + d z is an invariant quantity, in that theory, and defines the (Euclidian) geometry of the space. Is there anything which can tae its place in Einstein s space-time. The answer is yes. The Lorentz transformations do yield an invariant quantity. If we consider infinitesimal space-time displacements dx, dy, dz and dx t = c dt we have the result: So the quantity d x + d y + d z d x t = d x + d y + d z d x t (6.) d x + d y + d z d x t (6.) is invariant and can be used to define a distance. Unfortunately, unlie the corresponding quantity in Newton s space, this quantity is not necessarily positive. We need to distinguish three cases. (i) dx + dy + dz dx t > 0 In the case we call the displacement space-lie (because spatial displacements exceed the time displacement) and we can define the space-lie distance (ii) dx + dy + dz dx t = 0 ds = dx + dy + dz dx t (6.3) This is called a null displacement. It is a displacement that could be made by a photon (or beam of light). In this case we assign a distance zero. (iii) dx + dy + dz dx t < 0 3

32 6. Time Dilation and Length Contraction 3 This is called a time-lie displacement (because the time displacement exceeds the spatial displacements) and we can define the time-lie distance: dτ = dx t (dx + dy + dz ) (6.4) Even within this category there are two types of time-lie displacements: a forward timelie displacement in which dx t > 0 (that is one which moves forwards in time) and a bacward time-lie displacement in which dx t < 0 (that is one which moves bacwards in time) The most important type of displacement is the forward time-lie displacement: It is the type of displacement made in the motion of any physical object (apart from a photon). If a space-time displacement is null (or space-lie or time-lie) in one inertial reference frame then it is null (or space-lie or time-lie) in all inertial reference frames. A space-lie displacement with space-lie distance ds has the property that we can find one inertial reference frame F in which the displacement is purely spatial: dt = 0; dx + dy + dz = ds. Similarly a time-lie displacement with time-lie distance dτ has the property that we can find one inertial reference frame F in which the displacement is purely a displacement in time: dxt = dτ ; dx + dy + dz = 0. These definitions can all be extended to finite displacements. A forward time-lie path is one where all the infinitesimal components of the path are time-lie and so satisfy dx + dy + dz dx t < 0 and dx t > 0 The time-lie distance along such a path is τ (L) = L dτ = (dxt (dx + dy + dz )) L = c (( dt (v/c) ) (6.5) where the integral is over the infinitesimal displacements along the path L. τ(l)/c is called the proper time. This proper time is the time that would be measured by a cloc that moved along the time-lie path L. 6. Time Dilation and Length Contraction I want to loo again at time dilation and length contraction but now using the proper theory. For simplicity I shall consider space displacements only in the x direction.

33 6. Time Dilation and Length Contraction Time-Component Dilation Suppose two events, in a reference frame F, have a space-time separation ( x t, x). An event is jargon for anything which is specified by a time and a position. Suppose first the space-time separation is time-lie. That is, The magnitude of the separation is x t > x x t x and this, divided by c, is called the proper time separation. I can find a reference frame F 0, moving with speed V with respect to F, in which the space-time displacement is ( ) sign ( x t ) ( x t x ), 0 That is, the spatial component is zero. If I write the time component as x 0 t then I can obtain the relation x t = ( V ) x0 t Note the the time-displacement x t is larger in magnitude than x 0 t. Remember that x 0 t is the time-component of the space-time separation of the two events in a reference frame in which the spatial component is zero. These relations represent the formal statement of time dilation. You will have noticed that I have not specified what the velocity V is in the above equations: I will leave this as a problem. 6.. Space-Component Dilation Now suppose that the space-time separation of the two events is space-lie. That is, The magnitude of the separation is x > x t x x t

34 6. Time Dilation and Length Contraction 34 and this is called the proper space separation. I can find a reference frame F 0, moving with speed V with respect to F, in which the space-time displacement is ( ) 0, sign ( x) ( x x t ) That is, the time component is zero. If I write the space component as x 0 then I can obtain the relation x = ( V ) x0 Note the the space-displacement x is larger in magnitude than x 0. Remember that x 0 is the space-component of the space-time separation of the events in a reference frame in which the time component is zero. Notice that there is a complete symmetry with time dilation. Notice also that this is NOT related to the determination of the length of an object Length Contraction Consider an object which is moving with speed v x, in the x direction, in some reference frame F. In order to determine the length of the object we need to determine the positions of the two ends at the same time. The space-time separation is then (0, L) Suppose now I change to a reference frame F 0 in which this object is at rest. Clearly F 0 is moving with speed v x with respect F. Applying the Lorentz transformation gives the space-time displacement in F 0 as v x v x c, v x L Notice that in the reference in which the object is not moving we can determine the positions of the two ends at different times. Hence the length L 0 in this reference frame is L 0 = L v x

35 6. Straight Lines 35 L 0 is the proper length. I can write the above relation as L = v x L 0 That is the length of a moving object is smaller than its proper length. 6. Straight Lines Straight lines are important because they are the paths followed by free particles that is particles with no force acting upon them. Newton s first law still applies! In Newtonian space a straight line is the shortest distance between two particular (space) points. Consider the most important case of two space-time points separated by a forward timelie distance. Then from the set of forward time-lie paths connecting these two points the straight line path is the longest. Yes, the longest! x P O ct Figure 6.. Space-time diagrams The diagram shows some forward time-lie paths connecting the space-time points O and P. The straight line path OP is the longest such path. Example: Suppose that the displacement OP has the space-time co-ordinates, in a particular inertial reference frame, (c t, v t). This would be, for example, the displacement of a particle moving with speed v along in the x-direction. Suppose also that the displacement OQ has co-ordinates, in the same inertial reference frame, (c t/, (v+ v) t/). The displacement QP can be evaluated simply by subtracting OP OQ: QP = (c t/, (v v) t/). Now let us wor out the length of the path OP and of the path OQP (= OQ + QP ). In principle, I use integral (6.5) to wor out the time-lie distances on the three segments but since the velocities are constant the integrations are trivial:

36 6.3 Space-Time Diagrams 36 x Q P O ct Figure 6.. Shortest or longest? τ(op ) = c t v τ(qp ) = c t Hence, the lengths of the two paths are τ(oq) = c t (v v) (v + v) (6.6) τ(oqp ) = τ(oq) + τ(qp ) = c t τ(op ) = c t (v + v) + c t (v v) v (6.7) The time-lie length of OQP is just τ(oq) + τ(qp ) which is always less than τ(op ) Problem 5. Choose values for v/c and v/c and demonstrate the truth of the above statement. 6.3 Space-Time Diagrams Some care is required in the interpretation of space-time diagrams such as that shown above. Consider the space diagram (6.3) showing the x y plane. In this diagram all the lines OP, OQ, OR etc have the same length. (I have assumed that the time and z-components are zero). In fact all displacements starting from the origin and ending on the circle will have the same (space-lie) length This is the sort of diagram that loos familiar to us. Now consider the space-time diagram (6.4) showing the x t x plane (where x t =ct). In this diagram the curve is a hyperbola: x t x =constant. The straight lines OP, OQ, OR all have the same time-lie length. And in fact any straight line starting at the origin and ending on the curve will have the same time-lie length.

37 6.3 Space-Time Diagrams 37 y R Q P O x Figure 6.3. x - y plane x Q P O ct R Figure 6.4. x t - x plane The concept of length involved in this diagram needs getting used to!

38 Chapter 7 Space-Time Vectors Ordinary space vectors were introduced because they enabled us to express physical laws and equations in ways that are independent of the choice of reference frame. For example: is Newton s equation written in vector form. m d x = F (x, t) (7.) d t Now I want to introduce space-time vectors (or four vectors) for the same reason. I will denote space-time vectors in the form Ã. Ordinary space vectors are special examples of space-time vectors with the time part equal to zero. I denote the four basis space-time vectors by ẽ t ẽ x ẽ y ẽ z (7.) The three basis space vectors ẽ x, ẽ y, ẽ z are usually denoted by i, j,. The present notation is chosen in order to emphasize the space-time symmetry. ẽ x, ẽ y, ẽ z are unit space-lie vectors. That is, they are space-lie vectors with space-lie size equal to. The corresponding ẽ t is a time-lie vector with time-lie size equal to. This is a vector pointing in the forward time direction. Remember that time-lie distances are calculated differently to space-lie distances! A general space-time vector can be written as à = A t ẽ t + A x ẽ x + A y ẽ y + A z ẽ z (7.3) The scalar products of the basis vectors are defined to be: 38

39 7. Transformations 39 ẽ µ ẽ ν = 0 ẽ µ ẽ µ = µ ν µ = x, y, z (7.4) ẽ t ẽ t = The minus sign has to be introduced for the scalar product ẽ t ẽ t in order to produce the correct space-time invariant (6.). The scalar product of two vectors à and B is given by à B = A t B t + A x B x + A y B y + A z B z (7.5) If I use this to form the scalar product of the space-time displacement dx = dx t ẽ t + dxẽ x + dyẽ y + dzẽ z with itself I get dx dx = x t + dx + dy + dz (7.6) This is the space-time invariant (6.). The size of a space-time vector can be assigned in exactly the same way as we assigned a size to a space-time displacement in Chapter Time-Lie vectors If à is a time-lie vector then à à < 0 and its (time-lie) magnitude is à à 7.0. Space-Lie vectors If à is a space-lie vector then à à > 0 and its (space-lie) magnitude is à à 7. Transformations The components of any space-time vector transform in exactly the same way as the components of a space-time displacement. That is, we simply use the Lorentz transformation. If we consider two inertial reference frame F and F which are moving with relative speed V along the x-direction then the components of a general space-time vector à are related by

40 7. Problems 40 A t = A x = V V ( A t V ) c A x ( A x V ) c A t (7.7) A y = A y A z = A z This is just the Lorentz transformation (5.3) with the components of a space-time displacement replaced by the components of Ã. The corresponding inverse transformation is A t = A x = V V A y = A y A z = A z ( A t + V ) c A x ( A x + V ) c A t (7.8) 7. Problems Show the definitions of space-time vector magnitudes mae sense when applied to the basis vectors (7.).

41 Chapter 8 Doppler Effects Although, as we have seen, the speed of light is the same in all inertial reference frames this is not true for other properties of light. If light is emitted from a source that is stationary in one reference frame and if it is viewed by an observer in another frame then the observed frequency of the light will depend on the relative speed of the two frames. This is nown as the Doppler effect. A plane wave can be written as S (x, t) = A cos ( x ω t) (8.) where A is the amplitude of the wave; ω is the angular frequency; and is the wave vector. The direction of the (space) vector is the direction of propagation of the wave. The wavelength and period are given by λ = π (8.) T = π ω The speed of propagation of the wave is v = ω/. If the wave is light then this speed is, of course, c and so and ω are related by. = ω c (8.3) The term ( x ω t) is called the phase of the wave. This is an intrinsic property of the wave and its value should not depend on the choice of reference frame. That is. x ω t = x ω t (8.4) We can ensure that this is so by writing it as the scalar product of two space-time vectors, K and x 4

42 8. Longitudinal Doppler Shift 4 These two space-time vectors are defined as K = K t ẽ t + K x ẽ x + K y ẽ y + K z ẽ z K t = ω/c K x = x K y = y K z = z (8.5). x = x t ẽ t + xẽ x + yẽ y + zẽ z Then the scalar product, which is automatically an invariant quantity, is precisely the phase:. K x = x ω t (8.6) The advantage in expressing things this way is that we now how the components of a space-time vector transform in going from one reference frame to another. We simply have to use equations (7.7). K t = K x = V V K y = K y K z = K z ( K t V ) c K x ( K x V ) c K t The scalar product of K with itself is, of course, invariant but it also it is very simple: (8.7) I intend to consider two special cases: K K = 0 (8.8) 8. Longitudinal Doppler Shift In this case the direction of propagation of the light is parallel to the relative motion of the reference frames:. K x = x = ω/c K y = y = 0 K z = z = 0 K t = ω/c

43 8. Longitudinal Doppler Shift 43 I have chosen the direction of propagation of light to be along the positive x-axis. Using (7.7) to get the components in the other frame gives: ω /c = K x = V V (ω/c Vc ω/c ) ( K x V ) c K x (8.9) K y = 0 K z = 0 This yields the following equations for the frequency and wavelength: (c ) V ω = ω c + V (c ) (8.0) + V λ = λ c V 8.. Light from distant stars I now relate this to a physical situation: The observation of light from a distant star. If we tae the frame F to be that of the distant star. In this frame the source of light is (almost) stationary and then ω is the natural frequency of the emission spectrum of some atom (due to the difference in energy levels of the atom). Reference F is that of Earth. What I actually mean is that F is an inertial reference frame on Earth in which compensation has been made for the rotation. Light is propagating in the positive x-direction. That is, from the star towards Earth. In the above equations a positive value of V corresponds to the observer in frame F moving away from the source. This effect gives rise to the well-nown red-shift of light from distant stars. Because (we believe) the universe is expanding distant stars are moving away from us with a large speed (hence V is positive in the above equations). Light emitted by atoms on a distant star has the natural frequency ω of that atom because the atom is (almost) at rest in that reference frame. We observe the frequency ω which is smaller than ω. The shift is to a lower frequency and hence to a longer wavelength. If the universe were contracting distant stars would be moving towards us and V would be negative in formula (8.0) and

44 8. Transverse Doppler Shift 44 there would be a shift to shorter wavelengths. In the steady-state model of the universe a red-shift and a blue-shift would be equally liely. The formula (8.0) is used to determine the relative speed of distant stars: V = c λ λ λ + λ (8.) 8. Transverse Doppler Shift Suppose, that in reference F, the direction of propagation of the light is perpendicular to the relative motion and is in the y-direction: K x = x = 0: K z = z = 0; K y = y = ω/c; K t = ω/c. Using 8.7 the transformed components are: K x = K t = V V K t ( V ) c K t (8.) K y = K y K z = 0 The frequency shift in this case is much smaller. The direction of propagation the wave is however different in the two frames. In one frame it is moving in the y-direction whereas in the other frame it is moving at an angle θ to this direction where θ is given by tan (θ) = K x K y = V c V (8.3) 8.. Stellar Aberration An example of this transverse Doppler shift is stellar aberration. If a distant star is observed then its position (taing the rotation of earth into account!) will vary as the direction of motion of earth changes.

45 8.3 Problems 45 In the earth s orbit it will be transverse to the light twice per year. The angular difference between the star s position at these two times is twice that given by equation (8.3) that is θ = arctan V c V (8.4) Data for stellar aberration was collected by Bradley in 77. See Introduction to the Relativity Principle, Gabriel Barton, chapter Problems (i) The observed wavelength of a particular spectral line emitted from Hydra galaxy is 475 nm. The corresponding line in the laboratory has wavelength 394 nm. Is the galaxy moving towards or away from us? Determine the speed of the galaxy relative to the earth. (ii) In the previous question, how do you thin the corresponding line could be identified. Remember that the ratio ω /ω (or λ /λ) is the same for all frequencies (and wavelengths). (iii) Bradley s data on stellar aberration gave the angle θ (equation(8.4)) to be 39 seconds. Determine the speed of the earth.

46 Chapter 9 Particle Lifetimes One of the most tested aspects of relativity is that of the velocity dependence of the lifetimes of unstable particles. Suppose some unstable particle is created at time t = 0 in a particular inertial reference frame (the laboratory say) and moves with a constant speed v in the x-direction in this frame before it decays at time t L. The space-time displacement vector of the particle (from creation to decay) is The size of this time-lie displacement is x = c t L ẽ t + v t L ẽ x (9.) t L c v (9.) This quantity is invariant and, if I divide by c, is the proper lifetime τ L of the particle. The proper lifetime τ L is the lifetime in a reference frame in which the particle is not moving. The relation between this proper lifetime τ L and the time measured in the laboratory reference frame is t L = τ L v (9.3) which says that the observed lifetime depends on the speed v of the particle. In fact t L increases as v increases. Note that τ L is an inherent property of the particle: the speed v is a property of the particular process used to create the particle. There are usually many different processes in which a particular particle can be created. There have been extensive tests of this relation with complete verification of the theory. Of course particle creation and decay is a quantum effect and as such is a random process. What experimenters have to measure is the average lifetime, determined by looing at the decays of many particles (of the same ind from the same process). 46