The Special Theory of Relativity

Transcription

1 The Special Theory of Relativity The views of space and time which I wish to lay before you have sprung from the soil of experimental physics, and therein lies their strength. They are radical. Henceforth space by itself, and time by itself, are doomed to fade away into mere shadows, and only a kind of union of the two will preserve an independent reality. Hermann Minkowski, Newton s laws of motion and gravity have been wildly successful in the everyday areas in which they are applicable. However, once we move past these everyday applications and begin to discuss systems moving at speeds approaching that of light, or strong gravitational fields, then Newton s laws have to be modified. In this chapter we want to discuss the modifications that Einstein made to Newton s laws of motion, obtaining his Special Theory of Relativity, leaving the changes to the gravitational law to be discussed later. We begin by giving Einstein s motivation for Special Relativity stemming from his investigations of Maxwell s equations for electrodynamics. The laws of electrodynamics predict that electromagnetic waves propagate through the vacuum at a precise speed, c, which seems to be independent of the relative motion of the observer. This was a tremendous puzzle, requiring new concepts which ultimately led to a fundamental new understanding of the nature of the Universe. Part of that understanding is developed in this chapter. We discuss the observations seen by two different observers in relative motion, deriving the Lorentz transformations. These transformations lead to some very unintuitive ideas, mixing together space and time into a single spacetime, the interesting properties of which we will study through various examples. In order to better understand the new concepts introduced by Special Relativity, we need to introduce some new mathematical ideas including four-vectors and tensors. These ideas generalize the simple concepts of scalars and vectors of elementary physics to be consistent with Special Relativity. We explore these mathematical ideas, which will be extremely important in our later work, once again using electrodynamics as a guide. We finish the chapter with a discussion of the Minkowski metric and gauge invariance, each of which will be crucial later. 1 Electrodynamics. Einstein was motivated to discover the modification to Newton s laws of motion by considering a puzzling aspect of the laws of electrodynamics, namely Maxwell s equations. These equations for the electric and magnetic fields, E and B, respectively, are E = ρ ɛ 0 E = B t B = 0 B = µ 0J + E µ0 ɛ 0. t (1) Here ρ is the charge density, and J is the current density. The constants ɛ 0 and µ 0 are the permittivity and permeability of free space, respectively. Together with the Lorentz force 1

2 law, ( ) F = q E + v B, (2) which gives the force on a charge q, moving with a velocity v through an electric and magnetic field, Eq. (1) gives a complete description of electricity and magnetism. As a specific example of these laws, let s look at the source-free (vacuum) Maxwell s equations, obtained from Eq. (1) by setting the sources ρ and J to zero. Then, Eq. (1) reads E = 0 E = B t B (3) = 0 B E = µ 0 ɛ 0 Let s play around with these a little bit. Take the curl of the second equation ( ( E ) ( = E ) 2 E ) B = = ( B t t ). But, from the first of Eq. (3), E = 0, and from the fourth equation, B = µ 0 ɛ 0 E t. Plugging these in gives 2 E µ0 ɛ 0 2 E t 2 = 0. This is precisely the equation for a wave traveling at a speed v if we identify the value (µ 0 ɛ 0 ) 1/2 as the velocity of the wave. Since these are constants, we find a constant velocity of v 1 µ0 ɛ 0 = 299, 792, 458 m/s, which is precisely the speed of light, c! Thus, the wave equation reads t. 2 E 1 c 2 2 E t 2 E = 0, (4) where = is the d Alembertian operator. Furthermore, starting from the fourth c 2 t 2 of Eq. (3) and performing the same calculations one finds 2 B 1 c 2 2 B t 2 B = 0. (5) Thus, both the electric and magnetic fields travel in waves in vacuum. Eqs. (4) and (5) showed that not only are the electric and magnetic fields part of the same electromagnetic field, but light is, as well! This was Maxwell s crowning achievement. However, Eqs. (4) and (5) contain a curious property. These equations clearly say that the electromagnetic field always travels at the speed of light; but, one might ask: at the speed of light with respect to what? Conventional intuition says that speeds are only defined with respect to something else. As an example, suppose that you are riding in a train at 50 miles per hour. Then you throw a ball to a friend at 30 miles per hour. You would see the 2

3 ball traveling at 30 miles per hour, but someone standing outside the train watching you would see the ball traveling at the speed of the ball plus the speed of the train, i.e., 80 miles per hour. This idea is completely reasonable and intuitive. We can make it more mathematically precise using the Galilean Transformations, which relate the motion of objects as seen by different observers, as seen in Figure 1. Here we have two observers, in inertial reference frames S and S, respectively, initially overlapping at time t = t = 0. The observer in frame S is moving to the right at speed V with respect to S. y y V S S x x Figure 1: The Galilean transformations relate the motion of an object moving nonrelativistically as seen by two different observers. The object in the figure has coordinates r = (x, y, z) in frame S, and r = (x, y, z ) in figure S. What we would like to do is give the relationship between the coordinates, which we can easily just read off of the figure. Clearly y = y, since the motion is along the horizontal axis, we similarly see that z = z. For this nonrelativistic motion, the two times measured by each observer will agree with each other, so t = t. Now, the differences in the x coordinates of each frame depends on the relative motion between the two frames, and how long the frames have been in relative motion. So, x x = V t = V t, since t = t, and so x = x V t. This gives us the full Galilean transformations, relating the primed coordinates to the unprimed coordinates, x = x V t y = y z (6) = z t = t. We could easily invert these equations to find the components of motion in the unprimed coordinates in terms of the primed coordinates. Knowing the relationship between the coordinates we can find the relationship for the velocity of the object in each frame. Supposing that the velocity of the object in the unprimed frame is v = d r, then taking the derivative of the expressions in Eq. (6) with respect to time dt 3

4 (and recalling that t = t so that v = d r dt = d r dt ) gives v x = v x V v y = v y v z = v z. Thus, we see that the velocities add intuitively, just as expected. Furthermore, taking one more derivative tells us that the accelerations (and therefore the forces) seen by each observer is the same. The result that light always travels at the same speed, regardless of the observer, seems to contradict the velocity addition rules in Eq. (7). However, the speed of light is very specifically predicted by Maxwell s equations, and as time went on these equations were found to always agree with the results of experiments. Thus it seems that, if nothing is wrong with the unintuitive result from Maxwell s equations, then something must be wrong with the intuitive relationship between the observers in Eq. (6)! Various ideas were put forth in trying to reconcile these two ideas; foremost among them was the concept of the aether (also called the luminiferous ether). The aether was supposed to be some sort of background medium, like a space-filling Jell-O, providing the reference background from which we could measure the speed of light. This solved the issue associated with Maxwell s equations, saying that the speed of light was measured with respect to this background medium. The aether was a purely theoretical concept, invented to explain the propagation of light, but it lacked any experimental basis up until that point. So various experiments were predicted to try to measure properties of this aether. The most notable of these was the Michelson-Morely experiment, which tried to measure the motion of the Earth through the aether, determining our speed through this background. However, when performed, the Michelson-Morely experiment suggested that our velocity through the aether was precisely zero! Clearly, the Earth moves through space (we are orbiting the Sun, after all), and these experimental results said that the Earth and aether are moving in exactly the same way, even through our elliptical orbit. Obviously, the aether concept needs some refinement; since all the experiments suggest that the aether doesn t produce any measurable effects, then we should conclude that the aether isn t there! Einstein finally realized that a new concept was required. Maxwell s equations say that light always travels at the same speed, c, but with respect to whom? Einstein decided that light travels at speed c, with respect to whomever cares to measure it! This means that every observer always sees light traveling at the same speed, independent of their own motion. This idea is completely unintuitive, disagreeing with the Galilean transformations in Eq. (7) (and so therefore also Eq. (6)), but it is what seems to be required by the results of experiment. Thus, Einstein arrived at the idea which would become one of the cornerstones of his Special Theory of Relativity, which is that light always moves at the same speed, relative to all observers. However, this isn t quite enough to get the full results of experiment; we also need one more idea. Not only, Einstein said, do all observers have to agree on the speed of light, but they also have to agree on all the laws of physics! We will actually see later that this will be vitally important for the formulation of Einstein s General Theory of Relativity. For now, though, let s see what these ideas say about the laws of mechanics. 4 (7)

5 2 Mechanics in Special Relativity. As we will see, the simple postulates Einstein proposed will lead to some very interesting effects, including (but not limited to) the merging of space and time! Let s begin with a simple example which demonstrates some of the interesting physics. Consider a moving train of height L, moving to the right with velocity v, as seen in Figure 2, with one person riding inside the train, and another standing outside watching the train go by. The train has mirrors at the top and bottom, and a beam of light fires from the bottom mirror to the top. v L Observer Inside Train Observer Outside Train Figure 2: The motion of light as seen by two different observers, one moving along with the light, and another moving relative to the light. The different observers will measure different amounts of time for the light to complete its journey. The two observers see different motions of the light. The person sitting inside the train sees the light fire straight up from the bottom to the top, with the light traveling for a time t = L/c. The observer outside sees something different; since the train is moving (relative to the person standing outside), the light travels along a different, diagonal path. The total distance that the light travels is not just the usual distance up, but also the distance the train moves along, which, if the outside observer sees the light hit the top mirror in a time t, then the distance the train moves is v t. So, the total distance the light goes is, from the Pythagorean theorem, D = L 2 + v 2 t 2, and so the total time is t = D/c = L 2 + v 2 t 2 /c. Now, before Einstein both observers would have had to agree on the amount of time (disagreeing on the speed of the light), but now, since the speed of light is fixed, we find a difference in times! Solving for the time t gives t = L c2 v 2. But, the observer inside the train saw that L/c = t, and so t = t 1 v2 /c 2, (8) which says that the two observers see the light taking different times! The person sitting outside sees the light taking a longer amount of time to reach the top than the person inside 5

6 does. The time measured by the person inside the train ( t) measures what is called the proper time, the time that passes moving along with the clock. The height of the train could be chosen (at least in principle) such that it takes one second for the light to go up to the top (or we could just bounce the light up and down a bunch of times until a second has passed). In this way, we ve made a light clock which will tick off the seconds. Then, the two different observers will see different seconds ticking off. What this means is that moving clocks run slow! The amount of time passing differs for different observers; time can be affected simply by moving. This idea is called time dilation, and is completely unexpected according to Newton, who believed that time is fixed and immutable. It s easy to see that the exact expression in Eq. (8) is important only when the clock is moving at speeds comparable to that of light, and when v c, then t t, and the two observers agree on the ticking rates of clocks. We ll see soon that time dilation is only the tip of the iceberg. Let s look at one more example before moving on. Consider a bar moving along with speed v, being watched by two observers, one moving with the bar, and the other remaining stationary, as in Figure 3. The person moving along with the bar measures its length L, called it s proper length. L v L t 1 t 1 t 2 t 2 Moving Observer Stationary Observer Figure 3: Different observers will measure different lengths for the moving bar, with stationary observer measuring a shorter bar. A flash of light is fired from the back end, reflects off the front end, back to the rear. Let s look at how the different observers see the flash of light, first from the point of view of the observer moving along with the bar (the first half of the figure). The flash of light takes a time t 1 = L/c to reach the front end, then the same amount of time t 2 = L/c to come back for a total time of t = 2L/c. Now let s look at the same process from the point of view of someone watching the bar move past with a speed v (the second half of the figure). This observer may not see the same length of the bar, so we ve labelled the length L instead. The bar takes a longer time t 1 = L /(c v), since the end of the bar is running away from the light, and a shorter time t 2 = L /(c + v) since the end of the bar is now moving towards the end of the bar. This gives a total time ( 1 t = L c v + 1 ) = 2L c + v c 1 c 2 v 2. Now, as we ve seen before the two times are related by Eq. (8), and so 2L c 1 t = c 2 v 2 1 v = 2 /c 2 6

7 2L 1 c 1 v 2 /c 2, and so comparing gives L = 1 v2 L, (9) c2 which says that the length of the bar is shorter as measured by the stationary observer! This idea is called Length Contraction, and is yet another interesting effect that is required by the constancy of the speed of light; not only do moving clocks run slow, but moving rods get short! Once again, we see that this effect is important only if the rod is moving close to the speed of light; in order to change the length of the rod by even one percent, the rod would need to be moving at more than 14% of the speed of light, or about meters per second! This is far outside our everyday range of experience. 2.1 Lorentz Transformations. We have seen some of the interesting effects of moving near the speed of light, including time dilation and also length contraction. Neither of these effects is predicted by the Galilean transformation equations, Eq. (6). So, we need to try to figure out how those equations in Eq. (6) need to change to accommodate Einstein s new ideas. The correct transformation laws should not be too very different than those in Eq. (6), since the correct laws must reduce down to the simple equations when the speeds are low compared to that of light. So, with this in mind, for relative motion along the x direction at speed v, then let s try a solution of the form x = γ (x vt), where γ is to be determined, and must have the property that γ 1 as v 0. expression has the inverse transformation This x = γ (x + vt ). Suppose that the origin of each observer overlaps at t = t = 0. Then we fire a laser from the origin. In the unprimed frame after a time t the light beam travels a distance x = ct, while in the primed frame it travels a distance x = ct. Thus, ct = γ (ct vt) = γ (c v) t ct = γ (ct + vt ) = γ (c + v) t. Solving the second for t and plugging into the first gives γ 2 = c 2 /(c 2 v 2 ), or γ 1, (10) 1 v2 c 2 which is the same factor as in Eqs. (8) and (9)! So, we have the correct transformation for the x components, but what about t? Start with x = γ (x + vt ), and plug in x = γ (x vt). Solving for t gives t = γ (t v ) c x. 2 7

8 Thus, we have the correct transformation laws, called the Lorentz Transformations x = γ (x vt) y = y z = z t = γ ( t v x ), c 2 where the directions perpendicular to the motion do not transform (we say they re invariant under these transformations), and it s easy to see that we get the old Galilean transformations, Eq. (6), in the limit that v c. Thus, we have the correct transformations between the two frames. We can easily see that these transformations give us the time dilation and length contraction effects in Eqs. (8) and (9). The Lorentz transformations are really quite spectacular. They mix together space and time! These equations say that one can change the amount of time that passes, or the space between objects, just by moving! This merging of space and time together into one spacetime is our first taste of the malleability of space and time, a concept that will command our attention in the chapters to follow. 2.2 The Doppler Shift. We can use ideas introduced above to obtain an important result about the motion of light when considered by two different observers. We have already introduced the idea of redshift in Chapter?? when we discussed the expansion of the Universe, but we did not give any details. It s time to remedy this. We ve all heard the change in pitch of a siren as a fire engine is passing by us. This familiar effect, known as the Doppler effect, comes from the shift in frequency of the sound heard by an observer from an emitter in relative motion. This is a property of waves, and so is true for light, as well. A source moving away from us will have it s wavelength stretched out, and so will appear redder, while a source moving toward us will have it s wavelength shortened and so will look bluer than its original color. This effect will arise for light, in part, because the emitter and receiver will disagree on the amount of time between waves, due to the time dilation effects discussed above. Let s figure out the expression for the frequency shift. Consider two observers, Jack sitting at rest at the origin, and the Jill, initially at a distance x 0 at time t = 0, moving away at a speed v. Jack has a flashlight which he flips on and off at intervals of T, so there is a frequency ν = T 1. What is the frequency seen by Jill as she s moving away? Suppose Jack pulses the light at time t = 0, he will see Jill receive the pulse at a distance x 1 = ct 1, where t 1 is the time it took the pulse to reach Jill. Since she s moving away, in a time t 1 Jill moves a distance vt 1, so x 1 = x 0 + vt 1 = ct 1, or t 1 = x 0. c v Now, a time T later, Jack pulses the light again, and sees Jill receive the pulse at a distance x 2 = x 0 + vt 2, where vt 2 is the distance away from her starting point that Jill could move in a time t 2. But, this distance isn t just ct 2, because the light hasn t been traveling for the full time - there was a delay of time T between pulses, and so the light has only been traveling for time t 2 T. Thus, x 2 = c (t 2 T ) = x 0 + vt 2, such that t 2 = x 0+cT. Then c v t 2 t 1 = 8 ct c v (11)

9 gives the difference in time measured by Jack sitting at the origin. Furthermore, just from their definitions, x 2 x 1 = v (t 2 t 1 ), or x 2 x 1 = cvt c v. Now, because Jill is moving relative to Jack they will disagree on the amount of time passing, and from Eq. (11) we see that t 2 t 1 = γ [ (t 2 t 2 ) v (x c 2 2 x 1 ) ] = γ [ ] ct But, since γ = ( 1 v2 c 2 ) 1/2, we find which can be simplified to = γ ct c v v cvt c v( c 2 c v ) 1 v2 c 2 t 2 t 1 = 1 1 v 1 v2 c T, c 2 t 2 t 1 =. 1 + v/c 1 v/c T. Since t 2 t 1 is the period, T, in Jill s frame, she sees the light pulses stretched out in time. Since the frequency is just the inverse of the period we find, for an observer moving away, 1 v/c ν obs = 1 + v/c ν source, (12) which says that the frequency is decreased for a receding source. Since the color of light depends on the frequency, light sources moving away appear redder, in agreement with our earlier assertions (as always, to get the frequency for an approaching source, you just flip the sign of the velocity v v). Since the wavelength of a light wave is just λ = c/ν, we can easily express Eq. (12) in terms of the wavelength, 1 + v/c λ obs = 1 v/c λ source, (13) We will return to these ideas, yet again, when we discuss the expansion of the Universe according to Einstein s Theory of General Relativity. 2.3 The Interval. The transformations in Eq. (11) have a very interesting property, which will be even more useful in discussing General Relativity. Recall what happens if we have a vector, A, having components A = (A x, A y ) in one frame, and then we rotate our coordinate system by an 9

10 y y A y A A x x A y φ A x x Figure 4: While the components of a rotated vector change, the length is always the same. angle φ about the z axis to a new frame, as in Figure 4. In the new frame the vector has components A = (A x, A y ). We can easily relate the components in the primed system to those in the unprimed system from the geometry in Fig. 4. In this case A x = A x cos φ + A y sin φ A y = A x sin φ + A y cos φ. (14) So, the components of the vector are different. But, the actual length of the arrow, A A = A 2 x + A 2 y = A 2 x + A 2 y does not, as you can easily check. The length of the vector, which would be the real physical part, independent of our choice of coordinates, is an invariant value. We can do something very similar with the transformation laws in Eq. (11). But, just squaring and adding the components doesn t work; and there s no reason to expect that it should. Even though the Lorentz transformations mix together time and space, time still sticks out a little bit. We can move in three directions of space, back and forth at will, but we seem to be stuck moving along always forward in time (although we can change the rate of passage of time by moving, as we ve seen). So, instead of squaring and adding together all the terms, let s square and add the space components (this way we still get the rotational invariance for a fixed time), and subtract the square of the time components. We define the interval, s 2, s 2 = c 2 t 2 + x 2 + y 2 + z 2 = c 2 t 2 + x 2 + y 2 + z 2, (15) which is invariant under the Lorentz transformations! This can be written in another way, recalling that r = xˆx + yŷ + zẑ, we can write s 2 = c 2 t 2 + r r. This form of the interval shows the rotational invariance of the spatial components explicitly (since the dot product is a scalar, and independent of the coordinate system). Before looking 10

11 at what the interval means, note that we can look at a tiny differential interval, obtained by moving only a small distance in space and time, ds 2 = c 2 dt 2 + dx 2 + dy 2 + dz 2. (16) This form will turn out to be the most useful one. The interval in Eq. (16) will turn out to have a very important application describing spacetime in the absence of gravity, which we will discuss in detail later. For now, however, we want to focus on a particular property. While the length of the vector in Figure 4 is always positive, the interval need not be. If the spatial components are less than the time component, then ds 2 < 0. This means that a particle moving along moves faster through time than it does through space (this is clearly the case if the particle is standing still), and is the case for any particles moving slower than the speed of light. Particles moving in this way are called timelike, and are said to move along timelike paths. On the other hand, suppose that the spatial components are bigger than the time component. In this case the particle can get to some distance faster than light can travel, and so the interval ds 2 > 0. These particles move along spacelike paths. Finally the interval can also be zero, ds 2 = 0, which says that (for a particle moving along x, say) dx/dt = c, which is just the equation for the velocity of light. This means that light travels along paths (called lightlike) of zero interval. We can include all of this information in a handy diagram, called a spacetime diagram, seen in Figure 5. ct timelike path timelike region lightlike path spacelike region Figure 5: A spacetime diagram including the light cone. The region inside the shaded triangles are in causal contact with the particle moving along on the timelike path, while spacelike regions are out of causal contact. x The diagram in Fig. 5 shows a particle moving at less than the speed of light along the timelike path on the x axis (we ve suppressed the y and z axes for simplicity). The particle 11

12 is currently at the intersection of the two wedges. The lines stretching out at 45 (where x = ct, or ds 2 = 0) are the possible paths of light moving straight out along the x axes, defining what is called the light cone. The region in the lower wedge is called the past light cone, and represents the maximum distance away for which anything could possibly affect the particle. Only particles inside this back wedge are close enough to have any kind of effect on the particle, while particles outside this wedge are too far away that light couldn t get to the particle in time to affect it. Similarly, the top wedge is called the future light cone, and represents the maximum possible distance that the particle could affect. Since information can only travel at a maximum speed of light, these wedges are bounded by the lightlike paths moving out at 45. Regions inside the light cones (in the timelike region) can be affected by (or can affect) the particle, and are said to be in causal contact with the particle. Regions outside the light cones (the spacelike regions) are too far away to affect (or be affected) by the particle (at least right now), and are said to be out of causal contact with the particle. The timelike path that the particle is moving along is also called its worldline, and traces out it s motion. Massive particles always have timelike worldlines, and massless particles (like photons) have lightlike worldlines. There are hypothetical particles, called tachyons, which are supposed to move faster than light; however, such particles have never been found experimentally, and in fact would result in potential instabilities of the vacuum. For now, we ll ignore the possibility of tachyons. The interval, ds 2, is also called the proper length of the path, an example of which is the rest length of the rod in the length contraction experiment. This interval can be related to the proper time, dτ 2, which is the time measured on a clock moving along with a particle simply by dτ 2 1 c 2 ds2. (17) As an example, consider a particle moving along the x axis with a constant speed v dx. dt Then, the proper time is ( dτ 2 = dt 2 1 c 2 dx2 = 1 1 ( ) ) 2 ) dx dt 2 = (1 v2 dt 2, c 2 dt c 2 which, after taking the square root gives dτ = 1 v2 dt. We can integrate both sides to c 2 find the time in both frames, the proper time, τ, measures the time passing along with the particle, while the time t measures the time passing with respect to a stationary observer. Integrating and solving for t gives τ t =, (18) 1 v2 c 2 which is precisely the equation for time dilation, Eq. (8)! We ll see later that both the proper length and proper time are very useful concepts. 2.4 Velocity Addition in Special Relativity. Galilean velocity additions were intuitive and simple in Eq. (7), but these can t be right since it wouldn t let everyone see the same speed of light. We need to figure out the correct 12

13 expressions for the addition of velocities now, and it happily turns out to be very simple. Suppose that the relative velocity of the two frames is V, and that the velocity of a particle in the stationary frame is v dx, while the velocity in the moving frame is dt v dx. Taking dt the differentials of the first and last of Eq. (11) gives Dividing the first of these by the second gives dx = γ (dx V dt) dt = γ ( dt V c 2 dx ) dx dt = dx V dt dt V c 2 dx = dx V dt 1 V dx c 2 dt. Now, replacing v = dx dt with an inverse transformation dx, and v =, then we find the law of velocity addition dt v = v V 1 V c 2 v, (19) v = v + V 1 + V c 2 v. (20) Let s consider a couple examples, first suppose a ball is moving with velocity v in one frame, that the relative motion between the two frames is V = 0, then both observers clearly agree on the velocity. Now, suppose that a particle is moving at speed c/2 in the moving frame, which is moving at a speed of c/2. Classically, the outside observer would measure a speed c, just by adding the velocities. Eq. (20) gives, instead, v = c/2 + c/2 1 + c c 2c 2 2 = c = 4 5 c, which is less than the speed of light. Now, suppose that the moving observer is looking at light, and is moving at speed V. What is the speed measured by the stationary observer? In this case, Eq. (20) gives v = c + V 1 + V c 2 c = c + V 1 + V c ( ) c + V = c = c, c + V independent of V, which shows that all observers agree on the speed of light! In the same way, we can find the relativistic velocities in the other directions, v y = dy and v z = dz, which give dt v y = v y γ(1 vxv c 2 ) dt, v z = v z γ(1 vxv c 2 ), (21) where we have used the fact that v y = dy and v dt z = dz. Thus, we have the full description of dt velocities. 13

14 2.5 Energy and Momentum in Special Relativity. We now have all of the equivalent expressions to the Galilean transformations, including the laws of velocity addition. We ve seen that these expressions lead to some interesting and counter-intuitive results, but we re not done yet. We now need to expand our analysis beyond just the coordinate transformations and velocity additions to include momentum and energy. We begin with the momentum. In the classical case the momentum of a particle of mass m moving along the x direction is just p = mv = m dx. The question is how to generalize this result to include Special dt Relativity. The momentum is the ratio of the distance traveled by the particle to the time that it takes to go that distance; but what time should we use? We only have one observer now, watching a particle zipping by and asking what its momentum is. So, we don t need to try to relate the values seen by two observers; instead, we should be looking at a time intrinsic to the particle, itself. This suggests that we use the proper time of the particle, given (for a constant velocity) by Eq. (18). Thus, we define p = m dx dτ = m dx 1 v2 /c 2 dt = γmv, (22) ( ) 1/2 where we have recalled that γ = 1 v2 c and v = dx. Eq. (22), which reduces to 2 dt p = mv when the velocity is much less than that of light, is the correct expression for the relativistic momentum. The expression, Eq. (22), is very interesting. We can still describe the force by finding F = dp, and say that the momentum of a particle is changed by the force acting on it. dτ However, according to Newton we can just keep pushing harder and accelerating faster to an arbitrarily high speed. But, Eq. (22) says that as we push, the particle picks up more momentum. The momentum asymptotically approaches infinity as the velocity of the particle approaches that of light. Changing this momentum by even a little bit requires a tremendous force, becoming infinite as v c. This demonstrates that no massive particle can travel faster than light, since it would take an infinite amount of energy to push it up to the speed of light! Now that we have the relativistic momentum, how do we get the relativistic energy? Let s go back to the expression for the proper time, moving only along x, dτ 2 = dt 2 1 dx 2. c 2 Now, let s divide by dτ 2 to find 1 = But, as we ve just seen, dx = p dt, while dτ m dτ ( ) 2 dt 1 dτ c 2 = γ, and so ( ) 2 dx. dτ Rewriting gives 1 = γ 2 p2 m 2 c 2 ( mc 2) 2 = ( γmc 2 ) 2 (pc) 2. ( γmc 2 ) 2 = (pc) 2 + ( mc 2) 2. 14

15 The question now becomes, how do we interpret these values, specifically the one on the left? To find out, let s consider the nonrelativistic case, where the velocity is small compared to the speed of light. Then p mv, and γmc 2 = p 2 c 2 + m 2 c 4 mc m2 v 2 m 2 c 2 mc ( v 2 2 = mc mv2, where we have used the binomial expansion for the third line. The last term is the nonrelativistic kinetic energy, while the first term is an energy associated with the mass of the particle. So, we should associate γmc 2 with the total energy of the particle, and so we should expect that the total relativistic energy is E = γmc 2 such that c 2 ) E 2 = p 2 c 2 + m 2 c 4. (23) This is the correct expression for the energy of the system. The value mc 2 is identified with the rest-mass energy of the particle, giving us Einstein s famous expression E = mc 2, (24) for a stationary particle. Thus, unlike the classical case, a free particle has an energy just from it s sheer existence! This is a very important result, saying that energy and mass are equivalent! This idea will lead to some very interesting and important results, for it says that, given sufficient energy, one can make mass! All of particle physics is predicated upon this result, allowing for the creation of different particles in accelerators, and will give us the origin of all structure in the Universe after the Big Bang! We will discuss this in more detail as we go along. We can see a couple more properties of the energy. First of all, subtracting away the rest mass energy from the total energy gives the kinetic energy of a free particle KE = E mc 2 = (γ 1) mc 2. (25) Finally, what is the energy of a particle with no mass? Since E = γmc 2, one might first expect that the energy would be zero. But this makes no sense, since we know that light carries energy (the Sun heats the Earth, for example). We also know that massless particles (like light) travel at the speed c. In that case the energy becomes E = mc2 0/0, which 1 v 2 /c 2 is completely useless to us. However, we also know that E 2 = p 2 c 2 + m 2 c 4, which says that E = pc for a massless particle. All of the energy is carried by the momentum of the particle (of course, the momentum isn t given by Eq. (22) anymore), which depends on the frequency of the light, as is familiar from quantum mechanics. Thus, we have the correct expressions for the relativistic energy and momentum of a particle, both of which must be conserved in any reaction. Let s consider an interesting example. Figure 6 shows a reaction producing anitprotons in two different reference frames. In the top reaction we are looking at the reaction in the center of mass frame, in which both 15

16 protons are moving towards each other at the same speed v cm. This produces antiprotons via the reaction P + P P + P + P + P, where P is the antiproton. You might wonder why we get so many products from this reaction. The answer is that this reaction must conserve a quantity called baryon number, meaning that the total number of baryons (protons) that we start with (2) has to be the same number that we end up with (since antiprotons count as 1, we get 3 1 = 2). Center of Mass Frame Initial Final v cm v cm v = 0 f Lab Frame Initial Final v lab v=0 v f Figure 6: Smashing together two protons with enough energy can produce antiprotons, as shown in the two figures above. The reaction is seen in two different reference frames, one in the center of mass frame, and another in the lab frame where the second proton is stationary. In the lower reaction, looked at in the lab frame, we have a single proton moving towards a stationary one. What we want to determine is how much energy does the incident proton have to have (in the lab frame) to produce these antiprotons. Let s look at the center of mass frame, first. The initial energy of each proton is the same, since they re moving at the same speed, and is E i = γ cm mc 2, where m is the mass of the proton, and γ cm is the gamma factor in the center of mass frame, which is different from the gamma factor in the lab frame! The initial momentum is zero in this frame. Supposing that all the energy went into making the proton and antiproton, then the final energy in the center of mass frame is just the mass energy of the four product particles, E f = 4mc 2, since the proton and antiproton have the same mass. Equating the energies gives γ CM = 2, which says that v CM = 3 c. 2 Now, we want the energy in the lab frame, so let s use the velocity addition equations, Eq. (19) to transform to the lab frame. To do so, we need to run along with the proton on the right at a speed V = v cm. Thus, we find the velocity of the moving proton to be 1 Now we can figure out γ lab = r v = v CM + v CM 1 + v2 CM c 2 1 v2 lab c 2 = c. = 7, meaning that the initial energy of the proton in 16

17 the lab frame is E i = 7mc 2. Since one of these mc 2 s is the mass energy of the proton, itself, this means that 6 are left for the kinetic energy. Thus, a proton needs a kinetic energy of 6mc 2 to just make this reaction happen. This was a fair amount of work, requiring us to transform between frames; we ll soon revisit this problem using a different method which is more useful for many problems. 2.6 Acceleration in Special Relativity. In everything that we ve discussed so far we have typically kept the velocities constant, for example in the equations for velocity addition, Eqs. (19). It is sometimes claimed that Special Relativity doesn t include accelerations, and that we need the complete general relativistic theory to describe it, but this is actually false. By the same analysis that led to Eq. (19) one can show that the acceleration transforms as a x = a x γ 3 ( 1 V v c 2 ) 3. (26) By considering the acceleration we can obtain a very interesting result. Suppose we have a rocket of height h with a laser at the bottom, pointing up. The laser fires a series of pulses, separated by time intervals T. At the top of the rocket we have an observer counting the pulses. The rocket is accelerating upwards at a rate g, which is the usual gravitational constant. We expect that the observer won t measure the same time between the pulses as the laser is sending out since the observer is running away from the light, as she accelerates. Let s figure out what she sees. Suppose that the rocket starts from rest. Then when the laser fires it reaches the top of the rocket in a time t 1. In that time the light travels a distance y 1 = ct 1, which is the initial height, h, plus the distance that the rocket travelled during that time, 1 2 gt2 1, so y 1 = h gt2 1 = ct 1. The laser sends out the next pulse at a time T later, and the light pulse now reaches the top at a time t 2, which is again the initial height, plus the distance the rocket moved during that time, y 2 = h gt2 2, but the light has travelled only for a time t 2 T, there was no light before the beam was sent out. So, y 2 = h gt2 2 = c (t 2 T ). Solving these expressions gives the time it takes for the light to reach the top as [ ] c t 1 = 1 1 2gh g c 2 t 2 = c g [ 1 ] 1 2g(h+cT ), c 2 where we have taken the minus sign in the quadratic equation solution to get the correct g 0 limit. Now, the laser sent out the light pulses with period T, but our observer sees them with period T = t 2 t 1, which is T = c g [ 1 2gh c 2 1 ] 2g (h + ct ). c 2 In general we don t expect the periods to agree. To see what the difference is, let s expand this result to second order in a Taylor series. In general, both 2gh and 2g (h + ct ) c 2, 17

18 and so we can use the binomial theorem to write [ ] T c 1 gh g c 1 g 2 h 2 g (h + ct ) g 2 (h + ct ) 2, 2 2 c 4 c 2 2 c 4 Canceling off the common terms and dropping the terms of order T 2 gives ( T = 1 + gh ) T, c 2 which says that the period of reception of the pulses is longer than the period of emission, as we should expect. Instead of expressing our result in terms of the period, we can instead express it in terms of the frequency, ν = 1/T. Calling ν obs the observed frequency, and ν source the emitted frequency, we have ν obs = ν source. 1 + gh c 2 Now, in general gh c 2, and so we can expand the result once again to find ( ν obs = 1 gh ) ν c 2 source. (27) Thus, we find that the acceleration produces a Doppler shift in the frequency (in fact this result could have been obtained from Eq. (12) by writing v = gt, where t = h/c is the time to reach the top of the rocket, and expanding the square root recalling that v c). This is actually not a surprising result. But, Eq. (27) does contain a surprising result. The frequency depends on the acceleration, which we have chosen to be a = g, the acceleration due to gravity on Earth. If we ignore the derivation of Eq. (27), and just look at the result, then we would be led to believe that a beam of light just traveling upwards in a gravitational field would lose frequency! This is, in fact, completely true. As we will discuss further later, light is affected by gravity, and as light tries to escape from a gravitational field it experiences a redshift, causing its frequency to decrease (hence becoming redder). We can think of this in another (classical) way. When we throw a ball up into the air, it slows down, using it s kinetic energy to do work against the force of gravity. Light has to do work against gravity, too, but it can t change it speed. Therefore it has to lose energy, not by losing speed, but by losing frequency, since the energy of light depends on its frequency. We ll return to this result later. As a final comment, notice that the gravitational potential energy of a mass near the surface of the Earth is U = mgh, and so gh = U/m, but we know that U/m Φ, the gravitational potential. Thus, we can write Eq. (27) as ν obs = (1 Φc ) ν 2 source. (28) This will turn out to be a useful form. convenient and important notation. For now, though, we want to introduce a very 18

19 3 Four Vectors and Tensors. As we ve seen, there is a deep connection between space and time. They are no longer the seperate, immutable backgrounds that Newton imagined, but rather a plastic, changable framework unified into a single spacetime, where time and space can be changed into each other simply by moving! The Lorentz transformations, Eqs. (11), mix together time and space in much the same way as the rotation of the coordinate axes mix together the compoents of a vector, as in Eqs (14). This gives us a hint suggesting that we look at the Lorentz transformations as a sort of generalized rotation of a vector containing both spatial and time components! This idea will turn out to be crucial to every aspect of our future work, so let s develop it in detail. First, let s examine Eq. (14) a bit more, rewriting the components in matrix notation, now including the z components (z = z since we ve rotated the coordinate system about the z axis), A x cos φ sin φ 0 A x A y = sin φ cos φ 0 A y. (29) A z A z The preceding equation can be written fully in matrix notation. If A is the column vector, and R is the rotation matrix, then Eq. (29) is simply A = R A. (30) In order to form the scalar product giving the length of A, we need to multiply the column vector A by a row vector on the left (this way we have a 1 3 matrix multiplying a 3 1 matrix, giving a 1 1 matrix, or just a number, which is precisely what we need). We can form a row vector by taking the transpose of A, defined by A T = (A x, A y, A z ), (31) such that A T A = A 2 x + A 2 y + A 2 z A 2 (note that A T A A A T, in general; in the first case we get a number, and in the second we get a 3 3 matrix). The transpose operation just flips the rows and columns about the diagonal. In the same way, the transpose of the rotation matrix in Eq. (29) is just cos φ sin φ 0 R T = sin φ cos φ 0. (32) Now, in order for the length of the vector to be an invariant, we need A T A = A T A. Plugging in the expression for A from Eq. (30), and remembering that for two matrices A and B, (AB) T = B T A T, A T A = A T R T R A = A T A. In order for this expression to be true for any rotations at all, we need R T R = 1, (33) 19

20 where 1 is the unit matrix, consisting of 1 in every entry along the diagonal and zero everywhere else. Eq. (33), which says that R T = R 1, is the orthonormality condition for the rotation matrix, and you can easily check that the R and R T in Eqs. (29) and (32) satisfy it. Matrices which satisfy Eq. (33) are called orthogonal. Instead of working in matrix notation, we can instead work entirely in terms of the components of the vector and matrix, which is often much more convenient. Denoting the components of the vector A by A i (where i = x, y, z), and the entries of the rotation matrix by R ij (R xx = cos φ, etc.), we can rewrite Eq. (29) as A i = 3 R ij A j. (34) j=1 Eq. (34), which is just Eq. (30) in component form, is, in fact, the definition of a vector. Any object that transforms upon the change of coordinates in this way is a vector. Because the length of the vector is an invariant, we know that 3 A ia i = i=1 3 A i A i. i=1 Plugging in Eq. (34) for A i on the left-hand side gives R ij A j R ik A k = i=1 j=1 k=1 3 3 A j A k j=1 k=1 3 R ij R ik = i=1 3 A i A i. i=1 Once again, in order for the equality to hold we need the rotation matrices to satisfy an orthonormality condition, 3 R ij R ik = δ jk, (35) where δ jk is the Kronecker delta symbol, defined such that i=1 δ ij = { 1 i = j 0 i j. (36) This is the component form of the orthonormality condition in Eq. (33), since R ij R ik = R T jir ik for proper matrix multiplication. Now, the length (squared) of the vector becomes 3 3 A j A k j=1 k=1 3 R ij R ik = i=1 3 3 A j A k δ jk = j=1 k=1 3 A j A j, j=1 since the Kronecer delta kills off every term of the summation over k which isn t j, and shows the invariance of the length of the vector under rotations. So, the lesson here is that we can work with either the abstract matrix notation, as in Eq. (30), or with the component notation, as in Eq. (34). It will turn out for our work that the component notation will be more convenient. 20

21 Let s return to Eq. (34) for a moment. Suppose that the vector A was just the displacement vector, r. Then, rotating the coordinate system about the z axis by an angle φ gives the components of the new displacement vector r as x y z = cos φ sin φ 0 sin φ cos φ Now, for a given angle, it s clear that since x = x cos φ + y sin φ, then x = cos φ, while x x = sin φ, and so on for the y y derivatives. Thus, the components of the rotation matrix can be written in terms of partial derivatives. In general we can write (letting x 1 = x, x 2 = y, and x 3 = z) 3 A x i i = A j, (37) x j j=1 x y z. which is a more general definition of a vector. We will return to these ideas, generalizing them to more dimensions, very soon Spacetime Four Vector. Now that we have a good grasp on ordinary vectors in three dimensions (called threevectors ), let s now try to generalize the idea to including the time component. Going back to Eq. (11) and rewriting it in matrix notation after a bit of manipulation we find ct x y z = γ γ v c 0 0 γ v c γ ct x y z. (38) Eq. (38) is very similar in form to Eq. (29), and so we ll pursue this analogy further. We define the spacetime four-vector as being the column vector with components x µ, where µ = ct, x, y, z, or µ = 0, 1, 2, 3 where by 0 we mean ct, etc. Instead of writing the abstract four vector, we ll now just refer to it in terms of its components, so x µ ct x y z ( ct r ). (39) As we ll see, writing the index µ as a superscript defines the coordinates as a vector, and is not meant to imply raising it to a power. We can further define the Lorentz transformation matrix in terms of its components, and write Λ µ ν γ γ v c 0 0 γ v c γ , (40)

22 where we require two indices on the matrix to label the rows and columns. We ll explain the placement of the indices in a moment. Thus, we can rewrite Eq. (38) completely in component notation as x µ = Λ µ νx ν, (41) ν where the sum over ν runs over ct, x, y, z. Eq. (41) says, for example, that x 0 = ct = ν Λ 0 νx ν = Λ 0 0x 0 + Λ 0 1x 1 + Λ 0 2x 2 + Λ 0 3x 3 = γct γ v c x, which is precisely correct. We ve rewritten the Lorentz transformations in terms of the components, defining the coordinate four-vector in process. The next thing to consider is the scalar product of fourvectors, generalizing the usual dot product. Remember that the dot product s defining characteristic was that it preserved the length of the vector in any coordinate system; we want the scalar product of the four-vectors to keep this property. A first guess might be simply to write µ xµ x µ, in analogy with the ordinary three-dimensional dot product. In fact, if the time was zero, then we would need the four-dimensional dot product to reduce to this case. However, recall that we have already defined the invariant quantity associated with space and time, namely the interval, s 2, in Eq. (15), which contains a minus sign on the time component. If we just took our interval to be the sum of the squares of the four-vector components then we wouldn t get back the proper form. Instead let s define a new vector, much like the transpose of the four-vector, x ν, with the index written as a subscript, x µ = (x 0, x 1, x 2, x 3 ) = ( ct, x, y, z) ( ct, r), (42) such that s 2 = µ x µ x µ = µ x µ x µ = c 2 t 2 + x 2 + y 2 + z 2 = c 2 t 2 + r 2, (43) which is invariant. We now see the importance of the index placement. Vectors with superscript indices, as in Eq. (39) are called contravariant, or just vectors. Vectors with subscript indices, as in Eq. (42) are called covariant or one-forms. The product of a vector and a one-form, summing over the common index, is a number. Eq. (43) can be generalized to differential lengths ds 2 as ds 2 = µ dx µ dx µ = µ dx µ dx µ = c 2 dt 2 + dx 2 + dy 2 + dz 2. (44) Just as we can generalize the idea of the three-dimensional displacement vector r to other vectors (such as velocity, forces, etc.), we can have different four-vectors. We ll see other specific examples, soon, but for now any four-vector is defined as a µ such that the components transform under the Lorentz transformations, a µ = ν Λ µ νx ν, (45) 22