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6.1Reference frames and Lorentz transformations

In physics experiments, we (the observers) record the time and position at which events something that happens at a particular point in space and time take place. To do that, we use a reference frame a set of co-ordinate axes and a set of clocks at every point in space. A reference frame is a set of co-ordinate axes and a set of clocks at every point in space. If this set is not accelerating, the frame is called an inertial reference frame a reference frame which is not accelerating (see Figure 6.1).

Figure 6.1: A two-dimensional reference frame. There are clocks at every point in space (only a few are shown here). All clocks are synchronised to show the same time.

CHECK YOURSELF 1

Which reference frames in this list are inertial frames?

aA sky diver falling freely

bA sky diver falling at terminal velocity

cA train moving with constant speed on a straight line

dA plane taking off

eThe earth as it rotates around the sun

If the event is a strike of a lightning, an observer will look at the reading of the clock at the point where lightning struck and record that reading as the time of the event. The co-ordinates of the strike point give the position of the event in space. In Figure 6.2, lightning strikes at time t = 3 s at position x = 60 m and y = 0. (We are ignoring the z co-ordinate.)

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Figure 6.2: In this frame of reference the observer measures that lightning struck at time t = 3 s at position x = 60 m.

The same event can also be viewed by another observer in a different reference frame. So, consider a situation involving one observer on the ground and another who is a passenger on a train travelling at a velocity of v = 15 m s−1 relative to the ground observer, as shown in Figure 6.3.

Figure 6.3: The origins of the two frames of reference coincide when clocks in both frames show zero. The origins then separate.

We can see things more clearly if we concentrate on the rulers (in metres) used by the ground and train observers. In Figure 6.4 we see the two rulers on top of each other. This happens at time zero.

Figure 6.4: The situation at time zero.

The train moves and its ruler moves along with it. Lightning strikes at the point shown (Figure 6.5). The ground and train observers assign different co-ordinates to this event.

Figure 6.5: The situation at time 3 s. Clearly, v t + x′ = x so that x′ = x − v t.

The ground observer uses the symbol x to denote the position of an event. The observer on the train uses the symbol x′. As shown in Figures 6.3 and 6.5, these are related by:

x′=x−v t

since in the time of t seconds the train moves forward a distance vt. The equation expresses the relationship between the space co-ordinates of the same event as viewed by two observers who are in relative motion. In pre-relativity physics, time was absolute; that is, all observers make the same measurements for time, so t′ = t.

KEY POINTS

The equations x′ = x − vt and t′ = t are two of the Galilean transformation equations x′=x−v t t′=t u′=u−v

Thus, to the event of Figures 6.3 and 6.5 (‘lightning strikes’) the ground observer assigns the co-ordinates x = 60 m and t = 3 s. The observer in the train assigns to this same event the co-ordinates t′ = 3 s and x′ = 60 − 15 × 3 = 15 m.

We are assuming here what we know from everyday experience (a guide that, as we will see, may not always be reliable):

Two observers always agree on what the time co-ordinates are; in other words, time is common to both observers.

Figure 6.6 shows Galileo Galilei in an oil painting by Justus Sustermans that can be seen in the Uffizi Gallery in Florence.

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Figure 6.6: Galileo Galilei (1564–1642).

Galilean relativity has an immediate consequence for the law of addition of velocities: consider a ball that rolls on the train floor. The train moves at 20 m s−1. At time zero the ball is at the origin of the train frame, as shown in Figure 6.7.

Figure 6.7: At time zero the ball is at the origin of the train frame. Its position is given by x = x′ = 0.

Two seconds later, the train has moved forward a distance of 40 m. During this time the ball has rolled on the train floor a distance of 20 m, as shown in Figure 6.8. So the velocity of the ball for the train observer is 20 2=10 m s−1.

Figure 6.8: Two seconds later the ball has moved a distance of 20 m along the train floor. The train moved forward a distance of 40 m according to the ground.

The ground observer measures that the ball moved a distance of 60 m in 2 s, and for the ground observer the ball’s velocity is:

60 2=30 m s−1

Notice that 30 m s−1 = 20 m s−1 + 10 m s−1.

In general, if the train observer measures a velocity u′ for the ball, the ground observer will measure a velocity u = u′ + v, where v is the velocity of the train.

The equation u = u′ + v or u′ = u − v is the final Galilean transformation equation.

Check yourself 2

A bird flies due south at 4 m s−1. A glider flies due north at 5 m s−1.

What is the velocity of the bird relative to the glider?

Worked Example 6.1

A ball rolls on the floor of a train at 2 m s−1 towards the right (with respect to the floor). The train moves with respect to the ground:

ato the right at 12 m s−1

bto the left at 12 m s−1.

What is the velocity of the ball relative to the ground?

Answer

a u=u′+v=2+12=14 m s−1.

b u=u′+v=2−12=−10 m s−1.

The (apparently correct) argument in the answer to worked example 6.1 presents problems if we replace the rolling ball in the train by a beam of light moving with velocity c = 3 × 108 m s−1 as measured by the train observer (see Figure 6.9).

Using the Galilean velocity addition formula implies that light would be travelling at a higher speed relative to the ground observer. At the end of the nineteenth century, considerable efforts were made to detect variations in the speed of light depending on the state of motion of the source of light. The experimental result was that no such variations were detected.

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Figure 6.9: An observer in the train measures the speed of light to be c. An observer on the ground would then measure a higher speed, c + v.

Einstein had solid theoretical reasons (based on Maxwell’s theory of electromagnetism) to believe that the speed of light should be the same for all inertial observers. It should not depend on the speed of its source. So, either Galileo’s velocity formula was wrong or Maxwell was wrong. Einstein trusted Maxwell. Doing so meant that the equations of Galilean relativity had to be revised.

The postulates of special relativity

Einstein (Figure 6.10) based his theory of relativity on two postulates 1. All the laws of physics are the same in all inertial frames; 2. the speed of light in vacuum is the same for all inertial observers. The first was the old postulate of Galileo which Einstein extended to all laws of physics, not just mechanics:

KEY POINTS

All the laws of physics are the same in all inertial frames.

The second was new:

KEY POINTS

The speed of light in vacuum is the same for all inertial observers.

Figure 6.10: Albert Einstein (1879–1955).

THEORY OF KNOWLEDGE

Einstein was so convinced of the constancy of the speed of light that he elevated it to one of the principles of relativity. But it was not until 1964 that conclusive experimental verification of this took place. In this experiment, at CERN, neutral pions moving at 0.99975c decayed into a pair of photons moving in different directions. The speed of the photons in both directions was measured to be c with extraordinary accuracy. The speed of light does not depend on the speed of its source. The theory of relativity has modified our fundamental ideas about space and time, ideas that went unchallenged for over two thousand years. Accepting the new ideas was not easy; the beauty of the theory of relativity lies in the amazingly simple postulates upon which it is based. It is interesting that at the time of the formulation of the theory little experimental evidence for the postulates was available. Yet, despite the simplicity of the postulates, their consequences are far reaching and have revolutionised physics.

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These two postulates of relativity have far-reaching consequences. Einstein showed, in 1905, that, following from his two postulates, the correct equations relating measurements in the two frames are:

Galileo | Einstein
x′ = x − vt | x′ = γ(x − v t) | x = γ(x′ + v t′)
t′ = t | t′=γ(t−v c 2 x) | t=γ(t′+v c 2 x′)

where the factor γ is called the gamma factor and equals γ=1 1−v 2 c 2.

The first column under Einstein gives x′ and t′ in terms of x and t, and the second gives x and t in terms of x′ and t′.

In 1890 the Dutch physicist H. Lorentz also derived these equations by demanding that the laws of electromagnetism should be the same in all inertial frames, and so these are known as Lorentz transformations. Einstein re-derived them from more general principles 15 years later. Notice right away that if v is very small compared to the speed of light, then γ ≈ 1 and v c 2 is negligibly small, and so the Lorentz equations reduce to the Galilean equations! In other words the effects of relativity become important only at speeds close to that of light. Notice also that these equations become meaningless for speeds equal to or greater than that of light. This is a sign that no material body can reach or exceed the speed of light.

These equations may be used to relate the co-ordinates of a single event in one inertial frame with those in another. We will denote with S some inertial frame. A frame that moves with speed v to the right relative to S will be called frame S′. These equations assume that when clocks in both frames show zero (i.e. t = t′ = 0) the origins of the two frames coincide, i.e. x = x′ = 0.

Note that γ > 1. A graph of the gamma factor against speed is shown in Figure 6.11. We see that the gamma factor is approximately 1 for velocities up to about half the speed of light but approaches infinity as the speed approaches the speed of light.

Figure 6.11: The gamma factor as a function of velocity. The value of γ stays essentially close to 1 for values of the velocity up to about half the speed of light but approaches infinity as the velocity approaches the speed of light.

Check yourself 3

You are on an island watching the mainland some 20 km away. The instant you see light from the headlights of a car on the mainland the clock strikes midnight. Is the time co-ordinate of the event ‘light is emitted from the headlights’ 12 o’clock midnight?

Worked example 6.2

Note: Unless stated otherwise, when clocks in both frames show zero, the origins of the two frames coincide.

Lightning strikes a point on the ground (frame S) at position x = 3500 m and time t = 5.0 s. Determine where and when the lightning struck according to a rocket that flies past the ground at speed 0.80 c to the right.

Answer

This is a straightforward application of the Lorentz formulas (the gamma factor is γ=1 1−0.80 2=5 3):

x′=γ(x−v t)=5 3(3500−0.80×3×10 8×5.0)=−2.0×10 9 m

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and

t′=γ(t−v x c 2)=5 3(5.0−0.80×3×10 8×3500(3×10 8)2)=8.3 s

The observers disagree on the co-ordinates of the event ‘lightning strikes’, but both are correct.

However, notice that in S the quantity c 2 t 2 − x 2 is equal to (3 × 108 × 5.0)2 − 35002 = 2.25 × 1018 m2. In the rocket frame c 2 t′ 2 − x′2 is equal to (3 × 108 × 8.3)2 − (−2.0 × 109)2 = 2.25 × 1018 m2 which is the same value as that in S. So both observers agree that this combination of time and space co-ordinates is the same in both frames. We will understand this in the next section.

Worked Example 6.3

A rocket is moving at 0.80c past a space station (Figure 6.12).

aWhere is the rocket according to the space station when the rocket clock shows 0.60 µs?

bWhat does the space station clock read at this position?

Figure 6.12: For worked example 6.3.

Answer

aWe examine the event E = rocket clock shows 0.60 µs. For this event x′ = 0 and t′ = 0.60 µs in the rocket frame.

Hence, from the Lorentz formula:

x=γ(x′+v t′)=5 3(0+0.80×3×10 8×0.60×10−6)=240 m

b t=γ(t′+v x c 2)=5 3(0.60×10 6+0)=1.0 μ s

(The quantity c 2 t 2 − x 2 in S is equal to (3 × 108 × 1.0 × 10−6)2 − 2402 = 32 400 m2. In the rocket frame c 2 t′2 − x′2 is equal to (3 × 108 × 0.60 × 10−6)2 − 02 = 32 400 m2.)

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Exam tip

It is important that you can use the Lorentz equations to go from frame S to frame S′ as well as from S′ to S.

Sometimes we will be interested in the difference in co-ordinates of a pair of events. So for events 1 and 2 we define Δ x′=x 2′−x 1′ and Δ t′=t 2′−t 1′ in S′ and Δ x=x 2−x 1 and Δ t=t 2−t 1 in S.

These differences are related by

Δ x′=γ(Δ x−v Δ t);Δ t′=γ(Δ t−v c 2 Δ x)

or, in reverse,

Δ x=γ(Δ x′+v Δ t′);Δ t=γ(Δ t′+v c 2 Δ x′)

Worked Example 6.4

Consider a rocket that moves relative to a laboratory with speed v = 0.80 c to the right, as shown in Figure 6.13. We call the inertial frame of the lab S and that of the rocket S′. The length of the rocket as measured by the rocket observers is 630 m. A photon is emitted from the back of the rocket towards the front. Calculate the time taken for the photon to reach the front end of the rocket according to observers in the rocket and in the laboratory.

Figure 6.13: For worked example 6.4.

Answer

Event 1 is the emission of the photon. Event 2 is the arrival of the photon at the front of the rocket:

Δx′ = x′arrival − x′emission = 630 m

This distance is covered at the speed of light, and so the time between emission and arrival (in S′) is

Δ t′=630 c=2.1 μ s

In frame S we use

Δ x=5 3(630+0.80 c×2.1×10−6)=1890 m

Δ t=5 3(2.1×10−6+0.80 c×630 c 2)=6.3 μ s

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The spacetime interval

We have seen that observers in different inertial frames assign different space and time co-ordinates to the same events. For example, frame S assigns (Δx, Δt) for the difference in co-ordinates of two events and S′ assigns (Δx′, Δt′). These are related by Lorentz transformations. Interestingly, however, there is a quantity that both observers will agree on as we have already indicated. This is called the spacetime interval(Δs)2 = (cΔt)2 − (Δx)2, has the same value in all inertial reference frames. It is given by (Δs)2 = (cΔt)2 − (Δx)2 in S. In S′ the spacetime interval is (Δs′)2 = (cΔt′)2 − (Δx′)2 and it has the same value as in S (this is left as an exercise in algebra for you). In other words

(c Δ t)2−(Δ x)2=(c Δ t′)2−(Δ x′)2.

In practice this means that if we know three of the quantities Δt, Δx, Δt′, Δx′ we can find the fourth from the spacetime interval. Consider an observer X at the origin of a frame S′ that moves past a frame S with speed of 0.80c. An observer in S measures that X turns on a light at t = 0 and turns it off at t = 4.0 s. For how long was the light on according to S′?

We know that Δt = 4.0 s, Δx = 0.80c × 4.0 = 9.6 × 108 m, Δx′ = 0. Hence

(c Δ t)2−(Δ x)2=(c Δ t′)2−(Δ x′)2

(3×10 8×4.0)2−(9.6×10 8)2=(c Δ t′)2−0 2

Hence Δt′ = 2.4 s.

Of course we could get the answer with a Lorentz transformation as well:

Δ t′=5 3×(4.0−0.80 c c 2×0.80 c×4.0)=2.4 s.

Check yourself 4

A photon travels from A to B on a straight line. What is the spacetime interval for these points?

theory of knowledge

Note that there is no question as to which observer is right and which is wrong when it comes to measuring time intervals. Both are right. Two inertial observers moving relative to each other at constant velocity both reach valid conclusions according to the principle of relativity.

TEST YOUR UNDERSTANDING

1Give two reasons why the ground on which you stand is not an inertial reference frame. Under what approximation is it an inertial frame?

2Imagine that you are travelling in a train at constant speed in a straight line and that you cannot look at or communicate with the outside. Think of the first experiment that comes to your mind that you could do to try to find out that you are indeed moving. Then analyse it carefully to see why it will not work.

3Outline an experiment you might perform in a train that is accelerating along a straight line that would convince you that it is accelerating. Discuss if you could also determine the direction of the acceleration.

4An inertial frame of reference S′ moves past another inertial frame S moving to the right with speed 15 m s–1. At time zero the origins of the two frames coincide.

aA phone rings at location x = 20 m and t = 5.0 s. Determine using Galilean transformation equations the location of this event in the S′ frame.

bA ball is measured to have velocity 5.0 m s–1 in frame S′. Calculate the velocity of the ball in frame S.

5An inertial frame of reference S′ moves past another inertial frame S moving to the left with speed 25 m s–1. At time zero the origins of the two frames coincide.

aA phone rings at location x′ = 24 m and t = 5.0 s. Determine using Galilean transformation equations the location of this event in the S frame.

bA ball is measured to have velocity –15 m s–1 in frame S. Calculate the velocity of the ball in frame S′.

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6A rocket moving at speed 0.80c relative to the ground emits a light signal in the direction of the velocity. What is the speed of the light signal relative to the ground according to Galileo?

In the following questions the frames S and S′ have their usual meaning, i.e. S′ moves past S with velocity v, and when the origins coincide clocks are set to zero. You are expected to use relativistic equations for these questions.

7 aCalculate the gamma factor for a speed of 0.850c and a velocity of –0.850c.

bCalculate the speed that gives a gamma factor of 4.00.

cIs there a speed that gives a gamma factor of 0.50?

8In frame S an event E has co-ordinates (x = 600 m, t = 2.0 μs). Frame S′ is moving at speed 0.75c in the positive direction. Determine the co-ordinates of E in S′.

9An event E has co-ordinates (x′ = 520 m, t′ = 4.0 μs) in a frame S′ that moves with velocity 0.60c relative to frame S. Determine the co-ordinates of E in S.

10Frame S′ moves with velocity 0.98c relative to S. An explosion occurs at the origin of frame S′ when the clocks in S′read 6.0 μs. Calculate where and when the explosion takes place according to frame S.

11Frame S′ moves with velocity 0.60c relative to S. Calculate the reading of the clocks in frame S′ as the origin of S′ passes point x = 120 m.

12Two events in frame S are such that Δx = x 2 − x 1 = 1200 m and Δt = t 2 − t 1 = 6.00 μs.

aFrame S′moves with velocity 0.600c relative to S. Calculate Δx′ = x′2 − x′1 and Δt′ = t′2 − t′1.

bDetermine if there is another frame S″ such that in this frame Δt″ = 0. Comment on your answer.

13A rocket moving at 0.80c leaves earth towards a planet that is a distance of 24 ly from earth as measured in the earth’s frame. By considering the events E1 = rocket leaves earth and E2 = rocket arrives at the planet, calculate the time it takes to get to the planet according to the rocket. (ly stands for ‘light year’; it is the distance travelled by light in one year. 1 ly = c × year.)

14A rocket travelling at 0.60c with respect to earth leaves earth towards a planet. After 4.0 years (as measured by the rocket clocks) a radio message is sent to earth. Calculate when it will arrive on earth as measured by:

aobservers on earth

bobservers on the rocket.

15In frame S event E(arly) occurs before event L(ate). The events are separated by a time interval Δt = t L – t E and a space interval Δx = x L – x E.

aIf E is the event that causes event L to happen, explain why Δ x Δ t≤c.

bIs there a frame in which L occurs before E?