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6.3Spacetime diagrams

We now introduce a pictorial representation of relativistic phenomena, first introduced into physics by one of Einstein’s friends, H. Minkowski.

Spacetime diagrams

In Chapter 1 we saw lots of motion graphs. In particular we saw graphs of position (vertical axis) versus time (horizontal axis). In relativity it is customary to show these graphs with the axes reversed, i.e. time is plotted on the vertical axis and position on the horizontal. These are called spacetime diagrams a graph in which the y-axis represents time (although it is more usually given as ct), and the x-axis represents position, x; single events are then shown by a dot on the diagram (sometimes called Minkowski diagrams). For reasons we will see shortly, it is convenient to instead plot ct on the vertical axis rather than time itself. The vertical axis then also has units of length. Since the speed of light is a constant everyone agrees on, knowing the value of ct allows us to find the value of t.

Figure 6.22: a The space and time co-ordinates of an event: x = 5.0 m and ct = 6.0 m. b The worldline of a particle at rest is the sequence of events showing the position of the particle at different times (blue line). A particle moving to the right has a worldline shown by the red line.

Figure 6.22a shows a spacetime diagram and an event (marked by the dot). The space and time co-ordinates of the event are read from the graph in the usual way: we draw lines through the dot parallel to the axes and see where the lines intersect the axes. This event has x = 5.0 m and ct = 6.0 m giving t=6.0 3.0×10 8=2.0×10−8 s.

Now consider a particle that is at rest at position x = a, Figure 6.22b. As time goes by we may think of a sequence of events showing the position of the particle (which does not change) at different times. This sequence of events traces the blue straight line on the spacetime diagram. This is called the worldline a line on a spacetime diagram relating a sequence of events, the angle of which is related to the speed at which something is moving; tan θ=x c t=v c of the particle. Another particle that starts from x = 0 at t = 0 and moves with constant positive velocity would have a worldline as shown in red. From the diagram:

tan θ=x c t=y c

The tangent of the angle between the worldline and the ct axis gives the speed divided by the speed of light.

This is why it is convenient to plot ct on the vertical axis: a photon moves at speed c, and so it makes an angle of θ=tan−1⁡c c=45° to both axes. Since nothing can exceed c, the worldline of any particle will always make an angle less than 45° with the ct axis. In Figure 6.23a, the red worldline represents a particle moving to the right. The green worldline belongs to a particle moving to the left. The blue worldline is impossible since it involves a speed greater than c. Figure 6.23b shows an event E at x = 0 and ct = 0. The region coloured yellow consists of all events that can be influenced by event E. For example, E could be an explosion. We see the glass of a window shatter later on, event L. Could the event E be responsible for L? The answer is that it could if L is in the yellow region but could not if L is outside this region. The fastest a signal can travel is at the speed of light so any signal from E to L has to be within the yellow region. Similarly, events that could have influenced E are within the region coloured blue.

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Exam tip

An event E can be the cause of an event L if the time separating E and L is greater than or equal to the travel time of a photon from the position of E to that of L.

Figure 6.23: a Various worldlines. The one in blue is impossible because it corresponds to a particle moving faster than light. b Future and past light-cones. Event E could be the cause of an event if the time separation is greater or equal to the photon travel time. This means the events E could cause lie in the yellow region.

To proceed we need a small digression. In mathematics we are used to working with axes that are at right angles to each other, Figure 6.24a. But they don’t have to be. The axes could be slanted, Figure 6.24b. To find the co-ordinates of a point on a diagram with slanted axes we simply draw lines parallel to the axes and see where they intersect the axes.

Figure 6.24: To find the co-ordinates of a point we draw lines from the point parallel to the axes and see where these lines intersect the axes.

Convince yourself that the x co-ordinate of point P and the ct co-ordinate of point Q in Figure 6.24 are both zero.

Now consider a second inertial reference frame that moves to the right with speed v. Let us assume that when clocks in both frames show zero the origins of the frames coincide. The origin of frame S′ moves to the right with speed v. Therefore, the worldline of the origin of S′will make an angle θ=tan−1⁡v c with the ct axis of S. But this worldline is just the collection of all events with x′ = 0 and so is the time axis of the frame S′. Where is the space axis? It has to make the same angle θ with the x-axis, Figure 6.25. To see why, consider a photon emitted at the origin (yellow worldline). In the frame S′ the distance travelled is L and the time taken is c t′=L⇒t′=L c. So the speed of light in S′ is L L τ=c which is precisely what we want: the speed of light must be the same in all inertial frames. Hence the axes must be slanted as the red axes in Figure 6.25.

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Figure 6.25: Two frames in relative velocity on the same spacetime diagram. The red axes represent a frame moving to the right.

Worked Example 6.12

Use the spacetime diagram in Figure 6.26 to estimate the speed of frame S′ relative to frame S.

Figure 6.26: For worked example 6.12.

Answer

The tangent of the angle θ is tan θ=3.0 5.0=0.60 and so v c=0.60⇒v=0.60 c.

Finding the scale on the primed axes

The diagram in Figure 6.27 shows the axes of a frame S, and the red axes are the axes of a frame S′ that moves with speed 0.60c relative to S. The important thing to notice about the red axes is that they have a different scale than those of the black axes! In other words we cannot use our ruler to find out how long a unit is on the x axis and then lay the ruler on the x′ axis in order to find a length of 1. So where is the 1 on the x′ axis and where is the 1 on the c t′ axis?

On the x′ axis we have a point with co-ordinates x′ = 1 and c t′ = 0. In the S frame this point has a co-ordinate that we can find using a Lorentz transformation (the gamma factor for a speed of 0.60c is 1.25):

x=γ(x′+v t′)=γ(1−0)=1.25

Similarly, to find the point with co-ordinates c t′ = 1 and x′ = 0 we use

c t=γ(c t′+v c x′)=γ(1−0)=1.25

Figure 6.27

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Worked Example 6.13

Find the scale on the red axes in Figure 6.28a.

Figure 6.28a The scale on the primed axes is different from that on the unprimed axes.

Answer

The speed of the red frame is about 0.75c. The gamma factor is then 1 1−0.75 2≈1.5. Where the blue lines intersect the red axes is where the ‘1’ is.

Figure 6.28b For worked example 6.13.

Length contraction and spacetime diagrams

Spacetime diagrams show length contraction in a very clear and direct way. Imagine a rod of proper length 1 m at rest in the S′ frame. This frame moves away from a frame S with some speed v. We expect that the observers in S will measure a length for the rod that will be less than 1 m. The left end of the rod is at x′ = 0 and the other end is at x′ = 1 m. The worldlines of the ends of the rod are shown in Figure 6.29 (red line for the left end, blue line for the right end).

Figure 6.29: Rod at rest in S′. The observers in S′ measure a length of 1 m for the rod. The blue line is the worldline of the right end of the rod. The observers in S measure a shorter length.

The worldlines of the two ends intersect the x axis at 0 and 0.8 m. These two intersection points represent the positions of the moving rod’s ends at the same time in frame S. Their difference therefore gives the length of the rod as measured by S. The length is 0.8 m, less than 1 m. The rod which moves according to observers in S has contracted in length.

What if we had a rod of proper length 1 m at rest in the S frame which was viewed by observers in the S′ frame, Figure 6.30?

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Figure 6.30: Rod at rest in S. The observers in S measure a length of 1 m for the rod. The blue lines are the worldlines of the ends of the rod. The observers in S′measure a shorter length.

Exam tip

If the rod is moving its length must be measured when the position of its ends are recorded at the same time.

It is only lengths in the direction of the velocity that are contracted.

The two blue lines represent the worldlines of the ends of the rod. The worldlines of the two ends intersect the x′ axis at 0 and 0.8 m. These two intersection points represent the positions of the rod’s ends at the same time in frame S′. Their difference therefore gives the length of the rod as measured by S′. The length is 0.8 m, less than 1 m.

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Exam tip

Do not be led astray by your knowledge of Euclidean geometry! The length in S′ ‘looks’ longer. But it isn’t. The scale on the two axes is not the same!

Time dilation and spacetime diagrams

Spacetime diagrams also show the phenomenon of time dilation very clearly. Figure 6.31 shows a spacetime diagram showing the standard frames S and S′. A clock at rest in the origin of S′ ticks at O and then again at P. The duration of the tick is c t′ = 1. What is the duration of the tick according to frame S? We have to draw a line parallel to the x axis through P. This intersects the S time axis at point Q. This interval is greater than 1.

Figure 6.31: The green line is the worldline of a clock at rest in frame S′. This clock shows c t′ = 1 at P.

Similarly, Figure 6.32 can be used to analyze a clock at rest at the origin of frame S. It ticks at O and then at P. The duration of the tick is such that c t = 1. From P we draw a line parallel to the axis x′. The line intersects the c t′ axis at Q. The interval OQ is longer than 1.

Figure 6.32: This clock shows ct = 1 at P.

Exam tip

Do not be led astray by your knowledge of Euclidean geometry! In triangle OPQ the hypotenuse is the side OP, and so it should have been the longest side of the triangle. But the geometry of spacetime diagrams is not Euclidean! The Pythagorean theorem does not hold.

Simultaneity and spacetime diagrams

Simultaneity is easy to see on spacetime diagrams. In Figure 6.33 events A and B are clearly simultaneous in frame S (ct =1.2 for both).

Figure 6.33: A and B are simultaneous in the S frame.

In frame S′, B occurs first. To find out how much earlier B occurs use a Lorentz transformation:

Δx = 1.0 − 0.5 = 0.5 m and Δt = 0. So

c Δ t′=c(t B′−t A′)=γ(c Δ t−v Δ x c)=1.25×(0−0.60×0.5)=−0.38 m

The negative sign indicates B occurs earlier as we see from the diagram.

Check yourself 6

A rocket moves past the ground. Two trees are hit by lightning the instant the rocket is halfway between the trees. The strikes happen at the same time as far the rocket is concerned. (See Figure 6.34.)

Figure 6.34: For check yourself question 6.

Use a spacetime diagram to determine which tree is hit fi rst according to the ground.

Does light from the two strikes reach the rocket at the same time?

Worked example 6.14

The spacetime diagram in Figure 6.35 shows a number of events. List the events from earliest to latest in each frame.

Figure 6.35: For worked example 6.14.

Answer

In S (blue frame) events A and C are simultaneous and occur after event B. (A and C are on the same line parallel to the x axis).

In S′ (red frame) events B and C are simultaneous and occur before event A. (B and C are on the same line parallel to the x′ axis).

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Worked example 6.15

A spacecraft is moving away from earth at speed 0.60c. At ct = 0, just as the spacecraft begins to move, a flash of light leaves a space station which is at rest 4.0 ly from earth, according to earth. Using a spacetime diagram, estimate the time the flash of light arrives at the spacecraft according to a earth and b the spacecraft.

Answer

We must draw the worldline of a photon leaving the position x = 4.0 ly and see where it intersects the worldline of the spacecraft. This is a time of ct = 2.5 ly for earth. To estimate the time for the spacecraft we must find the scale on the worldline of the spacecraft (red line). The gamma factor is 1.25. This allows us to find the ‘1’ on the worldline of the spacecraft. The photon worldline is at 45° as shown. By measuring on the worldline we find a time of c t′ = 2.0 ly. This is all shown on the spacetime diagram in Figure 6.36.

Figure 6.36: For worked example 6.15.

Check yourself 7

Verify the conclusion of worked example 6.15 using Lorentz transformations.

Worked example 6.16

Two stars, A at position –3 ly and B at position +3 ly, both explode at ct = 0. A spacecraft moving at velocity +0.60c is at x = 0 when ct = 0. Use a spacetime diagram to determine a which star exploded first according to the spacecraft observers and b from which star explosion light reaches the spacecraft observers first.

Answer

The red lines are the axes of the spacecraft frame, Figure 6.37. By drawing lines through the events representing the explosions parallel to the space axis of the spacecraft and see where they intersect the time axis we see that B occurred first. Photon lines at 45° from A and B show that light from B arrives first.

Figure 6.37: For worked example 6.16.

Science in context

The power of diagrams

Spacetime diagrams offer a simple and pictorial way of understanding relativity. They can be used to unambiguously resolve misunderstandings and ‘paradoxes’. Their power lies in their simplicity and their clarity. In physics, the use of diagrams, as well as appropriate notation, in describing situations has always helped understanding. Spacetime diagrams are especially useful in seeing what event can or cannot cause another. Feynman diagrams are another example of this, as is the use of vectors as opposed to components.

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TEST YOUR UNDERSTANDING

In the questions that follow, the spacetime diagrams represent two inertial frames. The black axes represent frame S. The red axes represent a frame S′ that moves past frame S with velocity v.

34Use the spacetime diagram (Figure 6.38) to calculate the velocity of frame S′ relative to S.

Figure 6.38: For question 34.

35 aOn the spacetime diagram in Figure 6.39 draw the worldline of a particle that is:

iat rest in frame S at x = 1.0 m.

iimoving with velocity –0.40 c as measured in S and is at x = 1.0 m at t = 0.

Figure 6.39: For question 35.

bA photon is emitted from position x = 1.5 m at t = 0. Use Figure 6.39 to estimate when the photon arrives at the eye of an observer at rest at the origin of frame S′ according to:

i S

ii S′

36 a S′represents an alien attack cruise ship. Use the diagram in Figure 6.40 to determine the speed of the cruise ship relative to S.

bAt t = 0, a laser beam moving at the speed of light is launched from x = 2 ly towards the cruise ship. By drawing the worldline of the laser beam, estimate the time when the beam hits the cruise ship according to:

i S

ii S′.

Figure 6.40: For question 36.

37The diagram in Figure 6.41 shows two lamps at x = ±0.5 m that turn on at time zero in frame S.

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Figure 6.41: For question 37.

aDetermine which lamp turns on first in frame S′.

bDraw the worldlines of the photons emitted from the two lamps towards an observer at rest at the origin of S′.

cState the lamp whose light reaches an observer at the origin of frame S′ first.

38 aLook at Figure 6.42. The dashed blue line is parallel to the primed time axis and intersects the primed x axis at P. The S′ frame moves with speed v relative to S. Find v and, by using Lorentz transformations, find the co-ordinates of P in the S′ frame. Hence label the event with co-ordinates x′ = 1, c t′ = 0.

Figure 6.42: For question 38.

bRepeat part a where now the speed v is arbitrary to find the space co-ordinate of P. Express your answer in terms of the gamma factor.

39By any suitable method, locate the events (x′ = 1 m, c t′ = 0) and (x′ = 0, c t′ = 1 m) on the spacetime diagram in Figure 6.43.

Figure 6.43: For question 39.

40The purple line in the diagram in Figure 6.44 represents a rod of length 1.0 m at t = 0 as measured in the S frame.

Figure 6.44: For question 40.

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aBy drawing appropriate lines, show that the length of the rod measured in frame S′ is less than 1 m.

b iDraw a line to represent a rod of length 1.0 m as measured in the S′ frame.

iiBy drawing appropriate lines, show that the length of the rod measured in frame S is less than 1 m.

41A clock is at rest at the origin of frame S′ as shown in Figure 6.45. Its first tick occurs at c t′ = 0. The second tick of the clock occurs at c t′ = 1 m as measured in S′.

Figure 6.45: For question 41.

aMark these events on the spacetime diagram.

bBy drawing appropriate lines estimate the time in between ticks as measured in S.

42A rocket leaves earth at speed 0.80 c on its way to a planet a distance 20 ly as measured by earth. When the rocket gets to the planet it sends a light signal to earth.

aWhen does the signal get to earth according to (i) earth and (ii) rocket?

bIllustrate your answers with a spacetime diagram.

cAt t = 0, another rocket leaves the planet returning to earth with speed 0.80 c relative to earth. As the rocket sets out it sends a light signal to earth. When does the light signal arrive at earth according to (i) earth and (ii) rocket clocks?

dIllustrate your answers on a spacetime diagram.

Links

The formulas of special relativity make no sense when the speed of a material object reaches or exceeds the speed of light. This implies that the speed of light is a limiting speed. But the laws of Newtonian mechanics say that as long as we provide energy to a body, its speed will increase indefinitely. This implies that the laws of Newtonian mechanics also need to be modified! This has serious consequences implying that mass and energy are equivalent and can be converted into each other—see Unit E.