minkowski
TRANSCRIPT
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Einstein for EveryoneMichel Janssen
STR AND MINKOWSKI SPACE-TIME
This handout will give you the diagrams that I’ll be showing on transparancies in class. I have triedto keep the explanatory text short and concise. The best way to get a handle on this material is toredraw the diagrams for yourself and to do the exercises.
1. SPACE-TIME DIAGRAMS
timeaccordingto Tom
Lightbulb 1 Lightbulb 2
Michel
FIGURE 1
One of the difficulties of learning STR is that you have to visualize systems in motion. SupposeMichel is standing in the middle of a railroad car with lightbulbs attached to both ends, zooming byTom standing by the tracks. Both lightbulbs flash once and the two flashes reach Michel at thesame moment. As we saw earlier, this means that, from Tom’s point of view, the lightbulb at the rearend of the railroad car flashed before the lightbulb at the front end. In order to visualize whathappens, according to Tom, we have to go through a series of “snapshots” such as the ones shownin Fig. 1. Consider the sequence from bottom to top. The first snapshot shows the flashing oflightbulb 1. In the following snapshots we see a photon (a light particle) emitted by lightbulb 1catching up with Michel, who is rushing away from the photon. The fifth snapshot shows theflashing of lightbulb 2. Michel is rushing towards the photon emitted by lightbulb 2. In the finalsnapshot, both photons reach Michel.
In Fig. 2 a so-called space-time diagram or Minkowski diagram is shown for this collection ofevents. Such diagrams were first introduced in a famous lecture entitled “Space and time” by theGöttingen mathematician Hermann Minkowski (1864–1909) at the annual meeting of GermanNatural Scientists and Physicians in Cologne in 1908. The lecture was published post-humously inPhysikalische Zeitschrift 10 (1909): 104–111. It is reprinted in the Dover anthology The Principleof relativity.
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Worldline oflightbulb 1
Worldline of lightbulb 2
O
P
Q
Tom
's tim
e ax
is
Tom's space axis
R
S
FIGURE 2
You can think of a space-time diagram as a “stripped-down” version of a series of snapshots suchas given in Fig. 1. Leaving out all unnecessary details in Fig. 1 and retaining only the crucialinformation, you get Fig. 2. Every horizontal slice of Fig. 2 corresponds to a snapshot of the sortshown in Fig. 1. Notice how we can trace the position of all elements we are interested in (such asthe two lightbulbs, and the photons emitted by these lightbulbs) through time. The trajectories theseelements trace out in space and time are called world lines.
Fig. 2 also shows “Tom’s time axis” and “Tom’s space axis.” A time axis is a line in the“time direction.” On such lines all points (events) happen at the same place. For Tom, these arethe vertical lines in Fig. 2. A space axis is a line in the “space direction.” On such lines all points(events) happen at the same time. For Tom, these are the horizontal lines in Fig. 2. When the eventO is chosen as the origin (the zero point for both time and distance), we get the axes shown in Fig.2. With the help of these two axes, Tom can read off the time and place of any event happening inspace and time. For instance, The event Q (“the photons reach Michel”) happens some time ORafter event O at a distance OS—measured with clocks and rods at rest with respect to Tom.
2. RELATIVITY OF SIMULTANEITY
Worldline oflightbulb 1
Worldline oflightbulb 2
P
Q
O
T
U
Michel's space axis
Mic
hel's
tim
e ax
is
FIGURE 3
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Fig. 3 shows the same events in space and time as Fig. 2 but instead of Tom’s space and time axes,it shows Michel’s time and space axes. Michel’s axes are tilted compared to Tom’s. Why is this?(Answer below; skip Fig. 4 for the time being.)
Worldlinelightbulb 1
Worldlinelightbulb 2
Q
O P
T
R
S
U
Mic
hel's
tim
e ax
is
Michel's space axis
Tom's tim
e axis
Tom's space axis
FIGURE 4
Consider the time axis in Fig. 3 first. Recall that all events on a time axis happen at the same place.Hence, according to Michel, the time axis is parallel to the world lines of anything at rest withrespect to him (such as the two lightbulbs). Now consider the space axis. All events on a space axishappen at the same time. According to Michel, events O (“lightbulb 1 flashes”) and P (“lightbulb2 flashes”) happen at the same time.* Hence, according to Michel, the lines parallel to OP are thelines connecting points (events) that happen at the same time (in other words: according to Michel,lines parallel to OP are lines of simultaneity). Hence, Michel’s space axis has to be parallel to OP.When the event O is chosen as the zero point for time and distance again, we get the axes shown inFig. 3. With the help of these two axes, Michel can read off the time and place of any eventhappening in space and time. For instance, The event Q (“the photons reach Michel”) happenssome time OT after event O at a distance OU—measured with clocks and rods at rest with respectto Michel.‡
Of course, there is nothing sacred about representing Tom’s space and time axes by horizontaland vertical lines. We are perfectly free to represent Michel’s space and time axes by horizontal andvertical lines. This is what I have done in Fig. 4. The same set of events is pictured in this diagram.Events O and P are the flashing of the lightbulbs 1 and 2, event Q is the encounter of two photonsemitted by these lightbulbs with Michel. From Fig. 4, we once again read off that for Michel O andP are simultaneous, whereas for Tom P is later than O.
* Look back at Fig. 1. Michel is standing in the middle of the railroad car. The flashes from the two lightbulbs reachhim at the same moment. Hence, he concludes, that lightbulbs 1 and 2 flashed simultaneously.‡ Notice that, according to Michel, the same time OT elapses between the events O and Q and the events P and Q.Further notice that the spatial distance OU between the events O and Q is the same as the spatial distance UPbetween the events P and Q. So, the figure shows is that Michel will claim that the photons he receives fromlightbulbs 1 and 2 covered the same distance in the same time.
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EXERCISE 1: (pertaining to Fig. 4)a. Draw the point V on Tom’s time axis that, according to Tom, is simultaneous to P
(HINT: according to Tom, O is simultaneous to S, and Q is simultaneous to R).b. Draw the point W on Tom’s space axis that, according to Tom, corresponds to an event
happening at the same place as P (HINT: according to Tom, Q happens at the same placeas S, and R happens at the same place as O).
c. Which line segments on Tom’s space axis give the distances that the flashes from lightbulbs1 and 2 have to cover to get to Michel according to Tom? (HINT: according to Michel, theseline segments are OU and UP). Are these line segments equally long? Explain.
d. Which line segments on Tom’s time axis give the times it takes the flashes to get fromlightbulbs 1 and 2 to Michel according to Tom? (HINT: according to Michel, the line segmentis OT in both cases). Are these line segments equally long? Explain.
3. TIME DILATION
A moving clock ticks at a slower rate than an identical clock at rest. This can be read off directlyfrom Fig. 5. For Tom, the time between successive ticks on his clock (OA, AA’, A’A”) is shorterthan the time between successive ticks on Michel’s clocks (OC, CC’, C’C”). For Michel, the timebetween successive ticks on his clocks (OB, BB’, B’B”) is shorter than the time betweensuccessive ticks on Tom’s clocks (OD, DD’, D’D”).
Worldline ofTom's clock Worldline of
Michel's clock
A
A'
A"
B
B'
B"
C
C'
C"
D
D'
D"
events A, A', A": ticks on Tom's clock
events B, B' , B": ticks on Michel's clock
events C, C', C" simultaneous to B, B', B" according to Tom
events D, D', D" simultaneous to A, A', A" according to Michel
O
FIGURE 5: time dilation (OA < OC, OB < OD)
EXERCISE 2: Redraw Fig. 5, now representing Michel’s space and time axes by a horizontal and avertical line, respectively (cf. Fig. 4).
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4. LENGTH CONTRACTION
A moving rod is shorter than an identical rod at rest. This can be read off directly from Fig. 6:
For Tom, the length of his own rod (OA) is longer than the length of Michel’s (OC).
For Michel, the length of his own rod (OB) is longer than the length of Tom’s (OD).
AC
D B
O
Worldline of shadedend of Tom's rod
Worldline of whiteend of Tom's rod
Worldline of shadedend of Michel's rod
Worldline of whiteend of Michel's rod
OA = length of Tom's rod according to Tom
OB = length of Michel's rod according to Michel
OC = length of Michel's rod according to Tom
OD = length of Tom's rod according to Michel
FIGURE 6: length contraction (OC < OA, OD < OB)
EXERCISE 3: Redraw Fig. 6, now representing Michel’s space and time axes by a horizontal and avertical line, respectively (cf. Fig. 4 and exercise 2).
EXERCISE 4:a. The event “shaded ends meet” in Fig. 6 is labeled O. Label the event “white ends meet” E.b. Draw a line segment EF showing that, according to Tom, event E happens after event O (cf.
Fig. 5).c. Draw a line segment EG showing that, according to Michel, event E happens before event O
(cf. Fig. 5).d. Draw the time and space axes for some observer John for whom events E and O happen
simultaneously.e. According to John, which line segment represents the length of both Tom’s and Michel’s
rod?
5. CONSTANCY OF THE SPEED OF LIGHT AND THE LORENTZTRANSFORMATION
The space-time diagrams drawn so far have the following features:
(1) The world lines of photons or light signals are always at 45o;
(2) The angle the tilted time axis makes with the vertical is the same as the angle the tiltedspace-axis makes with the horizontal.
This is true for all Minkowski diagrams. In this section, we will see why that is. It is a directconsequence of the convention we use for choosing units on the space and time axes.
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In the internationally accepted system of units (SI units), we measure distance in meters andtime in seconds (and velocity accordingly in meters per second).* As we will see below, we want tocompute distances in space-time by adding distances in space to distances in time. In order to do so,we need to use the same units for distances in space and distances in time. Adding seconds tometers (or miles to hours) would be like adding apples and oranges. The way one typically getsaround this problem in relativity theory is to multiply all times (measured in seconds) by c = 3× 108 (the velocity of light in meters per second), thus turning seconds into meters. That way we, ineffect, introduce a new unit of time which can be called—although this term does not seem to becommonly used—the light meter. One light meter is defined as the time it takes for light to travelone meter. This means that 3 × 108 light meters go into one ordinary second. Note that the lightmeter is thus a much smaller unit of time than the ordinary second.†
We can now understand where features (1) and (2) mentioned above come from. We start withfeature (1) [world lines of photons are 45o lines]. Look back at Fig. 2. The line segment OQ is at45o: the component OS along the space axis is equal to the component OR along the time axis. Inother words:
distancec × time
= OSOR
= 1.
This means that the velocity of whatever is following the world line OQ is given by:
distancetime
= c.
So, with our convention of using meters as our unit on the space axis and light meters as our uniton the time axis, world lines of photons are represented by 45o lines.††
We now turn to feature (2) [the space axis is tilted at the same angle as the time axis]. Look atthe tilted space and time axes in Figs. 3 and 4. Obviously, the components of the line segment OQread off on those space and time axes—(OU,OT) in Fig. 3 and (OS, OR) in Fig. 4—should also beequal, just as the components read off on the horizontal and vertical axes. After all, the observerswhose space and time axes are represented by those tilted lines must also find that light moves atthe velocity c. You will see that OU is indeed equal to OT in Fig. 3 and that OS is indeed equal toOR in Fig. 4. Notice that this requires the tilted space and time axes to be tilted at the same anglewith respect to the horizontal and vertical axes, respectively.
Consider an arbitrary point or event in space-time. To be specific, consider the point Q in Fig. 4,but we could have chosen any other point. Tom and Michel will give different answers to the
* In principle, we could also use miles and hours, of course, in which case we would measure velocity in miles perhour.† Instead of using meters for distances in time as well as for distances in space, we could, of course, also avoid everhaving to add apples to oranges (or meters to seconds) by using seconds for both types of distances. In that case, we
divide all distances (measured in meters) by c = 3 × 108 (the velocity of light in meters per second) thus turningmeters into seconds. That way we, in effect, introduce a new unit which can be called the light second. One lightsecond is defined as the distance that light travels in one second. This means that 3 × 108 ordinary meters go intoone light second. The light second is thus a much larger unit of distance than the ordinary meter. The concept of alight second is fully analogous, of course, to the concept of a light year, the distance travelled by light in one year, auseful unit of distance in cosmology. A light year is clearly an even bigger unit of distance than a light second (thereare about 365 × 24 × 60 × 60 light seconds in one light year). The term “light meter” that I introduced above was,in fact, inspired by the term “light year,” with which everybody who has ever watched shows like “Star Trek” willpresumably be familiar.†† Once you grasp the notion of a light meter, there is a much simpler way to see this. If you plot the distancetravelled in meters against the time elapsed in light meters for a photon or a light signal, you get a 45o line (in 1light meter, light by definition travels 1 meter, so in 2, 3, 4, ... light meters light travels 2, 3, 4, ... meters).
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questions where and when the event Q happened. Michel will say that Q happened some time OT**
after the event O at a distance OU from O, where time and distance are measured in seconds andmeters on clocks and rods at rest with respect to Michel. Tom, on the other hand, will say that Qhappened some time OR after the event O at a distance OS from O, where time and distance aremeasured in seconds and meters on clocks and rods at rest with respect to Tom. So, Tom andMichel will assign a different time and place to the same event Q.
The time and place assigned to an event are called the space-time coordinates of an event. Oncewe have the coordinates of an event for one observer, say Michel, we can easily figure out what thecoordinates of that event would be for any other observer, e.g., Tom. Draw a space-time diagramwhere Michel’s space and time axes are horizontal and vertical. Draw the event at issue in thespace-time diagram on the basis of its coordinates with respect to Michel’s space and time axes.Now draw Tom’s space and time axes in the figure. Finally, read off the coordinates of the eventson Tom’s space and time axes. Such conversions from coordinates for one observer to coordinatesfor another observer are called coordinate transformations. Typically, physicists will not gothrough the geometrical construction described above to perform such coordinate transformations.They just use the coordinates assigned to a point by one observer to compute the coordinatesassigned to that point by some other observer. The geometrical construction gives you a goodpicture though of what is going on.
We tacitly assumed that we are only considering inertial observers, i.e., observers for whom thevelocity of light has the same constant value c. Start with one such observer, say Michel, whosespace and time axes we will draw as horizontal and vertical lines. If we further stipulate that allobservers take the same event O that Michel takes as his zero point in space and time—i.e., as theorigin of his space-time coordinate system—as their zero point in space and time as well, we caneasily make a full inventory of space-time coordinate systems used by inertial observers. The timeand space axes can be tilted at any angle less than 45o with respect to Michel’s space and timeaxes. The closer to 45o, the faster the observer using these tilted axes is moving with respect toMichel. There is another set of possible coordinate systems. To see this, we need to leave our 1+1dimensional space-time for a while and look at the full 3+1 dimensional space-time. Michel nowneeds three space axes instead of one. We let Michel pick any three axes he wants, the onlyrestriction being that they are mutually perpendicular. Other observers, however, are allowed torotate this set of axes in any way they please. So, on top of space and time axes corresponding todifferent states of motion, we also have space and time axes corresponding to different spatialorientations. A coordinate transformation that gets us from one such coordinate system to anotheris called a Lorentz transformation.
6. THE LIGHTCONE; TIME-LIKE, SPACE-LIKE AND LIGHT-LIKE LINES
Future lightcone
Inside the future lightcone
Out
side
the
light
cone
Outside the lightcone
Inside the past lightcone
Past lightcone
P
FIGURE 7: 1+1 dimensional Minkowski space-time
** More accurately: OT divided by the speed of light c. For convenience, we will treat OT as measuring timedirectly. This is a common and totally harmless simplification.
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Suppose the event O correspond to here and now for some observer. The rest of space-time nownaturally breaks up into different parts, no matter what his/her state of motion is at O. This isillustrated in Fig. 7 for a 1+1 dimensional space-time and in Fig. 8 for a 2+1 dimensional space-time. The observer can only move at velocities less than the velocity of light (the full answer to thequestion why this is the case will be given later). This means that wherever his world line takeshim/her from O, s/he is confined by the 45o lines representing the world lines of light. These worldlines are called the lightcone. The reason for this name will be clear from looking at the 2+1dimensional case in Fig. 8.
FIGURE 8: lightcone, world lines and simultaneity slicesin a 2+1 dimensional Minkowski space-time
Return to our observer. Just as his/her future is confined by the top half of the lightcone, the so-called future lightcone, his/her past is confined by the bottom half of the light cone, the so-calledpast lightcone. So, the observer’s total world line lies inside the lightcone (and always has an anglewith the vertical of less than 45o). Therefore the areas “inside the future lightcone” and “inside thepast lightcone” are also called the absolute future and the absolute past of O. In Fig. 7, a number ofpoints are drawn inside the future lightcone. If you draw a line through P and one of those points,you get a possible world line for some inertial observer. For this observer the two points happen atthe same place some time after each other. The world line of this inertial observer can be used as histime axis. If you draw the corresponding space axis, you will see that the space axis goes throughone of the points drawn outside the lightcone. For this observer this point is an event that happensat the same time as event O.
These considerations can also be expressed as follows. For any point P inside the lightcone ofsome event O there is an inertial observer for whom O and P happen at the same place at differenttimes. Similarly, for any point Q outside the lightcone of some event O there is an inertial observerfor whom O and Q happen at the same time at different locations. Finally, a line through O and anypoint R on the lightcone represents a possible world line of a photon. Points P inside the lightconeare said to be timelike connected to O, points Q outside the lightcone are said to be spacelikeconnected to O, and points R are said to be lightlike connected to O. Similarly, line segments suchas OP are called timelike, line segments such as OQ are called spacelike and line segments such asOR are called lightlike.
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7. EUCLIDEAN GEOMETRY AND THE GEOMETRY OF MINKOWSKI SPACE-TIME
Tom's rod
Michel's rod
Tom's rod
Michel's rod
Event P:shaded endsmeet
Event Q:white ends meet
Tom
Michel
space
space
time
time
P Q
FIGURE 9
In Fig. 9, two events P and Q are drawn in a space-time diagram (you will recognize the situationfrom the first handout, section 4). Tom and Michel are in different (inertial) states of motion, sothey will use different space and time axes. Both Tom and Michel choose the event P as their zeropoint in both space and time. As you can read off directly from the space-time diagram, Tom andMichel disagree about the order in which the events took place. According to Tom, P happensbefore Q; according to Michel, P happens after Q.
This situation in space-time is closely analogous to the following situation in ordinary space. Asis shown in Fig. 10, Tom and Michel are looking at points P and Q. Because they are looking at thesituation from different angles, they have a different sense of “back and front” and “left andright.” On the right in Fig. 10, the “front-back” and “left-right” axes used by Tom and Michelare shown. Both Tom and Michel choose the point P as their zero point for “front-back distance”and “left-right distance.” Because of their different perpectives on the situation, Tom and Michelwill disagree about whether P is in front of or behind Q. According to Tom, P is behind Q;according to Michel, P is in front of Q.
Tom
Michel
P
Q
Tom
Michel
left-right
front-
back
left-right
fron
t-ba
ck
P
Q
FIGURE 10
In Fig. 11, the diagrams on the right of Figs. 9 and 10 are shown next to each other. Look at Fig.11a first. With the help of the line segments from Q to the various axes, we can read off the “front-back distance” and the “left-right distance” of Q to P. According to Tom, Q is QR in front of P
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and PR to the right of P; according to Michel, Q is QS behind P and PS to the right of P. The totaldistance between P and Q is a combination of “left-right distance” and “front-back distance.” Tocompute the total distance from the “left-right distance” and the “front-back distance,” we use thePythagorean theorem. Tom will apply the theorem to the right angled triangle PQR:
PQ2 = RQ2 + PR2.
Michel will apply the Pythagorean theorem to the right angled triangle PQS:
PQ2 = QS2 + PS2.
These are just two equivalent ways of computing the same distance PQ. So, Michel and Tom agreeon the total distance between P and Q, they only disagree about how this total distance breaks upinto “front-back distance” and “left-right distance.”
space
space
time
time
P Q
Tom
Michel
left-right
front-
back
left-right
fron
t-ba
ck
P
Q
S
R
R
S
(a) 2-dim. Euclidean space (b) 1+1-dim. Minkowski space-timeFIGURE 11
A similar result holds in Minkowski space-time, as is illustrated in Fig 11b. Tom and Michelwill agree on the “distance” between P and Q in space-time. However, according to Tom, thespace-time distance PQ breaks up into a distance RQ in time and a distance PR in space, whereas,according to Michel, the space-time distance PQ breaks up into a distance QS in time and a distancePS in space.
How do we compute the distance in space and time between P and Q? Can we use thePythagorean theorem that we used above? The answer to this last question is: NO! The geometry ofMinkowski space-time is different from the geometry of ordinary Euclidean space. It is preciselyfor this reason that we cannot represent all features of Minkowski space-time in a drawing on aEuclidean piece of paper! Here are two such features:
(1) space and time directions are always perpendicular, just as the “front-back direction” isalways perpendicular to the “left-right direction.” In our drawings, however, the space and timedirections only look perpendicular for one observer (for Tom in Fig. 11b and most other figures,for Michel in Fig. 4).
(2) The units on different sets of space and time axes are always the same (all inertial observersuse the same clocks and measuring rods). In the drawings, however, the units on the “tilted” axeslook larger than the units on the horizontal and vertical axes (see Figs. 5 and 6).
Given these differences between Minkowski space-time and ordinary Euclidean space, it neednot surprise us that we cannot compute distances in space-time the same way as we computedistances in Euclidean space. It turns out, however, that the rule for computing distances in space-time is very similar to the Pythagorean theorem. The rule is:
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(distance in space-time)2 = (c × distance in time)2 – (distance in space)2.
Looking back at Fig. 11b, you will see that applying this rule from Tom’s and Michel’s perpective,respectively, gives:
PQ2 = RQ2 – PR2 = QS2 – PS2.
Hence the only difference with the Pythagorean theorem is that we have a minus sign instead ofplus sign. This minus sign does lead to some important differences between Minkowski space-timeand ordinary Euclidean space. For one thing, the space-time distance—or interval as it is usuallycalled—can be negative! This happens when the space distance between two events is larger thanthe time distance, i.e., for all spacelike line segments. Moreover, when the interval between twopoints equals zero, this does not mean that the two points are the same! If the space distance isequal to the time distance, the space-time distance equals zero. Hence, the interval between twopoints on the world line of a photon is always zero.
Figs. 12 and 13 show more differences between distance in ordinary Euclidean space anddistance in Minkowski space-time. In Fig. 12a, for ordinary Euclidean space, I have drawn the setof all points at some fixed distance, say 1 m, from the point in the center. This set of points form acircle. In Figs. 12b and 12c, for 1+1 dimensional Minkowski space-time, I have drawn the set of allpoints at some fixed space-time distance from the point in the center. The two hyperbolae in Fig.12b represent all points at fixed positive distance (i.e., a fixed number of seconds); the twohyperbolas in Fig. 12c represent all points at fixed negative distance (i.e., a fixed number ofmeters). In our Euclidean drawing of the situation in space-time it does not look as if these pointsare at equal distances from the center at all!
(a) (b) (c)
FIGURE 12
In Fig. 13a, two sets of “left-right” and “front-back” axes are drawn. We see that the units onthese axes are the same. In Fig. 13b, two sets of space and time axes are drawn. Again, the units onthese axes are the same, even though it looks as if the units on the tilted axes are larger than theunits on the horizontal and vertical axes. Notice that this drawing of the units on space and timeaxes on the basis of the definition of space-time distance reproduces what we already found insections 3 and 4 (see Figs. 5 and 6).
(a) Euclidean space (b) Minkowski space-time
FIGURE 13
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The striking similarity between situations (a) and (b) in Fig. 13 led Minkowski to an importantconclusion. The fact that the laws of physics stay the same when we switch to a different inertialstate of motion is no more surprising than the fact that the laws of physics stay the same when weswitch to a rotated set of space axes. As he pointed out, in Newtonian mechanics the statement thatuniform motion does not make a difference for the laws of physics is something completelydifferent from the statement that orientation in space does not make a difference for the laws ofphysics. In STR, these two statements turn out to be intimately connected. Switching to a differentinertial state and switching to a rotated set of space axes can both be seen as “rotations” of youraxes in space-time.
8. THE ANALOGUE OF THE PYTHAGOREAN THEOREM IN MINKOWSKISPACE-TIME
In this section I will show that the rule for computing distances in space-time that I gave in section 7is a straightforward generalization of the Pythagorean theorem in ordinary Euclidean space. To thisend, I will go over a simple geometrical construction used to prove the Pythagorean theorem inEuclidean space and show that essentially the same construction can be used to prove the analogueof the Pythagorean theorem (with the plus sign replaced by a minus sign) in Minkowski space-time. I will use some math in this section, though nothing more than high school geometry.
A B
CD
•• ••
• •
FIGURE 14
Fig. 14 shows a standard construction that is used to prove the Pythagorean theorem in ordinaryEuclidean space. We start with the right angled triangle ABC. We want to prove that AC2 = AB2 +BC2. We proceed as follows. First draw a line through B that is perpendicular to AC. In this wayyou get BD. The Pythagorean theorem follows directly from the similarity of the triangles ABC,ADB, and BDC:
A
similar to
B
C
•
••
A B
D
•••
⇒ ABAC
= ADAB
⇒ AB2 = AC × AD.
Likewise, we get:
A B
C
•
••
similar to
DC
B
•
••
⇒ BCAC
= DCBC
⇒ BC2 = AC × DC.
Now, notice that AD + DC = AC (see Fig. 14). Hence, adding the second relation to the first gives:
AB2 + BC2 = AC × (AD + DC) = AC2,which is what we wanted to prove.
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We will now go through the analogous argument in Minkowski space-time. Consider the rightangled triangle ABC in Fig. 15, made up of the spacelike line segments AB and AC and the timelikeline segment BC.‡
A B
C
D
FIGURE 15
As in Fig. 14, we construct a line segment BD perpendicular to AC. In Fig. 15, BD does not looklike it is perpendicular to AC, but that is because the geometry of the paper we are using to draw thediagram is different from the geometry of Minkowski space-time. In space-time, the direction BD isthe time direction that goes with the space direction AC. Hence, BD and AC are, in fact,perpendicular.
As in Fig. 14, the triangles ABC, ADB, and BDC are similar. We can therefore write down theexact same relations we had in the Euclidean case:
similar to
A B
C
•
••
⇒ ABAC
= ADAB
⇒ AB2 = AC × AD.•••A B
D
A B
C
•
••
similar to ⇒ BCAC
= DCBC
⇒ BC2 = AC × DC.•
B
CD
••
Now, notice that AD – CD = AC (see Fig. 15): a minus sign instead of the plus sign we had in theEuclidean case! Hence, subtracting the second relation from the first, we get:
AB2 – BC2 = AC × (AD – CD) = AC2.
Since AC is a spacelike line segment, the space-time distance—the interval—between A and C is –AC2. For this reason, we rewrite the result above as:
– AC2 = BC2 – AB2.
This is indeed the rule for computing distances in Minkowski space-time: space–time distancesquared equals time distance squared minus space distance squared. As the argument in this sectionshows, this rule is the direct space-time analogue of the Pythagorean theorem in ordinary space.
‡ A completely analogous argument can be given for the case of a right angled triangle made up of two timelikelines and one spacelike line.
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9. THE TWIN PARADOX
Consider Fig. 16. For the time being, ignore the shaded lines and the points P and Q, and focus onthe lines SUR and SAR, the world lines of the identical twin sisters Suzy and Sara. Up to the pointS, the world lines of Suzy and Sara coincide. At S, the twins get separated. Suzy remains in thesame state of inertial motion she has been in all along; Sara, on the other hand, boards a spaceshipthat takes her away from her twin sister at 75% the speed of light. At A, the spaceship abruptlyturns around and starts moving back to Suzy, again at 75% the speed of light. At R the sisters arereunited. For Suzy, 18 years have passed (see the marks on Suzy’s world line SUR). For Sara,only 12 years have passed (see the marks on Sara’s world line SAR). The phenomenon is knownas the twin paradox.
S
R
U A
Suzy Sara
Tim
Tom
P
Q
FIGURE 16
As the phrase “twin paradox” suggests, many people have been baffled by this prediction ofSpecial Relativity (it has successfully been tested using identical atomic clocks instead of twins).Here is how people tend to get confused. They argue as follows: “Suzy and Sara are two equivalentinertial observers who will both find that the clocks of the other are ticking at a slower rate than herown clocks, and, similarly, that the other should age more slowly than she herself. So, according toSuzy, Sara should be younger at their reunion, whereas, according to Sara, Suzy should be youngerat their reunion. Clearly, they cannot both be younger than the other. Hence, there must be someother factor that ensures that they are equally old when reunited.” As we just saw, Special Relativitytells us that Sara will be younger than Suzy. So, what is wrong with the argument we just gave? It isthe very first step that is wrong. Suzy and Sara are not two equivalent inertial observers. Suzy is aninertial observer, but Sara is not. At A she switches from one inertial state of motion to another; inother words: she accelerates.
But, you might object, just as Sara accelerates with respect to Suzy, Suzy accelerates withrespect to Sara. Why can’t they both claim that the other is accelerating. The answer is thatacceleration, unlike velocity, is an absolute, not a relative affair. Perhaps the easiest way to see this isto compare how Suzy and Sara experience the moment that they start to approach each other again.For Suzy—apart perhaps from some strange mixture of excitement and disappointment about hersister’s return—there is nothing special about this moment. Sara, on the other hand, will feel it inher stomach, especially if the spaceship turns around as abruptly as is indicated in Fig. 16! So,Suzy and Sara are not equivalent observers. Sara accelerates and Suzy does not. Therefore, it neednot surprise us that Suzy and Sara age a different number of years.
There has been a lot of confusion about the role of the acceleration at A. Some people have heldthe view that Sara ages 6 years during the acceleration so that she ends up being as old as Suzywhen the twins are finally reunited. Others have pointed to the acceleration as causing the difference
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in aging of the twins. Both views are erroneous. Here is a simple argument showing that theacceleration cannot be causing the difference in aging of the twins. Let Sara make a second journey.Again she takes off at 75% the speed of light, but now she keeps going for 12 rather than 6 yearsbefore reversing her course. If it were the acceleration that is causing the difference in aging, Sarashould gain the same number of years on Suzy on both trips. After all, the acceleration part in bothtrips is exactly the same. Special Relativity tells us, however, that Sara gains 6 years during the firsttrip and 12 years during the second. Clearly then, the acceleration does not cause the difference inaging. But then again, what does?
The answer is that it is simply a matter of the space-time trajectory SAR between S and R beingshorter than the trajectory SUR. The situation is completely analogous to the following situation inordinary space. Suppose Suzy and Sara drive from Pittsburgh to Minneapolis. Suppose there aretwo routes they can take. Suppose one route is one long straight highway from Pittsburgh toMinneapolis going through Chicago. This route is the analogue of the route SUR in space-time.Suppose the other route is a terrible detour consisting of a straight highway from Pittsburgh toDallas and another straight highway from Dallas to Minneapolis. This second route corresponds tothe route SAR in space-time. On this second route, you will not drive straight all the time. You needto make a right turn in Dallas. This right turn corresponds to the acceleration at A. The second routeis clearly longer than the first. This is a minor disanalogy with the situation in space-time: inordinary space a straight line is the shortest route between two points, in space-time a straight line isthe longest route between two points. Suppose you don’t know at first whether Suzy or Sara tookthe detour. Now, you are told that Sara took a right turn whereas Suzy drove straight all the way.This tells you immediately that Sara went from Pittsburgh to Minneapolis the “long way.”Similarly, knowing that Sara accelerated at some point in her space-time journey, whereas Suzy didnot, tells you that Sara went from separation to reunion the “short way.” So the acceleration servesto pick out which one of the twins took the shortest trip in space-time, just as “making a right turn”serves to pick out which one of the twins took the longest trip in space. But you would not invokethe right turn in Dallas to explain the difference between Suzy and Sara’s odometers upon arrival inMinneapolis! You would just point out that when you go from Pittsburgh to Minneapolis via Dallasyou cover a greater distance than when you go via Chicago. Likewise you are not going to invokethe acceleration at A to explain the age difference between Suzy and Sara upon their reunion. It isjust that the space-time journey SAR is shorter than SUR.
To show that the acceleration really does not play a crucial role, we will change the logistics ofthe experiment in such a way that we can establish the age difference between Suzy and Sara at Ravoiding the acceleration at A altogether. To this end we introduce the inertial observers Tom andTim whose world lines are also shown in Fig. 16. Tom’s world line coincides with Sara’s on thefirst half of her journey (SA), Tim’s world line coincides with Sara’s on the second half of herjourney (AR). Tom times the first half of the journey, Tim times the second. Tom and Tim will bothfind that they accompanied Sara for 6 years. So, the space-time journey SAR takes 12 years, thespace-time journey SUR takes 18 years. Notice that we reached this conclusion using inertialobservers only.
This maneuver of, in effect, replacing Sara with the inertial observers Tim and Tom seems toleave us with another puzzle though. Look at the first half of the journey. According to Suzy, Tomages only 6 years between S and U, which for her is 9 years. Now, Tom and Suzy are two fullyequivalent inertial observers. Hence, if Suzy finds that Tom ages at 2/3 her own rate, Tom shouldfind that Suzy ages at 2/3 his rate. So, according to Tom, Suzy only ages 4 years between S and A,which for him is 6 years. A similar result is found if we compare Suzy and Tim on the second halfof the journey. According to Tim, Suzy ages only 4 years between A and R, which for Tim is 6years. Combining Tom’s and Tim’s conclusions, it looks as if Suzy only ages 8 years between Sand R. However, as we saw earlier, Suzy ages 18 years! What is going on?
Fig. 16 gives the answer to this question. Tom and Tim fail to take into account 10 years ofSuzy’s life between S and R because of their disagreements about simultaneity. According to Tom,A is simultaneous with P, whereas, according to Tim, A is simultaneous with Q. This observationrestores the consistency of the whole story. For Tom, Suzy aged 4 years (between S and P) duringthe 6 years he accompanied Sara from S to A. Likewise, for Tim, Suzy aged 4 years (between Qand R) during the 6 years he accompanied Sara from A to R. The 10 years that Suzy aged between
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P and Q should be added to the 4 years counted by Tom and the 4 years counted by Tim, giving atotal of 18 years for Suzy’s aging between S and R.
Remember: there is nothing mysterious about the twin paradox. All there is to it is that differentjourneys between two points in space-time will generally take a different amount of time, just asdifferent journeys between two points in space will generally not be equally long.
10. THE TROUBLE WITH TACHYONS
A tachyon is a hypothetical particle travelling at a velocity greater than the velocity of light. Suchvelocities are called superluminal. Suppose a particle is moving at a superluminal speed withrespect to Tom (see Fig. 17).
Tom
Michel
space
time
time
Tachyon
FIGURE 17
The world line of this tachyon makes an angle between 45o and 90o with Tom’s time axis.Consider this world line from Michel’s point of view. You will notice that, according to Michel, thetachyon is moving backwards in time!
To see that tachyons—if they exist at all—could indeed get us into trouble, look at Fig. 18. Thisfigure shows the world lines of two spaceships and the world lines of two tachyon torpedos thesespaceships exchange. To determine the sequence of events and the nature of the various world lines(timelike or spacelike), the figure also shows the lightcones for the relevant events. Here is whathappens.
O
A
C
B
Worldline ofspaceship 1
Worldline ofspaceship 2
FIGURE 18
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Event O: Spaceship 2 passes by spaceship 1. Spaceship 1 warns spaceship 2: “You are violatingour territory. Return immediately or we will use our new tachyon torpedos against you.”Spaceship 2 replies: “If you use tachyon torpedos against us, we will use tachyon torpedosagainst you.”
Event A: Spaceship 1 fires a tachyon torpedo at spaceship 2.Event B: The tachyon torpedo from spaceship 1 hits spaceship 2. Spaceship 2 immediately
retaliates and fires a tachyon torpedo at spaceship 1.Event C: The tachyon torpedo from spaceship 2 hits spaceship 1.
Notice that spaceship 1 gets hit (at C) by the retaliation torpedo before it has even fired its own(at A). Here is a possible scenario for what went on on the bridge of spaceship 1 between eventsO and A:Between events O and C: There is a staff meeting to decide whether tachyon torpedos will be used
against the intruder. The captain is worried about retaliation, but his technical expert argues: “Ifthey retaliate, we would feel the hit before we would even fire. We have not been hit so far, so ifwe fire now, we can be sure that they will not retaliate.”
Event C: The tachyon torpedo from the intruder hits.Between events C and A: The captain decides to fire: “We might as well fire since we have already
taken the retaliation.”Event A: The captain gives the command to fire the tachyon torpedos.
This is a consistent scenario but minor modifications would leave us with flat contradictions.Suppose the decisions aboard spaceships 1 and 2 would be as follows. Spaceship 2 fires if andonly if it is hit, i.e. it only uses its tachyon torpedos for retaliation purposes. Spaceship 1, on theother hand, fires if and only if it is not hit (rationale: the technical expert’s argument). This leads toa flat contradiction. Suppose, 1 does not get hit between O and A. Hence, 1 fires. But then 2retaliates and 1 does get hit between O and A. So, “1 does not get hit” implies “1 gets hit.”Similarly, “1 gets hit” implies “1 does not get hit.” Here is the argument. Suppose 1 gets hit.Hence, 1 does not fire. But then 2 does not retaliate. Hence, 1 does not get hit.
Moral: causal loops, such as ABCA in Fig. 18, put severe restrictions on the decisions that canbe made at points in the loop. Somehow, all decisions leading to contradictions should be madeimpossible (such as the combination of decisions on the bridges of spaceships 1 and 2 leading tothe contradiction that 1 is and is not hit). In general this will require elaborate conspiracy schemes.Believing in such conspiracy schemes is the price you pay for believing in the possibility oftachyons (or more generally: for believing in the possibility of causal loops). For most people thisprice is too high.
Durkacs HD:mhp files:Teaching:Minkowski