Lecture 5 The Lorentz Transformation -...

EPGY Special and General Relativity

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Lecture 5

The Lorentz Transformation

We have learned so far about how rates of time vary in different IRFs in motion with respect to each otherand also how lengths appear shorter when in motion. What we want to do now is to develop a set of equations that willexplicitly relate events in one IRF to a second IRF. This will allow us to quantify more complex events betweenuniformly moving frames.

The Galilean TransformationBefore proceeding to the relativistic transformation we review the classical, Galilean, transformations for two

reference frames in motion with respect to each other. In both the following and the relativistic scenario we will dealwith references frames that are in the standard configuration, that is, the origins coincide at time t = t’ = 0 and all axesare collinear (x is collinear with x’, y with y’, and z and z’).

Two reference frames, S and S’, are in relative motion at constant velocity. From the perspective of frame S,frame S’ moves in the positive x direction at speed v. (This same scenario will be used for the relativistic derivation).First, the perpendicular directions, y and z, are unaffected by the relative motion since the two frames are at rest withrespect to each other along these directions. Thus, we have that y = y’ and z = z’. In classical physics there is auniversal time that ticks at the same rate for all observers. Thus in standard configuration we have t = t’.

For the remaining relation we consider an event, 1, in frame S’ given by the coordinates (x’, y’, z’, t’). At thetime t = t’ = 0, the position of x and x’ coincide and as the frame S’ moves off to the right the origin is measured to beat position x = vt when the event occurs and additional displacement is that observed at time t = 0, i.e. x’. Thus whenevent 1 occurs the position as measured by S is x = x’ + vt. To complete the connection we note that t = t’ and thisyields the Galilean transformation equations:

x = x’ + vt’y = y’z = z’t = t’

Thus given a point in the frame S’ these equations will give the position in S if the frames are in standardconfiguration (if they are not it is easy to translate the two frames).

If we have the inverse scenario – given a point in S and want to find it in frame S’, we use the inverseGalilean transformation. This is easily found by setting v to –v in the above equations and switching primed andunprimed coordinates:

x’ = x - vty’ = yz’ = zt’ = t

Now that the classical result is understood we proceed to the relativistic case.

The Lorentz Transformation

The problem is the following, given a point in frame S’, specified by the coordinates x’ and t’, how does thispoint map into the frame S. That is, what are the coordinates; x(t’,x’) and t(t’,x’)? (Again noting that perpendicularcoordinates are unaffected, y = y’, z = z’). To begin, we will examine a spacetime diagram representing the question athand in frames S (on the left) and S’ (on the right). The goal is to express the point P, where we are given the x’ andct’ coordinates, in the S frame.

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peci

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Combining these results we have the full Lorentz transformation,

x = (x’ + vt’)y = y’ The Lorentz Transformationz = z’ (given point in S’, these give the point in S)t = (t’ + vx’/c2)

And just as before, the inverse transformation (if given a point in S) is found by setting v = -v.

x’ = (x - vt)y’ = y The Inverse Lorentz Transformationz’ = z (given point in S, these give the point in S’)t’ = (t - vx/c2)

To Boldly Go Where No Object Can Ever Go.

You are no doubt familiar with the old television and movie series “Star Trek” in which the starshipEnterprise hurtles through space at speeds far exceeding that of light. We now want to show why this will alwaysremain purely fiction unless we give up a basic philosophical tenet concerning cause and effect.

There are several hints we’ve already seen telling us that c is the highest speed that can be achieved byanything. Lets examine these results.

We saw that time is observed to pass more slowly for moving objects, Explicitly t’ = t, and if an objecttravels at c with respect to frame S we see that time appears to stop for the moving object t’ = t = . It takesan infinite amount of time for the moving clock to tick.

Now what happens if v > c? Well then,

= 1 v2

c2

1/ 2

= 1 v2

c21

1/2

, v2

c2>1

the factor g becomes imaginary. Imaginary numbers can not represent real physical quantities so that t’ ~ i has nomeaning. Similarly the observed length becomes imaginary as well. So we see that special relativity can notincorporate objects which move faster than light. This does not forbid objects traveling faster than c but just says thatthe theory of special relativity can not describe them. Such objects are called tachyons.

In order to prove that objects can’t move faster than light we will have to show that it would cause problems.To do this lets devise another gedankenexperiment, this one is a pretty farcical one but will prove the point. (See BoxL-1 of your text to see an equivalent scenario involving a “Star Trek” theme).

Our gedankenexperiment posits a alien race of beings who are very large and inhabit deep space. They alsoenjoy sports, especially American Football which they’ve picked up on broadcasts. Owing to their immense size, theyhave developed their own version with some modifications. First, the size of the playing field is 4 light years long.They also require a large number of referees to establish reference frames to view the action. These creatures all travel


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and throw the ball at speeds less than the speed of light. One year a young recruit is drafted for one of the teams as aquarterback. He is a very talented player who can throw the football at speeds faster than light, up to 3c. Thinking thiswill ensure an easy route to the Universe Bowl the team pays the young recruit lots of money. However, on his firstplay from scrimmage there is a problem, and the young recruit never manages to score a touchdown and becomeswashed up within the first year. Let’s see if we can determine why he is forced out.

Let’s examine one play where this team is at the 0 ly line. The young gun receives the hike of the ball, waitsan appropriate amount of time, and then hurtles a bomb to the streaking receiver who runs at 0.6 c. The receivercatches the ball at the 3 ly line, runs into the end-zone and scores. The team celebrates until they see one referee throwa flag indicating a penalty. The referee states that more than one ball was used in the play, a 1 ly penalty, in this case,half the distance to the goal.

To analyze this play let’s plot the worldlines of the quarterback, the receiver, and the ball. Also, let’sdetermine beforehand what the value of the dilation factor will be for the reference frames of the quarterback and thereceiver,

= 1

1 0.62=

10.64

=1.25 .

Let’s plot the events in the quarterback’s frame.

Let’s list the events in both frames via the Lorentztransformation equations,

QB WR

x(ly) t(y) x’(ly) t’(y)

Start of play: 0 -4 3 -5Throw ball: 0 0 0 0Catch ball: 3 1 3 -1

With these values we can now plot the play in the widereceiver’s frame.

Notice the worldline of the football. It travelsbackwards in this frame! Notice at time t’ = -0.5 y thereare 3 balls! For this is after the catch in the WR frame,before the QB throws the ball, AND the ball is en routeto the receiver.


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Notice that to make a call on this play, there needs to be a series of referees all running at the same speed asthe receiver. Each referee sees a ball near him, and when they later confer they all say that the ball was in front of themat one particular time. This violates conservation of energy, (there are now 3 balls in existence in this frame asopposed to one in the QB frame). This also violates causality, the result (the WR catching the ball), occurs before theQB throws it. Imagine if this was some kind of gun which shot bullets which traveled faster than c. The WR would getshot before the gun was fired in one frame and after in another. Cause and effect no longer holds. Hence, to preservecausality, no thing (not even information), can travel faster than c.

Spacetime Maps

In a previous lecture we examined objects traveling faster than c by viewing events in different referenceframes. Now we are going to examine spacetime maps in more detail. What we will find is a new, easy, way toexamine events in many different reference frames. This method will also display the strange hyperbolic geometry ofLorentz spacetime.

To begin, we go back to the example of the rocket ship moving at 0.4c with the light clock in it. Rememberthat in the rocket frame the emission and absorption of a light pulse occurs at the same spatial location. In thelaboratory frame we concluded that the time must run slower in the rocket because of the path traced out by the lightpulse is longer. We show the two spacetime diagrams here and remind you that the interval between the two events isthe same.


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Alice’s frame Bob’s frame

s2 =1m2 = (ct)2 (x)2

=(ct)2[1 (x)2

(ct)2]

=(ct)2[1 v2

c2] = (ct)2 0.84

(t)2 =c2 (3.6 ns)2

s2 =1m2 = (ct ')2 = c2 (3.3ns)2

Now consider a third frame, the super rocket frame, which is traveling in the same direction as the rocket butat 0.8c. In this frame the light pulse of the rocket and laboratory will move in the opposite direction. This being an IRFas well, the interval between these two events (emission and absorption) will be the same. We plot this frame as well.

s2 =1m2 = (ct '')2 (x '')2

=(ct '')2[1vsr

2

c2] = (ct '')2 0.36

(t '')2 =c2 (5.6 ns)2

By placing these three diagrams within one diagram, since they have the same origin, we begin to notice a pattern,


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We can also plot these events in any other IRF we can think of which moves along the x axis and shares the sameorigin. Again the interval is 1 m in all of these,

By filling these in a line emerges,

This line traces out all points which have the interval equaling 1 m with respect to the origin. This line is in the shapeof, (the upper half), a hyperbola.

The equation for a hyperbola is a2 = x2 – y2 , where a is the point of closest approach to the origin. (Tocompare a2 = x2 + y2 is the equation of a circle and 1 = x2/a2+ y2/b2 is the equation for an ellipse).


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What we see is that to describe two events in any frame we can draw the invariant hyperbola to find theinterval or event in any other IRF. This plot tells us is what the (x,t) event location will be in various frames. Note thatthe frame in which the two events occur in the same place, the proper frame, has the shortest time interval between twoevents. In all other frames, clocks appear to run slower, giving a larger time interval.

Just to be complete, if all of these frames are coincident at the reception of the light pulse we could describethese events in all frames using the lower branch of the hyperbola.

Now we want to discuss worldlines in more detail. We will stick toobjects moving in one dimension for simplicity.

Note that at no time does the particle go beyond 45 degrees of slope, that would correspond to faster than light travel.We examine now our light clock traveling in spacetime starting at the origin, heading out some distance, and

then returning to the x = 0 point. This is similar to the discussion of the twin paradox before, but now we want to gointo more detail about the proper time indicated on the light clock.

Each mark indicates one roundtrip of a light pulse for each light clock. In thelaboratory frame, the interval between events (two receptions) in the rocket canbe found easily as s2 = (ct )2 – (x)2. But since the two events occur at thesame place in the rocket we see that this interval is the same as the proper timemeasured by the rocket’s light clock, (or one tick of that clock). Note that therocket is not always a free float frame, (IRF), since it accelerates at times, (itspath is curved). We need to be careful with neighboring events not joined by astraight line; in this interval the frame is accelerating and the principles we’veset up do not need to hold. To get around this we will assume that all of thereceptions are close enough together such that they are all connected bystraight worldlines. We cut up the path into small segments.

When the rocket returns home at event B the light clock will have readoff the total proper time elapsed during its journey. To other reference frames,like the lab, this proper time will equal the cumulative interval of all the ticksalong the path, i.e. taking the interval between 0-1, 1-2, 2-3, etc. and addingthem up. All IRFs will agree on this cumulative interval which equals theelapsed proper time. The only way the lab frame finds the total proper time isto add up these segments separately. Finding the interval between O and Bdirectly will not work.

An analogy might suffice here. Consider a car travelingbetween two cities. The distance traveled by the car can easily be readoff of the odometer. To an observer not in the car to measure thedistance is not so easy. Simply measuring the distance between the


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cities will not give the correct result since the road bends and twists. The only way for the observer tomeasure the distance the car has traveled is to cut up the path into small segments and add up the distance ofthe segments. Then the observer will get the same result as the car’s odometer.

We see that finding the interval between two events can vary if the path or worldline between the two eventsis curved. Before we only examined IRFs and we saw that there was only one IRF which can have the events occur atthe same location. (Assuming that the events are timelike separated). But if we have a worldline which is not straight

we will get a different result for the proper time. The proper time for thecurved worldline will not equal the proper time for an IRF connecting thesame two points.

This is the same as in flat space (not spacetime), the shortest distancebetween two points is a straight line but there are infinitely many longercurved paths which can connect the two points. The same occurs here exceptthe metric throws in a twist. What we see is that the straight worldline has thelongest proper time. s2 = (ct )2 . Any other frame will require summing the

segments, s2 = (cti )2

i=start

end

(xi )2 . Since we are subtracting off the

spatial part we see that it will be less than the straight worldline.Again, the straight worldline has the longest proper time. It is important to note that the total proper time along aworldline is an invariant. It has the same value in all IRFs, (since it is essentially the interval).

What we see is that if we have a set of identical siblings and all but one sets off on a space journey and theyall meet later, the one with the straight worldline, (the one who stays at home), will be the oldest.

Consider a particle in deep space and picture its worldline between two events. If no outside forces act on thisparticle, (it’s a free particle) it will not accelerate. Hence in special relativity it will have a straight worldline. Hence itwill follow a worldline of maximal aging. This is raised to a principle, The Principle of Maximal Aging. This is notonly true in deep space (in an IRF) but also in a region near a gravitating mass. In such regions we will need to useGeneral Relativity to examine systems. This principle provides a bridge between SR and GR.

The proper time (cumulative interval) is a fundamental method of comparing different worldlines betweenevents.

One last point to make is regarding the stretch factor and IRFs. Remember that we found that the time dilatesbetween two uniform moving frames.

t = t’Now if one of these is the proper frame (indicating the proper time by ) we see we can define as follows,

=t

=1

1 v2

c2

=timein frameproper time

(Again this is for IRFs, if the worldline is curved we will need to cut up the worldline into segments where eachsegment is approximately an IRF). We can easily find the stretch factor if we know the proper time. Say a rocket goesby and you see 1 second go by on its clock while yours ticks off 3 seconds. Well then = 3s/1s = 3, from which youcan invert to find the speed of the other clock.

13= 1 2 1 / 9 =1 2 v

c=

89= 0.94 . Hence can be used as a measure of speed.


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Transforming Between Frames, Spacelike Separated Events

We continue our discussion of spacetime maps and translating betweenreference frames. Examine the following plot for IRFs all observing thesame interval between two events. This line gives all of the timelikeseparated frames.[The following refers back to the discussion in Box L-1 of your text, pg108].Remember back to our discussion of the object moving faster than the speedof light, discussed in Box L-1 of your text. The squared interval between thethrow and reception is negative,

Lab frame.x = 0 x’ = 3 lyt = 0 t’ = 1 ly from which s2 = c2 (1 y) 2 – (3 ly)2= -8 ly2.

This was the same in the shuttle’s frame (traveling at 0.6 c),

x = 0 x’ = 3 lyt = 0 t’ = -1 y from which s2 = - 8 ly2.

And now consider a frame traveling at 0.3c then,

x = 0 x’ = 8 lyt = 0 t’ = 0 y from which s2 = - 8 ly2. The events occursimultaneously in this frame.

We see that for spacelike separated events there is an invariant hyperbolalying outside the light cone along the x axis. And note now thattransforming between these frame the temporal order between events canswitch. (And for timelike separated events the spatial order can switch).

For timelike and spacelike separated events there is a barrier, thespeed of light, or the lightlike events.

So that the type of separation between events is invariant. (timelike for one,timelike for all).

Timelike (interval)2 +Spacelike (interval)2 -Lightlike (interval)2 0

In three dimensions the lightcone creates a barrier, a naturalpartitioning of spacetime.

s2 = c2t2 - x2 -y2 -z2.Note that spacelike separated events can not be causally related, (no information can pass from one event to the other).

Lecture 5 The Lorentz Transformation -...

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