1. Advanced EM - Master in Physics 2011-2012 1 Motion in special relativity We have seen “the composition of velocities” in various forms, perhaps without having clearly explained what it is about: it is about “which velocity sees an observer in a certain IRF (Inertial Reference Frame) of an object moving with velocity v’ in an IRF’ moving with velocity β wrt IRF. We have obtained it in a simple way via the LT and again in a more complicated way using only the invariance of “ C ” in changing IRF. We have looked at both cases of the two velocities being parallel and having an angle. The formulas we found are For the case of parallel velocities For orthogonal velocities Some consequences of the formula for parallel velocities: Case 1) v’ : =1. The object moving in IRF’ is a light pulse, moving with velocity=1. Then the resulting velocity is: We “find” back as a result the condition we used as a postulate at the beginning of the derivations: the speed of light in vacuum “ C ” is the same in all IRFs.

2. Advanced EM - Master in Physics 2011-2012 2 Case 2: large velocities, very near to 1 (parallel velocities) We shall write them like this: Then: In words, adding two velocities very near – but not equal to – the speed of light we obtain a velocity nearer – but not equal to that of light. Case 3: Velocities very low wrt the velocity of light. Then: And we find again, as an approximation for velocities much, much smaller than “ C ”, the good, old formula for addition of velocities of the galilean relativity.

3. Advanced EM - Master in Physics 2011-2012 3 Motion in special relativity and… the Paradoxes Let us now consider motion within the special relativity. As we have seen, the main effect of special relativity – the composition of velocities – tends to the galilean relativity for low speeds. The effects of special relativity become apparent only at velocities tending to the speed of light, or at least not many orders of magnitude below it. It is therefore natural to study movement in the context of interstellar trips… large distances and, necessarily, large velocities. So, let us consider a star at large distance from the Earth: Canopus at 99 light years. A spaceship leaves the Earth towards Canopus at velocity of 0,9802 (this also means that we have adopted a unit system in which ‘ c ’=1. Since we use for lengths the unit “light-years”, times are measured in years). With these values for distance and velocity, the trip will take 101 years, as measured in the reference system of the Earth, of course. The LI is then 400 years 2, and γ =5.05. And, since the LI is then 400 years 2, the proper time, which is its root, is 20 years. Which is the trip duration as measured in a reference frame in which the spacecraft is at rest. The reference frames of choice to describe this trip are therefore 2: the IRF of the spacecraft and that of the Earth – and, for that matter, also of Canopus, which we assume to have no movement relative to the Earth. Then to describe the motion we need some numbers: the distance (101 years), and the velocity ( β or γ ) or, the LI between the two events: departure from the Earth and arrival on Canopus.

4. Advanced EM - Master in Physics 2011-2012 4 A word on the “infrastructure”, i.e. the reference system, with all its coordinate axes and all its clocks: this physical system we can consider as a linear system, in the sense that the trajectory of the spacecraft is a straight line, which is also covered at constant speed. We do not need then all this system of tridimensional coordinate axes, we only need one axis, equipped with its clocks of course, but one axis; or, better, two axes, one for each reference frame (Earth and spacecraft), the two axes moving with speed 0.9802 wrt each other. This, of course, for the trip Earth to Canopus. For the return trip, Canopus to Earth, the spacecraft frame will be different from the previous one, since the direction of motion is reversed. With a little imagination it is not difficult to imagine the three rigid bars sliding fast next to each other, each equipped with a clock every so many light-years. In fact, a good analogy is with two trains, each many wagons long, crossing at “high” speed. Because, after arriving at Canopus, the spacecraft does a few revolutions around the star, at non-relativistic speed (we can approximate β ≈ 0), the crew take pictures of the star, of whatever else is there, souvenir photos. And then they head for the Earth, accelerate briefly to γ =5.05 and - 20 years later on their clocks – they are back and land on Earth. For them – who saw the system Earth-Canopus travelling at speed 0.9802 for 20 years, the Earth-Canopus distance is 19.6 light-years. They land on Earth – 40 years after leaving it. They left the Earth when they were 30 years old. They are eventually back – and they are 70. But this is nothing compared to what they find: on Earth, on the Earth clocks, since they left 202 years ago, it is another world. This is actually the Twin Paradox.

5. Advanced EM - Master in Physics 2011-2012 5 The Twin Paradox. Two twins were born the same day, but one of them took the same sort of relativistic trip to somewhere and back that has been just described, and when they meet again on Earth the travelled twin is much younger than the twin who always stayed home. Now, let us consider the following argument which, rather than an argument is a description of how the whole process is seen by a member of the crew: at departure, the Earth starts travelling in the direction opposite to Canopus and at the same time Canopus is noticeably starting to approach the spacecraft. The trip time as measured on the spacecraft is 20 years but, measured on Earth, should be 4 years. Then, another long, boring 20 years pass with nothing happening beside the Earth and Canopus moving until the Earth eventually arrives at the spacecraft. The crew reckons that the proper time seen by people on the Earth should be in total 8 years. Therefore, at reunion time of the crew with their old friends, the crew should be about 32 years older than their friends; and not 162 years younger…. So, at this point, the problem is: which of the two possible solutions is true; and why? Well, the fact that this happens in the space-time complicates things. Let us take a three-dimensional example. Let suppose a car A goes from Ferrara to Paris along a virtual freeway which is a straight line between the two cities, while another car B, more conventional, follows the existing highway passing through the Monginevro tunnel. The first car will have to travel for 1000km, while the second, straightening a bit its trajectory, will cover 600 km until the Monginevro pass, make a turn there and then cover another 800km to arrive to Paris. The second car has travelled 400km more than the second one.

6. Advanced EM - Master in Physics 2011-2012 6 Only notice that, considering only the trips in space of the two cars, the driver of A has seen car B going off for a while and then coming back; and this is the same in reverse that B has seen of car A. Superficially, it seems that the trip of B as seen by A is just the opposite than the trip of A as seen by B. So B may think that, since while he was driving straight to Paris A took a detour right he has run a longer distance. But the fact is that the wheels will have made more turns in one well defined case, there is no confusion possible: the car which has made more kilometers is the one which went through the existing roads. And he ran a longer distance because it is this car who made a turn, and therefore covered two straight stretches; its trip is NOT an INERTIAL frame. The same happens for the Canopus trip: it is the spacecraft who made a turn, not the Earth-Canopus system: there is not only the relative motion to take into account, but also – in the Canopus case – that one has to consider motion in an Inertial Reference Frame. And why then the people on the spacecraft, who actually made the turn, are at the end of the trip younger instead of older? In the car example the car who took the turn covered a longer distance. Well, this is a direct consequence of that MINUS SIGN in the Lorentz invariant! That issue settled (or at least it seems) there is still another problem: if we take only one trip into account, p. ex. the trip to Canopus, we have found that when the spacecraft arrives to Canopus its clock shows 20 years. But reversing the point of view, the observer on the spacecraft sees the system Earth-Canopus moving at constant speed. Therefore, he sees the clocks in the Earth IRF indicate a time a factor 5.05 shorter than the 20 years his own clock shows, i.e. 3.96 years! Then if we repeat the same argument for the return trip, we come to the conclusion that we have a situation worse than a paradox: The Earth observer (E) sees the whole trip last 202 years, while the Spacecraft crew (S) sees it last 40 years but, according to them, for the Earth observer the 2 stretches last 3.96 years each…. Well, the issue is complicated. We put it aside for now, and discuss first various aspects of space travel.

7. Advanced EM - Master in Physics 2011-2012 7 Another example: Caesar’s death The data: Julius Caesar died in 44 A.C., in Rome. Let us choose that place and date as the origin of our “Earth” IRF. Coordinates: [0,0] NOW, year 1956, is a student on Earth doing this exercise. This is event A [t A,x A ]=[2000, 0]. NOW, event B: a spacecraft travels in the Andromeda galaxy, 210 6 light-years away: B = [2000, 210 6 ] Units are obviously years and light–years. The question: How fast must the Spacecraft travel (in the Earth IRF) so that B is simultaneous (in the spacecraft’s IRF’) with Caesar’s death? The question is really paradoxical. No wonder, simultaneity is involved: and especially… simultaneity of events which are very far away in space. Well, first we remind that we consider IRF and IRF’ to have the same origin, in time and space: the event at the origin is the death of Caesar on the Earth. Then we request We have the two events A and B defined. What we have to find is the β of IRF’ (the spacecraft) wrt IRF that makes

8. Advanced EM - Master in Physics 2011-2012 8 We know the coordinate of event B in IRF (Earth’s). We need those of event B in IRF’ (spacecraft’s). Then we write the Lorentz transformation from IRF to IRF’ with IRF’ moving away from IRF in the direction of positive x. From these equations we will obtain the value of β and its sign (is the spacecraft travelling towards the Earth or away from it?). The condition we have is LT. From it we obtain: The events whose coordinates in IRF satisfy this condition - which is obviously a straight line in the space-time plot, passing through the origin – are all simultaneous (in IRF’) with the death of Caesar.! Now, we do know the coordinatesand replace their values [2000, 210 6 ] in the previous equation to find So, this is the answer: the spacecraft is moving away from the Earth – because β has turned out to be positive – and has the value 10 -3. Now, here we have met the “line of simultaneity” -in IRF’!- with event B plotted in IRF spacetime plot.

9. Advanced EM - Master in Physics 2011-2012 9 Now, here we have met the “line of simultaneity” -in IRF!- with event B plotted in IRF spacetime plot. Events which are simultaneous with any event have this property – we have learnt that – only in one IRF, in other IRFs they are no longer simultaneous. The simplest thing is to plot them in the same IRF plot in which they are simultaneous: the simultaneous events are those who have the same time and any space coordinate. In the spacetime plot they are a straight line parallel to the origin. They also obviously are spacelike points (more precisely, any pair of points on that line is a spacelike pair). In another IRF these events are no longer simultaneous, and therefore do not lay on a straight line parallel to the space axis. But they do lay on a line passing through the event they are simultaneous with in IRF. In our case, since that event is the origin, which is also the origin of IRF’, (the two systems have the same origin in space and time), that line satisfies the equation: In this case β obviously is not a velocity, just the slope of a simultaneity curve whose points, as was to be expected, are all spacelike. Event A: Caesar’s death time space Event B Line of simultaneity in Earth’s frame Line of simultaneity in spacecraft’s frame

10. Advanced EM - Master in Physics 2011-2012 10 The spacetime plot So far we know that events happen in the spacetime. Motion in the spacetime plot – a two-dimensional plot in which on the abscissa is indicated a space coordinate, usually the most significant for the system under study, and on the ordinate the time. Motion of an object – at least motion in one direction only – is indicated on this plot as a line; if this line is straight, the motion is uniform. But before studying motion with the spacetime plot, let us study the representation of the Lorentz Invariant LI. For a given (timelike) pair of events it takes the value LI = τ², where τ is the proper time of event pair. When we change frame, the coordinates of the distance between those points change, but stay on a line of constant ( t 2 –x 2 ) which is a hyperbola. It is obvious that the frame in which the object lays on the time axis and does not move has the least value of time (proper time). X (m) Time (m) τ This plot shows how change the coordinates of an event (x,t) when we change the frame from which we observe it.

11. Advanced EM - Master in Physics 2011-2012 11 In a spacetime plot we can draw the line which represents the motion of an object… With the limitation that what is represented here is the motion along one coordinate only. The slope of the curve (see plot) is the velocity β of the motion of new IRF wrt the old one. The motion of an object being limited to speeds lower than 1, the line is limited to slopes lower than 1: a slope of precisely 1 indicates the speed of light. In the figure some straight lines are shown: they indicate objects which travel with uniform motion. Note that all the motion lines have slopes (|dx/dt|, NB!!) smaller than 1. The lines at ±45°, which correspond to the speed of light and mark the separation between timelike and spacelike events, are shown as thicker lines. X (m) Time (m) τ ψ

12. Advanced EM - Master in Physics 2011-2012 12 X (m) Time (m) O B In this plot 2 possible routes from event O (the origin) to event B are shown. The blue line represent an object which does not move, just lets the time go by: we have in fact chosen to plot the coordinates in an IRF in which B and O are in the same position. The red line represents an object which moves: accelerates along some direction, then slows down and eventually goes back to B. Note that, in order to satisfy the requirement of an object never to reach the velocity of light in vacuum, it is not sufficient that the line be contained in the timelike region: what is necessary is that the slope never exceeds 1. These lines (like the blue and the red) are called worldlines. Now, imagine such a plot to represent the x and y coordinates of an object moving. The 2-dimensional distance (distance in space) of the red curve is obviously “longer” than that of the blue curve: in ordinary space, the straight line is the shortest line between two points. This is not true –in fact, the opposite is true- in spacetime (quasi-euclidean space), because of that Minus sign in the Lorentz Invariant.