1. Advanced EM - Master in Physics 2011-2012 1 SPACE TRAVEL REVISITED After we got ourselves familiar with the spacetime plot and the trips in spacetime, let us go back to the mysteries of the Canopus trip. We understood some of the values of time and distance covered that were calculated for the outbound trip: An observer on Earth knew that the distance Earth- Canopus was 99 light-years (l-y), and with his clock measured a time of the trip of 101 years. An observer on the spaceship measured a travel time of 20 years. This time, combined with a velocity of.9802, made a distance Earth-Canopus –in his IRF’!!- of 19.6 l-y But then, paradoxes arise when we consider the outbound trip as seen by the observers on the spacecraft. In particular, what is odd is the following: at departure from Earth an observer on the spacecraft sees the Earth go away and, at the same time, Canopus, at a distance of 19.6 starts to move at β = 0.9802 towards him. The observer on the spacecraft, who knows that Canopus is at a distance of 19.6 quickly calculates that a man on Earth (which goes at a high speed) should see the spacecraft arriving at Canopus at a time of 20/ γ = 3.96 years. Then, time passes and, when he arrives on Canopus, he looks at what time do the Earth clock show: the time is, as he expected, 3.96 years. But! But, but…. But, we started the example saying that the watches on Earth indicate a time of 101 years when the spaceship arrives on Canopus, because the spaceship covers 99 l-y with a speed of 0.98: what is wrong in this argument? Well; the first thing is that we must understand which is the event we want to measure now. We must specify its time and location.

2. Advanced EM - Master in Physics 2011-2012 2 We must define the event whose time we want to know. And, remember, we can use an indirect definition, as in the example of Caesar’s death: at which speed must the spacecraft go so that the time an observer on the spacecraft IRF’ in Andromeda at a time t=2000 (in IRF) sees Caesar’s death simultaneous with him in IRF’…. Which as the definition of an event is pretty complicated. So, let us focus on the trip to Canopus, and look at its spacetime plot, as seen in the Earth (and Canopus) IRF. The problem we have is the compatibility of the arrival to Canopus at t =101 with the statement that an observer in IRF’ (the spacecraft) looks –upon arrival at Canopus- at the clock on Earth and reads 20/ γ = 3.96 years Well, we have already treated this case,when (in lesson 14) we have looked at simultaneity in some detail. Now... the clock is on Earth: so, we are looking at the time in IRF; and on Earth. The next piece of information is: an observer in IRF’ upon arrival at Canopus. This is tricky, but: the taking of the Earth clock reading is simultaneous (for the observer on the spacecraft, then in IRF’!!!!!) with the arrival at Canopus. Then, we know how to deal with it: The event has the location on Earth and lays on the line of simultaneity (in IRF’) with the arrival at Canopus. We know how to draw that line in the spacetime plot in IRF: it goes through the event B (arrival at Canopus ) a has slope β. This line is shown in red, dashed line. Its equation is: O (Earth) 99 (Canopus) 101 202 time B C

3. Advanced EM - Master in Physics 2011-2012 3 Let us use again the model we had for the coordinate system of an IRF, with all the coordinate axes in space and the clocks over them. In these cases of interstellar trips, in which the travel is along one direction, we can use a much simpler representation of the IRF: one single bar, of length in principle infinite but in practice as long as is needed to contain the distances studied, with clocks at regular distances. If there are various velocities and two directions of motion we can use as many of these bars as needed, all moving wrt each other. To have a better model for such bars, let us imagine a train station, with a very long train standing still, and other trains going in the two directions with different velocities. In the middle of each wagon there is a clock. The distance L between two successive clocks is the same in all trains (in their own reference system). All the clocks in a train are, of course, synchronized between themselves. Now an event happens: the engine of train B, at its front, and which has IRF frame, arrives at a switch. At that moment (for IRF) all the clocks in train B show the same time. All observers in each wagon of train B “look at” the clocks on the train A at rest and, beside finding them not at distance L but L γ, they find them showing different times, staggered by βγ L. Now, in the spacecraft’s travel to Canopus and back, three IRFs are involved: one in which Earth – and Canopus – are at rest; one moving with the spacecraft with β =0.9802 towards Canopus; and a third IRF with the same speed β =0.9802 but direction from Canopus to the Earth. And let us figure the three IRFs as trains. If two events at different locations δ l are simultaneous in the train at rest, they will happen at times different by βγδ l in a moving train (i.e, IRF’). Let us draw again the spacetime plot of the Canopus trip.

4. Advanced EM - Master in Physics 2011-2012 4 O (Earth) 99 (Canopus) 101 202 time B C D In the previous slide we discussed the various ways to look at the travel to Canopus. Of course, there is also the trip back. Starting from Canopus (point B in the plot) and returning to Earth, at (its) time 202 l-y. The trip, as seen in IRF, is the same as the outward journey, but with velocity reversed: in the formulae, with a minus sign, in the plot, its mirror image wrt the line (101 – B). In such case too, an observer on the spacecraft sees the Earth suddenly moving towards him with speed 0.982 from a distance of 19.6 l-y. He therefore concludes that, since it will take the Earth 20 years to get to him, for an observer on the Earth the trip duration will be 20/ γ = 3.96 years. The duration... The duration is the difference between two times: the arrival at Earth minus the departure from Canopus – both seen on an Earth clock, but by a spacecraft observer. Well, when he is on Earth, he reads its time. But when he is on Canopus, at departure, when he reads the time on Earth…. He is actually reading on a simultaneity line (shown again in red in the plot). And this line is the mirror image of the red dashed line we drew before. So the value he reads is 202-3.96= 198.04years! Let us see that better: when he arrives at Canopus, he looks the Earth clock: it shows 3.96 years. Then he turns around and starts immediately towards Earth, and he sees that the time on Earth has jumped to 198.04 years. Why is that? That is because he has jumped another IRF”, which differs by 2 β from the previous IRF’.

5. Advanced EM - Master in Physics 2011-2012 5 More on SpaceTime Spacetime is a 4-dimensional space, defined as the extension of ordinary 3-dimensional space by addition of the dimension “time”. This because the units of time turned out to be compatible with being the same as those of space. There is a difference of course, a mathematical difference, beside those of our perception of time and space: the different sign in the expression of the Lorentz Invariant. In a linear space the points (called events in our case) are identified by a set of coordinates (4 real numbers), which are the components of the vector “point position” along the unitary vectors m i of the coordinate axes. Ordinary 3-dimensional space is called “Euclidean space”. Its peculiarity is the existence of a quadratic form between points which is defined as the (distance) 2. In the 4- dimensional spacetime an invariant quadratic form is defined, similar to that of the 3-dimensional space, with the important difference that the signs of the squares of spaces coordinates are different from those of the time one. For this reason the space is called “pseudo-Euclidean”. The choice of the coordinate axes is arbitrary, as long as they are orthogonal. Given a reference frame, if I want to change it for another I will have to change the coordinates of each event: this can be done with a linear combination of the old coordinates. In 3-dimensional space the change from a reference system to another corresponds to a Rotation of the coordinate axes. And, here we come to the point, in a ref. system we can plot where the new axes are.

6. Advanced EM - Master in Physics 2011-2012 6 x y y’ X’ This is the representation of the new coordinate axes in the old reference frame: they have equations y’=0 (new axis x’) x’=0 (new axis y’) In the spacetime however this simple geometric representation is lost: in large rotations it would lead to situations in which time runs backwards! But the linear combination between time and space coordinates is still valid, as we have seen with the LT. And it corresponds to a change from a IRF to another in uniform motion wrt the previous frame. Which leads us to say that a LT represents the rotation between two inertial frames in uniform relative motion. And…… the LT has coefficients which are very different from the usual sines and cosines of an ordinary rotation, and it contains also coefficients larger than 1. Now, if we refer to spacetime, which are the new axes t’ and x’ in the x,t spacetime plot? They still are the two axes x’=0 and t’=0. And, given the form of the LT:

7. Advanced EM - Master in Physics 2011-2012 7 x y x’ t’ Light cone (β=1) X’’ t’’ In this figure are shown – in a spacetime plot in IRF – the coordinate axes x’ and t’ in IRF’, which moves with positive β wrt IRF; and the inverse transformation (x”, t”). The transformation seems odd, since the two axis t’ and x’ are NOT orthogonal. Well, they are! But… but not in this page, in the Minkowski space instead! We can calculate the scalar product of the vectors (Origin to a point on x’, with coordinates [ βx and x] ) and (Origin to a point in t’, with coordinates [ t and βt] ). The scalar product of the two is: The scalar product of the vector to any point on axis t’ with the vector to any point on axis x ’ is null. Therefore, (NB: with this definition of scalar product!) the two axes are orthogonal. We shall now discuss the handling of non-orthogonal coordinate axes, axes such that at least for some i, j the following Eq. is valid: Where the m i are the unitary vectors of the axes. α tg α=β

8. Advanced EM - Master in Physics 2011-2012 8 Oblique axes Since some scalar products between the unitary vectors along the coordinate axes are not zero, we shall use them to build a matrix The elements g i,j define a Symmetrical Tensor which is called Metric tensor. It gives a representation of the obliquity of the reference frame. Now, there are two ways –once we have defined a reference system- to define in it the components of a vector A. First definition: the components defined with parallel projections on coordinate axes are called contravariant components A I. x1x1 x2x2 The scalar products of a vector A with the unitary vectors m i are the covariant components of the vector. A1A1 A2A2

9. Advanced EM - Master in Physics 2011-2012 9 This is a formula that allows to calculate the covariant coordinates of a vector once the contravariant coordinates of that vector are known. After that, we can define the inverse tensor [g i,j ]: From this equation we can obtain the formula for the change from covariant to contravariant components of a given vector. The scalar product We can use this formula to calculate the norm of a vector: N.B.: if the axes (be they oblique or orthogonal) are straight, the g ij are constant; if the axes are also orthogonal, then And the covariant components are equal to the contravariant ones.

10. Advanced EM - Master in Physics 2011-2012 10 After the generalities on oblique coordinates, let us see how they apply to our 4-dimensional space or “Minkowski space”. A point in space has 3 coordinates x, y, z. The event which we obtain by considering what happens at point [x,y,z] at time t has the following contravariant coordinates: [x 0 =t, x 1 =x, x 2 = y, x 3 = z]. The metric vector has therefore the form As a consequence, the covariant coordinates are: [x 0 =t, x 1 =-x, x 2 =-y, x 3 =-z] These two definitions are necessary in order to have the quadratic form which is the invariant in Minkowski space.

11. Advanced EM - Master in Physics 2011-2012 11 Much in the same way, we can convert well known vectors in 3- dimensional space into so-called four-vectors in spacetime. We start with a 3-vector A [A x, A y, A z ] and rewrite as a 4-vector with components: contravariant [ A i ]= [ A 0, A 1 =A x, A 2 =A y, A 3 =A z ] and covariant [ A i ]= [ A 0, A 1 =-A x, A 2 =-A y, A 3 =-A z ] The change of coordinate axes from IRF’ to IRF can be written in matrix form: We have applied the change of coordinates only to the new coordinates: of course, it also applies to the components of a quadrivector (in the formula above, replace the coordinates x with the components B i of vector B ).

12. Advanced EM - Master in Physics 2011-2012 12 One last remark before embarking into rewriting the physics formulas to incorporate spacetime and all that? So far we have used scalars, vectors, tensors in ordinary space. It is clear that the objects that are invariant in spacetime are scalars -and that creates no problems- quadrivectors, quadritensors,… Here comes the principle of relativity, as restated by Einstein: “the equations must have the same form in all reference systems”. It implies that the transformation properties must be the same on both sides of the ‘equal’ sign, and both have well defined transformation properties, such as are satisfied by the objects of special relativity i.e. scalars, quadrivectors etc. So, if we have an ordinary vector with three components, how can we create a quadrivector out of it? In the previous slide we have talked about such a “creation”: there is no problem for the 3 space components, they are precisely the components of the 3-vector, with maybe the ‘minus’ sign in front, it depends on our components being contravariant or covariant. But the time components? Well, we will have to find out what “ A 0 ” is case by case, from the physics that the quadrivector represents. In fact, this is how was found that the 3-vector “position in space” is actually a quadrivector, and that “time” is its “ x 0 ”.