Changing coordinates - the Lorentz transformation

Transcription

1 Changing coordinates - the Lorentz transformation In figure 5, we have two observers, A and B, and the two photons passing through the event E are shown. Each observer assigns coordinates to the event E. How are they related? If we label t a as the midpoint of the indicated segment, then, as mentioned above, the photons intersect at t a +x a and t a x a. Similarly for B. The lower triangle gives t b x b = k(t a x a ), while the upper one gives t a + x a = k(t b + x b ). Rearranging, we get Now t b = x b = ( ) ( ) + k 2 k 2 t a + x a ( ) ( ) k 2 + k 2 t a + x a k 2 = + v v k2 + = v 2,

2 so we get, after a llittle algebra, t b = v 2 (t a vx a ) and similarly, x b = v 2 (x a vt a ). This linear transformation between the coordinates of A and B is called a Lorentz transformation. Note that, when v is small (relative to c), these read, respectively, t b t a x b x a vt a, the Newtonian approximation. (Remember that when we go back to normal units, both v and x get divided by c, so t a vx a becomes t a (vx a )/c 2 t a, when v << c. Note that observer B s world line, given by x b = 0, has the equation x a = vt a according to A (which is correct). And the constant time surface t b = 0 has the equation x a = (/v)t a. The constant time surfaces of any observer are obtained by reflecting the observer s world line through the light rays! This is how we see it from our 2-dimensional Eucliean viewpoint. In fact, the angle made by a world line and a constant time surface is not an observable, so it s physically meaningless - it s just an artifact of the non-physical Euclidean viewpoint. 2

3 Proper time and the invariant interval Given 2 events, let the coordinate differences between them be ( t a, x a ) for observer A, and similarly for observer B. Using the Lorentz transformation,we compute t 2 b = x 2 b = v 2 ( t2 a 2v t a x a + v 2 x 2 a) v 2 ( x2 a 2v t a x a + v 2 t 2 a). Subtracting and simplifying gives t 2 b x2 b = t2 a x 2 a. This quantity, often written as τ 2 is invariant in the following sense: Every observer computes τ 2 for E and E 2 in his own coordinate system (inertial frame). Everyone gets the same answer. It is the relativistic analog of the quantity s 2 computed in Cartesian coordinates in Euclidean space. It has a simple physical meaning: Suppose the two events E and E 2 lie on the world line of observer A. Then for these two events, x a = 0, and t a = τ. That is, the time between the two events, measured by an observer present at both events, is τ, which is also called the proper time. Exercise: Show that an observer B, who is not present at both events, and who therefore has a nonzero velocity relative to A, will necessarily find the time interval between them to be strictly greater than τ. Notice that there s a rather prominent minus sign in the expression for τ 2, so in spite of the way we write it, this quantity may be positive, negative or zero. The geometry described by a space with this sort of invariant is decidedly non-euclidean. 3

4 Definition: A vector in space time is called Timelike if τ 2 > 0 Null if τ 2 = 0, and Spacelike if τ 2 < 0 This is an invariant notion in SR, since it is defined in terms of the invariant τ 2. And this exhibits an immediate difference between spacetime and Euclidean space - there is nothing like spherical symmetry about the origin. Exercise: Pick an origin and an observer A. Find the locus of all events with τ =, and t a > 0. How far is it possible to go (relative to A) if you re going really fast? Example - the observation of µ-mesons at the surface of the earth We can make elementary particles called µ-mesons in various nuclear experiments on earth. These particles (muons ) are unstable, and decay into electrons and neutrinos with a half-life of seconds. This means that half of any existing population of muons decays during any interval of duration seconds. Muons are also created by collisions of cosmic rays with gas nuclei in the upper atmosphere; the energy of the cosmic rays is substantial and the newly minted muons move at speeds just fractionally less than the speed of light. They are observed on the earth s surface by the tracks they make in bubble chambers. The muons are created at an altitude of approximately 60 km. Moving at about the speed 4

5 of light, they will take 60 km/300,000 km/sec = seconds to reach the earth s surface. Approximately what fraction of these muons make it to the earth s surface? Ignoring relativity, we note that in the seconds that pass in the laboratory frame between the creation of the muons and their observation in the lab, 00 = (2 0 4 )/(2 0 6 ) half lives have passed. Therefore, the fraction of the created muons arriving at the earth s surface is ( ) Thisi is a ridiculously small number; according to this computation, we shouldn t observe any muons! But of course we do, and the reason is that the time delay between the two events E (creation of the muon at a height of 60 km) and E 2 (observation of the muon at the earth s surface) is different when measured in the rest frame of the muon instead of the rest frame of the laboratory. And the muon is going to live its life in its own rest frame. In the laboratory frame on earth, we have x e = 2 0 4, t e = x e /v, where v is the velocity of the muon relative to the lab. In the muon s rest frame, we have x µ = 0, and therefore, t µ = τ, and so by invariance of the proper time interval, we can write τ = t µ = ( t e ) 2 ( x e ) 2 = x e v v 2, where we ve made the substitution t e = x e /v. We ask: at what velocity v must the muons be moving in order that /8 of the muons created actually make it to the surface of the earth? We ll have the fraction /8 5

6 remaining if the time measured in the muon rest frame is 3 half-lives or less, So we want to find the value of v such that t µ Since we know x e /v 2 0 4, we have v 2, and we find that v This is well within the range of velocities measured for muons arriving at the earth s surface, so it all works out correctly using special relativity. The twin "paradox" Example: A and B are twins; A stays in the solar system, while B heads off to Alpha Centauri (the nearest star, 4 light-years away) at a speed of.95c. Once there, having forgotten his towel, B immediately turns around, and returns to the solar system at the same speed. How much time has passed for each of the twins? Solution: The spacetime diagram for this setup is shown in the figure. Observer A s world line is approximately that of the Sun which is also (approximately) the closest thing we have in the solar system to an "inertial observer". The very small deviations are immaterial. The events E and F, denoting the departure and subsequent arrival of B are on A s world line. In this problem, rather than centimeters, we ll use light years as the unit of time. So A sees B going at v =.95 for a distance of x = 4; this takes, according to A, the time t = 4/.95 = 4.2 years, and the return trip takes the same time. According to A, his twin has been gone a total of 8.42 years. 6

7 However, if G is the event "B arrives at Alpha Centauri", then the coordinates of G in A s frame are (assuming E to be the origin) (t a, x a ) = (4.2, 4.0). But the line joining E to G is B s world line, so according to B, the time that s passed when he arrives there is τ = t 2 a x 2 a = =.724 years. So when B arrives back only years have passed for him. A is nearly 6 years older! Is there something wrong with this calculation? No. But in prehistoric times, this was known as the "twin paradox". It was thought to be a paradox because while it s true that B is in motion relative to A, it s also true that A is in motion relative to B; therefore, their situations should be symmetric, and they should both age by the same amount. What s wrong with this argument? (Exercise!) Here s one of the concrete examples of the fact that we re dealing with non-euclidean geometry: in spacetime, a timelike straight line is the longest distance between any 7

8 two of its points. To phrase this precisely, suppose E and E 2 are two events such that the vector joining them is timelike. Then there s an observer present at both events, say observer A. We use A s coordinate system in the following. We suppose the coordinates of E, E 2 are (t, 0), (t 2, 0). Let γ(t) be any piecewise smooth curve such that (a) the vector dγ/dt is timelike, and (b) γ(t ) = E, γ(t 2 ) = E 2. That is γ is the world line of an observer (in general not an inertial observer) present at both E and E 2. Now, using calculus, the proper time observed by γ is obtained by first approximating the curve by a polygonal line, adding up the proper times for each segment to get an approximate total time (a Riemann sum), and then taking a limit (Riemann integral) to get ( τ) γ = γ dτ = γ t2 dt 2 dx 2 = ( (dx/dt) 2 dt. t (This is just like the computations for arc length ds in Euclidean space, except for the minus sign.) Now ( (dx/dt) 2 = v 2. Here v depends on t, of course, but it s still true that v 2 < unless v = 0. So we have ( τ) γ = t2 t2 v 2 dt dt = t 2 t = ( τ) A, t t with equality holding only if v = 0 everywhere along γ. That is, equality holds only if γ is the world line of A. 8