Special Relativity - QMII - Mechina

Special Relativity - QMII - Mechina 2016-17 Daniel Aloni Disclaimer This notes should not replace a course in special relativity, but should serve as a reminder. I tried to cover as many important topics as I could but clearly not everything. If topics are unfamiliar it is recommended to read one of the references below or any other relevant literature. 0 References There are thousands of references about the subject. I found that many books about General relativity include good explanations in their first chapters. Other sources are advanced books on mechanics and electromagnetism. Here is a list of some good examples (to my opinion) which covers the subject from those different points of view. In each of them look for the relevant chapters. 1. Classical Mechanics, H. Goldstein. 2. Classical Electrodynamics, J. D. Jackson. 3. Gravitation and cosmology, S. Weinberg. 1 Motivation One of the main topics of the course QMII is to develop a (special) relativistic treatment of QM, which is done in the framework of quantum field theories. We will learn how to quantize (relativistic) scalar and fermionic fields, and about their interactions. For this end a basic knowledge in special relativity is needed. 1

2 Indexology We will find that index notation is the most convenient way to deal with vectors, matrices and tensors. For a vector v = (v 1 v 2 v 3... v n ) T we denote the i s component by v i. m 11 m 12... For a matrix M = m 21 m 22... we denote the (ij) s entry by M ij...... Note that M ij M ji, while M ij = (M T ) ji. When we multiply a vector by a matrix from the left we get a new vector v = M v. The i s component of the new vector is given by v i = (M v) i, where (M v) i is just the following sum M ij v j. j From now on we will use Einstein summation convention 1. If an index appears twice we sum over it M ij v j j M ij v j. 2. Index will never appear more then twice. What about multiplying by a matrix from the right v T M? Again we get a new vector v T = v T M. In component notation [ ] v T = [ v T M ] = i i j ( vt ) j (M T summation convention ) ji = v j (M T ) ji = v j M ij = M ij v j = v i. We see that the i s component is insensitive to the transpose operation. Here is an example why this is so useful. exercise - Prove that ( A) = ( A) 2 A. proof - by using component notation [ ( A) ] = ɛ ijk j ( A) k = ɛ ijk ɛ klm j l A m = ɛ ijk ɛ lmk j l A m i = (δ il δ jm δ im δ jl ) j l A m = i m A m 2 A i = [ ( A) ] 2 A. 2 i

3 Fast introduction to special relativity (SR) The theory of Special relativity (SR) tells us how to relate observables seen by different inertial observes, namely it is a set of transformation rules. Both Galilean transformations and SR act on space-time. This is a vector space which can be labeled as (t, x). An event is a point, and a path of a particle is a curve called world-line. Figure 1: World lines In Fig. 1 we show world-lines in Newtonian dynamics in the left panel and SR in the right panel. The differences are crucial: Galilei: Time is universal and time slicing is well-defined. All inertial observers agree on the same time slicing. A trajectory must cross each slice once (and only once), but the crossing point can be at any point, depends on the observer. SR: The speed of light c is finite and maximal, and all inertial observers measure the same velocity of a light wave front. The velocity of a massive particle is limited, v < c in all frames of reference. At each point of the trajectory one can draw a light-cone, and all inertial observers will agree that the trajectory is within this light-cone. 3

3.1 Transformation rules Let us derive the transformation rules of SR Imagine two systems O, O with respective velocity v between them. Figure 2: Two systems We choose x = x = 0 at t = t = 0. Assume the same wave front in both systems x 2 = ct 2, x 2 = ct 2 (1) We take the transformation to be linear x = γ[x ṽt], where γ and ṽ will be found below. The physical reason is that we want O T1 O T2 O to be identical to O T1 +T 2 O (you can compare it to rotations) 1. The origin x of O in the O system is given by x = vt and hence: 0 = γ [v ṽ] t ṽ = v x = γ[x vt]. (2) We can find the opposite transformation by changing the sign of v, namely x = γ [x + vt ]. Plugging x into x gives: t = γt + (1 γ2 )x. (3) γv 1 We will make this statement more precise once we will study group theory. 4

Now we demand that a wave front in O i.e. x = ct is also a wave front in O, i.e. x = ct (Here we demand that the speed of light is the same for all observers). By using Eqs. (2),(3) we get: [ x = ct γ[ct vt] = c γt + (1 ] γ2 )ct γv [ γ[c v] = c γ + (1 ] γ2 )c γ 2 v 2 = (1 γ 2 )c 2 γv ) (1 v2 γ 2 = 1. (4) c 2 So we conclude that γ = 1. (5) 1 v2 c 2 Using this transformation rules we can show that while the different inertial observer will determine different world-lines for the same particle, they will measure and agree that (from now on we set c = 1) t 2 x 2 = t 2 x 2. (6) Now we can make it infinitesimal dt 2 d x 2 = dt 2 d x 2 ds 2. (7) ds is called the interval and it is a Lorentz scalar (i.e. remain the same in all inertial frames of reference). 5

4 The metric, inner product, raising & lowering indices 2 Four vectors - In special relativity we put space ( x) and time (t) in the same four vector x µ = (ct, x). From now on c = 1. Metric - We define the metric as symmetric function which takes two vectors into R g(v µ 1, v2) ν = g(v2, ν v µ 1 ) R. (8) Note that we did not define g to be positive definite. In special relativity the metric is given by η µν = diag(1, 1, 1, 1). Some authors are using η µν = diag( 1, 1, 1, 1). The metric with upper indices is identical η µν = η µν = diag(1, 1, 1, 1). Summation convention: 1. If an index appears twice, once as a lower and once as an upper index, we sum over it. 2. An index will never appear twice as a lower/upper index. 3. An index will never appear more then twice. We raise and lower indices by using the metric v µ = η µν v ν If x µ = (t, x) x µ = (t, x). We define the interval as the following scalar ds 2 = η µν dx µ dx ν (9) The modern way to define SR is the other way around, namely we start from the interval, and the transformation rules between different frames of reference are those which leaves the interval invariant. 2 Some of the definitions in this section can be generalized, but we neglect it for the sake of simplicity. 6

5 Interval, rotations, boosts and all of that... The allowed transformations x µ x µ = Λ µ νx ν are those that leaves the interval ds 2 invariant η µν dx µ dx ν = η µν Λ µ µ dx µ Λ ν ν = ( Λ T ηλ ) dx dxν µ dx ν (10) µ ν So Lorentz transformations are given by all transformations that preserve the metric. Lets count d.o.f: Λ is a 4 4 real matrix 16 parameters. η µν = ( Λ T ηλ ) µ = ν gives 4 eq. µν µ ν gives 6 eq. 16 4 6 = 6 so in general we have 6 continuous parameters. How to we interpret them? cosh(η x ) sinh(η x ) 0 0 Three boosts, e.g. Λ = sinh(η x ) cosh(η x ) 0 0 0 0 1 0 0 0 0 1 1 0 0 0 Three rotations, e.g. Λ = 0 1 0 0 0 0 cos(θ x ) sin(θ x ) 0 0 sin(θ x ) cos(θ x ) This gives Λ(η i, θ i ). We will come back to that when we will get to the subject of Group theory. Defining tanh(η x ) = vx, we get the same transformation between two c inertial observer that we developed in the old-fashion point of view. 7

6 Example - The pole and the Barn paradox Consider the following (Fig. 3): A pole, which within its rest frame (O ) has a length L pole = 10 m. A barn, which within its rest frame (O) has a length L barn = 5 m. We throw the pole into the barn with a velocity β = 3/2. γ = (1 β 2 ) 1/2 = 2. Figure 3: Pole and barn. The first question is how do we measure the length of the pole in the barn rest frame. The way we measure length is to determine the x coordinate of the head and tail of the pole at the same time. We choose the origin of O and O to determine the head of the pole( and ) 0 the entrance of the barn at t = t = 0. This is the point y = y = 0 of Fig. 3. The point x µ (see figure) is the length of the pole in the barn rest ( frame. ) 0 By definition its time component is set to zero, so x µ =. l pole Here l pole stands for the yet unknown length of the pole in the barn rest frame. 8

Now we determine the same point in the pole rest frame. Since both systems share the same origin by definition the space component of the tail of the pole is minus its length (recall ( that the ) pole does not move within its rest frame...). Therefore x µ = is yet unknown. t 10. Note that the time Using our transformation rules from previous section x µ = 0 l pole 0 = t 10 ( ) 3 2 x µ = t l pole = 20 x µ 0 = 3t 5 10, and l pole = 5. We see that the pole is twice shorter in the barn system. It means that at time t f = 5m we can shut the door and lock the pole within the barn. β The paradox - We can repeat the exercise in the pole rest frame and find that the barn is twice shorter in this frame. How come that the pole can enter into a barn which is four times shorter from it? Solution - In the pole system the event of the head of the pole hitting the end of the barn occurs at t f = 2.5/β. On the other hand, the event of ( ) closing the door is given in the barn rest frame by z µ = 5/β 0 (see figure). At the pole rest frame this event is given by [ ] z µ = Λ 1 µνz ν γ 5 β 0 ( ) β [ ] 10/β not the same time =. γ 0 β 5 10 right place β We see that in the pole rest frame we shut the door at 4 t f. It means that the tail of the pole has precisely enough time to reach the door of the barn. 9

7 Hyparbolic structure of space time Recall that ds 2 = dt 2 dx 2, and all inertial observers measure the same value of ds 2. We can plot curves of constant values of ds 2. A curve determine the same event as observed by different inertial observers which have the same origin. We distinguish between three different scenarios: light-like - For light ds 2 = 0 dx = ±1. Determine the dashed curves dt of Fig. 4. All inertial observers with same origin will determine the same curve. time-like - ds 2 > 0. Determine the top and bottom parts of Fig. 4. Here causal order is well defined, i.e. all observers will agree which event happened before another event. space-like - ds 2 < 0. Determine the right and left parts of Fig. 4. Here causal order is not well defined, namely events which are space like cannot affect each other. t x Figure 4: Minkowski space-time As an example consider two dwarfs sitting in rest on the x axis, one in x = 1, and one in x = 2. They set their clocks in advance and decided to 10

push a button at t = 0. Another dwarf which sit at rest at the origin will say that they pushed the button at the same time and mark the blue squares of Fig. 4. A dwarf in a spacecraft which is flying to the right direction (and pass through x = 0 at t = 0) will say that the left dwarf (x = 1 in rest frame) pushed the button before the right dwarf (x = 2 in rest frame). She will mark the green squares of Fif. 4. Finally a dwarf in a spacecraft which is flying to the left direction (and pass through x = 0 at t = 0) will say that the left dwarf (x = 1 in rest frame) pushed the button after the right dwarf (x = 2 in rest frame). She will mark the red squares of Fif. 4. 8 Lorentz scalars, more 4-vectors and a bit of electromagnetism In order to get scalars we contract all indices by using the metric. A B η µν A µ B ν = A µ B µ A B = η µν Λ ν ν Λµ µ A µ B ν = ( Λ T ηλ ) µ ν A µ B ν = A B. (11) Let us consider few important examples related to important four vectors. 8.1 Four-momentum The four momentum is given by p µ = (E, p), (12) where E is the energy and p is the three momentum. The scalar which is obtained by taking the square is the mass square of the particle p µ p µ = m 2. (13) 11

The four-momentum is related to the four velocity by p µ = mu µ, where U µ dxµ ds = γ(1, v). Note that for light γ, but m 0, so the four 2 momentum is well defined and E = p. 8.2 Derivatives, currents and EM (Jackson 11.6,11.9) Another important four-vector is the four derivative α It transforms as α = = Λ β α β, as expected (see Jackson). x α = xβ x α x β x α = ( t, ). The four dimensional Laplacian α α scalar, and hence also the wave equation. = 2 t 2 is a Lorentz Recall the continuity equation ρ t + J = 0, which must hold at any frame of reference. We can define the four-current J µ = (ρ, J). Using this definition the continuity equation is clearly satisfied at all frames and can be written as µ J µ = 0. 8.2.1 Electromagnetism In Lorentz gauge the wave eqs. for scalar and vector potentials are 2 A 2 t A = 4πJ, 2 2 φ 2 φ = 4πρ, t 2 Gauge: t φ A = 0. It is just natural to define the four potential A µ = (φ, A). Then A µ = 4πJ µ wave eq.. µ A µ = 0 Gauge E = A The electric and magnetic fields are given by φ t B = A. 12

You can check that the electromagnetic tensor is given by 0 E x E y E z F αβ α A β β A α = E x 0 B z B y E y B z 0 B x. (14) E z B y B x 0 If you want, you can do the following exercises 1. Find F αβ = η αµ η βν F µν. 2. Convince yourself that F µν F µν is a Lorentz scalar. 3. (annoying) Find the transformation rules of E and B by using F µν = Λ µ µ Λ ν ν F µ ν. Most of the time in relativistic theory we are using A µ and F µν, and not E and B. 8.3 Fourier transform Just as last comment, we will make use of the relativistic Fourier transform e ixµ p µ, where the exponent x µ p µ is another useful Lorentz scalar. 13