10 Lorentz Group and Special Relativity

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1 Physics 129 Lecture 16 Caltech, 02/27/18 Reference: Jones, Groups, Representations, and Physics, Chapter Lorentz Group and Special Relativity Special relativity says, physics laws should look the same for different oservers in different inertial reference frames. In the non-relativistic setting, the coordinates of different reference frames are related y the Euclidean transformation. In particular, if two reference frames S and S coincide at t = 0 and are moving with relative velocity v = (v, 0, 0). Then the relation etween the coordinates of an event in the two reference frames is t = t, r = r vt ( x = x vt, y = y, z = z ) (1) Under such a transformation the spatial distance etween two points remains invariant ( r 1 r 2) 2 = ( r 1 r 2 ) 2 (2) In special relativity however, we need to use the Lorentz transformation and the replace the aove relation with t = γ(t xv/c 2 ), x = γ(x vt), y = y, z = z (3) where γ = c c. This particular transformation induced y a relative velocity is called a oost. 2 v 2 The transformation may look complicated, ut it is designed so that the velocity of light remains invariant in all inertial reference frames. Suppose that we send out a light signal from the origin at t = 0 in the x direction. In reference frame S, at a later time t, the signal has travelled to point x = ct, y = 0, z = 0. Transformed to the reference frame of S, we find that the coordinate of the corresponding signal is c v c v t = t c + v, x = ct c + v, y = 0, z = 0 (4) Therefore the velocity of light in the frame of S is also c. This is just one particular example of the whole group of Lorentz transformation, which we are going to study in detail elow Coordinate four vector The fact that time and space gets mixed together under Lorentz transformation motivates the definition of the space-time four vector x µ, µ = 0, 1, 2, 3 for the Minkowski space x 0 = ct, x 1 = x, x 2 = y, x 3 = z (5) Notice that the components of the four vector all have the dimension of length. 1

2 While usual Euclidean transformation leaves the length of the three vector (x, y, z) invariant, Lorentz transformation leaves the interval x 2 = c 2 t 2 x 2 y 2 z 2 (6) invariant. (This ensures that the speed of light remains invariant in all reference frames.) We define a metric tensor, g µν such that g µν = 0 if µ ν and g 00 = g 11 = g 22 = g 33 = 1. That is, g = (7) Then we can define the length of the four vector as x 2 = x T g x (8) This is very different from the metric we are used to. In the usual Euclidean space, g = is positive definite. The metric for the Minkowski space is not positive definite and will result in some special properties of the Lorentz group. Suppose that under a Lorentz transformation, the four vector transforms as Then the invariance of the length of the vector requires that Because this is true for all x, we have Λ T gλ = g. x = Λ x (9) x 2 = x T g x = xλ T gλ x = x T g x = x 2 (10) Notice that if g = I 3, this condition reduces to the orthogonality condition of three dimensional rotation transformations which form the group SO(3). The Lorentz group can e thought of as the group of orthogonal transformations on a space with metric g = diag( 1, 1, 1, 1) and it is denoted as SO(3, 1) Lorentz Transformations What kind of Λ satisfies the aove condition? First, any spatial rotation involving x 1, x 2, x 3 keeps the length of the four vector invariant. Therefore, the spatial rotation transformations SO(3) forms a sugroup of the Lorentz group. The transformation matrices take the form Λ n r (θ) = 0 0 R n (θ) 0 (11) 2

3 where R n (θ) is the three dimensional special orthogonal matrix representing the spatial rotation. This sugroup of transformations is generated y X 1 = i, X 2 = i , X 3 = 0 0 i 0 0 i 0 0 (12) 0 0 i 0 0 i A different kind of Lorentz transformations which do involve time are the oosts. Boost in the x direction gives rise to the transformation γ γv/c 0 0 Λ x(v) = γv/c γ (13) c where γ =. If we define γ = cosh ζ = eζ +e ζ c 2 v 2 2, so that tanh ζ = sinh ζ cosh ζ = v c, then Λ x(v) can e re-written as cosh ζ sinh ζ 0 0 Λ x(ζ) = sinh ζ cosh ζ (14) One can explicitly check that ( Λ x(ζ) ) T gλ x (ζ) = g. The infinitesimal generator for x direction oost is 0 i 0 0 Y 1 = i dλ x(ζ) = i dζ ζ= (15) By exponentiating Y 1, we can recover Λ x(ζ) Λ x(ζ) = e iζy 1 (16) However, notice one important difference with the generator for SO(3): ζ is not ounded. As v approaches c, ζ approaches +. Therefore, the Lorentz group SO(3, 1) is not compact. This has a series of consequence. One of them eing that the finite dimensional representations of SO(3, 1) are no longer unitary. Similarly, oosts in y and z directions are generated y 0 0 i i Y 2 = i 0 0 0, Y 3 = (17) i Boosts in an aritrary direction n can e otained first y rotating n to x axis, applying the oost in x direction and then rotating ack. In general, an aritrary Lorentz transformation contains oth spatial rotation and oost. The whole group is generated from X 1, X 2, X 3 and Y 1, Y 2, Y 3. The Lie algera of SO(3, 1) is a six dimensional real vector space with commutators [X a, X ] = iɛ ac X c, [X a, Y ] = iɛ ac Y c, [Y a, Y ] = iɛ ac X c (18) 3

4 Comments: (1) While X 1, X 2, X 3 is closed under commutation, Y 1, Y 2, Y 3 are not. Therefore, the oosts do not form a sugroup. (2) While X 1, X 2, X 3 are Hermitian, Y 1, Y 2, Y 3 are anti-hermitian. Therefore, the representation is not unitary (the oost transformation is not unitary). (3) The first and second commutation relation says that X 1, X 2, X 3 transform as a vector under SO(3), so do Y 1, Y 2, Y 3. (4) We can make linear cominations etween X and Y In terms of X ± a, the commutation relations ecome [X a, X ] = iɛ ac X c X (±) a = 1 2 (X a ± iy a ) (19), [X ( ) a ] = iɛ ac X ( ) c, [X a ] = 0 (20) That is, the set of six generators reak up into two susets, such that each suset is equivalent to the Lie algera of SU(2) and the two susets are independent of each other Irreducile representations The four dimensional matrices Λ provides one possile representation of SO(3, 1) while the group is astractly defined as that satisfying the same composition rule as the Λ s. In terms of Lie algera, the group is defined as that with a six dimensional Lie algera, satisfying the commutation relation [X a, X ] = iɛ ac X c, [X a, Y ] = iɛ ac Y c, [Y a, Y ] = iɛ ac X c (21) or [X a, X ] = iɛ ac X c, [X ( ) a ] = iɛ ac X c ( ), [X a ] = 0 (22) Now we can try to see what irreducile representations do SO(3, 1) have. Following our analysis of SO(3), in order to find irreps for a Lie group, we can try to find the irreps for its Lie algera, ut with the danger that we get the irrep of the covering group (SU(2) for SO(3)). Things work in a very similar way for SO(3, 1). We see that the Lie algera of SO(3, 1) contains two SU(2) part. Therefore, its irrep can e laelled y (j 1, j 2 ), where j 1, j 2 are integer or half-integer. The representation is then (2j 1 + 1)(2j 2 + 1) dimensional. The generators are From which we get X + a = J j 1 a I 2j2 +1, X a = I 2j1 +1 J j 2 a (23) X a = J j 1 a I 2j I 2j1 +1 J j 2 a, Y a = i(j j 1 a I 2j2 +1 I 2j1 +1 J j 2 a ) (24) If we then take the exponential, we can recover the group (or its covering group). 4

5 Let s see some example irreps. (1) j 1 = 0, j 2 = 0 This is the trivial representation. It is one dimensional, with all the generators eing 0 and all the group elements eing represented y 1. In quantum field theory, this representation is carried y a relativistic scalar field (e.g. Higgs field). (2) j 1 = 1/2, j 2 = 0 This is called a spinor representation. It is two dimensional. Correspondingly X + 1 = σ x, X + 2 = σ y, X + 3 = σ z, X 1 = 0, X 2 = 0, X 3 = 0 (25) X 1 = σ x, X 2 = σ y, X 3 = σ z, Y 1 = iσ x, Y 2 = iσ y, Y 3 = iσ z (26) The Lorentz transformations are then parameterized y six real numers θ 1, θ 2, θ 3, φ 1, φ 2, φ 3 Λ( θ, φ) = e i( θ X+ φ Y ) = e i( θ i φ) σ (27) Note that this is different from the SU(2) group which contains matrices e i θ σ. In fact, the set of matrices generated are the group of special (determinant 1) linear (invertile) matrices of dimension 2, SL(2, C). SL(2, C) is a covering group of the SO(3, 1) group as 2π spatial rotation results in I instead of I. In quantum field theory, this representation is carried y the Weyl fermion. (3) j 1 = 0, j 2 = 1/2 This is another spinor representation with generators X 1 = σ x, X 2 = σ y, X 3 = σ z, Y 1 = iσ x, Y 2 = iσ y, Y 3 = iσ z (28) This is inequivalent to the previous representation ecause they cannot e related y a asis transformation. In quantum field theory, the direct sum of the j 1 = 1/2, j 2 = 0 representation and the j 1 = 0, j 2 = 1/2 representation is carried y the Dirac fermion. (4) j 1 = 1/2, j 2 = 1/2 This is a four dimensional representation. Actually, it is exactly the four dimensional representation which we used to define SO(3, 1). That is, the space time four vector transforms with this representation. 5