Part 1: Boosts and rapidity For this part, we will need the definitions of the hyperbolic trig functions, sinh (pronounced "sinsh"), cosh (pronounced like it sounds), and tanh (pronounced "tansh"): enter sinh d = cosh d = tanh d = sinh o 2 (1) 2 cosh d epte 1.1. The hyperbolic trig functions obey similar identities to the ordinary trig functions, but with some extra minus signs. Prove the identities cosh' o- sinh = 1, cosh $+ sin = cosh(20). (2) 1.2. Define the rapidity @ in terms of the velocity 8 = v/c as follows: tanh o = . (3) (In other words, = tanh8 where tanhis the inverse function to tanh, just like tan is the inverse function to the ordinary trig function tan.) Show that in terms of rapidity, we can express the Lorentz transformation parameters y and ~ as y = coshd, 75 = sinho. (4) The easiest way to do this is just to write tanho in terms of exponentials. 1.3. The Lorentz transformation matrix for a boost along the x-axis can therefore be written cosh o sinho sinh of cosh o ) ( #). (5) Note how similar this looks to a rotation matrix! The essential difference is the absence of a minus sign and the replacement of ordinary trig functions with their hyperbolic cousins. Using the first identity you derived in question 1.1, show that this Lorentz boost matrix preserves the invariant interval: (6)1.4. Another extremely convenient fact about rapidities is that they make the velocity addition formula look easy. Show that if 8=les its: then 8 = tanh(@1 + $2) with #1 = tanho and 32 = tanh 02. In other words, velocity addition can be equivalently stated as rapidities add: =01 + $2. This is exactly the same thing that happens with 2-dimensional rotations! 1.5. Finally, rapidities are useful for understanding Minkowski diagrams. Show using the second of your identities derived in 1.1 that the angle # between the ct and ct' axes (and also the or and r' axes) satisfies tand = tanho, and that the length of a unit on the S' axes is Vizz = vcosh 20. Use the Taylor expansions of e and v1 + r to show that, to linear order in o, units in S are the same length as units in S'. Part 2: Commutation relations for Lorentz boosts Now that we understand boosts along the r-axis, let's bring back the boosts along the other coordinate axes as well. Keeping with the analogy to rotations, let's write these boosts in terms of rapidities: cosho sinho 0 0 cosho 0 sinho 0 \\ cosho 0 0 sinhd As(d) = sinh d cosh $ 0 0 0 , Ay(g) = 0 1 0 sinh $ 0 cosho 0 As(0) = 0 1 1 0 sinh d 0 0 coshd (7) (You should compare A, and Ay to the analogous matrices in Discussion 5 and make sure you understand where the nonzero entries go.) 2.1. In exact analogy to Part 2 of last week's Discussion 6, Taylor-expand the entries of A,((), Ay(e), and A,(e) to find A,(e) = I teke, Ay(e) = It cky, and A(e) = It EKs, where I is now the 4 x 4 identity matrix and 10 0 0 0 0 0 01 K, = 00 0 0 0 0 00 0 Ky K. = (8) 0 0 0 0 0 0 0 These are pretty easy to remember: the only nonzero entries are in the zeroth row and column, in the position labeled by the subscript (K, has the 1's in the r component, Ky has the 1's in the y component, and K, has the I's in the s component). To do the Taylor expansion, you can simply expand the exponentials in the definition of sinh and cosh as you did in question 1.5 of Part 1 above. 2.2. Compute the commutator [K2, Ky]. Is it equal to K-? If not, this means that boosts along different ares do not form a group! ! `Strictly speaking, we should actually check that [K, Ky] is not equal to a linear combination of any of the three generators Ke, Ky, K=. This is indeed the case.2.3. Rotations of the spatial coordinates do not change the time coordinate, by definition. This means that the rotation matrices from last week (which are 3 x 3 matrices) can be embedded in 4 x 4 matrices, for example 0 R (0) = 0 cos e - sine 0 sin e (9) cos # 0 Let this matrix act on a vector (of', s', y', ?') and show that the transformation is cf = off = 'cos 0 - y'sind, y = s' sind + y' cos0, = = 2', as expected for a rotation around the z-axis. 2.4. Compute the 4 x 4 analogue of the infinitesimal rotation /: from Discussion 6 by Taylor- expanding the 4 x 4 matrix R, above, and show that [Kr, Ky] as computed in question 2.2 is equal to -J. The interpretation is that two boosts along two different ares involve a rotation of the coordinate systems. You can't have Lorentz transformations without rotations! As with the rotation group, you can check the rest of the commutators to show that the commutator of a K and a J gives something proportional to K. This means that any combination of infinitesimal rotations (about any axis) and boosts (along any axis) is also a combination of a rotation and a boost. The combination of boosts and rotations is called the Lorentz group. The rotation group SO(3) is a subgroup of the full Lorentz group. Part 3: The Lorentz group SO (3,1) Recall in lecture that we introduced the metric 9= (10) which lets us encode the invariant interval as As' = guer"Ar". Just as orthogonal matrices pre- serve the length of a vector, Lorentz matrices preserve the invariant interval, which can be thought of as the relativistic "length" of the displacement 4-vector Ar". The mathematical expression of this fact is the equation 9 = A gA (11) which can be taken as a definition of the Lorentz group. If we replaced g with the 4 x 4 identity matrix, we would simply get / = A'A, which is satisfied by an orthogonal matrix. But since g has 3 minus signs and 1 plus sign, we call the Lorentz group SO(3,1). (Sometimes you'll see this written as SO(1,3); this is convention-dependent exactly like the sign of the metric.) 3.1. Show that the Lorentz matrix A, satisfies Eq. (11). 3.2. Compute Ar(1)Ay($2) in the rapidity parameterization and show that it also satisfies Eq. (11). At last, we have an answer to our question from Discussion 5: AyAy is indeed a member of the Lorentz group because it preserves the invariant interval, but it is not a boost: it is a combination of a boost and a rotation