Homework Assignment #1, due Fri. Jan 18, 2002.

Read Chapters 1 and 2 of MTW.

1. Exercise 2.5 in MTW.

a) -p.u is a scalar, with the same value in all frames. In the rest frame of the observer u^mu = (1,0,0,0), u_mu = (-1,0,0,0), so -p.u = p^0 = E in the observer's frame.

b) Since p.p is a scalar, m^2 = -p.p is the rest mass in any frame, including the observer's frame.

c) p.p = p^2 - E^2 in any frame. In the observer's frame (p.u)^2 = E^2. Therefore, [(p.u)^2 + p.p] = p^2 in the observer's frame.

d) Since p = gamma*m*v and E = gamma*m, v = p/E.

e) In the observer's frame p + (p.u)u = (E, p_x,p_y,p_z) + (-E)(1,0,0,0) = (0,p_x,p_y,p_z). Dividing by (-p.u) = E gives (0,v_x,v_y,v_z).

2. Exercise 2.7 in MTW.

a) Do the matrix multiplication.

b) An observer at rest in the primed frame has t' varying while x^j' is constant. The inverse Lorentz transformation gives t = gam*t' + beta*gam*n^jx^j', and x^j = beta*gam*n^j + ((gam-1)n^in^j + del^jk)x^k'. Then v^j = dx^j/dt = beta*gam*n^j/gam = beta*n^j.

c) Similar to b), but using the direct Lorentz transformation.

d) n^jn^k = 1 for j = k = 3, and zero otherwise. Thus Lambda^3'_3 = gam - 1 + 1 = gam, etc.

3. A large number N of observers are spread uniformly around a ring which is revolving with angular velocity \Omega about an axis through the center of the ring and perpendicular to the plane of the ring. In the inertial frame in which the center of the ring is at rest, the circumference of the ring is 2*pi*r.

a) Are these observers able to synchronize their clocks? Explain your answer.

The standard method of clock synchronization in special relativity is for Observer A to send a light signal to Observer B at time t1. The light signal is reflected by B and received back at A at time t2 measured by A's clock. B's clock is synchronized with A's if B's clock reads (t1+t2)/2 when the signal is reflected. Now consider the sequence of observers A1, A2, ..., AN evenly spaced around the ring in the direction of rotation, with spacing d = 2*pi*r/N in the inertial frame. Starting at A1 with t = t' = 0, where t is the inertial frame time, A2's clock (the leading clock) reads t' = 0 at time t = vd in the inertial frame. Continuing around the ring and synchronizing each pair of clocks in turn AN's clock reads t' = 0 at time (N-1)vd in the inertial frame and is clearly NOT synchronized with A1's clock. So while each pair of observers can synchronize their clocks in the standard way, the entire set of observers cannot. Of course, the clocks can be consistently synchronized in the inertial frame of the center of the ring, since they all run at the same time-dilated rate in the inertial frame.

b) Suppose one of these observers measures the times t+ and t- for light signals to propagate around the ring and back to him in the forward and backward directions (relative to the direction of rotation). Show how he can calculate \Omega and r from t+ and t-.

The light signal in forward direction takes a time 2*pi*r/(1-v) to return to the observer as calculated in the inertial frame (taking into account the the extra distance it has to go to catch up to the observer, where v = Omega*r. The observer's clock runs slow by sqrt(1-v^2) relative to the inertial frame, so t+ = 2*pi*r*sqrt((1+v)/(1-v)). Similarly, t- = 2*pi*r*sqrt((1-v)/(1+v)). Then 2*pi*r = sqrt(t+*t-). Since t+/t- = (1+v)/(1-v), v = (t+ - t-)/(t+ + t-) = Omega*r can be solved for Omega.

c) Is the ring Lorentz contracted? Discuss.

The PROPER distance between each PAIR of rotating observers is d/sqrt(1-v^2), as measured in the instantaneous comoving Lorentz frame of the pair. But this is based on standard clock synchronization, which cannot be maintained consistently all the way around the ring. Therefore, one can't really say that the circumference of the ring is Lorentz-contracted. The entire circumference is only meaningfully measurable in the global inertial frame.

d) Find the acceleration 4-vector for a point on the ring.

The 4-velocity of a point on the ring is (1,-v*sin(omega*t), v*cos(omega*t),0)/sqrt(1-v^2). The four-acceleration is the derivative of 4-velocity with respect to proper time tau = sqrt(1-v^2)*t. This is (0,-cos(omega*t),-sin(omega*t),0)(v^2/r)/(1-v^2).

4. Show that all vectors orthogonal to a non-zero null vector must be either spacelike or a multiple of the same null vector.

Transform to a frame in which the null vector reduces to (k,k,0,0). The dot product with an arbitrary 4-vector (a^t,a^x,a^y,a^z) is k*(a^x-a^t). This is zero only if a^x = a^t. But then a.a = (a^x)^2 + (a^y)^2 + (a^z)^2 - (a^t)^2 = (a^y)^2 + (a^z)^2 which is space-like (> 0) unless a^y = a^z = 0. Since a^t = a^x = a, a = (a/k) times the null vector if a is null.