Problem 68
Many of the stars in the sky are actually \(binary\space stars\), in which two stars orbit about their common center of mass. If the orbital speeds of the stars are high enough, the motion of the stars can be detected by the Doppler shifts of the light they emit. Stars for which this is the case are called \(spectroscopic\space binary\space stars. \textbf{Figure P37.68}\) shows the simplest case of a spectroscopic binary star: two identical stars, each with mass \(m\), orbiting their center of mass in a circle of radius \(R\). The plane of the stars' orbits is edge-on to the line of sight of an observer on the earth. (a) The light produced by heated hydrogen gas in a laboratory on the earth has a frequency of 4.568110 \(\times\) 10\(^{14}\) Hz. In the light received from the stars by a telescope on the earth, hydrogen light is observed to vary in frequency between 4.567710 \(\times\) 10\(^{14}\) Hz and 4.568910 \(\times\) 10\(^{14}\) Hz. Determine whether the binary star system as a whole is moving toward or away from the earth, the speed of this motion, and the orbital speeds of the stars. (\(Hint\): The speeds involved are much less than \(c\), so you may use the approximate result \(\Delta f/f = u/c\) given in Section 37.6.) (b) The light from each star in the binary system varies from its maximum frequency to its minimum frequency and back again in 11.0 days. Determine the orbital radius \(R\) and the mass \(m\) of each star. Give your answer for \(m\) in kilograms and as a multiple of the mass of the sun, 1.99 \(\times\) 10\(^{30}\) kg. Compare the value of \(R\) to the distance from the earth to the sun, 1.50 \(\times\) 10\(^{11}\) m. (This technique is actually used in astronomy to determine the masses of stars. In practice, the problem is more complicated because the two stars in a binary system are usually not identical, the orbits are usually not circular, and the plane of the orbits is usually tilted with respect to the line of sight from the earth.)
Problem 69
In high-energy physics, new particles can be created by collisions of fast- moving projectile particles with stationary particles. Some of the kinetic energy of the incident particle is used to create the mass of the new particle. A proton-proton collision can result in the creation of a negative kaon (\(K^-\)) and a positive kaon (\(K^+\)): $$p + p \rightarrow p + p + K^- + K^+$$ (a) Calculate the minimum kinetic energy of the incident proton that will allow this reaction to occur if the second (target) proton is initially at rest. The rest energy of each kaon is 493.7 MeV, and the rest energy of each proton is 938.3 MeV. (\(Hint\): It is useful here to work in the frame in which the total momentum is zero. But note that the Lorentz transformation must be used to relate the velocities in the laboratory frame to those in the zero-totalmomentum frame.) (b) How does this calculated minimum kinetic energy compare with the total rest mass energy of the created kaons? (c) Suppose that instead the two protons are both in motion with velocities of equal magnitude and opposite direction. Find the minimum combined kinetic energy of the two protons that will allow the reaction to occur. How does this calculated minimum kinetic energy compare with the total rest mass energy of the created kaons? (This example shows that when colliding beams of particles are used instead of a stationary target, the energy requirements for producing new particles are reduced substantially.)
Problem 70
(a) Consider the Galilean transformation along the \(x\)-direction : \(x' = x - vt\) and \(t' = t\). In frame \(S\) the wave equation for electromagnetic waves in a vacuum is $${\partial^2E(x, t)\over \partial x^2} - {1 \over c^2} {\partial^2E(x, t)\over \partial t^2} = 0$$ where \(E\) represents the electric field in the wave. Show that by using the Galilean transformation the wave equation in frame \(S'\) is found to be $$(1 - {v^2 \over c^2} ) {\partial^2E(x', t') \over \partial{x'^2}} + {2v \over c^2} {\partial^2E(x', t') \over \partial{x'} \partial{t'}} - {1 \over c^2} {\partial^2E(x', t') \over \partial{t'^2}} = 0$$ This has a different form than the wave equation in \(S\). Hence the Galilean transformation \(violates\) the first relativity postulate that all physical laws have the same form in all inertial reference frames. (\(Hint\): Express the derivatives \(\partial/\partial{x}\) and \(\partial/\partial{t}\) in terms of \(\partial/\partial{x'}\) and \(\partial/\partial{t'}\) by use of the chain rule.) (b) Repeat the analysis of part (a), but use the Lorentz coordinate transformations, Eqs. (37.21), and show that in frame \(S'\) the wave equation has the same form as in frame \(S\): $${\partial^2E(x', t') \over \partial{x'}^2} - {1 \over c^2} {\partial^2E(x', t') \over \partial{t'}^2} = 0$$ Explain why this shows that the speed of light in vacuum is c in both frames \(S\) and \(S'\).
Problem 71
An airplane has a length of 60 m when measured at rest. When the airplane is moving at 180 m/s (400 mph) in the alternate universe, how long would the plane appear to be to a stationary observer? (a) 24 m; (b) 36 m; (c) 48 m; (d) 60 m; (e) 75 m.
