Problem

Problems are listed in approximate order of difficulty. A single dot (•) indicates straigh...

Problems are listed in approximate order of difficulty. A single dot (•) indicates straightforward problems involving just one main concept and sometimes requiring no more than substitution of numbers in the appropriate formula. Two dots (••) identify problems that are slightly more challenging and usually involve more than one concept. Three dots (•••) indicate problems that are distinctly more challenging, either because they are intrinsically difficult or involve lengthy calculations. Needless to say, these distinctions are hard to draw and are only approximate.

••• In Problem 1 the minimum (or threshold) energy Ei for a proton incident on a stationary proton to produce an antiproton in the reaction (18.46) is shown to be 7mc2. A more sophisticated and slicker way to derive the same result uses the Lorentz transformation for energy of a particle. This was derived in Problem 4, and the relevant part of the result is this: If a particle has energy E and momentum p in a frame S, then its energy in a second frame S′ (traveling with velocity v along the x axis) is

E′ = γ(E − vpx)

Consider the process (18.46) as seen in a frame S in which the incident protons approach one another along the x axis with equal but opposite momenta. (This frame is called the center of momentum — or CM — frame.) In this CM frame it is obvious that the minimum total incident energy is 4mc2, for which all of the final particles are at rest. (a) Write down the energy and momentum of each of the incident protons in this frame. (b) The experiment is usually carried out in the lab frame S′ where one of the incident protons is at rest. Using the Lorentz transformation for energy, find the energies of the two incident protons in the lab frame, showing that one is mc2 and the other 7mc2.

Problem 1

•• The first antiprotons to be observed were produced in the reaction (18.46) (Problem 2), with one of the initial protons at rest. The minimum value of the incident kinetic energy Ki needed to induce the reaction is called the threshold energy for the reaction and is surprisingly large compared to 2mc2. To calculate this threshold energy, note that the minimum of Ki occurs when the final kinetic energy is as small as possible, to maximize the fraction of Ki available to create particles. This requires that all final particles have the same velocity, so that there is no kinetic energy “wasted” in relative motion. Write down the Pythagorean relation for the total energy and momentum of the final particles. (Use the result of Problem 3.) Then use conservation of energy and momentum to rewrite this relation in terms of the energy and momentum, Ei and pi of the incident projectile. Finally, use the Pythagorean relation for Ei and pi to eliminate pi. Show that Ei = 7mc2 and hence Ki = 6mc2.

Problem 2

•• The first antiprotons to be observed were produced in the reaction

which required a minimum kinetic energy of 6 GeV. Below are listed three reactions that would require less energy, if they were possible. Explain why none of them is possible.

(a)
(b)
(c)

Problem 3

• Suppose that n particles all have the same rest mass m and all move with the same velocity u. Prove that their total energy and total momentum satisfy the Pythagorean relation Etot2 = (ptotc)2 + (Mc2)2, appropriate to a single particle of rest mass M = nm.

Problem 4

••• (a) Suppose that a mass m has momentum p and energy E, as measured in a frame S. Use the relations (2.60) and (2.61) (Problem 5) and the known transformation of dr and dt to find the values of p′ and E′ as measured in a second frame S′ traveling with speed υ along Ox. (Notice that apart from some factors of c, the quantities p and E transform just like r and t. Remember that dt0 has the same value for all observers.) (b) Use the results of part (a) to prove the following important result: If the total momentum and energy of a system are conserved as measured in one inertial frame S, the same is true in any other inertial frame S′.

Problem 5

We saw that the relativistic momentum p = γmu of a mass m can also be expressed as

where dt0 denotes the proper time between two neighboring points on the body’s path and has the same value for all observers. Show that the relativistic energy E = γmc2 can similarly be rewritten as