ETHAN N. CARRAGHER AND ANTHONY G. WILLIAMS

Solutions Manual for

Introduction to Quantum Field Theory:

Classical Mechanics to Gauge Field Theories

ETHAN N. CARRAGHER AND ANTHONY G. WILLIAMS 1 / 4

• Lorentz and Poincar´e Invariance

1.1 Problem 1

Problem:

A muon is a more massive version of an electron and has a mass of 105.7 MeV/c 2 . The dominant decay mode of the muon is to an electron, an electron antineutrino and a muon neutrino,µ − →e −

• ¯νe+νµ. If we haveN(t) un-

stable particles at timetthen the fraction of particles decaying per unit time is a constant, i.e., we havedN/N=−(1/τ)dtfor some constantτ. This gives dN/dt=−(1/τ)N, which has the solutionN(t) =N0e −t/τ where we haveN0un- stable particles att= 0. The fraction of particles decaying in the intervalttot+dt is−dN/N0= (−dN/dt)dt/N0= (1/τ)e −t/τ dt. So themean lifetime(orlifetime) is R ∞

t(−dN/dt)dt/N0=τ R ∞

xe −x dx=τ, wherex=t/τ. Thehalf-life,t 1/2, is the time taken for half the particles to decay,e −t

1/2/τ

= 1 2 , which means that t 1/2=τln 2. Thedecay rate, Γ, is defined as the probability per unit time that a particle will decay, i.e., Γ = (−dN/dt)/N= 1/τis the inverse mean lifetime. A muon at rest has a lifetime ofτ= 2.197×10 −6 s.Cosmic raysare high-energy par- ticles that have traveled enormous distances from outside our solar system.Primary cosmic raysare particles that have been accelerated by some extreme astrophysi- cal event andsecondary cosmic raysare those resulting from collisions of primary cosmic rays with interstellar gas or with our atmosphere. Most cosmic rays reach- ing our atmosphere will be stable particles such as photons, neutrinos, electrons, protons and stable atomic nuclei (mostly helium nuclei). Muon cosmic rays there- fore are secondary cosmic rays produced when primary or secondary cosmic rays collide with out atmosphere. A typical height in the atmosphere for the production of cosmic ray muons is∼15 km. What is the minimum velocity that this muon be produced with in order that it have a 50% chance of reaching the surface of the Earth before decaying?

Solution:

A stationary observer on Earth will see the time experienced by a muon traveling at speedvto be dilated by a factor ofγ= (1−v 2 /c 2 )

−1/2

compared to their own.So according to the observer, the half-life of the muon will beγ t 1/2. If the muon travels at a speed that allows it to traverse theL= 15 km of the atmosphere in this time, it will therefore have a 50% chance of reaching the Earth’s surface. That

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2Lorentz and Poincar´e Invariance is, the speed of the muon must satisfy v= L γ t 1/2 = q 1− v 2 c 2L t 1/2 (1.1) to have a 50% chance of reaching the surface before decaying. This can be rearranged to give v= L q t 2 1/2 + L 2 c 2

.(1.2)

Now from the given information, the half-life of the muon is t 1/2=τln 2 = (2.197×10 −6

• ln 2 = 1.523×10

−6 s, (1.3) so the necessary speed is v=

15×10

3 m q

(1.523×10

−6 s) 2 +

(15×10

3 m) 2

(3×10

8 ms −1 ) 2 = 299,653,700 ms −1 = 0.9995c.(1.4) An inertial observer traveling with the muon will see the height of the atmosphere contracted by a factor ofγ, which at this speed gives a height of L γ = r 1− v 2 c 2 L = p

1−0.9995

2

(15×10

3 m) = 456 m.(1.5) In other words, from the muon’s point of view, it only needs to travel 456 m to reach the Earth’s surface.

1.2 Problem 2

Problem:

The diameter of our Milky Way spiral galaxy is approximately 100,000 - 180,000 light years and our solar system is approximately 25,000 light years from the center of our galaxy. Recalling the effects of time dilation, approximately how fast would you have to travel to reach the center of the galaxy in your lifetime? Estimate how much energy would it take to accelerate your body to this speed.

Solution: 3 / 4

3Problem 3 From the solution to Problem 1.1, the speed needed for an observer to travel a proper distanceLin a timeTis v= L q T 2 + L 2 c 2

.(1.6)

So for a human with a lifetime of, say,T= 80 years, to travel to the center of the Milky WayL= 25,000 light years = 25,000cyears away, a speed of v= 25,000cyears p (80 years) 2

• (25,000 years)

2 (1.7) = 0.999995c(1.8) would be required. The kinetic energy possessed by a human of massm= 70 kg at this speed is Ekin= (γ−1)mc 2 = ` 1 √

1−0.999995

2 −1 ´ (70 kg)(3×10 8 ms −1 )

= 2×10

21

J.(1.9)

For reference, it would take humanity slightly more than 3 years to consume this much energy at current rates.

1.3 Problem 3

Problem:

Consider two events that occur at the same spatial point in the frame of some inertial observerO. Explain why the two events occur in the same temporal order in every inertial frame connected to it by a Lorentz transformation that does not invert time. Show that the time separation between the two events is a minimum

in the frame ofO. (Hint: Consider Figs. 1.2 and 1.5.)

Solution:

Since the two eventsE1andE2occur at the same spatial point in frameO, the displacement vectorE2−E1between them will point entirely along the time axis in this frame, in the positive direction, say. Under any given Lorentz transformation, displacement vectors will be moved along hyperbolae in a spacetime diagram, as illustrated in Fig. 1.1. Further, Lorentz transformations that do not invert time will only move displacement vectors alongbranchesof those hyperbolae. Our vector of interestE2−E1lies on a positive-time branch, so under an orthochronus Lorentz transformation it will remain on that positive-time branch. That is, the temporal order of the events remains the same. And the point on such a positive-time branch that has the smallest time component is the one that lies on the time axis, asE2−E1

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