Summary Classical Field Theory (Instructor Solution Manual, Solutions) Series: Author(s): Joel Franklin.

Classical Field Theory Solution Manual 1st Edition
Joel Franklin




Chapter 1

Special Relativity

Problem 1.1
From the definitions,
1 2η
cosh2 η = e + 2 + e−2 η


4 (1.1)
1 2η
2
e − 2 + e−2 η ,


sinh η =
4
so that cosh2 η − sinh2 η = 1.

Problem 1.2
The only difference between L and L̄ is the direction of the relative motion –
from L̄’s point of view, L is moving to the left with speed v, so we can just
take v → −v in the Lorentz boost:
 v 
c t = γ c t̄ + x̄ x = γ (v t̄ + x̄) . (1.2)
c


Problem 1.3
From tanh η = −v/c, and using cosh2 η − sinh2 η = 1, we have:
sinh η v
p =− , (1.3)
2
1 + sinh η c

and solving for sinh η gives
v
sinh η = ± q c = ±β γ. (1.4)
v2
1− c2


5

,6 CHAPTER 1. SPECIAL RELATIVITY

Using this we can find cosh η:
1
q
cosh η = 1 − sinh2 η = q = γ. (1.5)
v2
1− c2



Problem 1.4
Using the transformation directly, we have: c t̄ = γ (c t − β x) and x̄ = γ (−β c t + x),
so that
−c2 t̄2 + x̄2 = γ 2 −c2 t2 − β 2 x2 + 2 v t x + β 2 c2 t2 + x2 − 2 x v t


(1.6)
= γ 2 1 − β 2 −c2 t2 + x2 = −c2 t2 + x2 .




Now taking t̄ = t, x̄ = x cos θ + y sin θ and ȳ = y cos θ − x sin θ, we have:

−c2 t̄2 + x̄2 + ȳ 2 = −c2 t2 + x2 cos2 θ + 2 x y sin θ cos θ + y 2 sin2 θ



+ y 2 cos2 θ − 2 x y cos θ sin θ + x2 sin2 θ


(1.7)
= −c2 t2 + x2 + y 2 .


Problem 1.5
Length contraction says that ∆x = γ −1 ∆x̄, and we are given,
∆x 12
= = γ −1 , (1.8)
∆x̄ 13
5
so that v = 13 c is the speed of L̄.

Problem 1.6
In L, the trip took: ∆t = ∆x/v = 5 m/(12c/13) ≈ 1.81 × 10−8 s. In L̄, the
rest frame of the clock, we have:
h v i
c∆t = γ c ∆t̄ + ∆x̄ (1.9)
c
with ∆x̄ = 0, so that ∆t̄ = γ −1 ∆t, and
s  2
−8 12
≈ 6.96 × 10−9 s.


∆t̄ ≈ 1.81 × 10 s 1− (1.10)
13


Problem 1.7

, 7

From time dilation: ∆t = γ ∆t̄, and γ = 5/4, so that for ∆t̄ = 2 years, we
have ∆t = 2.5 years

Problem 1.8
We know that the interval: ∆s2 = −c2 ∆t2 + ∆x2 has the same value in all
frames related by a Lorentz boost, so that if ∆s2 ≤ 0 in L, we also have
∆s̄2 ≤ 0 in L̄, but that means that:

−c2 ∆t̄2 + ∆x̄2 ≤ 0 −→ c2 (t̄2 − t̄1 )2 ≥ (x̄1 − x̄2 )2 , (1.11)

and the two points are causally related in L̄.

Problem 1.9
From the inverse Lorentz transformation, we have:

t1 = 0
t2 = γ (β x̄2 )
(1.12)
x1 = 0
x2 = γ x̄2 ,

and γ = 5/3, β = 4/5, so c t2 = 4/3 d and x2 = 5/3 d.

c t̂



event 2
d



x̂
event 1 d

Figure 1.1: Event 1 is at the origin – event 2 is at x2 = 5/3 d, c t2 = 4/3 d.


Problem 1.10
We can take 45◦ lines emanating from each event, the events that fall within
that forward “light cone” are causally related. The events, with light cone, are

, 8 CHAPTER 1. SPECIAL RELATIVITY


c t̂



C
2


D
1
A
B
x̂
1 1 2




Figure 1.2: Events with light cones in place.


shown in Figure 1.2. From that figure, we see that A can cause nothing, B can
cause C, C causes nothing, and D could cause C.

Problem 1.11
p
For x = α t2 we have c t = c x/α, and a sketch of this curve is shown
in Figure 1.3. In that figure, a 45◦ line has been placed at the first tangent
location, indicating that the particle has reached speed c.

c t̂




x̂


Figure 1.3: The star represents the first point at which the curve’s slope is
= c, beyond this point, the particle is moving faster than c.



Problem 1.12
Referring to Figure 1.4, anything in the t > 0 shaded section (between the two
“light” lines) could be caused by an event at the origin. Anything in the t < 0
shaded section could cause the event at the origin.