Phys. 506
Electricity and Magnetism
Prof. G. Raithel
Problem Set 7
Total 40 Points
Winter 2004
1. Problem 11.3
10 Points
Since the boosts are in parallel directions, we may consider - without loss of generality - two Lorentz
transformations in the x-direction with velocities v1 and v2 , described by matrices

γ1
 −β1 γ1
A1 = 

0
0
where βi =
vi
c
and γi = √
1
1−βi2
−β1 γ1
γ1
0
0

0
0 

0 
1
0
0
1
0

γ2
 −β2 γ2
A2 = 

0
0
and

0 0
0 0 

1 0 
0 1
−β2 γ2
γ2
0
0
with i=1,2. The net transformation then is


0 0
0 0 

1 0 
0 1
γ1 γ2 (1 + β1 β2 ) −γ1 γ2 (β1 + β2 )
 −γ1 γ2 (β1 + β2 ) γ1 γ2 (1 + β1 β2 )
A2 ◦ A1 = 

0
0
0
0
.
The 6= 1 diagonal terms can be written as
γ1 γ2 (1 + β1 β2 ) =
p
=
q
1 + β1 β2
(1 −
v1 +v2
v v
1+ 12 2
=q
1
(1−β12 )(1−β22 )
(1+β1 β2 )2
1
µ
1
c2
1−
v1 +v2
v v
1+ 12 2
³
β1 +β2
1+β1 β2
1−
1
¶2 =: q
1
=r
(1+β1 β2 )2 −(1−β12 )(1−β22 )
(1+β1 β2 )2
s
1−
with v3 =
−
β22 )
1
1−
=
β12 )(1
v32
c2
1
=: p
1 − β32
´2
=: γ3
c
. Then, the non-zero off-diagonal terms can be written in the form
c
−γ1 γ2 (β1 + β2 ) = −
Thus,
v3
β1 + β2
[γ1 γ2 (1 + β1 β2 )] = − γ3 = −β3 γ3
1 + β1 β2
c

γ3
 −β3 γ3
A2 ◦ A1 = 

0
0
−β3 γ3
γ3
0
0
This is a net Lorentz transformation with boost velocity v3 =
0
0
1
0
v1 +v2
v v
1+ 12 2
c

0
0 

0 
1
, q.e.d.
2. Problem 11.5
10 Points
Inertial frame K 0 moves wrt. frame K with a constant velocity v. Observed in K 0 , a particle has a trajectory
(ct0 , x0 (t0 )). The trajectory observed in K, (ct, x(t)), is the inverse Lorentz transform of (ct0 , x0 (t0 )). The
0 0
components of the particle velocities parallel to v in K 0 and K, u0|| = dxdt(t0 ) and u|| = dx(t)
dt , are related as
shown in Eq. 11.31 of Jackson. The longitudinal acceleration components observed in K 0 and K, denoted
by a0|| and a|| , are then related via


0
du||
d  u|| + v 
a|| (t) =
=
0
dt
dt 1 + u|| v
c2
"
#
0
u0|| v du0||
v du||
1
0
=
− (u|| + v) 2
(1 + 2 )
u0 v
c
dt
c dt
(1 + c||2 )2
"
#
¸ 0
·
u0|| v u0|| v v 2
1
dt0 du||
=
1+ 2 − 2 − 2
u0 v
c
c
c
dt dt0
(1 + c||2 )2
· ¸
1
1 dt0 0
a
=
2
u0 v
dt ||
(1 + c||2 )2 γ
· µ
¶¸
a0||
βγx|| (t)
d
=
γt −
u0 v
c
γ 2 (1 + c||2 )2 dt
¸
·
a0||
v dx||
=
1− 2
u0|| v
c
dt
γ(1 + c2 )2
h
a0||
v i
=
1 − 2 u||
u0|| v
c
γ(1 + c2 )2



0
a0||
u
+
v
1 − v  || 0 
=
u0|| v
c2 1 + u|| v
2
γ(1 + c2 )
c2
"
#
0
0
a||
u|| v v(u0|| + v)
=
1+ 2 −
u0 v
c
c2
γ(1 + c||2 )3
a0||
=
u0 v
γ 3 (1 + c||2 )3
a0||
and thus, written as a vector, a|| (t) =
q.e.d.
3
γ 3 (1 + u·v
c2 )
1
(Note that we use γ = γv = p
0
in K and K, denoted
a0⊥
2
1− v2
and a⊥
a⊥ (t)
.) Also, beginning with Eq. 11.31 we find for the transverse accelerations
c
=
=
=


1 d  u0⊥ 
0
γ dt 1 + u|| v
c2
"
#
u0|| v d 0
u0|| v
1
0 d
(1 + 2 ) u⊥ − u⊥ (1 + 2 )
u0 v
c dt
dt
c
γ(1 + c||2 )2
"
#
· 0¸
u0|| v d 0
1
dt
0 v d 0
(1 + 2 ) 0 u⊥ − u⊥ 2 0 u||
u0 v
c dt
c dt
γ(1 + || )2 dt
c2
=
=
=
=
=
=
=
=
"
#
u0|| v d 0
0 v d 0
(1 + 2 ) 0 u⊥ − u⊥ 2 0 u||
u0 v
c dt
c dt
γ 2 (1 + c||2 )3
"
#
u0|| v 0
1
0 v 0
(1 + 2 )a⊥ − u⊥ 2 a||
u0 v
c
c
γ 2 (1 + c||2 )3
"
#
u0|| v
1
0
0
0 v 0
a⊥ + a⊥ 2 − u⊥ 2 a||
u0 v
c
c
γ 2 (1 + c||2 )3
"
#
´ u0 v
³
1
||
0
0
0
0 v 0
a⊥ + a − a|| v̂
− u⊥ 2 a||
u0 v
c2
c
γ 2 (1 + c||2 )3
"
#
´
u0 v va0|| ³ 0
1
0
0 ||
0
a⊥ + a 2 + 2 −u⊥ − v̂u||
u0 v
c
c
γ 2 (1 + c||2 )3
·
´¸
1 ³ 0 0
1
0
0 0
a
+
a
u
v
−
va
u
⊥
||
||
u0 v
c2
γ 2 (1 + c||2 )3
·
¸
1
1 0 0
0
0
0
a⊥ + 2 (a (u · v) − (a · v)u )
u0 v
c
γ 2 (1 + c||2 )3
·
¸
1
1
0
0
0
a⊥ + 2 v × (a × u )
q.e.d.
0
c
γ 2 (1 + uc2·v )3
1
There, we have used the unit vector v̂ =
v
v.
3. Problem 11.6
10 Points
a): To calculate the time interval in the earth frame K along the first acceleration leg,
Z
Z
5a
T1 =
5a
γ(τ )dτ =
0
0
1
q
dτ
2
)
1 − u c(τ
2
,
(1)
we require u(τ ). To find u(τ ), we use the parallel-component result of Problem 11.5 for the case that K 0 is
the instantaneously co-moving frame of the rocket and K is the earth frame. Then, dt0 = dτ and u0|| = 0,
and
du
1 du0
1 du0
g
= 3
= 3
= 3
dt
γu dt
γu dτ
γu
(2)
Also, due to time delation between the earth and the instantaneously co-moving rocket frame it is dt = γu dτ ,
and therefore
du
dt
Z
γu2 du
u(τ )
u=0
1
du
1 − u2 /c2
u(τ )
c
du
g
= 3
γu dτ
γu
=
= gdτ
Z τ
=
gdτ = gτ
τ =0
= tanh
gτ
c
Insertion into Eq. 1 allows us to calculate the travel time of the first leg observed in K,
Z
5a
1
q
T1 =
τ =0
1 − tanh2
Z
¡ gτ ¢ dτ =
c
5a
cosh
³ gτ ´
c
τ =0
c
dτ = sinh
g
µ
g × 5a
c
¶
= 84a
By symmetry, all other legs give the same result, yielding a travel time observed in K of T = 4 T1 = 336a.
Thus, the year of return to earth is 2436.
Note that in the analysis the instantaneously co-moving frame (ICMF) of the rocket K 0 is an inertial frame;
for that reason the presented analysis is valid. As the rocket moves along, it marks the origins of an infinite
sequence of different ICMFs. The rocket itself is not an inertial frame, of course, but the rocket frame never
enters in our analysis.
b): The travel distance in K along the first leg
Z
Z
T1
L1 =
84a
u(t)dt =
t=0
u(t)dt
t=0
requires knowledge of the rocket velocity u(t) observed in K. From Eq. 2 it follows
γ 3 (u)du
Z
u(t)
u=0
q
1
u2
c2
1−
3 du
1
u(t) q
2
1 − u(t)
c2
u(t)
= gdt
Z t
=
gdt
0
= gt
=
gt
q
2 2
1 + gc2t
Insertion into the previous equation yields
c2
L1 =
g
Since
gT1
c
Z
T1 ×g/c
q
t=0
c2
d(gt/c) =
g
g 2 t2
gt/c
1+
"r
#
g2 T 2
1 + 21 − 1
c
c2
À 1, the square-root can be developed, yielding with T1 = 84a
L1 ≈ cT1 +
c2
g
µ
¶
c
−1
2gT1
≈ cT1 −
c2
= (84 − 0.969) lightyears
g
Since all legs have, for symmetry, the same length, we find a total travel distance of L = 2 L1 =
166 lightyears, which is almost as far as a beam of light would travel (which would be 168 lightyears).
4. Problem 11.9
10 Points
The objective of the problem is to find the generators of the Lorentz transformation in a simplified way
(compared with Chapter 11.7 of Jackson). The simplification is achieved by considering finite Lorentz
transformations as a sequence of infinitesimal ones. The generators of the latter are obtained in the following.
a): We consider the effect of the consecutive application of the transformation given in the problem and its
inverse,
x00α
=
(g αβ + ²0αβ ) x0β
=
(g αβ + ²0αβ ) gβγ x0γ
=
(g αβ + ²0αβ ) gβγ (g γδ + ²γδ ) xδ
=
(g αβ + ²0αβ ) gβγ (g γδ + ²γδ ) gδη xη
Since the two consecutive transformations are the inverse of each other, it also is x00α = δ αη xη for all x. By
comparison with the previous equation, we can write
δ αη
= (g αβ + ²0αβ ) gβγ (g γδ + ²γδ ) gδη
£
¤
= δ αγ (g γδ + ²γδ ) + ²0αβ gβγ (g γδ + ²γδ ) gδη
¤
£ αδ
=
g + ²αδ + ²0αβ (δβδ + gβγ ²γδ ) gδη
£ αδ
¤
=
g + ²αδ + ²0αδ + ²0αβ gβγ ²γδ gδη
£ αδ
¤
=
g + ²αδ + ²0αδ gδη
= δ αη + gδη (²αδ + ²0αδ ) = δ αη + (²αη + ²0αη )
Note that due to the infinitesimal character of the elements of the ²-tensors, we were allowed to drop terms
quadratic in them. From the last line it follows that ²αδ = −²0αδ , q.e.d.
b): Beginning with norm conservation, we find by application of the infinitesimal transformation law specified
in the problem,
xα xα
=
x0α x0α
= (g αβ + ²αβ ) xβ x0α = (g αβ + ²αβ ) xβ gαγ x0γ
= (g αβ + ²αβ ) xβ gαγ (g γδ + ²γδ ) xδ
£
¤
= (g αβ + ²αβ ) gαγ (g γδ + ²γδ ) gδη xβ xη
= (g αβ + ²αβ ) (δαδ + gαγ ²γδ ) gδη xβ xη
£
¤
= (g αβ + ²αβ ) (δαδ + gαγ ²γδ ) gδη xβ xη
= (g αβ + ²αβ ) (gαη + gαγ ²γδ gδη ) xβ xη
= (δ βη + δ βγ ²γδ gδη + ²αβ gαη ) xβ xη
=
xβ xβ + (²βδ gδη + ²αβ gαη ) xβ xη
=
xα xα + (²βα gαη + ²αβ gαη ) xβ xη
=
xα xα + (²βα + ²αβ ) gαη xβ xη
=
xα xα + (²βα + ²αβ ) gαη xβ xη
=
xα xα + (²βα + ²αβ ) xα xβ
Again, due to the infinitesimal character of the elements of the ²-tensors we were allowed to drop terms
quadratic in them. Since the result must be valid for all x, it follows that ²αβ = −²βα , q.e.d.
c): It is
x0α = (g αβ + ²αβ ) xβ = (g αβ + ²αβ ) gβγ xγ = (δ αγ + ²αβ gβγ ) xγ
Also, an infinitesimal Lorentz transformation matrix with generator L is of the form A = exp(L) = 1 + L.
In index notation, the effect of such a transformation is
x0α = Aαγ xγ = (δ αγ + Lαγ )xγ
Comparison of the last two equations shows
Lαγ = ²αβ gβγ
which is equivalent to Lαγ g γδ = ²αβ gβγ g γδ = ²αβ δβδ = ²αδ . Written in matrix form, L ◦ g = ² with an
antisymmetric matrix ². This is equivalent to Eq. 11.89 of Jackson, which directly leads to Eqs. 11.90-11.93.