1
Physics 445
Cornell University
Solution for homework 2
(40 points)
I.
Fall 2004
Steve Drasco
HARTLE CHAPTER 3, PROBLEM 5 (8 POINTS)
Under a variation in the path x(t) → x(t) + δx(t), the functional
S[x(t)] =
Z
T
dt (ẋ2 + x2 ),
(1.1)
0
becomes
S[x(t) + δx(t)] = S[x(t)] +
Z
T
dt 2(x − ẍ)δx +
0
Z
T
˙ 2 + (δx)2 ],
dt [(δx)
(1.2)
0
where I have used integration by parts on the O(δx) term. The function x(t) which makes S[x(t)] an extremum must
cause the O(δx) term to vanish. Since the variation δx(t) is arbitrary, except for the condition δx(0) = δx(T ) = 0,
this function must satisfy1 ẍ = x. The general solution is of the form
x(t) = A sinh t + B cosh t.
(1.3)
The boundary condition x(0) = 0 requires that B = 0, and x(T ) = 0 requires that A = 1/ sinh T
x(t) =
sinh t
.
sinh T
(1.4)
Substituting (1.4) into (1.1) gives the extremal value of the action S[sinh t/ sinh T ] = coth T . Lastly, since the O[(δx) 2 ]
term in Eq. (1.2) is always positive, (1.4) is a minimum of S[x(t)].
II.
HARTLE CHAPTER 3, PROBLEM 6 (8 POINTS)
The gravitational self-energy of the moon is the energy required to take it apart and move all of it’s pieces off to
spatial infinity against the resistance of gravity alone. Consider the outermost lunar shell of radial thickness dr. The
mass of this shell is dm = 4πr 2 ρdr, where r is the radial location of the shell and ρ is the lunar density, which we will
approximate as a constant. The work we do in pulling this shell out to spatial infinity is roughly
dw = dm[Φ(∞) − Φ(r)] = (4πr 2 ρdr)
16
G(4πr 3 ρ/3)
=
Gπ 2 ρ2 r4 dr
r
3
(2.1)
where we have used Φ(r) = −Gmenclosed /r. The total self-energy is then
Eself =
Z
dw =
16 2 2
π Gρ
3
Z
Rmoon
dr r4 =
0
2
16 2 2 5
3 GMmoon
π Gρ Rmoon =
,
15
5 Rmoon
(2.2)
3
where we have used ρ = 3Mmoon /(4πRmoon
). Since the moon’s rest-energy Erest = Mmoon c2 , we have Eself /Erest =
−11
1.88 × 10 . This is roughly two orders of magnitude larger than the fractional error of lunar laser ranging tests
1.5 × 10−13 (see box 2.1 in Hartle).
III.
HARTLE CHAPTER 4, PROBLEM 3 (8 POINTS)
A qualitative solution to the pole in the barn paradox is given in Figs. 1 and 2. For an observer in the “barn-frame”
the pole is inside the barn with both doors closed (instantaneously at time t AB on the left, and for a finite time tB −tA
on the right). From the runner’s point of view, the door at A is closed (t0 = t0A ) after the door at B has opened
1
This is the equation one gets by enforcing the Euler-Lagrange equation.
2
x’
Q
P
x
A
B
FIG. 1: Layout of the pole in the barn paradox.
barn
Q
le
P
po
B
barn
po
le
A
t’A
t’B
tB
tA
t AB
t’A
Q
t’B
P
A
B
FIG. 2: Spacetime diagrams for the pole in the barn paradox. Worldlines for Q, P , A, and B, are drawn in bold. The worldlines
for the doors at A and B are solid when the doors are closed and dashed when the doors are shut. The fine dotted lines are
lines of constant t and t0 . The left panel corresponds to the scenario in Hartle. The right panel corresponds to a scenario where
a barn-frame observer sees the pole enclosed in the barn for a finite time.
(t0 = t0B ). By examining the diagrams see if you can answer the following questions: Does one runner go faster than
the other (as measured in the barn frame)? Are the rest-lengths of the poles or barns the same in both panels?
Spacetime diagrams can be a powerful tool. However one must be careful to obey the rules when drawing them.
For an excellent review see David Mermin’s notes on special relativity [1] (from which the diagrams in Fig. 2 were
adapted).
We now give a quantitative explanation for Hartle’s scenario. Consider the two spacetime events “worldline of P
intersects worldline of B” and “worldline of Q intersects worldline of A”. In the barn-frame these events are are
simultaneous ∆t = 0, and separated in space by the length of the barn ∆x = 10 m. The invariant spacetime interval
between these two events is
∆s2 = 100 m2 .
(3.1)
From the runner’s point of view the events are separated in space by the length of the pole ∆x 0 = 20 m, so we have
∆s2 = −(c∆t0 )2 + 400 m2 .
(3.2)
Equating (3.1) and (3.2) shows that from the runner’s perspective the time between the events is ∆t 0 = 5.8 × 10−8 s.
IV.
HARTLE CHAPTER 4, PROBLEM 13 (8 POINTS)
Consider two inertial frames with coordinates (t, x) and (t0 , x0 ) related by a Lorentz boost in the x-direction. In
the unprimed frame, event A and is at (tA , xA ), and event B is at (tA , xA + ∆x), where ∆x = 3 m. In the primed
frame A is at (t0A , x0A ), and B is at (t0A + ∆t0 , x0B ), where ∆t0 = 10−8 s. We wish to determine
∆x0 = x0B − x0A .
(4.1)
3
We could use the Lorentz transformation relating two frames [Eqs. (4.21) from Hartle]
t0A
x0A
t0A + ∆t0
x0B
=
=
=
=
γ(tA − vxA /c2 ),
γ(xA − vtA ),
γ[tA − v(xA + ∆x)/c2 ],
γ(xA + ∆x − vtA ),
(4.2)
(4.3)
(4.4)
(4.5)
where γ = (1 − v 2 /c2 )−1/2 . The equations for x0A and x0B give ∆x0 = γ∆x. The remaining equations give
µ
¶2
c∆t0
β2
=
,
∆x
1 − β2
(4.6)
where β = v/c. Solving this, we find β 2 = 0.499654, γ = 1.41372, and lastly
∆x0 = 4.2412 m.
(4.7)
Since we all hate Lorentz transformations, we now solve this problem by invoking the invariance of the spacetime
interval between the events (∆x)2 = −(c∆t0 )2 + (∆x0 )2 . By solving for ∆x0 , we recover (4.7).
V.
EXAMPLE OF LOCAL INERTIAL REFERENCE FRAME (8 POINTS)
As indicated by the potential
"
µ
¶2 #−1/2
r
2z
GM⊕
Φ = −p
,
1+
+
=−
R⊕
R⊕
R⊕
x2 + y 2 + (z + R⊕ )2
GM⊕
(5.1)
where r 2 = x2 + y 2 + z 2 , this is a Newtonian problem. Keeping only the leading order part of each component, the
gradient of the potential is
¶
µ
µ
¶
GM⊕
xz
x
Φ,x =
+
O
,
(5.2)
2
2
R⊕
R⊕
R⊕
¶
µ
µ
¶
yz
y
GM⊕
+
O
Φ,y =
,
(5.3)
2
2
R⊕
R⊕
R⊕
µ
¶
z
GM⊕
+O
.
(5.4)
Φ,z =
2
R⊕
R⊕
We will assume the falling observer has no initial velocity, and that there is no initial displacement in x or y. After
the jump, the only nonzero component of the observer’s acceleration a is then
¶
µ
GM⊕
z
z
a = −Φ,z = − 2 + O
.
(5.5)
R⊕
R⊕
The path along which the observer falls is then
z(t) = −
GM⊕ 2
2 t +O
2R⊕
µ
¶
GM⊕ 2
t
,
3
R⊕
(5.6)
where we have written O (z/R⊕ ) in terms of the leading order value for z. If the observer’s local inertial frame (t 0 , x0 )
is to be aligned with (t, x) at the moment of the initial jump, the frames must be related by
x0 = x,
y 0 = y,
z0 = z +
GM⊕ 2
2 t ,
2R⊕
t0 = t.
(5.7)
3
up to corrections of order GM⊕ t2 /R⊕
.
Consider now another inertial (free falling) observer who has some initial velocity in x and y, but is otherwise
identical to the first observer at the time of the jump. If we drop terms of order r/R ⊕ , there will still be no
acceleration along x or y, and we have
µ
¶
¶
µ
GM⊕ 2
GM⊕ 2
y
t
,
y(t)
=
v
t
,
(5.8)
x(t) = v0x t + O
t
+
O
0
3
3
R⊕
R⊕
4
while the z-motion will be the same as that of the first observer.
In the coordinates of the first observer, this second observer then moves along a straight line at constant velocity
µ
¶
GM⊕ 2
0 x 0 y 0
(t , v0 t , v0 t , 0) + O
.
(5.9)
3 t
R⊕
Suppose that we were to keep higher order terms when solving for the x and y motion 2 . Then there would be
acceleration in the x and y directions
µ
¶
µ
¶
µ
¶
µ
¶
x
xz
y
yz
GM⊕
GM⊕
y
x
+O
,
a =− 2
+O
.
(5.10)
a =− 2
2
2
R⊕
R⊕
R⊕
R⊕
R⊕
R⊕
Solving the (simple harmonic oscillator) equations of motion for the given boundary conditions gives
s
s
Ãs
!
Ãs
!
3
3
R
R⊕
GM
GM⊕
⊕
⊕
x
x
x(t) = v0
sin t
,
y(t) = v0
sin t
.
3
3
GM⊕
R⊕
GM⊕
R⊕
For times t ¿
q
(5.11)
3 /(GM ) (about 13 minutes), the motion will be the same as (5.9). This is of course the same
R⊕
⊕
3
¿ 1, or rather |z| ¿ R⊕ .
condition as GM⊕ t2 /R⊕
[1] N. D. Mermin, Introductory notes on relativity, part 10, http://www.lassp.cornell.edu/c̃ew2/P209/P209 home.html .
2
This is somewhat inconsistent of course since we should solve each equation of motion to the same accuracy.