Chapter 2 Geometry as Physics 2-1. (pv-10) [B] (a) In a plane, show that a light ray incident from any angle on a right angle corner reflector returns in the same direction from whence it came. (b) Show the same thing in three dimensions with a cubical corner reflector.
Solution:
B
i i
i i O
A
a) Snell’s law of reflection is that the angle the incident ray makes with the normal to the surface is the same as the angle the reflected ray makes with the normal as shown above. Since the sum of the interior angles of the 1
2
CHAPTER 2. GEOMETRY AS PHYSICS right triangle OAB is π, this implies ı′ =
π −i . 2
Equivalently the angle the incident ray makes with OB is π/2 minus the angle the reflected ray makes with OA. The reflected ray is thus parallel to the incident ray. b) Snell’s law may be stated vectorially as follows. Let ~k be a unit vector along a ray incident on a surface with normal ~n. ~k can be divided into a component along ~n and a component perpendicular to ~n as follows ~k = ~k · ~n ~n + ~k − ~k · ~n ~n .
On reflection the component along ~n changes sign while the perpendicular component remains unchanged. Thus, ~k ′ after reflection is ~k ′ = − ~k · ~n ~n + ~k − ~k · ~n ~n (1) ~k ′ = ~k − 2 ~k · ~n ~n .
Consider a ray which reflects off of all three faces of the corner reflector with orthogonal normals ~n1 , ~n2 , ~n3 respectively. Using (1) at each of the three reflections, the output of the previous reflection being the input to the next, one finds for the exiting ray ~kex in terms of the incident ray ~kin ~kex = ~kin − 2 ~kin · ~n1 ~n1 − 2 ~kin · ~n2 ~n2 − 2 ~kin · ~n3 ~n3
But ~n1 , ~n2 , ~n3 are three orthogonal vectors that form a basis. Thus, ~kex = ~kin − 2~kin = −~kin . so the exit ray leaves in the direction opposite to the incident one.
2-2. (pii-10) [S] The center of the Sun is much further way from a terrestrial measurement of angles than the center of the Earth is. But it is also much 2
PROBLEM 2.3
3
more massive. Using (2.1), estimate which would have the greatest effect on a measurement of angles such as is attributed to Gauss. Solution: Gauss’ triangle was located near the surface of the Earth. The relevant radius in the expression (2.1) is the distance of the triangle from the center of attraction. Eq (2.1) gives the approximate size of the effect of the Earth where R⊕ = 6378 km is the radius of the Earth. However for Sun the relevant radius is the distance of the triangle from the center of the Sun, which approximately the size of the Earth’s orbit r⊕ ≈ 1.4 × 108 km. Then, from (2.1) the ratio of the effect of the Sun to that for the Earth can be written 3 2 R⊕ c GM⊙ . (ratio)SuntoEarth ∼ c2 GM⊕ r⊕ For the Earth GM⊕ /c2 = .443 cm and for the Sun GM⊙ /c2 = 1.48 km. For the ratio we get (ratio)SuntoEarth ∼ 10−19 (!) . The effect of the Sun is therefore much smaller than the effect of the Earth.
2-3. (pii-5) [C] (a) Verify the relation (2.4) between the sum of the interior angles of a spherical triangle and its area when two of the angles are right angles. (b) Prove the relation generally.
Solution: a) Such a triangle can be bounded by the equator and two lines of longitude differing by an angle α. The area A is (α/2π) × (area of a hemisphere) = αa2 . A (αa2 ) sum of the = π + . =π+α =π+ interior angles a2 a2 b) The three great circles that bound a spherical triangle divide the sphere up into eight triangles. Any two circles divide the sphere into wedges whose opening angle is one of the interior angles of a triangle, and whose area is 3
4
CHAPTER 2. GEOMETRY AS PHYSICS the sum of the areas of two of the triangles. This gives a set of relations of the form 1 interior ′ · 4πa2 A+A = 2π angle
which could be solved for the areas of the triangles in terms of their interior angles.
However, it is not necessary to carry out this solution. Arguments of symmetry and some special cases are enough to find the result. The above relations show that the area of a spherical triangle are linearly related to the three interior angles: α, β, γ. Since, in a general triangle, no one of these angles is preferred over any other, the area must be related linearly and symmetrically to the angles by a relation of the form A = c(α + β + γ) + d with constants c and d depending on a to be determined. The special case considered in (a) with β = γ = π/2 gives A = c(π + α) + d = αa2 holding for arbitrary α. Thus, c = a2 and d = −a2 π, giving A = a2 (α + β + γ − π) or, what is the same thing: α+β+γ =π+
A . a2
2-4. (pii-1) Draw examples of a triangle on the surface of a sphere for which: a) the sum of whose interior angles is just slightly greater than π. b) the sum of whose angles is equal to 2π. c) What is the maximum the sum of angles of a triangle on a sphere can be according to (2.4)? Can you exhibit a triangle where the sum achieves this value? 4
PROBLEM 2.5
5
Solution: α
Consider the triangle contained within the equator and two lines of longitude differing by an angle α. That triangle has two right interior angles at the equator and the interior angle α at the pole. The sum of the interior angles is π + α. By taking α near zero, one has a triangle whose sum of angles is slightly bigger than π. By taking α = π, one has a triangle whose sum of angles is 2π. From (2.4), the maximum sum of interior angles occurs when A = 4πR2 — the area of the whole sphere — and is 5π. A triangle which nearly realizes this bound is the complement of a small equilateral triangle. The three interior angles are each (2π − π/3) and add up to 5π. 2-5. (pii-2) Calculate the area of a circle of radius r (distance from center to circumference) in the two - dimensional geometry which is the surface of a sphere of radius a. Show that this reduces to πr 2 when r ≪ a. Solution: Refer to Fig.2.6 for the geometry. Consider an element of area at (θ, φ) spanned by coordinate intervals (dθ, dφ). The length of the edge of size dθ is adθ, the length of the edge of size dφ in a sin θdφ. Since the coordinate lines are orthogonal the area is (adθ)(a sin θdφ) . The circle of radius r lies at θ = r/a. Integrating the element of area above Z r/a Z 2π A= dθ dφ a2 sin θdθdφ 0
0
5
6
CHAPTER 2. GEOMETRY AS PHYSICS
gives the result: A = 2πa2 [1 − cos(r/a)] .
For small r/a cos
r a
A = πr 2
r 4 1 r 2 = 1− +O 2 a a r 4 . + O a
2-6. (pii-12) [B] Consider a sphere of radius a and on it a segment of length s of a line of latitude that is a distance d from the north pole measured on the sphere. What is the angle between the lines of longitude that this segment spans? Is this angle greater or smaller than the angle the segment would subtend at the same distance on a flat plane? Solution: This problem is solved in the same way that the ratio of the circumference to radius of a “circle” on the sphere was calculated in (2.17) — (2.19). The answer is d s = ∆φ a sin . a The subtended angle is therefore
−1 d ∆φ = s a sin a since sin x < x this is more than the angle ∆φ = s/d that would be subtended geometry were flat.
2-7. (pii-4) Consider the following coordinate transformation from familiar rectangular coordinates (x, y) labeling points in the plane to a new set of coordinates (µ, ν) x = µν 1 2 µ − ν2 y = 2
a) Sketch the curves of constant µ and constant ν in the (x, y) plane. 6
PROBLEM 2.7
7
b) Transform the line element dS 2 = dx2 + dy 2 into (µ, ν) coordinates. c) Do the curves of constant µ and constant ν intersect at right angles? d) Find the equation of a circle of radius r centered at the origin in terms of µ and ν. e) Calculate the ratio of the circumference to the diameter of a circle using (µ, ν) coordinates. Do you get the correct answer?
Solution: a) y
x
b) dS 2 = dx2 + dy 2 = (µdν + νdµ)2 + (µdµ − νdν)2 dS 2 = (µ2 + ν 2 ) (dµ2 + dν 2 )
c) The curves intersect at right angles because there are no cross terms dµdν in the metric. 7
8
CHAPTER 2. GEOMETRY AS PHYSICS
d) The equation of a circle is x2 + y 2 = r 2 which becomes 2 1 2 µ − ν2 = r2 µ2 ν 2 + 4 2 1 2 µ + ν2 = r2 4 µ2 + ν 2 = 2r e) The circumference C is I I C = dS =
1 1 µ2 + ν 2 2 dµ2 + dν 2 2
C = (2r)
= (2r) = 2πr
1 2
1 2
I
dµ 1 +
√ + Z 2r
√ − 2r
dµ 1 +
dν dµ
2 ! 12
µ2 (2r − µ2 )
12
This is, of course, the correct answer. The key step in the above evaluation √ is recognizing √ that the whole circle is covered by the coordinate range µ = − 2r to µ = 2r.
2-8. (pii-6) [A] The surface of an egg is an axisymmetric geometry to a good approximation. In the line element for two-dimensional axisymmetric geometries (2.21), pick an f (θ) such that the resulting surface would resemble that of an egg. Calculate the ratio of the biggest circle around the axis to the distance from pole to pole. Solution: There are many solutions to this problem corresponding to the different choices of f (θ) that make a surface like an egg which is smaller near one pole than the other. A simple linear choice is f (θ) = 1 − θ/2π which varies between 1 at θ = 0 (the larger end) and 1/2 at θ = π (the smaller end). The circumference of a circle around the axis at θ is C(θ) = 2π a f (θ). 8
PROBLEM 2.9
9
The maximum circumference occurs at θ = 0: C(0) = 2π a. The distance d from pole to pole is Z π d = dθ a 0
= πa
The ratio C/d is thus 2 for this example.
2-9. (pii-7) The surface of the Earth is not a perfect sphere. One quarter of the circumference aound a great circle passing through the poles is 9, 985.16 km. This is slightly less than one quarter the of equatorial circumference, 10, 0018.75 km, meaning the Earth is slightly squashed. Suppose the surface of the Earth is modeled by an axisymmetric surface with a line element of the kind in (2.21) with f (θ) = sin θ(1 + ǫ sin2 θ) for some small ǫ. What values of a and ǫ would best fit reproduce the known polar and equatorial circumferences? Comment: It is not an accident that one quarter of the polar circumference is almost exactly ten million meters. That was the original definition of the meter.
Solution: The line element (2.21) with the given f (θ) depends on two parameters a and ǫ. We can determine these by fitting to the circumferences of the equator and a great circle through the polar axis. One quarter of the circumference of the equator θ = π/2 from (2.21) is Z 1 π 1 2π π af (π/2)dφ = af (π/2) = (1 + ǫ)a . Ceq = 4 4 0 2 2 This must be 10018.7 km. The circumference of the great circle φ = 0 from (2.21) is Z 1 π 1 π adθ = a . Cpolar = 4 4 0 2 This must be 9985.16 km. Thus a = 6357 km, 9
ǫ = .003 .
10
CHAPTER 2. GEOMETRY AS PHYSICS
The data in this problem were taken from Allen’s Astrophysical Quantities, 4th ed., ed, by A. N. Cox, (Springer, 2000). 2-10. (pii-8) [B] (Equal Area Projections.) An equal area map projection is one for which there is a constant proportionality between areas on the map and areas on the surface of the globe. Given x = Lφ/2π, what function y(λ) would make an equal area map? [Hint: If an infinitesimal area dxdy has the same constant of proportionality to the corresponding infinitesimal area on the sphere wherever it is located, bigger areas will be also proportional.]
Solution: The metric on the sphere (2.24) can be written in the form (2.28) in terms of x = (Lφ)/2π and arbitrary y = y(λ). The area on the sphere bounded by a small rectangle of coordinate length dx and height dy is thus 2π dλ a cos λ(y) dx a dy L dy If the area dx dy on the map is to be proportional to this, then the coefficient of dx dy above must be constant. Choosing a convenient constant if proportionally, we have dλ cos λ = 1 . dy Integrating this and choosing y = 0 to be the equator λ = 0, we find y(λ) = sin λ or λ(y) = sin−1 (λ)
2-11. (pii-9) [B] (Conical Projections.) Conical projections map points on the globe into polar coordinates (r, ψ) in the plane of the map. (We use ψ to avoid confusion with the coordinate φ on the sphere.) Thus, in general r = r(λ, φ) and ψ = ψ(λ, φ). A particularly simple class of conical projections uses the north pole as the origin of the polar coordinates and has r = r(λ) and ψ = φ. For this simple class a) express the line element on the sphere in terms of r and ψ. 10
PROBLEM 2.11
11
b) find the function r(λ) which makes this an equal area projection in which there is a constant proportionality between each area on map and the corresponding area on the sphere. [Hint: See the hint for the previous problem.]
Solution: (a) dS 2 = a2 dλ2 + cos2 λdφ2 " # 2 dλ = a2 dr 2 + cos2 λdψ 2 . dr (b) The length on the sphere of a line of coordinate length dr extending in the r direction is a(dλ/dr)dr. The length of a line of coordinate length dψ extending in the ψ direction is a cos λdψ. Since r and ψ are orthogonal coordinates the area spanned by dr and dψ is dλ dr [a cos λdψ] . a dr The area of the corresponding element in the plane is (dr)(rdψ) . If these are related by a constant of proportionality L, we must have a2
dλ cos λ = −Lr . dr
(The constant of proportionality must be negative since latitude decreases as r increases from the north pole.) Integrating both sides and choosing the constant so r = 0 is λ = π/2 (the north pole) we find 1 a2 (sin λ − 1) = − Lr 2 2 r 2a2 r(λ) = (1 − sin λ). L 11
12
CHAPTER 2. GEOMETRY AS PHYSICS
2-12. (pii-11) [B,N]Your Personal World Map The maps in Box 2.3 were made with the Mathematica program WorldPlot. Make your own projection centered on your home city that uses a radial coordinate that represents your view of the importance of the rest of the world. Solution: The world map below is a projection which emphasizes the US and, in fact, the neighborhood of New York over other places. It was constructed using the projection x = φ′ /(1 + |φ′|/6000)1.2 y = λ′ /(1 + |λ′ |/1000)1.2 with φ′ = φ + 74◦ , and λ′ = λ − 41◦ expressed in minutes of arc.
There are probably many more elegant ways of solving this problem and certainly many more candidates for the most important city.
12
Chapter 3 Space, Time and Gravity in Newtonian Physics 3-1. (piia-4) A free particle is moving in an inertial frame (x, y, z) in the xy -plane on a trajectory x = d, y = vt where d and v are constants in time. Consider a rectangular frame (x′ , y ′, z ′ ) rotating with respect to the inertial frame with an angular velocity ω about a common z-axis (z ′ = z). What are the equations of motion obeyed by x′ (t), y ′ (t) and z ′ (t) in the rotating frame? Sketch the trajectory of the particle in the x′ y ′-plane and show explicitly that it satisfies these equations of motion.
Solution: Deriving the equations of motion in a rotating frame is a standard topic in Newtonian mechanics which can be found in almost any text on the subject. If ~x′ (t) is the vector with components (x′ (t), y ′ (t), z ′ (t)) in the rotating frame and V~ ′ (t) is its time derivative, the equation of motion is: ~′ dV ~ ′ × ~ω + ~ω × (~x′ × ~ω ) . = 2V dt where ~ω is the angular velocity of the rotating frame. The first term in this expression is the Coriolis force and the second the centrifugal force. The explicit equations for the components V x′ and V y′ for an angular velocity of 13
14
CHAPTER 3. NEWTONIAN PHYSICS
magnitude ω pointing along the z−axis are: dV x′ = +2ωV y′ + ω 2 x′ dt dV y′ = −2ωV x′ + ω 2 y ′ . dt The given trajectory in the inertial frame x(t) = d ,
y(t) = vt
becomes in the inertial frame x′ (t) = +x(t) cos(ωt) + y(t) sin(ωt) , y ′(t) = −x(t) sin(ωt) + y(t) cos(ωt) . A plot of the orbit for three periods of rotation with v = 1 and d = 1 is shown below: y’ 3
2
1
-3
-2
-1
1
2
3
x’
-1
-2
-3
′
′
Substituting the x (t) and y (t) given above into the equations of motion verifies that they are satisfied.
3-2. (piii-7) Show that Newton’s laws of motion are not invariant under a transformation to a frame that is uniformly accelerated with respect to an inertial frames of Newtonian mechanics. What are the equations of motion in the accelerated frame? 14
PROBLEM 3.3
15
Solution: Let (t, x) be the coordinates of an inertial frame and (t′ , x′ ) the coordinates of a frame accelerating along the x-axis with acceleration g. Then x = x′ + t = t′ .
1 ′2 gt , 2
Newton’s equation of motion for a free particle d2 x/dt2 = 0 implies d2 x′ = −g dt′2 which is not the form it takes in an inertial frame.
3-3. (piia-3) [B,S] How many degrees per hour does the Foucault pendulum described in Box 3.2 precess?
Solution: In this problem it is important to distinguish the Earth’s rotation, or revolution about its axis, from its orbital motion around the Sun. The Earth makes one complete revolution with respect to the Sun in 24 hr. The center of the Earth is not fixed in an inertial frame, but orbiting around the Sun. Thus the Earth would make on complete revolution with respect to the Sun in 365 days even if it were not rotating in an inertial frame in which the distant galaxies were at rest. The rotation period of the Earth with respect to inertial frame of the distant galaxies is therefore (24 − 24/365) hr = 23.93 hr or 23 hr, 56 min, and 4 s approximately. This is called the sidereal rotation period. (There are also negligible corrections for the Sun’s rotation around the center of the galaxy, etc. ) The plane of the pendulum makes one complete rotation in 23.93 hr so the angular rate is 360/23.93 = 15.04◦ /hr.
3-4. (piia-1) Find the gravitational potential inside and outside of a sphere of uniform mass density having a radius R and a total mass M. Normalize the potential so that it vanishes at infinity.
15
16
CHAPTER 3. NEWTONIAN PHYSICS
Solution: The mass density in the sphere is µ=
3M M = = const. (4πR3 /3) 4πR3
(1)
The spherical symmetry of the problem implies that Φ is a function only of the radius r. Poisson’s equation (3.18) for the gravitational potential is 1 1 2 dΦ r = 4πGµ. (2) r 2 dr dr The general solution inside (r < R) which does not diverge at r = 0 is Φ(r) = Ar 2 + B
(3)
where A and B are constants. A is determined from (2) and (1) to be 2π 1 GM 1 A= Gµ = . 3 2 R R2
(4)
B is determined by matching to the exterior solution Φ(r) = −GM/r at r = R to be B = −3GM/(2R). Thus, GM r 2 −3 , r<R (5) Φ(r) = 2R R GM , r > R. (6) = − r
3-5. (piv-5) Consider the functional Z " T
S[x(t)] =
0
dx(t) dt
2
2
#
+ x (t) dt .
Find the curve x(t) satisfying the conditions x(0) = 0,
x(T ) = 1 ,
which makes S[x(t)] an extremum. What is the extremum value of S[x(t)]? Is it a maximum or minimum? 16
PROBLEM 3.5
17
Solution: Lagrange’s equations ∂L d ∂L + =0 − dt ∂ ẋ ∂x are the necessary condition for an extremum to functionals of the form Z S [x(t)] = dt L (ẋ, x) .
(1)
(2)
In the present case, L = ẋ2 + x2 and the Lagrange equation (1) is ẍcl = xcl whose general solution is a linear combination of sinh t and cosh t. The solution satisfying x(0) = 0, x(T ) = 1 is xcl (t) =
sinh t . sinh T
The value of the action at this extremum can be found by doing the integral directly, but is most easily computed by integrating (2) by parts to give Z T T dt x(t) [−ẍ(t) + x(t)] . S[x(t)] = ẋ(t) x(t) + 0
0
The second term vanishes because of Lagrange’s equation, so S [xcl (t)] = coth T
(3)
for the extremal path. The argument of the action is positive for any choice of x(t) and can be made arbitrarily big by choosing a wiggly path with big ẋ. The extremum, therefore, cannot be a maximum but must be a minimum. For example, the simple path x∗ (t) =
t T
satisfies the boundary conditions, and 1 S [x∗ (t)] = T 17
T2 1+ 3
18
CHAPTER 3. NEWTONIAN PHYSICS
which is greater than (3). 3-6. (pv-8) [B,E,C] Estimate the gravitational self-energy of the Moon as a fraction of the Moon’s rest mass energy. Is this ratio larger or smaller than the few parts in 1013 accuracy of the Lunar laser ranging test of the equality of gravitational and inertial mass?
Solution: The gravitational self energy is of order Eself ∼
2 GMmoon Rmoon
and the ratio to the rest energy is Eself GMmoon (6.67 × 10−8 )(7.35 × 1025 ) ∼ ∼ ∼ 3 × 10−11 Erest Rmoon c2 (1.7 × 108 )(3 × 1010 )2 which is within the 10−13 accuracy of lunar laser ranging.
18
Chapter 4 The Principles of Special Relativity 4-1. (piv-1) [B,S] Today a TGV train (train à grande vitesse) leaves Paris (Gare de Lyon) at 8:00 and arrives at Lyon (Part Dieu) at 10:04 (using a 24hr clock). Assuming the train makes no intermediate stops, plot the world line of the train on a copy of the railway spacetime diagram on p. 71. If the distance between Paris and Lyon is 472 km, how fast is the train traveling on average?
Solution:
The average velocity is approximately 228 km/hr. Evidently, from the smaller slope of its world line, this TGV is faster than late nineteenth century trains.
19
20
CHAPTER 4. THE PRINCIPLES OF SPECIAL RELATIVITY
4-2. (piii-3) A rocket ship of proper length L leaves the Earth vertically at speed (4/5)c. A light signal is sent vertically after it which arrives at the rocket’s tail at t = 0 according to both rocket and Earth based clocks. When does the signal reach the nose of the rocket according to (a) the rocket clocks; (b) the Earth clocks?
Solution: a) In the rocket frame, as in all inertial frames, the velocity of light is c, so the time to traverse the proper length L is t′ = L/c. b) There are at least two instructive ways of doing this problem: Direct calculation in the earth frame; Let t be the time as read on earth clocks that the light signal reaches the nose of the rocket. The signal has traveled a distance equal to the contracted length of the rocket plus the distance (4/5)ct it traveled in the time t. Thus, 1/2 ct = L 1 − (4/5)2 + (4/5)ct
Solving for t, one finds
t = 3L/c Transforming back from the rocket frame; The event of the signal reaching the nose occurs at t′ = L/c, z ′ = L in the rocket frame if c is the vertical direction. Therefore in the earth frame t′ + (4/5)c z ′ 3L . t= p = 2 c 1 − (4/5) 4-3. (piii-5) A 20 m pole is carried so fast in the direction of its length that it appears to be only 10 m meters long in the laboratory frame. The runner carries the pole through the front door of a barn 10 m long. Just at the instant the head of the pole reaches the closed rear door, the front door can be closed, enclosing the pole within the 10 m barn for an instant. The rear door opens and the runner goes through. From the runner’s point of view, however, the pole is 20 m long and the barn is only 5 m! Thus the pole can never be enclosed 20
PROBLEM 4.3
21
in the barn. Explain, quantitatively and by means of spacetime diagrams, the apparent paradox.
10 m
Solution: t t
t * x F
R
x
Shown above is a spacetime diagram in the frame where the barn is at rest and the pole is moving. The solid, vertical lines are the world lines of the front and rear barn doors, the heavy parts indicating when the doors are shut. The dotted lines are the world lines of the end of the pole. At the moment t∗ the pole is in the barn and both doors are simultaneously shut. The coordinates (t′ , x′ ) of the frame in which the pole is stationary and the barn is moving are also indicated as well as some lines of constant t′ . 21
22
CHAPTER 4. THE PRINCIPLES OF SPECIAL RELATIVITY
From this spacetime diagram it is evident that the closing of the front door and the opening of the rear door are not simultaneous in the pole frame. Rather, the front door closes after the rear door opens. This allows the shorter barn time to pass over the longer pole as we now demonstrate quantitatively. In the pole frame the time difference between the two events simultaneous in the barn frame in [cf. (4.24)] is ∆t′ = γvL∗ /c2 where L∗ = 10 m is the proper width of the barn and v is its velocity. This velocity is such √ that the 20 m pole is contracted to 10 m in the barn frame, i.e., γ = 2, v = ( 3 /2)c. At the time the rear door opens, 5 m of√the pole is within the contracted 5 m length of the barn. Another v∆t′ = 2 · ( 3/2)2 · 10 = 15 m can pass through before the front door closes. The total makes the full 20 m length of the pole. There is no contradiction.
4-4. (piii-2) A satellite orbits the Earth in a circular orbit above the equator a distance of 200 km from the surface. By how many seconds per day will a clock on such a satellite run slow compared to a clock on the Earth? (Compute just the special relativistic effects.)
Solution: Neglecting the earth’s orbital motion, we can think of the earth as rotating about an axis in an inertial frame. The speed Vs of the satellite is related to the distance rs from the earth’s center by GM⊕ Vs2 = rs rs2 where M⊕ is the mass of the earth. Thus, Vs = c
GM⊕ c2 rs
12
= 2.6 × 10−5
for rs = r⊕ +200km, where r⊕ = 6378km. The speed of the earth at its surface is 2πr⊕ 1 Vsurf = 1.5 × 10−6 = c c 24hrs 22
PROBLEM 4.5
23
The satellite clock is moving faster in the inertial frame and will run slower compared with the clock on the surface. The ratio of rates is 1 2 2 (1 − Vs2 /c2 ) 2 1 Vs (Rate of sat clock) 1 Vsurf = − 1 ≈ 1 + 2 (Rate of surf clock) 2 c 2 c (1 − Vsurf /c2 ) 2 since both velocities are small compared to c. The ratio is thus 1 − 3.4 × 10−10
So, in one day the clocks will differ by (3.4 × 10−10 ) × (8.6 × 104 s) = 29 µs. 4-5. (piii-17) [B,E] The radio source 3C345 is participating in the expansion of the universe and its distance can be determined from the redshift arising from its recession velocity and assumptions about our universe. (Work Problem Chapter 19.1 when you have studied a little cosmology.) However, a rough idea of the distance can be obtained from Hubble’s law relating distance d to observed recession velocity V : V = H0 d where H0 ≈ 72 (km/s)/Mpc is the Hubble constant. (Look at the endpapers for astronomical units like the megaparsec (Mpc).) V for 3C345 is about .6c. Use these facts together with the data in Box 4.3 to roughly estimate the velocity of the cloud C2 assuming (contrary to fact) that it is moving transverse to the line of sight. Solution: From the figure in Box 4.3 we can roughly estimate that the cloud C2 moves about 2 mas (milliarcseconds) in 4.7 yr. To find out how far it moves we need the distance to 3C345. Hubble’s law gives .6c = .6(3 × 105 km/s) = H0 d
which with the value of H0 specified gives (1 pc = 3 × 1018 cm) d ≈ 2.5 × 103 Mpc ≈ 7.5 × 1022 km.
Assuming the cloud is moving transversely to the line of sight, the distance it travels is s = θd where θ is 2 mas in radians. Thus, 2π × 2 × 10−3 × 7.5 × 1022 km s ≈ 360 × 60 × 60 ≈ 7 × 1014 km. 23
24
CHAPTER 4. THE PRINCIPLES OF SPECIAL RELATIVITY
Since 4.7 yr ≈ 1 × 108 s, the transverse velocity VT is VT ≈ 7 × 106 km/s ≈ 24c.
This is just an estimate but the velocity is larger than c. A detailed calculation presented in the solution to Problem 19.1 gives about 13c.
4-6. (piii-6) Example 2 showed how time dilation in a moving clock could be understood in terms of the working of a model clock consisting of two mirrors oriented along the direction of motion. Show that the same result can be derived using a similar clock oriented perpendicular to the direction of motion.
Solution:
A V C
B
The mirrors are a distance L apart in their rest frame. A light signal starts from A, is reflected at B, and returns at C. We calculate the elapsed time for that in the frame where the mirrors are moving with speed V and compare with the elapsed in 2L/c in the rest frame. With an appropriate choice of origins of x and t, the positions of the left (L) and right (R) mirrors as a function of time are xL (t) = V t "
2 # 12 V xR (t) = L 1 − +Vt c
(1) (2)
1
where L[1 − (V /c)2 ] 2 is the Lorentz contract separation between the mirrors. Let tA = 0 be the time the light ray is emitted, tB the time it is reflected, and tC the time it returns. Since light travels with speed c, ctB = xR (tB ) − xL (0) c(tC − tB ) = xR (tB ) − xL (tC ) 24
(3) (4)
PROBLEM 4.7
25
give the relation between time and distance of the outgoing and reflected ray. Using (1) and (2) for solving (3) and (4) gives L tB = c
1 + V /c 1 − V /c
12
.
(5)
Substituting this in (4) and solving for tC gives 2L tC = c
"
2 #− 12 V . 1− c
(6)
Since ∆τ ≡ 2L/c is the interval between ticks in the rest frame, and ∆t ≡ tC is the interval in the moving clock frame, s 2 V ∆τ = ∆t 1 − . (7) c This is (4.15) — time dilation. 4-7. (piii-14) [S,P] In (4.4) we deduced a travel time ∆t′ for a pulse of light traveling between two mirrors that were moving with a speed V . This time was different from the travel time ∆t in the frame in which the mirrors are at rest, (4.3). In Newtonian physics, with its absolute time, these times would necessarily agree. Carry out the analysis that led to ∆t′ in (4.4) using the principles of Newtonian physics and show that this is the case, assuming that the rest frame of the mirrors is the rest frame of the ether. Solution: Maxwell’s equations which govern the propagation of light are valid only in the rest frame of the ether. Suppose this is the frame in which the two ~ | = (0, c, 0). In the mirrors are at rest. The velocity of the light signal is |V ~ ′ | = (v, c, 0) from the frame moving with speed −v along the x-axis, it is |V Newtonian addition of velocities (4.2). The key point is that the component V y is the same in both frames so ∆t′ = 2L/c = ∆t.
4-8. (piii-21) [S] Calculate the hyperbolic angle between the sides AC and AB of the triangle ABC illustrated in Figure 4.8. 25
26
CHAPTER 4. THE PRINCIPLES OF SPECIAL RELATIVITY
Solution: From (4.10), the point (ct, x) = (3, 5) makes a hyperbolic angle with the t = 0 line of ct tanh θ = = .6 , x so θ = tanh−1 (.6) = .69.
4-9. (piii-1) Consider two twins Joe and Ed. Joe goes off in a straight line traveling at a speed of (24/25)c for seven years as measured on his clock, then reverses and returns at half the speed. Ed remains at home. Make a spacetime diagram showing the motion of Joe and Ed from Ed’s point of view. When they return what is the difference in ages between Joe and Ed?
Solution: ct
τ2
2 t1
Ed
Joe
τ1
t1
x
The time t1 to the turn around point as measured by Ed is related to Joe’s proper time, τ1 = 7 yr to the same point by "
t1 = τ1 1 −
24 25
2 #− 12
= 25 yr
Since the return velocity is half the outbound velocity, it takes twice as long for the return trip (50 yr) according to Ed. Ed has therefore aged by a total 26
PROBLEM 4.10
27
of 75 yr. Joe aged τ1 = 7 yr on the outbound trip, and "
τ2 = 2t1 1 −
12 25
2 # 21
= 44 yr
on the return. The total for Joe is 51 years, so he is younger by 24 years on return.
4-10. (piii-12) In the novel “Return from the Stars” by S. Lem which is concerned with the problems a returning twin in the twin paradox situation might face, there is the following passage: “Her eyes were shining and attentive. ‘... I was thirty then’. The expedition ... ‘I was a pilot on the expedition to Fomalhaut. That’s twenty-three light years away. We flew there and back in a hundred and twenty years ship time. Four days ago we returned ... The Prometheus — my ship — remained on Luna. I came from there today. That’s all.’ ”1 Assuming that all accelerations are instantaneous and the the velocity of the Prometheus was constant in between, with what speed did it travel from the Earth to Fomalhaut?
Solution: We’ll present two ways to arrive at the answer. First version: Let V be the speed of the Prometheus, d the distance it traveled (23 ly), and τ the total proper time traveled (120 yr). The elapsed time in the earth based frame is τ T =√ 1−V2
and the speed is V = (2d)/T . Thus, V = 1
2d √ 1−V2 . τ
S. Lem, Return from the Stars, Harcourt Brace Jovanovich, San Diego, 1989.
27
28
CHAPTER 4. THE PRINCIPLES OF SPECIAL RELATIVITY
Solving for V gives, V
= q
2d/τ 1 + (2d/τ )2
2d = 2 × 23 ly = 4.35 × 1014 km τ = 120 yr = 3.78 × 109 s V = .36 = .36c. Second version: The spacetime interval is an invariant. For one leg, the twin on the ship measures ∆tship = 120 yr and ∆xship = 0. The twin on earth measures ∆xearth = 23 ly. Invariance of the interval supplies an equation for ∆tearth : −∆t2ship = −∆t2earth + ∆x2earth which gives ∆tearth = 64 yr. The speed in the earth frame is V = ∆xearth /∆tearth = .36 c.
4-11. (piii-20) [C] Alice and Bob are moving in opposite directions around a circular ring of radius R which is at rest in an inertial frame. Both move with constant speeds V as measured in that frame. Each carries a clock which they synchronize to zero time at a moment when they are at the same position on the ring. Bob predicts that when next they meet Alice’s clock will read less than his because of the time dilation arising because she has been moving with respect to him. Alice predicts that Bob’s clock will read less with the same reasoning. They both can’t be right. What’s wrong with their arguments? What will the clock’s really read? Solution: The problem is most easily analyzed in the inertial frame in which the ring is at rest. In that frame, the time to go once around the ring is T = 2πR/V . The proper time elapsed for both Alice and Bob is, from (4.14), ∆τonce around =
2πR √ 1−V2 . V
Alice and Bob agree and their clocks will thus be synchronized when next they meet. Their arguments about moving clocks running slow do not apply 28
PROBLEM 4.12
29
because neither Alice nor Bob, nor their clocks, are at rest in any inertial frame.
4-12. (piii-16) (a) Show explicitly that the straight line path between any two points in flat three-dimensional space (dS 2 = dx2 + dy 2 + dx2 ) is the shortest distance between them. (b) Is the straight line path between two spacelike separated points in flat spacetime the shortest distance between them? Solution:
a) Orient Cartesian coordinates (x, y, z) so that one point is at the origin and the other is a distance L away on the x-axis. Any curve connecting the two points can be specified by giving y(x) and z(x). The distance along such a curve is
S=
Z
ds =
Z
dx2 + dy 2 + dz
1 2 2
=
Z L
"
dx 1 +
0
dy dx
2
+
dz dx
2 # 12
.
(1) The distance is smallest when dy/dx = dz/dx = 0. But that is the straight line path along the x-axis.
b) In four dimensions, the generalization of (1) would be, from (4.6)
s=
Z L 0
"
dx −c2
dt dx
2
+1+
dy dx
2
+
dz dx
2 # 12
.
(2)
An argument as in (a) cannot be made because of the minus sign. Indeed the following path has zero distance between the points at x = 0 and x = L along the x−axis. 29
30
CHAPTER 4. THE PRINCIPLES OF SPECIAL RELATIVITY
ct
L
x
4-13. (piii-4) In an inertial frame two events occur simultaneously at a distance of 3 meters apart. In a frame moving with respect to the laboratory frame, one event occurs later than the other by 10−8 s. By what spatial distance are the two events separated in the moving frame? Solve this problem in two ways: first by finding the Lorentz boost that connects the two frames, and second by making use of the invariance of the spacetime distance between the two events.
Solution: The interval between the two events (∆s)2 = −(c∆t)2 + (∆x)2 must be the same in both frames. In the laboratory frame (∆s)2 = 02 + (3 m)2 = 9 m2 . In the moving frame
which gives
2 9 m2 = − 3 × 108 m/s · 10−8 s + (∆x)2 ∆x =
√
18 m2 = 4.24 m .
Let (t, x) be coordinates of the inertial frame in which the events are simultaneous, and (t′ , x′ ) coordinates of a frame moving with respect to this one along the x−axis. The Lorentz boost connecting the two frames implies i h v ′ ′ t2 − t1 = γ (t2 − t1 ) − 2 (x2 − x1 ) c 30
PROBLEM 4.14
31
In the unprimed frame, the two events are simultaneous (∆t ≡ t2 − t1 = 0) and separated by ∆x ≡ x2 − x1 = 3 m. Then v ∆t′ ≡ t′2 − t′1 = −γ 2 ∆x = 10−8 s c and solving for γ 2 gives 2
γ =1+c
2
∆t′ ∆x
2
= 2.
Therefore the separation of the two events in the moving frame is √ ∆x′ = γ∆x = 2(3) = 4.24 m.
4-14. (piii-22) [C] This problem concerns the toy model satellite location system discussed in the example on Example 4. Suppose you simultaneously receive broadcasts from two neighboring satellites A and B that report their locations x′A and x′B as well as their times of broadcast t′A and t′B which are equal t′A = t′B . The times and positions are in the rest frame of the satellites to which their clocks are all synchronized. Derive a condition that determines your position in x. Evaluate it to find your deviation from the midpoint between the satellites to first order in V /c where V is the speed of the satellites.
Solution: Two reference frames are relevant for this problem: The (t′ , x′ , y ′) rest frame of the satellites that is moving with velocity V with respect to the rest frame (t, x, y) of the observer. (The z-direction is irrelevant for this problem.) The satellites broadcast their location and the times of the emissions of their signals in their rest frame. Let (t′A , x′A , h) and (t′B , x′B , h) be the coordinates of the emissions of the two signals that are received simultaneously by the observer in her frame at (t, x, 0). (The problem states t′A = t′B , but let’s keep this general for a moment.) The coordinates of the two events of emission in the observer’s rest frame can be found from a Lorentz boost, e. g. (4.1a) tA = γ t′A + V x′A /c2 ′ ′ xA = γ (xA + V tA ) (4.1b) 31
32
CHAPTER 4. THE PRINCIPLES OF SPECIAL RELATIVITY
and similarly for (tB , xB ). The events will be received simultaneously by the observer if 1 (4.2a) (x − xA )2 + h2 2 = c(t − tA ) 1 (xB − x)2 + h2 2 = c(t − tB ) (4.2b) Subtracting gives the condition
1 1 (xB − x)2 + h2 2 − (x − xA )2 + h2 2 = c(tB − tA )
(3)
where tA , xA , tB , xB can be expressed in terms of t′A , x′A , t′B , x′B by (1). The condition (3) determines x. In particular, using t′A = t′B V V ′ ′ tB − tA = γ 2 (xB − xA ) = γ 2 L∗ (4) c c where L∗ is the proper distance between the satellites. If tB − tA = 0, the solution to (3) would have the observer at the position x̄ ≡ (xA + xB )/2 equidistant from xA and xB . But because of the relativity of simultaneity, tA − tB 6= 0 and the observer is closer to one satellite than to the other. It’s messy to solve (3) for x, but for V /c ≪ 1 we can write x = x̄ + δx and solve for δx to first order in V /c. The condition (3) becomes #− 21 " 2 V L 2 =γ L∗ +h δx γ L∗ 2 c where L = L∗ γ. The result for δx is # 12 " 2 V γL∗ δx = . + h2 c 2 4-15. (piii-9) Show that the addition of velocities (4.28) implies that (a) if 32
PROBLEM 4.15
33
~ | < c in one inertial frame then |V ~ | < c in any other inertial frame, (b) if |V ~ | = c in one inertial frame then |V ~ | = c in any other inertial frame, and that |V ~ | > c in any inertial frame then |V ~ | > c in any other inertial frame. (c) if |V Solution: Orient coordinates so that the relative velocity between the two frames is along the x-axis with magnitude v. The y-axis can be oriented so that V~ has only x and y components with V x = V cos ψ,
V y = V sin ψ.
(1)
~ ′ in the second There are then only x′ and y ′ components of the velocity V inertial frame related to the components in the first frame by the addition of velocities formulae (4.28). The picture below summarizes many algebraic demonstations. It shows the magnitude V ′ plotted against the magnitude V for v = .5 and various angles ψ. V¢ c 2 1.75 1.5 1.25 1 0.75 0.5 0.25
0.25 0.5 0.75
1
1.25 1.5 1.75
V 2 c
Starting from the bottom for small V the curves correspond to ψ = 0, π/4, π/2, 3π/4, π. In all cases V /c = 1 implies V ′ /c = 1, V /c > 1 implies V ′ /c > 1 and V /c < 1 implies V ′ /c < 1 which is what the problem asks for. The same result can be demonstrated algebraically. For convenience use units where c = 1. Then with a little algebra the addition of velocity formulas (4.28) together with (1) implies the following formula for the magnitude V ′ as a function of V , v, and ψ: 33
34
CHAPTER 4. THE PRINCIPLES OF SPECIAL RELATIVITY
V ′2 = (V ′x )2 + (V ′y )2 = 1 −
1 (1 − V 2 )(1 − v 2 ) F2
(2)
where F ≡ 1 − vV x = 1 − vV cos ψ
(3)
The relative velocity v between the two inertial frames is always less than 1. The relation (2) shows that if V = 1 then V ′ = 1, if V < 1 then V ′ < 1, and if V > 1 then V ′ > 1. The algebra in the problem can be simplified by assuming that ~v and V~ are colinear.
4-16. (piii-13) Lengths Perpendicular to Relative Motion are Unchanged v
x
Imagine two meter sticks, one at rest, the other moving along an axis perpendicular to the first and perpendicular to its own length, as shown above. There is an observer riding at the center of each meter stick. a) Argue that the symmetry about the x-axis implies that both observers will see the ends of the meter sticks cross simultaneously and that both observers will therefore agree if one meter stick is longer than the other. b) Argue that the lengths cannot be different without violating the principle of relativity.
Solution: a) If either observer saw one end of the other meter stick cross his or hers first that would violate the evident symmetry about the x-axis. Both ends must therefore cross simultaneously for both observers. b) The situation with regard to measuring the length of the moving meter stick is completely symmetric between the two observers. If one measured 34
PROBLEM 4.17
35
a shorter length than the other it would distinguish his or her inertial frame from the other one. That would violate the principle of relativity.
4-17. (piii-15) Another derivation of Lorentz contraction. Example 2 showed how the operation of a model clock was consistent with time dilation. This problem aims at showing how Lorentz contraction is consistent with ideal ways of measuring lengths. O
O v O
v
O
The length of a rod moving with speed V can be determined from the time it takes to move at speed V past a fixed point (left hand figure above). The length of a stationary rod can also be determined by measuring the time it takes a fixed object to move from end to end at speed V (right hand figure above). Taking account of the time dilation between the two frames, show that the length of the moving rod determined in this way is Lorentz contracted from its stationary length.
Solution: Consider, for example, a rod moving along its own length with speed v past observer O as in the above figure. As measured by Observer O, the length will be will be L = v∆τ where ∆τ is the time interval between when the nose of the rod coincides with O’s position and the time when the tail of the rod is coincident. An observer O ′ riding on the rod sees observer O moving in the opposite direction with speed v, as illustrated above. The time ∆t for the observer O to traverse the rod will be ∆t = L∗ /v where L∗ is the length measured by O ′. L∗ is the proper length of the rod since it is measured in its rest frame. ∆τ is a proper time interval on the clock of O and ∆t is the corresponding interval in a frame in which that clock is moving with speed v. Using the above result for ∆t and eliminating ∆τ from 35
36
CHAPTER 4. THE PRINCIPLES OF SPECIAL RELATIVITY
(4.14) one finds L = L∗
p 1 − v 2 /c2 .
(1)
The moving rod is contracted in the direction of its length.
4-18. (piii-19) [S] Show that for two timelike separated events there is some inertial frame in which ∆t 6= 0, ∆~x = 0. Show that for two spacelike separated events there is an inertial frame where ∆t = 0, ∆~x 6= 0 Solution: Two timelike separated events A and B, have ∆s2 < 0. Construct rectangular coordinates by using the straight line through A and B as the time axis and align the spatial axes along three orthogonal spacelike directions. The result is a rectangular system in which evidently ∆t′ 6= 0, ∆~x′ = 0.
One can also start with an inertial frame in which none of the ∆xα are zero and make a Lorentz transformation to a new frame where ∆t′ 6= 0, ∆~x′ = 0. Suppose, for simplicity, ∆y = ∆z = 0. The required Lorentz transformation is the boost along the x-axis such that 0 = ∆x′ = γ(∆x − v∆t) . The condition that the events are timelike separated ∆t > ∆x guarantees that this can be solved with v < 1. The spacelike case is exactly analogous.
4-19. (piii-11) [C] If a photograph is taken of an object moving uniformly with a speed approaching the speed of light parallel to the plane of the film, it does not appear contracted in the photograph, but rather rotated. Explain why. (Assume the object subtends a small angle from the camera lens.)
Solution: 36
PROBLEM 4.19
37
V
θ
Vb
a 1−V2
Consider a rectangular object moving parallel to the plane of the film with speed V as shown above. Suppose the long side has a rest length a and the short side a rest length b. Because of Lorentz contraction, the image of the √ 2 long side will have a length a 1 − V . The light from the far side takes a time b (c = 1 units) longer to get to the film than the near side. A photo taken at one instant will therefore, show the near side and the far side as it was a time b earlier when it was a distance V b to the left, as shown. That’s just the same as if the object were rotated by an angle θ with V = sin θ, since √ a cos θ = a 1 − V 2 b sin θ = bV .
37
38
CHAPTER 4. THE PRINCIPLES OF SPECIAL RELATIVITY
38
Chapter 5 The Spacetime of Special Relativity 5-1. (piv-2) [S] Consider two four-vectors a and b whose components are given by aα = (−2, 0, 0, 1) bα = (5, 0, 3, 4) . a) Is a timelike, spacelike, or null? Is b timelike, spacelike, or null? b) Compute a − 5b. c) Compute a · b.
Solution: a) a · a = −2 · 2 + 0 · 0 + 0 · 0 + 1 · 1 = −3 b · b = −5 · 5 + 0 · 0 + 3 · 3 + 4 · 4 = 0 Thus, a is timelike, and b is null. b) a − 5b = (−2, 0, 0, 1) + (−25, 0, −15, −20) = (−27, 0, −15, −19). 39
40
CHAPTER 5. THE SPACETIME OF SPECIAL RELATIVITY
c) a · b = −(−2 · 5) + 0 · 0 + 0 · 3 + 1 · 4 = 14 .
5-2. (piv-6) The scalar product between two three-vectors can be written as ~a · ~b = ab cos θab where a and b are the lengths of ~a and ~b respectively and θab is the angle between them. Show that an analogous formula holds for two timelike fourvectors a and b: a · b = −ab coshθab
where a = (−a· a)1/2 , b = (−b· b)1/2 and θab is the parameter defined in (4.18) that describes the Lorentz boost between the frame where an observer whose world line points along a is at rest, and the frame where an observer whose world line points along b is at rest.
Solution: Work in the frame A of the observer whose four-velocity is pointing along a, and orient the spatial coordinates so that x points along b. Then a = (a, 0, 0, 0)
(1)
Similarly in the frame B of an observer whose four-velocity points along b, b = (b, 0, 0, 0). Suppose B is moving with respect to A with a relative rapidity θab . Making a Lorentz transformation from the B to the A frame, the components of B are [cf (4.18)] b = (b cosh θab , b sinh θab , 0, 0) in the A frame. Taking the inner product with (2) in the A frame: a · b = −ab cosh θab . Another way of doing the problem is a follows: Since a1 and a2 are timelike, we can choose a frame so that a1 lies purely in the time direction a1 = (at1 , 0, 0, 0) 40