1. Newtonian Gravity, Geometry as Physics, and Special Relativity • 1.1) We’ll start with a disk whose center is at the north pole of a sphere of radius a, the disk’s boundary is at a polar angle θ0 , in the standard spherical coordinates (a,φ,θ) we have for the disk’s area A... Z Z 2π Z θ0 A = dA = a2 sin(θ)dθdφ 0 0 Z 2π

a2 cos(θ0 ) − 1 dφ = 2πa2 1 − cos(θ0 ) (1) = − 0

The “radius”, r, of the circle drawn on the sphere is actually an arc length on the sphere in terms of θ0 and a, namely r = aθ0 . So we have   r  A = 2πa2 1 − cos (2) a To leading order in ar  1 we can use the expansion cos(x) = 1 − 12 x2 + · · · to obtain     r 2 2 A = πr 1 + O (3) a • 1.2) This problem was discussed in class. There was confusion about the use of polar coordinates and we discussed that although θ is not the standard polar coordinate it parameterizes arc length along the peanut as defined by the metric and it is chosen to go from [0, π]. Our metric is ds22D = a2 dθ2 + a2 f (θ)2 dφ2 The waist of the peanut, has circumference, Z Z C(π/2) = ds = af (π/2)dφ = 2πaf (π/2)

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We are given that f (θ) = sin(θ)(1 − (3/4) sin(θ)2 ) and that C(π/2) = 1, so we find a = 2/π. We also know that the radius, ρ, of the peanut at any θ is just C(θ)/(2π). and (θ) dθ so dρ = a dfdθ So we make a small triangle on the surface at any point theta. Looking at the legs of the right triangle, we see that (adθ)2 = dh2 + dρ2 where dh is an infinitesimal of height. Integrating... s  2 Z Z π df H = dh = a 1− dθ = 2.578 a = 1.648 m dθ 0

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• 1.3) Light travels from the back of the spaceship to the front, while moving v = (4/5)c relative to another frame (earth frame). We must find the time it takes to traverse the ship in both frames. In the spaceship’s frame, the light moves as speed 1

2

c (as in all frames), and travels the proper length of the ship, because the ship is at rest with respect to it’s own frame. vτ = ∆x. τship = l0 /c

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Now, in the earth frame, the light must traverse the contracted length of the ship, l, but in addition the ship is moving at speed v. So in the earth frame, ∆x = xf − xi , where xf = xi + l + (4/5)cτearth (9) where τearth is the time it takes for the traverse to occur. Since, cτearth = ∆x, and the relation between contracted and proper length is γl = l0 , we have τearth = ∆x/c = l0 /(cγ) + (4/5)τearth

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and solving for τearth and noting γ = 5/3 we obtain τearth = 3l0 /c .

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Adding either of these elapsed times to the initial time (when the light passed/emitted by the back of the ship), ti = 0, gets you the final time in either frame, tf . • 1.4) t

2t1

Lisa inbound !in !

Louise at home

t1

Lisa outbound !out

x

Figure 1. Spacetime diagram of Lisa trip Treat each section separately (the outward journey and the return journey). Treat Lisa’s turn around as an instantaneous change of velocity that would kill any *real* human piloting the ship. In fact, an instantaneous switch from 24/25c in +x to 12/25c in −x direction would surely kill the ship itself. But lets not worry about that. On the outward journey, Lisa’s clock (and thus proper time as her watch is moving with her) reads τout = 7 years. During this time Louise’s clock reads t1 = τout γout . Since v = 24/25c, we have p t1 = γout τout = 7/ 1 − (24/25)2 yrs = 25 yrs (12)

3

For the incoming trip, Lisa is moving half as fast in the Lousie’s frame. Therefore, for Louise, the incoming trip should take twice as long. So the inward journey on Louise’s clock took 50 years. According to her clock she had aged 75 years since Lisa left. The inward journey on Lisa clock was is the proper time, but now her speed is 12/25c, and thus γin = 1.13. Thus Lisa’s clock on her return shows a time interval τin = 2t1 /γ = 43.86 yrs .

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Thus during the journey Lisa’s clock indicates she aged by 50.86 yrs. she almost 24 years younger than Louise. 2. Gravity as Geometry • 2.1) The equivalence principle can be stated as Experiments in a sufficiently small freely falling laboratory, over a sufficiently short time, give results that are indistinguishable from those of the same experiments in an inertial frame in empty space. How can we use this to show that light must fall in a gravitational field? The classic story goes as follows: Suppose you are in a small rectangular lab (inertial frame) in space, and you shoot a beam of photons parallel to the floor from the wall some height above the floor. The photons will arrive at the opposite wall, at the same height above the floor. If you now perform this same experiment, where your lab is now freely falling towards earth, and you must get the same result via the equivalence principle. Since you are falling, this is only possible if the photon beam falls as well. Thus, light must fall in a gravitational field. There were two caveats in the equivalence principle. Sufficiently small laboratory, and sufficiently short time. So we take the limit that the opposing walls in our laboratory have a separation distance that goes to less than any finite ε, and our argument is still the same. With the walls getting closer, the experiment will last over a shorter and shorter amount of time, less than any required δt, and this, as well, in no way affects the argument. • 2.2) For this problem, we merely want to know how time elapses differently in different gravitational potentials. So we ignore how the earth actually forms, (surely the middle rocks got there first?), and assume that the planet is just born all at approximately the same time. The rocks at the center will experience time elapse at a different rate than those at the surface. For spherical symmetric shells of mass, at any distance from the center, one only feels the effect of the mass of all the shells closer to the center and located at the center (In 3D only!). We are to assume that the earth has a uniform density, i.e. ρ = M/( 34 πR3 ) Then, at any radius inside the earth, the mass beneath the shell of said radius is M (r) = ρ 34 πr3 .

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Therefore we have, for test mass m, Gρ4πr F ~(r) GM (r) rˆ = − =− rˆ (14) m r2 3 Where M (R) is the mass of the earth. To find the potential difference (per unit mass) between r = 0 and r = R, where R is the radius of the earth , we have R

Z ∆V = −

(− 0

GM (r) rˆ · d~r) = r2

Z

R

( 0

Gρ4πr GM (R) )dr = 1/2( ) 3 R

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From Hartle (6.23), we have that two observers experience durations related via their gravitational potential difference. ∆ΦBA )∆τA (16) c2 where ∆ΦBA is our ∆V from before. The problem does not say whether it is the surface or the center that is 5 billion years old. Let us assume it is the surface that is 5 billion years old. Then ∆τB = (1 +

GM ∆Φ )∆τcenter = (1 + )∆τcenter 2 c 2Rc2 and so, with GM/(Rc2 ) = 7 × 10−10 we have ∆τsurface = (1 +

∆τcenter = (1 + 3.5 × 10−10 )−1 (5 × 109 )

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and the difference ∆τcenter − ∆τsurface ≈ −1.7 So the surface rocks are about 1.7 years older.

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3. Description of Curved Spacetime 3.1.

a) Proper distance. dt, dφ, dθ = 0 Z Z Z R √ grr dr = s = ds = 0

R

(1 − Ar2 )dr = R −

0

b) Area of sphere of radius r = R, dr, dt = 0 Z Z 2π Z π q A= gφφ (r = R, θ)gθθ (r = R) dθdφ = 0

0

0

2π

Z

A 3 R 3

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π

R2 sin[θ]dθdφ = 4πR2

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0

c) Since dt = 0, so on the sphere, the volume is then Z q V3 = grr (r)gφφ (r, φ)gθθ (r)drdθdφ Z R2 1 = (1 − Ar2 )r2 sin[θ]drdθdφ = 4πR3 ( − A ) 3 5

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d) Since the signature on the 4D metric is minus, Z

RZ T

2π

Z

Z

V4 = 0

Z

0 0 RZ T

= 4π 0

0

π

q −gtt (r)grr (r)gφφ (r, φ)gθθ (r)dφdθdtdr

0

4 6R2 3R4 (1 − Ar2 )2 r2 dtdr = πT R3 (1 − A + A2 ) 3 5 7

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3.2. a) Using the relations between coordinates X, Y, Z, W and χ, θ, φ we find that X 2 + Y 2 + Z2 + W 2 = (R sin[χ] sin[θ] cos[φ])2 + (R sin[χ] sin[θ] sin[φ])2 + (R sin[χ] cos[θ])2 + (R cos[χ])2 = (R sin[χ] sin[θ])2 (cos[φ]2 + sin[φ])2 ) + (R sin[χ] cos[θ])2 + (R cos[χ])2 = (R sin[χ])(sin[θ]2 + cos[θ]2 ) + (R cos[χ])2 = R(sin[χ]2 + cos[χ]2 ) = R2

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identically for any {χ, θ, φ}. Thus any point on the surface X 2 + Y 2 + Z 2 + W 2 = R2 labelled by X, Y, Z, W can be accessed by varying χ, θ, φ. b) Using the coordinate relations we can calculate the differentials dX = R (− sin[χ] sin[θ] sin[φ] dφ + sin[χ] cos[θ] cos[φ] dθ + cos[χ] sin[θ] cos[φ] dχ) + sin[χ] sin[θ] cos[φ] dR . (25) However since we want to know the metric restricted to the surface of the sphere we set dR = 0 to obtain dX dY dZ dW

= = = =

R(− sin[χ] sin[θ] sin[φ] dφ + sin[χ] cos[θ] cos[φ] dθ + cos[χ] sin[θ] cos[φ] dχ) (, 26) R(sin[χ] sin[θ] cos[φ] dφ + sin[χ] cos[θ] sin[φ] dθ + cos[χ] sin[θ] sin[φ] dχ) , (27) R(− sin[χ] sin[θ] dθ + cos[χ] cos[θ] dχ) , (28) −R(sin[χ]) dχ . (29)

Plug these into the metric for euclidean 4D space, dS 2 = dX 2 + dY 2 + dZ 2 + dW 2

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dS 2 = R2 (dχ2 + sin[χ]2 (dθ2 + sin[θ]2 dφ2 )

(31a)

to get

after noticing that cross-terms cancel and using the trigonometric relation sin2 + cos2 = 1.

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3.3. We start with orthonormality, eα · eβ = ηαβ . Since eα are the components of a basis, we can use a general linear transformation to change the basis, i.e., ˜γ = Aαγ eα e

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˜γ , are orthonormal, i.e. ηγδ = e ˜γ · e ˜δ now, if we still demand that these new basis vectors, e then we have restricted set of general linear transformations to something smaller, Λµν , which are a subset of the original Aµν . Then, we have ˜γ · e ˜δ = Λαγ Λβδ eα · eβ = Λαγ Λβδ ηαβ ηγδ = e (33) µ which means that these restricted transformations, Λν are precisely the set of matrices that preserve the Minkowski metric — the Lorentz transformations. 4. Geodesics and Schwarzschild Orbits 4.1. Circular orbits occur at the extrema of the effective potential. For the phonon the effective potential in Schwarzschild coordinates is   2M 1 (34) Wef f (r) = 2 1 − r r and solving dWef f (r)/dr = 0 we obtain that radius of the circular orbit R = 3M . Choosing a coordinate system such that the orbit is in the equatorial plane with θ = π/2 have dθ = 0, and because it is a circular orbit dr = 0. Thus, along the photon trajectory (using geometerized units) we have   2M 2 dt2 + r2 dφ2 , (35) ds = 0 = − 1 − r and this implies that   2M −1/2 dt = 1 − r dφ . (36) r The coordinate time interval ∆t for one orbit at r = R is obtained by setting dφ = 2π and is given by   √ 2M −1/2 2πR = 3 6πM (37) ∆t = 1 − R where we set R = 3M to obtain the second equation. The time Kirk measures is his proper time   2M 1/2 ∆τKirk = 1 − ∆t = 6πM (38) R Now we reverting back to physical unit we can write the above equation as GM c∆τKirk = 6π 2 , (39) c and use the fact that τKirk = 30µs and cτKirk ' 9 km to obtain GM M S 6π 2 = 3π R ' 9km . (40) c M

7 S ' 3 km we obtain Since the Schwarzschild radius of the sun is R

1 M (41) π Since the orbit is outside the star we require that RN S < (3/π)M or in physical units RN S < 1.41 km. Far away, on earth M’Kkshi’s √ metric can be assumed to be flat, and the time interval he measures is ∆τM = ∆t = 6π 3 M ' 52µs . M'

4.2. Spherical symmetry allows us to orient the z axis in arbitrary directions and we shall use this freedom to solve this problem. To begin we note that the time evolution of the radial coordinate will be the same in any two spherical coordinate systems labelled ˜ φ) ˜ which only differ by the orientation of their z-axes. (t, r, θ, φ) and (t, r, θ, Consider a particle at an instant in coordinate time t = 0 and orient the axes to ensure that uφ = dφ/dτ = 0. The killing vector η = (0, 0, 0, 1) associated with the φ translation symmetry implies that η · u = constant. Since uφ = 0 at t = 0, it remains zero at all time and this proves that the orbit is confined to a plane. Reorient the z-axis of the (t, r, θ, φ) coordinate system to ensure the orbit is in the plane with θ = π/2 and uθ = 0. In this coordinate system the constants of motion are η · u = gφφ uφ = r2 uφ = pφ (42)     GM dt −ξ · u = 1− =e (43) r dτ    2      2 GM dt GM −1 dr 2 dφ 2 u·u = − 1− + 1− +r = −1 r dτ r dτ dτ   2      dt GM −1 dr 2 1 GM (44) + 1− + 2 p2φ = −1 = − 1− r dτ r dτ r ˜ φ), ˜ the constants of motion are In an alternate coordinate system (t, r, θ, ˜ ˜ η · u = gφ˜φ˜ uφ = r2 sin2 θ˜ uφ     GM dt −ξ · u = 1− =e r dτ   2      GM dt GM −1 dr 2 + 1− u·u = − 1− r dτ r dτ !2 !2 dθ˜ dφ˜ + r2 + r2 sin2 θ = −1 dτ dτ    2     GM dt GM −1 dr 2 + 1− = − 1− r dτ r dτ ! 2 pφ˜ 1 2 + p + = −1 ˜ θ r2 sin2 θ˜

(45) (46)

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(48)

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Comparing these equations we can deduce that for the time evolution of the radial coordinate to be the same in both coordinate systems we require that p2θ˜ +

p2φ˜ sin2 θ˜

= p2φ = constant ,

since pφ is constant in time . If we expand the LHS of Eq. 49 about the point θ = π/2 we obtain !2 dδ θ˜ 4 + p2φ δ θ˜2 = constant , r dτ

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where δ θ˜ = π/2 − θ. Since p2φ > 0, it shows that the effective potential for motion along θ has a minimum at θ = π/2 and consequently the orbit is stable with respect to small perturbations in θ. 5. GR in Astrophysics 5.1. Positing that the photon propagates in the plane defined by θ = π/2, its momentum four vector uα = (ut , ur , 0, uφ ) where ut = d(ct)/dλ, ur = dr/dλ and uφ = dφ/dλ. We can use the following constants of motion    2  2 d(ct) 2 dr α β 2 dφ u · u = gαβ u u = −A(r) + B(r) +r = 0, (51) dλ dλ dλ   d(ct) −ξ · u = A(r) = e, (52) dλ   2 dφ −η · u = r = l, (53) dλ to obtain

√   dφ B 1 1 −1/2 = 2 − , dr r b2 A r2

where b = l/e is the impact parameter. The deflection angle √   Z ∞ B 1 1 −1/2 ∆φ = 2 dr 2 − , r b2 A r2 r1 which can be recast in the new variable w = b/r as  −1/2 Z w1 √ 1 2 ∆φ = 2 dw B −w . A 0

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At this stage it is useful to recognize that x = GM/bc2 is small parameter because the impact parameter b & R where R is the radius of the star and for most stars (neutron stars

9

being the only exception) R >> GM/c2 . Expanding √ GM B = 1 + γ 2 w + O[x2 ] , bc 1 2GM = 1+ w + O[x2 ] A bc2 we find that to linear order in x we have   −1/2  Z w1 2GM GM 2 1+ w−w . ∆φ = 2 dw 1 + γ 2 w bc bc2 0

(57) (58)

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This integral h can be done using mathematica o √
−1 n Integrate 2(1 + γxw) 1 + 2xw − w2 2 , w, 0, x + x2 + 1 , Assumptions → {γ > 0, x > 0}] π + 2ArcTan[x] + xγ(2 + πx + 2xArcTan[x]) Series[%, {x, 0, 1}] π + (2 + 2γ)x + O[x]2 You can can also be look this up in Abramowitz and Stegun, which is now available online at http://people.math.sfu.ca/~cbm/aands/frameindex.htm. The answer is GM + O[x2 ] (60) bc2 Setting b = R and M = M we find that 4GM /R c2 = 1.7 arc seconds. To measure γ to an accuracy of 1%, we need to be able to measure the angle to an accuracy of 0.0085 arc seconds. ∆φ = π + 2(1 + γ)

6. Vectors, Dual Vectors 6.1. Call the spherical basis primed, and the rectangular unprimed. The basis for our tangent space dxµ , if we want to change basis to spherical we have, ∂xµ µ0 dx ∂xµ0 likewise, the basis of the cotangent space (or dual space) is related via dxµ =

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0

∂xµ ∂µ = ∂µ0 ∂xµ Therefore, components of a dual vector in the cotangent space are related by aµ =

∂xµ µ0 a . ∂xµ0

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10

Calculating

∂xµ ∂xµ0

from the coordinate relations given in the problem we have ax = sin θ cos φar + r cos θ cos φaθ − r sin θ sin φaφ , a

y

a

z

r

θ

φ

= sin θ sin φa + r cos θ sin φa + r sin θ cos φa , r

θ

= cos θa − r sin θa .

(64) (65) (66)

To find the components in the dual space, use the metric gij to lower the indices. For the rectangular components this is trial because the metric gij = diag(−1, 1, 1, 1) and at = −at , ax = ax , ay = ay and az = az . The spherical metric in flat space is gij = diag(−1, 1, r2 , r2 sin2 θ)gives ar = ar ,

aθ = r2 aθ ,

aφ = r2 sin2 θ aφ

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Plug this into the above relations. 6.2. Consider an arbitrary basis set in the tangent space, who’s elements are labeled by a, e(a) . Each of which has components in terms of any other basic, including dxµ , e(a) = (e(a) )µ dxµ

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Likewise, for the dual space, we have the basis vectors of the cotangent space e(a) , which can also be expanded in any general basis e(a) = (e(a) )µ ∂µ

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Now, to say that e(a) is dual to e(a) means that e(a) · e(b) = (e(a) )µ eµ(b) = δba

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In particular, for a coordinate basis, we have defined that e(a) · e(b) = gαβ

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e(a) = Mab e(b) → e(a) · e(c) = Mab (e(b) · e(c) ) → gac = Mab δcb

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Using these we have where M is just some change of basis matrix. so e(a) = gab e(b)

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e(a) = M ab e(b) → e(a) · e(c) = M ab e(b) · e(c) = δca = M ab gbc

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e(a) = g ab e(b)

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e(a) · e(b) = g ac g bd e(c) · e(d) = g ac g bd gcd = g ab

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Likewise so Finally