PHZ 6607 Fall 2004 Homework #4, Due Friday, October 22, 2004

Transcription

1 Read Chapters 9, 10 and 20. PHZ 6607 Fall 2004 Homework #4, Due Friday, October 22, The usual metric of four-dimensional flat Minkowski-space in spherical-polar coordinates is ds 2 = dt 2 + dr 2 + r 2 dθ 2 + r 2 sin 2 θdφ 2. Find g, the components of g ab and all non-zero components of the Christoffel symbols. g ab ( ) 2 ( ) 2 ( ) 2 ( ) 2 x a x = + + r 2 + r 2 sin 2 θ b t r θ φ g = r 4 sin 2 θ g = r 2 sin θ Γ r θθ = r Γ r φφ = r sin 2 θ Γ θ θr = Γ θ rθ = Γ φ φr = Γφ rφ = r 1 Γ θ φφ = sin θ cos θ Γ φ φθ = Γ φ φθ = cot θ 2. Assume that Ψ(t, r, θ, φ) = R(r)e iωt Y lm (θ, φ), where R is a function only of r, ω is a constant, and Y lm (θ, φ) is a spherical harmonic function. Evaluate and simplify g ab a b Ψ in Minkowski space using the spherical polar coordinates of the previous question. g ab a b Ψ = g ab a b Ψ + g ab Γ c ab c Ψ ( 1 gg ab = ) g x a x Ψ b = 2 t Ψ + 1 (r 2 r ) 2 r 2 r Ψ + 1 (sin θ θ ) r 2 sin θ θ Ψ + 1r 2 2 φ Ψ { 2 1 d = (r 2 ddr ) [ ] } r 2 dr R + ω 2 l(l + 1) R e iωt Y r 2 lm (θ, φ)

2 3. Consider the two-dimensional spacetime with the line element ds 2 = X 2 dt 2 + dx 2. Find the shapes X(T ) of all timelike geodesics in this spacetime. T a = / x a = / T is a Killing vector. Let u a = (dt/ds, dx/ds) be the tangent to the geodesic, so u a u a = 1 = X 2 (dt/ds) 2 + (dx/ds) 2 from the normalization of the worldline and T a u a E = X 2 dt/ds where E is constant because T a is a Killing vector. Divide the first equation by (dt/ds) 2 and substitute from the second to obtain (dx/dt ) 2 = X 2 X 4 /E 2, which is the equation of motion. It is clear that X 2 E 2 always, and that for X 2 E 2, X(T ) e ±XT. One consequence is that X cannot cross X = 0. Doing the integral, and choosing initial conditions and/or constants of integration properly ultimately yields This is easy to visualize by noting that X = E/ cosh(t). X = e XT for T 1 and X = e XT for T 1 follow from the differential equation. Also X(T ) has a turning point when X 2 = E 2. Sketch a smooth curve which has these properties, and you will see what the worldline looks like. 4. Consider a medium with an index of refraction n(x i ) that is a function of position. The velocity of light in the medium varies with position and is c/n(x i ). Fermat s principle states that light rays follow paths between two points in space (not spacetime!) that take the least travel time. (a) Show that the path s of least time are geodesics in the the three-dimensional space with the line element ds 2 Fermat = n 2 (x i )ds 2 = n 2 (x i ) [ dx 2 + dy 2 + dz 2]. (b) Write out the geodesic equations for the extremal paths in (x, y, z) coordinates. Page 2

3 (a) The infinitesimal travel time between two point in this medium is given by dt 2 = [n(x a )/c] 2 (dx 2 + dy 2 + dz 2 ) Thus the path of least travel time minimizes the distance traveled in a geometry given by dl 2 = [n(x a )] 2 (dx 2 + dy 2 + dz 2 ). (b) The Christoffel symbols for this geometry are Γ a bc = n 1 ( δ a c b n + δ a b c n δ bc δ ad d n ) where δ ab = δ a b = δb a = the Kronecker delta. The acceleration of a worldline is u b b u a = dxb dl dx a b dl = d2 x a dl + dx b dx c 2 Γa bc dl dl + ( n 1 δc a b n + δb a c n δ bc δ ad d n ) dx b dx c dl dl + 2 dn dx a n 1 dx b dx c δ bc n dl dl dl dl a n [ ) ] d = n (n 2 2 dxa n 1 a n = 0, (1) dl dl = d2 x a dl 2 = d2 x a dl 2 where we have used the normalization of the worldline of a geodesic, with the metric being n 2 δ ab, in the last step. Finally it might be convenient to change the parametrization of the curve to distance in real, flat space ds, rather than distance in the artificial geometry dl above. So we use dl = nds to obtain ) d (n dxa a n = 0, ds ds for the path of a light ray through a medium. 5. An observer is stationed at fixed radius R in the Schwarzschild geometry of mass M. A proton moving radially outward from the star traverses the observer s laboratory. Its energy E and momentum P are measured. (a) What is the relationship between E and P? (b) What are the components of the four-momentum of the proton in Schwarzschild coordinates in terms of E, P and R? (a) Let the observer have four-velocity v a and the proton have four-velocity u a. Then mu a v a = E, Page 3

4 and Thus p 2 = m 2 u a u b (g ab + v a v b ) = m 2 + m 2 (u a v a ) 2. p 2 = E 2 m 2. (b) For the observer v a = [dt/dτ, 0, 0, 0] and the normalization v a v a = 1 implies that dt/dτ = (1 2M/R) 1/2. Then, mu a v a = E implies that mu t = E/(1 2M/R) 1/2. Further, u a u b g ab = 1 implies that (u r ) 2 = (1 2M/R)(u t ) M/R. This results in mu r = (1 2M/R) 1/2 (E 2 m 2 ) 1/2 = (1 2M/R) 1/2 p. Together, these give mu a = [E/(1 2M/R) 1/2, (1 2M/R) 1/2 p, 0, 0] 6. A spaceship is moving without power in a circular orbit about a Schwarzschild black hole of mass M. The radius of the orbit in Schwarzschild coordinates is 7M. (a) What is the period of the orbit as measured by an observer at infinity? (b) What is the period of the orbit as measured by a clock in the spaceship? Here is a summary of useful formulae for the remaining problems. For a general worldline in equatorial plane of the Schwarzschild geometry: u a = [E/(1 2m/r), ṙ, 0, J/r 2 ] u a u a (1 2m/r) = E 2 + ṙ 2 + (1 2m/r)J 2 /r 2 Ω dφ/dt = J(r 2m)/r 3 E For a geodesic: E and J are constants of the motion. If E > 1 the orbit is unbound and the impact parameter is given by For a circular geodesic: b 2 = J 2 /(E 2 1). E 2 = (1 2m/r) 2 /(1 3m/r) ṙ = 0 J 2 = mr 2 /(r 3m) Ω 2 = m/r 3 Ω 2 r = m(r 6m)/r 4 Page 4

5 (a) ΩT = 2π defines the period T, thus T = 2π r 3 /m = 2πm 343 = 14πm 7. (b) The proper time τ in the spaceship runs with dt/dτ = 1/ 1 3m/r. Thus, r 3 T ss = T/(dt/dτ) = 2π 1 3m/r = 28πm. m 7. A comet starts at infinity, goes around a Schwarzschild black hole of mass M and goes out to infinity. The impact parameter at infinity is b. The radius at closest approach is R, in Schwarzschild coordinates. What is the speed of the comet at closest approach as measured by a stationary observer at that point? At the position of closest approach ṙ = 0, and the normalization condition gives E 2 = (1 2m/R)(1 + J 2 /R 2 ), where R is the value of r at closest approach. Also, the impact parameter is given by b 2 = J 2 /(E 2 1). These two equations may be solved for E and J in terms of b and R, E 2 1 2m/R = R 2 b 2 R 2 b 2 (1 2m/R) J 2 = 2mRb 2 b 2 (1 2m/R) R 2. Let the speed of the comet at closest approach as measured by an observer v a, who is static at the point of closest approach, be u. Also v a = [1/ (1 2m/R), 0, 0, 0], from the normalization v a v b = 1. Then, and These last two equations give u 2 1 u 2 = ua u b (g ab + v a v a ) = 1 + (u a v a ) 2 u a v a = E/ 1 2m/R. u 2 = 1 1 2m/R E 2, which, after a substitution for E 2 in terms of b, implies that u 2 = 1 R2 b 2 (1 2m/R) R 2 b 2 = 2mb 2 R(b 2 R 2 ), Page 5

6 8. A beam of photons with a circular cross section of radius a is aimed toward a Schwarzschild black hole of mass M from far away. The center of the beam is aimed at the center of the black hole. What is the largest radius a = a max of the beam such that all of the photons are captured by the black hole? The capture cross section is πa 2 max. Rather than a, Let me use b for the impact parameter of a photon then b = J/E. Let r = R at closest approach, where ṙ = 0. Thus, E 2 (1 2m/R)J 2 /R 2 = 0. In principle, we can solve these last two equations for R in terms of b from the roots of When R 3 b 2 ((R 2m) = 0. b = 3 3m the three roots are R = (3, 3, 6). For b < 3 3m there is only one real root and it is negative. For b > 3 3m there are three real roots: one greater than 3 3m, one positive but less than 3 3m, and one negative. Close examination shows that, for b > 3 3m, the photon passes the black hole and returns to infinity, for b = 3 3m the photon orbits the hole and spirals in toward (but never reaches) the photon orbit at 3m, and for b < 3 3m the photon is ultimately swallowed up by the hole. Thus the cross-section is σ = π(3 3m) 2 = 27πm (a) What is the speed of a particle in the smallest possible circular orbit in the Schwarzschild geometry as measured by a stationary observer at that orbit? Note: The orbit in question happens to be unstable. (b) What is the relationship between this orbit and the unstable circular orbit of a photon in the Schwarzschild geometry? (a) A circular orbit at r has an orbital frequency, measured at infinity, given by Ω 2 = m/r 3. A static observer at r has four-velocity v a = [1/ 1 2m/r, 0, 0, 0] and measures the speed u of the particle to be given by As in problem 7, u 2 1 u 2 = ua u b (g ab + v a v b ) = (v a u a ) 2 1. (u a v a ) 2 = E 2 /(1 2m/R) = (1 2m/R)/(1 3m/R) for a circular orbit. Page 6

7 Thus, which yields u 2 = (1 2m/R)/(1 3m/R) 1, 1 u2 m u = r 2m, and we see that u = 1 in the limit that the orbit approaches r = 3m. (b) This is the same radius as the photon orbit, r = 3m. Page 7