Homework 7 Solutions

Transcription

1 Homewok 7 olutions Phys 4 Octobe 3, 208. Let s talk about a space monkey. As the space monkey is oiginally obiting in a cicula obit and is massive, its tajectoy satisfies m mon 2 G m mon + L 2 2m mon 2 G L 2 m mon 3 = E2 2m mon. ) To wite this expession, we have multiplied the enegy consevation equation by the mass of the monkey m mon. Now, E is just the monkey s enegy and L is its angula momentum, and not the espective quantities pe unit mass. Futhe, a cicula obit can only occu whee the potential is flat which poduces an additional constaint: G N Mm 2 mon 2 L 2 + 3G N ML 2 = 0 2) Knowing the monkey s mass m mon and adius m mon, one can use these two equations to solve fo the monkey s angula momentum, L. Now, when the coconut is dopped in, it has exclusively adial velocity, and so has no angula momentum. That is, befoe and afte the monkey catches the coconut, it has the same angula momentum. By catching the coconut, howeve, the mass of the monkey and coconut system is lage than just the monkey the monkey. o, it s possible fo the monkey-coconut to still be in a cicula obit of adius if G N Mm mon + m coco ) 2 2 L 2 + 3G N ML 2 = 0 3) Howeve, this equation may o may not have a solution fo the adius, depending on the mass of the monkey and coconut. If a cicula obit is not possible, then thee ae thee possibilities: elliptical obit, open obit slingshot to infinity), o sucked into the black hole. Which of these possibilities is ealized depends on the shape of the potential afte the monkey catches the coconut. If the monkey was initially in a minimum of the potential, the potential still has a minimum, and the monkey does not gain enough enegy in catching the coconut, then the obit will be elliptical. If the monkey had a sufficiently lage initial angula momentum so thee was a baie to fall in the black hole and the monkey gained a significant enegy in catching the coconut but not enough to go ove the baie),

2 then the monkey is slingshot off to infinity. If the monkey was initially at an unstable cicula obit o the mass of the coconut was significantly lage to decease the angula momentum pe unit mass, then the monkey would get sucked into the black hole. I guess the evil baboon would know befoe he thew the coconut in. : 2. Now, we e asked to detemine the minimum ate at which the adial coodinate must decease when you e inside the event hoizon of a chwazschild black hole. We ae also told that this doesn t necessaily need to be a geodesic; that is, the paticle could have a little ocket stapped to it. Nevetheless, its tajectoy is still descibed by the space-time inteval: ds 2 = 2 = 2G ) 2 + 2G ) d dω 2. 4) We ve intoduced the pope time τ hee. We can solve fo d/ as ) 2 d = 2G ) + 2G Howeve, as we ae inside the event hoizon, we have that ) 2 ) 2 2 2G ) ) 2 dω. 5) < 2G N M, 6) so we can expess the ight-hand side as a sum of manifestly positive quantities: ) 2 d 2GN M = ) 2GN M + ) 2 ) 2 ) ) GN M dω. 7) The ight-most two tems ae always non-negative, so we establish the bound d 2GN M. 8) Then, we ae asked to calculate the maximum lifetime as the paticle tavels fom the even hoizon to the singulaity at = 0. Because the adius is deceasing, the maximum lifetime coesponds to d = 2GN M. 9) We can integate both sides: 0 2G N M d = 2G N M = τ. 0) 2

3 The integal ove can be done with actangents and such, but the final esult is just τ = πg c 3, ) whee we have einseted factos of the speed of light, c. If we say that M = αm, with the sola mass M = kg. 2) α is some constant the epesents the numbe of sola masses the black hole is. Then, plugging in Newton s constant and the speed of light, we find τ α s. 3) o, even fo the appoximately 4 million sola mass black hole at the cente of ou galaxy, you might have a minute in which to contemplate life befoe enteing the black hole. Finally, on a geodesic, / is popotional to the conseved enegy, so indeed if this is 0 you must be taveling adially dω = 0) and so the lifetime is maximized. 3. Now we e asked to analyze the motion of a beacon that is dopped fom a spatial coodinate, θ, φ ) on a adial geodesic of a black hole of mass M. The beacon emits light at a wavelength λ em in its est fame. a) Fist, we ae asked to calculate the coodinate velocity, d/, as a function of adius. This is easiest to do with the consevation equations. The enegy is E = 2G ). 4) When τ = 0, the beacon is stationay and at the location of the spaceship. Theefoe, we can calculate the initial / value and use this to detemine the value of E along this geodesic. We have the spacetime inteval: ds 2 = 2 = 2G ) 2, 5) because we stat at. Theefoe, because time tavels fowad, we have = τ=0 2G ) /2. 6) Plugging this into the expession fo the enegy, we have E = 2G ) = 2G. 7) τ=0 3

4 Theefoe, the ate that time ticks away is = R. 8) Next, on the adial geodesic, we have the total enegy consevation law ) 2 d = E 2 R ). 9) Plugging in the esult fo the total enegy fom ealie, we find ) 2 d = R + R = R R. 20) O, because the adius deceases on this geodesic, d = R Now, we just take the atio of d/ to /: R. 2) d = d/ / = R R /2 R ). 22) Note that the coodinate velocity goes to 0 as the beacon appoaches the event hoizon; that is, it neve eaches the event hoizon fom the pespective of the spaceship. b) Now, we want to calculate the pope speed of the beacon. Actually, we aleady have all of the ingedients fo this. We had identified the pope speed d/ as d = R R. 23) When = R = 2G N M, this pope speed is d = R, 24) =Rs which is some sensible finite velocity. If the initial adius, then this velocity becomes the speed of light. With a geneous intepetation, this is the consequence that once you ae at the event hoizon of a black hole, you have to tavel faste than the speed of light to escape off to infinity. 4

5 c) Now we e asked to calculate the wavelength of the emitted light λ em as obseved by the spaceship, λ obs. We now want to find the coodinate time at which the light is obseved at the spaceship. We can define the fequency of the light o coespondingly the invese wavelength as ω = λ = g µνu µ dxν. 25) Hee, U µ is the fou-velocity of the beacon and x ν is the null geodesic tajectoy of light as it tavels fom the beacon. The fou-velocity of the beacon is just defined fom its pope velocity as found in the pevious poblem: U µ = R R ), R R, 0, 0 ). 26) The geodesic of the light can be identified by demanding that it is null: dx µ dx ν 0 = g µν = R ) ) 2 + ) 2 d, 27) o that d = R ). 28) Now, plugging this into the expession fo the invese emitted wavelength we have = g µν U µ dxν λ em = R ) R = R + R R em R em ) 29) ) + R R R em em em. To detemine the emaining time deivative, we calculate the conseved enegy E as a poduct of the time-tanslation Killing vecto and the photon s tajectoy: E = g µν K µ dxν = R ). 30) Theefoe, the time deivative is = E. 3) Theefoe, the emitted wavelength is = E R ) R + R. λ em em em ) 5

6 We do the same execise at the spaceship. Now, the spaceship s fou-velocity is U µ = /2, 0, 0, 0), 32) because the spaceship is stationay, located at the adius. obseved by the spaceship is theefoe The wavelength = g µν U µ dxν λ obs = R ) /2. 33) We can eplace the time deivative by the conseved enegy E at the adius so that = E /2. 34) λ obs Putting it all togethe, the obseved wavelength is [ λ obs = R R ] R + R λ em. 35) em em d) Now, we want to detemine the time at which the light fom the beacon is obseved at the spaceship. Because the light tavels along a null geodesic, we have o that ds 2 = 0 = R ) ) 2 + R ) ) 2 d, 36) d = R ). 37) Then, we can integate both sides to detemine the elapsed coodinate time t obs : em d = R tobs 0 = t obs. 38) This can be integated with integation by pats, and the answe is t obs = em + R log R em R. 39) Again, note that if em R, the time to obseve the light is longe and longe. 6

7 e) Finally, we ae asked to calculate the edshift facto which is the atio of the wavelength of the obseved light to the emitted light. Fom ealie, we had found [ λ obs = R R ] R + R. 40) λ em em em At late times, the emitted adius appoaches the chwazschild adius, so this expession simplifies to λ obs 2 R ) 2 R R. 4) λ em em em R Fom the pevious pat, we can solve fo the atio of adii diffeences in tems of obseved time t obs : [ ] [ ] R tobs em ) tobs R ) = exp exp, 42) em R R whee we have taken the late-time expession on the ight. Theefoe, the atio of wavelengths is λ obs = 2 R [ ] tobs R ) exp. 43) λ em R It then follows that the time constant T = R = 2G N M. R 7