Physics 161 Fall 2011 Extra Credit 2 Investigating Black Holes - Solutions The Following is Worth 50 Points!!!

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1 Physics 161 Fall 011 Exta Cedit Investigating Black Holes - olutions The Following is Woth 50 Points!!! This exta cedit assignment will investigate vaious popeties of black holes that we didn t have time to investigate in lectue We will answe the following questions: 1 What is the closest obit that a massive body can have aound a black hole, and is it stable? Can light obit aound a black hole, and, if so, what is the distance and is it stable? et s begin with the chwazschild metic, expessed in tems of the pope time, dτ = dt 1c [ 1 R / + dθ + sin θdφ ), 1) whee R = G N M/c is the chwazschild adius Now, since we ae inteested in obits, which occu in a plane, we can choose coodinates such that θ = π/, which gives dτ = dt 1c [ 1 R / + dφ, ) Dividing though by dτ gives 1 = 1 R ) ) [ dt 1 d/dτ) dτ c 1 R / + ) dφ 3) dτ Now, the complete theoy of Geneal Relativity tells us that thee ae some constants of motion, E = 1 R mc = dφ, 4) m dτ whee E is the enegy of the body of mass m, and is the angula momentum Hee s whee you wok begins: ) dt dτ 1 Plug in above constants to Eq 3) to find c ) = E d m c dτ m Next, use the chain ule to wite d dτ = d dφ dφ dτ, and plug in again fo dφ/dτ to find c = E m c m 4 1 ) d 5) dφ m

2 The expession in Eq 5) is a mess, full of nonlineaities The standad way of handling this sot of obital expession is to make the substitution Using this, show that Eq 5) becomes m c 1 R u) = u = 1 ) E ) du u 1 R u) 6) dφ 3 Now, we could solve fo the deivative and integate, but a simple way of poceeding is to diffeentiate eveything with espect to φ, noting that u is the only vaiable Do this, and cancel off a common du/dφ to find a much simple equation, u dφ + u = 3 R u + R m c 7) 4 Detemine the adius of the constant obit, 0 = u 1 0 Note: you ll need to solve a quadatic equation Explain which sign fo the squae oot you need to take 5 We now want to investigate the stability of this obit, looking at small petubations about a constant obit, u φ) = u 0 + δu φ), 8) whee δu u 0 Plug Eq 8) into Eq 7), expand out and neglect tems of ode δu how that you get an equation of the fom dφ δu + ω 0δu = 0, 9) and detemine the hamonic oscillato fequency, ω 0 Detemine the smallest adius that keeps the obit stable to these small petubations What does the angula momentum need to be fo this obit? 6 The above analysis has been pefomed fo a massive body Now we want to look at the obit of light Fotunately, we can simply use Eq 7) setting m 0 What is the adius of the constant obit, 0 = u 1 0? 7 Finally, pefom the same stability analysis as fo the massive body, and show that thee is no stable obit fo light! olution

3 1 Plugging Eqs 4) to Eq 3) gives 1 = ) 1 R dt ) [ dτ 1 d/dτ) + ) c 1R / dφ dτ = ) 1 R 1 [ E 1 d/dτ) + ) m c 4 c 1R / m = ) 1 R 1 [ E 1 d/dτ) + m c 4 c 1R / m Multiplying though c 1 R /) gives c ) = E d m c dτ m Now, fom the chain ule d dτ = d dφ dφ d = d dφ m, whee we have plugged in fo dφ/dτ Inseting this esult into ou above expession gives c ) 1 R = E ) ) m c dτ m 1 R ) ) = E d m c m 4 dφ m 1 R, which is Eq 5) Making the substitution u = 1 says that Hence, Eq 5) becomes d dφ = d du du dφ = 1 du u dφ c 1 R u) = E m c u 4 m 1 = E m c du u 4 dφ du m dφ ) u 1 R m u) ) u 1 R m u) Now, getting the deivative tem alone by multiplying by m / gives ) m c ) E du 1 R u) = u 1 R u), dφ and is Eq 6) 3 Taking the deivative of Eq 6) gives m c R du s dφ = 0 du dφ Canceling off the the common du/dφ tem gives ) u dφ udu dφ + 3R u du dφ m c R = d u dφ u + 3R u 3

4 We can ewite this as giving us Eq 7) u dφ + u = 3 R u + R m c, 4 The constant adius 0 = u 1 0 has zeo deivative, so u 0 /dφ = 0, so Eq 7) educes to 3 R u 0 u 0 + R m c = 0 This is just a quadatic equation with solutions ) R mc 1 ± 1 3 3R Now, we need to detemine which sign to take In the limit that R 0 ie, the mass of the black hole goes to zeo), then the obital adius should go to infinity the obit is just a staight line), meaning that u 0 0 In this limit then u 0 1 3R [ 1 ± 1 3 ) ) R mc = 1 ± 1 3R 1 mc ) R The second tem vanishes when R 0 In ode fo the fist tem to vanish in this limit we need to choose the minus sign Thus, ) R mc 1 1 3, 3R such that the obital adius is 0 = 3R R ) 10) mc 5 Now, to detemine the stability of this obit we make the petubation, u = u 0 + δu, such that u dφ = d dφ δu, and so Eq 7) eads Expanding the squae gives dφ δu + u 0 + δu = 3 R u 0 + δu) + R m c dφ δu + u 0 + δu = 3 R u 0 + 3R u 0 δu + 3 R δu + R m c 4

5 Now, we can neglect the small δu tem, to find afte eaanging a bit) dφ δu + δu 3R u 0 δu = 3 R u 0 u 0 + R m c Now, because of the definition of u 0 the ight-hand side of this expession vanishes Collecting the δu tems we find which is Eq 9) with dφ δu + 1 3R u 0 ) δu = 0, ω 0 1 3R u 0 Now, the obit will be stable as long as ω 0 > 0 this means that the petubations will be hamonic oscillatos, stably oscillating aound the constant obit, u 0 ) Howeve, if ω 0 0 then the petubation solutions become exponentials, destoying the stability fo the gowing exponentials) o, the cutoff is when ω 0 = 0, so 3R 0 = 3R o, the minimum stable obit has to be just outside thee times the chwazschild adius Fom Eq 10) we see that the angula momentum would have to be such that 0 = 3R = 3R mc 6 Now we want to pefom the same analysis fo light etting the mass m = 0 in Eq 7) gives u dφ + u = 3 R u Again, looking fo a constant adius, u 0 we need to solve u 0 = 3 R u 0 1 u 0 = 0 = 3 R ight is able to obit moe closely than the massive body, as we should expect, at a distance of only 15R 7 Finally, petubing the light obit u = u 0 +δu and neglecting tems O δu ) as befoe gives dφ δu + u 0 + δu = 3 R u 0 + 3R u 0 δu d dφ δu 3R u 0 δu = 0 Now, since 3R u 0 is always geate than zeo, the solutions ae always exponentials, showing that the photon obit is always unstable! 5