Physics 161: Black Holes: Lectue 5: 22 Jan 2013 Pofesso: Kim Giest 5 Equivalence Pinciple, Gavitational Redshift and Geodesics of the Schwazschild Metic 5.1 Gavitational Redshift fom the Schwazschild metic We aleady discussed the cuvatue of time in the Schwazschild metic. Clocks un slowe in a gavitation field (at smalle values of ). This is one of the moe measuable effects of GR. Measuements ae possible because atomic tansitions act like small clocks which send messenges (photons) to us hee at Eath. Recall the pope time in the Schwazschild metic: dτ = dt 1 2GM c 2, whee dτ is tick of a clock measued at, and dt is the length of that same tick measued fa away ( ). Now conside some atomic tansition, fo example, the n = 3 to n = 2 tansition of atomic Hydogen. This is called the H-alpha tansition and esults in emission of a photon with enegy E γ = 13.6eV( 1 3 1 2 2 ) = 1.89 ev. Using λ = c/ν, and E 2 γ = hν, we find the coesponding wavelength Hα : 6563.5 Angstoms. Now in the fame at, the photon has a fixed fequency ν 0, and each oscillation is like a little clock ticking with peiod dτ 0 = 1/ν 0. Thus the time between cests in the outgoing photon measues dτ. That same photon tavels out to us at =. Then that same peiod is obseved as dt. Thus the dτ dτ obs = dt = dτ 0 / 1 2GM/c 2 is longe. Using the definitions above of ν and λ we see that the fequency deceases by the same facto and the moe easily obseved wavelength is longe at when obseved at infinity than the emitted wavelength at by λ obs = λ 0 λ = 0 1 2GM 1 S, c 2 whee we used the definition of the Schwazschild adius S. The edshift of an emission line is defined as z = λ = λ obs λ 0 = (1 S λ 0 λ 0 )1/2 1, whee λ 0 is the emitted wavelength in the est fame of the atom. Note fo small values of 2GM/c 2 = S /, this expession can be Taylo expanded as z GM/c 2 = 1 2 S/. 1
This gavitational edshift is totally diffeent fom eithe the nomal Dopple shift edshift caused by an atom moving away fom the obseve, o the cosmological edshift caused by the expanding Univese. Let s see how big the effect is by calculating it fo some typical astonomical objects. Fist how about emission fom the suface of the Sun? We expect the wavelengths ecieved by us on Eath to be longe than the emitted wavelengths since the emitted photons have to climb out of the potential well of the Sun whee clocks un slowe. Using the appoximation above z 1 2 3km/7 105 km 2 10 6. This is not a vey big effect; a clock on the suface of the Sun loses about 30 seconds/yea. Fo H-alpha the line edshift would be only 0.014 Angstoms. How about fo a white dwaf sta? The mass of a white dwaf is about 0.6M, and the adius is about 5500 km, nea to the Eath adius. Thus z 1 2 (.6)(3km)/5500km 1.6 104, and the wavelength of H- alpha emission ecieved on Eath would be 6563.5(1+z) = 6564.6 Angstoms, quite easily distinguishable fom local H-alpha lines. This esult agees well with the measuements fom the neaby white dwaf sta Siius B. A clock on the suface of Siius B would lose about 0.6 second/hou. Next conside the case of a neuton sta, with M = 1.4M and R = 10 km. Now S = (1.4)(3km) = 4.2 km, so eally should not use the appoximation, but calculate z = [1(4.2km/10km)] 1/2 1 = 0.313. This is big shift: the H-alpha line would appea at 8618.3 Angstoms and clocks would un 31% slowe. Finally what about light emitted fom the Schwazschild adius of a black hole? With = S, z! Light would be invisible since it would be edshifted to infinitely low enegies. Clocks sitting at that location would appea to have stopped! 5.2 Light bending and the Equivalence Pinciple To find pope answes in GR we calculate using the metic, in ou case the Schwazschild metic, but thee is an inteesting, moe intuitive way that uses the Equivalence Pinciple diectly. Let s look biefly at two examples. Imagine someone in a falling elevato. The equivalence pinciple says that expeiments done in this falling fame will give the same esults as expeiments done floating in fee space fa fom any massive object. Suppose the guy in the falling elevato shines a lase beam hoizonally fom one side of the elevato to the othe. That light must follow geodesics and theefoe go staight acoss the elevato; if the beam leaves fom a height of 3 feet above the elevato floo, it must aive on the othe side also thee feet above the floo. Fom the point of view of the elevato man, the light must tavel exactly hoizontally. But now think of someone watching though the (glass) walls of the elevato. The man shoots the lase fom a height of 3 feet above the floo, but then while the light is taveling acoss, the elevato continues to fall. Thus when the light hits the opposite wall 3 feet above the floo, the wall has moved down. Theefoe the light cannot go staight in the Eath fame. The light must also fall. This gedanken expeiment shows the light must fall in a gavitational field in the same way as eveything else. Of couse, fom the point of view of GR, light is just following the geodesic though a cuved spacetime; we will calculate those soon, but hee we fosee what that calculation will give. In paticula, light fom distant stas tavelling nea the limb of the Sun must be deflected towads the Sun. Si Autho Eddington did this expeiment in 1919, and it ganeed wold wide fame fo Albet Einstein, because it was one of the fist poofs of Einstein s GR theoy. In the homewok, I am asking you to calculate how fa the light falls in a gavitational field, that is how much it bends. 5.3 Gavitational Redshift again The equivalence pinciple can also be used to show that light must gavitationally edshift. Again conside ou man in a falling elevato. This time he shoots his lase fom the floo to the ceiling. 2
Fist think about it fom the point of view of someone outside watching though the glass walls. In the Eath fame the feely falling elevato picks up speed while the light is in tansit, and so thee is Dopple shift caused by the speeding elevato ceiling. The time the photons spend in tansit is t = h/c, whee h is the height of the elevato, and the speed incease of the elevato is v = gt = gh/c, whee g = GM Eath / 2 = 9.8m/s 2. We could find the expected Dopple shift fom the elativistically coect fomula ν obs = ν 0 (1 + v)/(1 v), but a Taylo expansion of this is good enough fo the slow speeds involved hee: ν obs ν 0 (1 + 1 2 v)(1 + 1 2 v) ν 0(1 + v), o ν = ν obs ν 0 = vν 0 Thus in the Eath fame we expect the light beam to be blue shifted by ν/ν 0 = v/c = gh/c 2, whee I put the c s back in. Howeve, the equivalence pinciple says that this cannot be! The man in the elevato can t tell whethe he is falling o floating feely in deep space, so he expects coectly that the fequency of light ecieved at the elevato s ceiling is the same as the fequency he sent fom the floo. Thus thee must be anothe effect to cancel this one! The light taveling up must be gavitationally edshifted to countebalance this effect. Thus thee must be a facto of ν/ν 0 = v/c = gh/c 2 = GMh/ 2 c 2. This is not quite the esult we got befoe fom the metic, since thee is a facto of 2 athe than 1, but befoe we found the edshift between = 0 and fa away ( = ) and hee we found it only between 0 and. So we can continue with this deivation by integating the small edshift we found z ν/ν 0 fom 0 to. z ν ν 0 dν ν 0 GM 2 c 2 d. This gives ln(ν /ν 0 ) GM/ 0 c 2, o λ = λ 0 exp (GM/ 0 c 2 ). This is still not the coect answe, but if we assume the facto in the exponential is small and Taylo expand the exponential using e x 1 + x + x 2 /2 +..., we do find λ /λ 0 1 + GM/c 2, o z = GM/c 2 = 1 2 S/, the Schwazschild esult fo small z. 5.4 Geodesics of Schwazschild metic fom Eule-Langange Let s etun to ou wok on geodesics and apply the Eule-Lagange equations to find the geodesics of the Schwazschild metic. Remembe that the Schwazschild metic is the unique metic aound stationay, spheically symmetic, unchaged objects, so what these geodesics do is tell us how thing move aound the Eath, aound the Sun, and aound unchaged, non-spinning black holes. These ae the Geneal Relativistic extension of Newton s and Keple s laws of motion. Recall the Schwazschild metic is ( ds 2 = 1 2GM ) ( c 2 dt 2 + 1 2GM ) 1 c 2 d 2 + 2 dθ 2 + 2 sin 2 θdφ 2. As above, we set c = 1, and take affine paamte λ = s = τ, and extemize s = Ldτ, with [ ( L = 1 = 1 2GM ) ] 1/2 ṫ 2 ṙ 2 1 2GM 2 θ2 2 sin 2 θ φ 2 The Eule-Lagange equation fo t is then d dτ ṫ t = 0, which since / t = 0, implies thee is a conseved quantity we will call enegy pe unit mass: ṫ E m. 3
Calculating, ṫ = 1 2 [ ]1/2 (1 2GM )2ṫ, whee we abbeviated L = [ ] 1/2. Using L = 1, we find ou t equation ( 1 2GM ) ṫ = E m. We don t yet know that the constant has anything to with enegy, but we call it E/m, because of ou expeience with the Minkowski metic. Fo, the Schwazschild metic goes to the Minkowski metic and fo the Minkowski metic ṫ = E/m. Next we find the φ equation: d = dτ φ φ = 0, so again we have a conseved quantity p φ = / φ l/m, whee we will call this conseved quantity the angula momentum pe unit mass. Recall Noethe s theoem which says if physics is unchanged by a otation then angula momentum is conseved. Since the metic does not depend explicitly on the angle φ, we get that esult hee. Doing the diffeentiation we find φ = 1 2 [ ]1/2 ( 2 sin 2 θ2 φ) = 2 sin 2 θ φ. Thus ou φ equation eads l m = 2 sin 2 θ φ. Note that it makes sense that we called the constant of motion l/m, since this matches the nomal definition of angula momentum, l = P, with v = sin θ φ. Next, we conside the θ equation. Hee we find that d dτ = θ θ 0, thus we do not have a conseved quantity fo this equation. We find θ = 1 2 [ ]1/2 ( 2 φ2 2 sin θ cos θ) = 2 φ2 sin θ cos θ, and θ = 1 2 [ ]1/2 ( 2 2 θ) = 2 θ. Thus ou θ equation eads: d dτ (2 θ) = 2 φ2 sin θ cos θ. Finally we come do the equation, which is kind of messy because of all the explicit dependence in the metic. Howeve, we don t have to do it, because we can get the fouth equation we need to specify the equations of motion fom ou definition of the Lagangian, L 2 = 1: 1 = (1 2GM )ṫ 2 ṙ 2 (1 2GM ) 2 θ2 2 sin 2 θ φ 2. 4
Since ou object and metic ae spheically symmetic we can simplify things by only consideing motion in the equatoial plane (θ = π/2, and θ = 0). Of couse we have to emembe that we made this assumption late when we use ou equations! If we ty to conside motion that has a changing θ, o which is not in this plane we would need to come back to the equation above. In this case then, the equation L = 1 becomes: 1 = (1 2GM )ṫ 2 ṙ 2 (1 2GM ) 2 φ2. Now eliminate vaiables othe than, using the constants of motion we have fom the above equations: l/m = 2 φ, and E/m = (1 2GM )ṫ, to get: 1 = E 2 m 2 (1 2GM ) ṙ2 (1 2GM ) l2 m 2 2. We can wite this in a nice fom by solving fo mṙ 2, ) mṙ 2 = (m E2 m + l2 m 2 (1 2GM ), Remembeing the definition of ṙ and using dimensional analysis to put back the c s, we can wite this as: ( ) 2 ( d m = E2 dτ mc 2 1 2GM ) ) (mc 2 c 2 + l2 m 2. Notice that, as I mentioned in my handwaving desciption, a step in pope time dτ foces a step d in the diection. Thus this geodesic equation shows that things fall due to the spacetime cuvatue of the metic. We will look at these geodesic equations in some detail, but fo now let s just take one limit of this last equation. Suppose the angula momentum l = 0, which we might expect fo adial infall towads a spheical mass. Ou equation then is: m( d ( dτ )2 = E2 mc 2 1 2GM ) c 2 mc 2 = 0. Next conside a case whee you stat at est fa fom the object so that at pope time τ = 0, m(d/dτ) 2 = 0, and. Ou equation becomes: E 2 2GM (1 mc2 )mc2 = 0, o E 2 /mc 2 = mc 2, o simply E = mc 2! So at τ = 0 the total enegy is just E = mc 2. In Newtonian mechanics the enegy at infinity is usually defined as E = 0. Isn t it nice how these impotant esults ae just built ight into the geneal elativistic view of spacetime. Also since enegy E is conseved along geodesics we know that E = m always. This E is not the Newtonian enegy; it is the conseved quantity, which is bette than the sum of 1 2 mv2 + V. Finally note that if we would have stated with some velocity at = then E > mc 2 but it still would have been conseved. At late times duing this adial infall fom est, ou equation becomes: m( d dτ )2 = E2 mc 2 mc2 + 2GM c 2 5 mc2 = 2GMm.
That is simply 1 d m( 2 dτ )2 GMm = 0, which looks emakably like the Newtonian case 1 2 mv2 GMm/ = 0! But this is not the same equation. It is fully elativistic and the time deivative is τ not t (emembe t = γτ in SR). In geneal you integate these 4 equations to get the complete pictue of motion nea the Eath, Sun, o Black Hole. We will come back to these soon. 6