Physics 161: Black Holes: Lecture 6: 14 Jan 2011

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1 Physics 161: Blck Holes: Lecture 6: 14 Jn 2011 Professor: Kim Griest 6 Geodesics: Moving in stright lines through curved spcetime We sw tht grvity curves time nd spce. A very importnt result is how things move in curved spce. The bsic principle is tht things move on pths tht re s stright s possible. These cn be defined s the pths which extremize the distnce between two points, nd re clled geodesics. In 3-D Euclidin geometry stright line is defined s the shortest distnce between two points, nd Newton s lw sy in the bsense of outside forces prticles move long such lines. It is similr in GR, but sometimes it is the mximum invrint intervl which is relevnt, not the minimum. Tht is why we sy extremizes the distnce insted of minimizes. To get n ide of this, consider the 2-D nlogy of n nt crwling on the surfce of n pple. The nt is forced to follow the curved 2-D spce of the pple skin, but suppose it wlks s stright s possible, i.e. not veering to the left or right. As it wlks from one side of the pple to the other coming ner the stem, the nt will be deflected nd come wy from the stem t different ngle thn it pproched the stem. If the nt is close to the stem it could even circle the stem continully while still following the geodesic. This should seem similr to the nlogy of the two surveyors. Now consider two points on opposite sides of the stem nd sk which pth is the shortest between them (the geodesic). Note tht without the curved surfce the shortest pth would be quite different. However, the metric requires us to sty on the surfce. If we wnt to clculte the motion of objects in GR, we need to be ble to find the geodesics. By finding them we will be ble to derive Newton s lws from the Schwrzschild metric, s well s Einsteinin corrections Newton s lws. We cn find out the rel nswer obeyed by ctul objects in the solr system. We will lso be ble to find out how objects move round blck holes. 6.1 Geodesics nd Clculus of Vritions GR sys tht the motion of prticle tht experience no externl forces is geodesic of the spcetime metric. One cn summrize GR in two sttements: 1. Mtter nd Energy tell spcetime how to curve. 2. Curved spcetime tells mtter nd energy how to move. In solving for the geodesics we re finding how mtter nd energy (light) move. In 3-D Euclidin spce the definition of geodesic, k stright line is the shortest distnce bewteen two points. Mthemticlly this cn be found from clculus of vritions on the metric distnce. The 1

2 differentil distnce ds = dx 2 + dy 2 + dz 2 cn be integrted between two end points to find s = ds = 1 + (dy/dx)2 + (dz/dx) 2 dx, where we fctored out dx so the integrl is done over the x-xis. Note now the integrnd contins the derivtives of the the functions y(x) nd z(x) which define the pth. How do find the pth (functions y(x) nd z(x)) tht minimize s? We need generl method. This method is clled the clculus of vritions nd you probbly lerned it in clculus. However, I will remind you of how it is done. If you hven t seen it before, tht s ok, since I ll show you ll you need. This is beutiful nd fun mth tht is used throughout dvnced physics. The bsic ide is to minimize ds the sme wy you minimize ny function is clculus: tke its derivtive, set it to zero, nd solve. This solution will be the geodesic, tht is the extreml pth. This method works on curved surfces since ds mesures the ctul distnce following the curved surfce. It works the sme wy in GR; you just dd the time prt of the metric. 6.2 Geodesics s equtions of motion A geodesic on pure sptil mnifold (e.g. curved surfce) is line. For exmple, stright lines on plne, or gret circles on sphere surfce. In GR the geodesics include time nd so re ctully the equtions of motion! You cn understnd this by remembering the distinction between time nd spce. One hs choice in sptil motion, but one is forced by nture to move forwrd in the time direction: 1 sec per sec. You will hit the yer 2011 no mtter wht you do. Now in flt spce possible geodesic is one in which you don t move in spce t ll: dx = dy = dz = 0. Then ds = dt is n extreml pth nd you just sit there getting older. However, if spcetime is curved, then motion in dt cn require motion in dx! Think of the surveyor nlogy, where motion north (z) required motion in the x nd y direction to sty on the surfce of the Erth. Thus since motion in t direction is forced, motion in the x (or other sptil) direction will lso be forced. Thus you hve the geodesic requiring dx/dt 0. dx/dt is velocity, so you cn t sty t rest nd be on the geodesics. This is how grvity ttrcts things nd why geodesics ner mss require bodies to fll towrds the mss center. Let s do some mth. 6.3 Euler-Lgrnge Equtions To get used to clculus of vritions, let s do clcultion some of you hve lredy done: Lgrnge s equtions in clssicl mechnics. Then we will do it on metrics. Doing it severl times is necessry nd I highly recommend you go over ll these clcultions t home severl times. We wnt to extremize the integrl s = ds, which cn be written s = (ds/dt)dt, where we hve fctored out t. We cn fctor out ny vrible we wnt so s to mke the integrting esier. As n exmple, consider the generl lest ction integrl S = Ldt, where L is clled the Lgrngin. In our exmple L = ds/dt, but in regulr clssicl mechnics the Lgrngin is L = T V = 1 2 mẋ2 V (x), where ẋ = dx/dt = v x, T is the kinetic energy, nd V (x) is the potentil energy. (Here we used only the x nd t dimensions.) Thus for 1-D clssicl mechnics S = ( 1 2 mẋ2 V (x))dt is clled the ction, nd the equtions of motion, F = m, with F = dv/dx 2

3 re found minimizing this ction (principle of lest ction ). In our problem we re trying to find the pth tht gives the shortest (or longest) distnce long the pth between two fixed points. Let s do it first in generl nd get the equtions. S = L(x, y, z, ẋ, ẏ, ż, λ)dλ, where we explicitly put in the velocities v x, v y, nd v z s vribles tht hve to be solved for, nd hve fctored out the generl vrible λ ( usully just time). This lmbd is clled the ffine prmeter nd is the vrible you use to trce long the geodesic pth. In the bove, the dot mens differentiting with respect to λ, i.e. ẋ = dx/dλ. Now we tke the vrition of this integrl δs using the chin rule nd set it to zero: δs = 0, nd solve. ( ) δs = δx(λ) + δẋ + x ẋ y δy + dλ = 0. Next note tht δ(ẋ) = δ dx dλδx, where δx(λ) is smll devition from the extreml pth. Now integrte by prts the term with the δẋ, by using: dλ = d b ẋ δẋdλ = ẋ udv = uv b d dλ vdu. b (δx)dλ = dẋ δx b d dλ dẋ δxdλ. We wnt pth defined by x(λ), y(λ), etc. tht goes from point to point b. Thus we wnt δx(λ), the vrition of the pth from the geodesic to be zero t the end points. Tht is δx(λ = ) = 0 nd δx(λ = b) = 0 (nd similrly for δy nd δz). Thus the first term on the right hnd side of bove eqution vnishes, nd the eqution becomes: ( 0 = x d ) δxdλ + y nd z prts. dλ ẋ The finl step in getting our Euler-Lgrnge equtions is to note tht the vrition in the pth δx(λ), δy(λ), nd δz(λ) is completely rbitrry. Thus for the integrl s whole to vnish, the integrnd itself must vnish everywhere. Tht is, to be true for every possible function δx(λ), the prts of the intregrnd multiplying δx, δy, nd δz must be zero. Thus we hve the Euler-Lgrnge equtions: x d dλ ẋ = 0. y d dλ ẏ = 0. z d dλ ż = 0. 3

4 6.4 First Exmple of Euler-Lgrnge equtions: clssicl mechnics As first exmple of the use of The Euler-Lgrnge equtions, let the Lgrngin be L = T V = 1 2 mẋ2 V (x) nd let λ = t. Then x = V x = F, where we used the norml definition of force s d the derivtive of the potentil energy. Since v = ẋ, ẋ = mẋ, nd dλ ẋ = d dtmẋ = m, where the ccelertion = dv/dt. Thus the Euler-Lgrnge equtions found by extremizing the ction is just F = m. This might seem like lot of work to get something you lredy know, but the beuty of the method is tht it works in difficult situtions nd in difficult coordinte systems. It is usully lot esier to write down the kinetic nd potentil energy thn it is to use the vector form of F = m in complicted situtions. 6.5 Second exmple of Euler-Lgrnge equtions: Flt spce geodesics Now let s extremize the 3-D flt spce metric to see if the shortest distnce between two points is indeed stright line! So s = ds, with ds = dx 2 + dy 2 + dz 2 = 1 + ẏ 2 + ż 2 dx, where we hve chosen λ = x, nd L = 1 + ẏ 2 + ẋ 2. The y Lgrngin eqution thus reds: or or d dx ẏ y = 0, ( ) d 1 dx 2 (1 + ẏ2 + ż 2 ) 1/2 2ẏ 0 = 0, ẏ 1 + ẏ2 + ż 2 = constnt. For the z eqution we find similrly, ż 1 + ẏ2 + ż 2 = constnt. Dividing the y eqution by the z eqution we get ẏ/ż = constnt, so ż = c 1 ẏ. Substituting this into the y eqution, we get ẏ/ 1 + ẏ 2 + c 2 1ẏ2 = constnt. Since the only vrible in this entire eqution is ẏ, solving this eqution for ẏ will give constnt. Thus we find the Euler-Lgrnge equtions for the extreml distnce between two points re dy/dx = m y, nd dz/dx = m z, where m y nd m z re some constnts found by the boundry condition. Thus y = m y x + b y nd z = m z x + b z, the eqution for stright line in 3-D. This gin might seem like lot of work to prove tht the shortest distnce between two points is stright line, but the method is generl. Before going on to extremize the invrint intervl in spcetime metric, I wnt to do the lst problem gin nd show you useful trick. In the bove we used x s the ffine prmeter. Insted we could hve used s itself. Thus we write our integrl s s = ds, with the Lgrngin L = 1! However we write this one in specil wy: ds = dx 2 + dy 2 + dz 2 = ( dx ds )2 + ( dy ds )2 + ( dx ds )2 ds = ẋ 2 + ẏ 2 + ż 2 ds, 4

5 where now L = 1 = ẋ 2 + ẏ 2 + ż 2. As before in the Euler-Lgrnge equtions the term / x = 0, nd similrly for the y nd z equtions, so they reduce to: ( ) d ẋ ẋ2 = 0, ds + ẏ 2 + ż 2 nd similrly for the y nd z equtions. But since L = 1 is constnt it comes out of the differentil, becoming ( ) d ẋ = dẋ/ds = 0. ds L This sys tht ẋ = dx/ds = constnt, or x = m x s + b x, nd similrly, y = m y s + b y, nd z = m z s + b z. Agin the eqution for stright line, with s being the distnce trveled long the line, but with less lgebr. 6.6 Geodesics in Minkowski spcetime Next we use the Euler-Lgrnge equtions to extremize the invrint intervl, ds 2 = dt 2 + dx 2 + dy 2 + dz 2, between two events in Minkowski spcetime. This will give us the equtions of motion of Specil Reltivity! Let s use proper time τ s the ffine prmeter: dτ = ds 2. We write s = dτ = dt2 dx 2 dy 2 dz 2 = ṫ2 ẋ 2 ẏ 2 ż 2 dτ, where ṫ = dt/dτ, ẋ = dx/dτ, etc. Since bsiclly τ = s, here gin the Lgrngin L = 1 = ṫ 2 ẋ 2 ẏ 2 ż 2. Noting tht / t = 0, nd similrly for / x = 0, etc. the Euler-Lgrnge equtions become d dτ ṫ = 0, nd d dτ ẋ = 0, nd similrly for y nd z. Thus s we move long the pth (vrying τ to move long the geodesic) we hve 4 conserved quntities, = constnt, ṫ ẋ = constnt, etc. Explicitly, these give / ṫ = ṫ/l = ṫ = constnt, nd / ẋ = ẋ = constnt, so our equtions re ṫ = c t, ẋ = c x, ẏ = c y, nd ż = c z, or t = c t τ + t 0, x = c x τ + x 0, etc., where the c t,c x, t 0, x 0, etc. re constnts. To find the vlues of the constnt let s simplify to only the x direction nd let time t strt with τ = 0, so t = c t τ. Then x = c x τ+x 0 = (c x /c t )t+x 0, nd we recognize the combintion c x /c t = v x the velocity in the x direction. Thus c x = c t v x. Now evlute the Lgrngin L = 1 = ṫ 2 ẋ 2 = c 2 t c 2 x = c 2 t c 2 t vx, 2 or 1 = c t 1 v 2 x. Thus c t = 1/ 1 vx 2 = γ, the Lorentz fctor, nd we hve s our geodesics the equtions of specil reltivity: motion in stright line with time diltion included: t = γτ, x = v x t+x 0, etc. Note tht it is possible to hve ll the constnts v x = v y = v z = 0, so just t = τ (stnding still ging) is geodesic. In summry, wht did here is extremize (in fct mximize) the proper time between two events to find the geodesics. Thus the geodesic is tht pth for which the mximum time psses on the wrist wtch of the observer trveling tht pth. [It is mximum, rther thn minium due to the minus sign in the metric.] Note tht this is bsiclly the nswer to twin prdox. The twin tht went out nd then bck 5

6 did not trvel geodesic, they ccelerted three times, while the sty t home twin did not ccelerte nd therefore followed geodesic. Thus we understnd why the sty t home twin ges more (in fct ges mximlly!). Whenever someone ccelertes, they leve their geodesic nd therefore re ging less! The solution to the twin prdox is lso esily understood using spcetime digrm. Both twins strt together t the origin. One twin stys on Erth (worldline is geodesic going stright up). The other ccelertes close to the speed of light (worldline close to 45 0 ). The proper time for the speedy twin is very smll; remember the hyperbol of constnt proper time. Hlfwy out, the speedy twin ccelertes gin (leves the geodesic gin), nd speeds home, meeting up with the first twin. From this it is cler tht if one minimized the proper time, it would require ccelerting to the speed of light for hlf the time nd then coming bck with totl proper time pproching 0. We see tht ctul geodesics mximize the proper time. 6