Schwarzschild metric: an application of Einstein s equations to stars

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1 Schwazschild metic: an application of Einstein s equations to stas 1. The metic outside a spheical non-otating mass M is given by (without poof) ds = (1 [ ] s d )d(ct) + (dθ + sin θdφ ), 1 s s GM c. Fo genealized coodinates q µ = (ct,, θ, φ) (check this), 3. the above fixes the components of the metic g µν, which has no off-diagonal components. 4. One can see that this metic is spheically isotopic in spheical angles θ and φ, and has a adial coodinate. 5. and static with the coodinate time t. 6. We will conside only the outside egime whee the adial coodinate > s, such that the coodinate time t and the adial coodinate ae popely defined with g tt > 0, g < The spheical aea of the same coodinate measues as π daea = gθθ dθ π π g φφ dφ = dθ [π sin(θ)] = 4π 0 so takes the meaning of the angula adius. 8. The distance between two obseves at the adial coodinates, e.g., = s and = π s, measues (check this) πs πs dl = g d = (1 s /) 1 d s s 9. 0 = s [ π(π 1) + ln( π + π 1) ] = s > (π 1) s so the adial distance is not the angula distance in the cuved space. g tt = 1 + Φ c = 1 g, Φ GM whee Φ is the Newtonian potential aound a point mass. At vey fa away, the space is nealy Minkowski, and is appoximately the adial distance. 0 1

2 10. A stationay clock B at infinity measues a time dτ = dt A clock O moving tangentially with a speed v measues a pope time ds dτ = c = dt g tt (dφ/cdt) Φ v = dt 1 + c < dt Let the obiting clock O flash with intinsic inteval dτ = 1 second at adius nea the sta. These flashes would each the distant stationay obseve B. They would appea to to be flashes with a dilated inteval dt > 1 second, which is longe than the flashing inteval dt = dτ = 1 second of a clock B local to obseve B (check this). Light signals fom O show up with a longe peiod o a lowe fequency (i.e., edshifted) by the Tansvese Dopple effect 1 + Φ c. 11. Calculate the Chistoffel symbol. 1 v c, and the Gavitational Redshift effect Γ a bc = 1 gaa [ g bc,a + g ac,b + g ab,c ] = 0 if a b c thanks to vanishing coss tems g ab etc. So non-zeo Γ equies that two of the thee indices must be identical. E.g., Still fo each i = 1,, 3 Γ i 00 = g [ g 00,] δ i1 = (1 s ) s δ i1 Γ 0 0i = g00 [g 00,i] = (1 s ) 1 s δ i1 Γ i 0i = 1 gii [g ii,0 ] = 0 Γ 0 ii = 1 [ g00 dg ] ii = 0 d(ct) because the metic and Chistoffel symbols ae all time-independant. Othe zeo and non-zeo symbols include (check these) 3 β=0 Γ 0 00 = Γ θ θθ = Γ φ φφ = 0. Γ β β = [ln ( g 00 g g θθ g φφ ) ] =

3 1. Now compute the Ricci tenso fom the Chistoffel symbols. we get (check this) R 00 = Γ 00, 0+Γ 00 R 00 = R α 0α0 = Γ α 00,α Γ α 0α,0 + Γ α 00Γ β αβ Γα 0βΓ β 0α 3 β=0 Γ β β ( Γ 00Γ Γ 0 0Γ ) d 00 = d ( Γ 00) s 4 = 0 One can pove that all Ricci tenso components ae zeo outside the sta (although Riemann tensos ae not all zeo). This satisfies the eaanged Einstein s equation (in the absence of Λ) R µν = κ(t µν T α α g µν) = 0 since the enegy-momentum stess tenso is zeo outside the sta. 13. A possible action fo a test paticle of mass m is S m Ldλ, whee the Lagangian is ds L(q µ, dq µ /dλ) = dλ = (1 [ s )(d(ct)/dλ) (1 ] s ) 1 (d/dλ) + (dθ/dλ) + sin θ(dφ/dλ) 14. The geodesic equation fo θ is One can check that d dθ dλ Ldλ = L θ = (L) 1 ( ) dφ sin θ cos θ dλ θ(λ) = π = const, dθ(λ) dλ = 0 is a solution, which makes l.h.s. = 0 =.h.s. because cos θ = 0. So we will conside only obit in the equatoial plane θ = π/ fom now on without loss of geneality. 15. The geodesic equation fo q 1 = ct is d L dλ (d(ct)/dλ) = L ct = 0 because the metic is independent of the coodinate time t. So we have an enegy consevation ( ) (1 s /) dct = p t E L dλ mc = const whee m is the mass of the test paticle ; not the point mass M. 3

4 16. The geodesic equation fo φ is d L dλ (dφ/dλ) = L φ = 0 because L does not explicitly depend on φ. So we can integate to find consevation of the angula momentum in the equatoial plane whee we also used sin θ = 1. dφ Ldλ = p φ J mc = const 17. The geodesic equation fo is (ty this) [ (d(ct) dp dλ = L = 1 ) s L dλ + p ( ) d s dλ ( s ) L d ( d dλ ) = dλ L(1 s /) ( ) ] dφ, dλ Howeve, a bette way to give the equivalent infomation is to use the definition of Lagangian L = (1 s /)(d(ct)/dλ) [ (d/dλ) (1 s /) 1 + (dφ/dλ) ]. This can be divided by L on both sides, and ewitten as 1 = E (mc ) gtt + p g + J m c gφφ µ p µ p ν g µν which takes the meaning of enegy-momentum equation. This leads to 1 d ( p = L(1 s /) dλ = ± Q, Q 1 ) s E (1 m c 4 + J ) ( m c 1 ) 1 s. 18. Combining equations p φ and p we get dφ = J mc( s )p d This allows computation of the obit shape φ() and the deflection angle. 19. Combining equations p t and p we get d(ct) = E mc p d This allows us to compute the time delay t(), and the ada echo time. ν 4

5 0. E.g., along the bent path of a photon, we can assume m 0 while keeping E and J finite. Thus ( (mcp ) 1 ) ( ) s E J ( c 1 ) 1 s. So the obit is given by (ty) cj(1 s /) J d E d(ct) = dφ = (1 s /)(mcp ) du, W W s b u (1 u) ( s min 3 s 3 min ) ( u u 3), b cj E, u s, whee b = R ; the adius of the Sun is the peicente whee the denominato W = 0, and b has meaning of the impact paamete at. du W 1. A cicula obit would imply u s, so in ode fo to be nondivegent, W must appoach zeo quadatically with W 0 dw/du = u 3u, hence u = 3, i.e., photon can be bent to a cicula obit at = 1.5 s = =. Likewise fo massive paticles: cicula obits b 3 satisfy Q 0 dq d, which elates obital E/m and J/m with.. Slight light deflection: in the limit u s 1, define f φ π+ dφ = 0 s du 1 + (1 f) 1 s u 0 s s min +u s +u min s, +u min fdu s u s with Taylo-expansion fo small f. Fo the Sun φ = 4GM c R = In contast Newtonian theoy pedicts only half as much deflection. This hallmak pediction of Einstein s GR is confimed by obseving the bending of the sta field duing sola eclipses. Is the deflection angle same fo adio/ultaviolet light? What about deflection by Sun s atmosphee? 3. Time delay: Fom the time pat s d d(ct) = b (1 s /) W The geometic delay fom adius to the closest appoach of and bounce-back is min, the exta delay is (ct) = d + 1 d (1 s ) 1 1 s 1 s 5

6 d [ 1 s + s ( + ) ] s ln( + min min ) + s + with Taylo-expansion fo small s. Eath-Sun-Venus ound tip ada echo has exta delay [ s ln E S R + s ] + [ s ln V S R + s ] 60 km. 4. An altenative/fast way to compute obits is to stat with p µ = g µµ p µ = g ds µµ ( dλ ) ( dqµ dλ p = d cdτ = ± whee we used Ldλ = ds = cdτ. Then the adial motion satisfies ) V (), p φ = dφ c cdτ = J mc, pt = d(ct) cdτ = E mc (1 s /), whee V (), the effective potential enegy pe unit mass, obeys V () c (1 + J ) ( m c 1 ) s ( E m c 4 ) We can use classical mechanics to ague that stable cicula obits satisfy V = 0, dv d = 0 and ω d V d > 0 (no stable cicles with < 3 s ), hence pedicting a adial oscillation fequency [ d V GM ω = d = 3 n, n 1 3 ] 1/ s < 1 fo a nealy cicula obit of adius. The slowe adial oscillation implies a fowad peihelion pocession of φ = π(1 n) 3π s = 6π GM c = 0.43 /yea fo the peicente of Mecucy, as fist calculated by Einstein. 5. Anothe altenative (faste) way to computation of obits is to use the total action S. Assume (without poof) that S = Et + Jφ + S, S mcp d E.g., fo sub-elativisitic paticles, we define the classical enegy Ẽ E mc and keep only the lowest ode of c and s = GM c. Thus GM S = mcp d m d (1 + 4Ẽ mc ) + Ẽ m (1 + Ẽ mc ) J m This means the action is the simila to the classical Kepleian obit except fo a petubation of total enegy and the value of the point mass. The obit would look like = A(1+e cos(n(φ φ 0 ))) 1, whee n [ ] 1/ 1 3s < 1. This gives pecession φ = S J 6 3π s.

7 6. To model Gavitational Lensing the metic outside a point mass can also be witten in genealized coodinates q µ = (ct, γ, θ, φ) as (check this) ds = (1 ( m γ ) ) (1 + m γ )4 d(ct) (1 + m γ )4 dl, m s = GM c whee the spatial pat of the line-element dl = [ dγ + γ (dθ + sin θdφ ) ] = dx + dy + dz, The coodinate γ elates to the Schwazschild by a tansfomation = γ(1 + m γ ), d = (1 m (γ) )dγ and is elated to the Catesian coodinates (x,y,z) by γ = x + y + z 7. The metic g µν is puely diagonal. But unlike the spheical Schwazschild metic, this epesentation is homogeneous in the spatial coodinates (x,y,z). 8. In the limit fa fom the sta, define a Newtonian potential Φ = GM/γ, then (check this) ds (1 + Φ c )d(ct) (1 Φ c )dl, Φ c GM/γ c 1 9. Photon path is given by ds = 0. The gavitational potential fom the sta bends the photon path slightly, such that the photon aival time (1 Φ c ) Φ ct = d(ct) = ndl, n = (1 + Φ (1 c ) c ) = 1 + GM/γ c So the time delay has two pats, a geometic pat, γ=dl γ=dls dl + d γ + ξ Dl + ξ + D ls + ξ [D l + (ξ ξ s) ]+[D ls + (ξ ξ s) ] D l D ls whee ξ and ξ s is the closest appoach of a possible light path and the unbent path espectively, and a gavitational pat GM/γ. c dl GM c [ln D l ξ + ln D ls ] = s ln ξ + const ξ 30. The aival time is extema fo tue path (Femat s pinciple), so 0 = ξ ct = α s ξ, α = (ξ ξ s)(d 1 l + D 1 ls ) This eq. poves the deflection angle α = s ξ, and fixes the closest appoach ξ of the tue path as function of the unbent appoach ξ s. Moe eadings, cf. lensing fomalism#femat Suface. 7

8 Tutoial 3 solutions 1. Solution of Q1: We can stick to ou oiginal sign convention (and switch signs at the end if needed). We have fo a pessueless (P = 0) fluid of density ρ with fou velocity u = (c, 0, 0, 0) the stess tenso T µν = u µ u ν (ρ + 0) 0g µν = ρc δ µ 0 δν 0 has only one non-zeo tem. So the Einstein teno has only non-zeo G 00 = G 00 = κt 00 = κρc. (used g 00 = 1 when loweing the indices). Einstein s eq. says 0 = κt ii = G ii = R ii Rg ii / and take the expession fo Ricci tenso fom the lectue note (aound Eq.3), we get 0 = G ii = k + aä/c + ȧ /c a g ii [ 6ä/c 6ȧ /c a a 6k ] a g ii / = k a g ii 6k 0 0 a g ii / = k a g ii so we pove that ka = 0, i.e., the cuvatue paamete k = 0 (flat Minkowski metic) fo a static pessueless univese. Take fom the lectue note (aound Eq.3) the expession fo G 00 = 3 ȧ /c +k a = 3(0 + k/a ), (whee a is the scale facto instead of R, and has no time deivative fo a static univese.), we have G 00 = G 00 = 3k a = κρc. So the matte density ρ = 0, too since k = 0. I.e. it is an empty static flat univese. To ewite things to the othe sign convention, one can simply eplace G ii with G ii fo i = 1,, 3 espectively, because the new definition of Ricci cuvatue diffes by a sign by a contaction of the fist and last indices (instead of the fist and the thid indices) of the Riemann tenso. Likewise G 00 is eplaced with with G 44 because 0 is eplaced by 4.. Solution of Q: We manipulate Einstein s equation (without cosmological constant) R ab g ab R = κt ab = 0 whee we note that outside the sta thee is no enegy-momentum tenso T ab. Contact both side with g ab, we have So the Ricci scala R = 0. R 4R = 0 8

9 3. Solution to Q3: The thee obseves aive at the same point. Thee intenal clocks egistes a time (fom lectue note, eq 1.) dcτ = dt (1 m )c (1 m ) 1 v vφ, v d dt, v φ dφ dt, m GM c The bench mak is the clock at infinity whee dcτ = dct. Fo each fixed ding-dong intenal click dτ = 1 of each of the clocks, the thee clocks egistes with a diffeent coodinate time dt when the light eaches the bench mak obseve at infinity. We can compae dt to see which clock appeas slowest. Diffeent clocks Fo the stationay obseve S at the adius v = v φ = 0, dcτ = dt S (1 m/)c Fo the obseve F falling fom infinity v = v F dcτ = dt F (1 m/)c (1 m/) 1 v F Fo the obseve C ciculating v φ = v C dcτ = dt C (1 m/)c v C In the classical limit, when m/ is small, classical mechanics suggests that the v F GM/ > v C GM/ > 0 because the v F is like the escape speed, the highest speed fo a bound obit, and v C is only the cicula speed. So compaing the factos in font of the dt in above equations, and note that (1 m/)c (1 m/) 1 vf is the smallest, and c is the biggest, and (1 m/)c vc is in between, we can ague that dτ = dt < dt S < dt C < dt F **A moe elaboate answe ** is to compute v F and v C in full GR. Fom the lectue note on enegy consevation (eq 15. note we substitute Ldλ dcτ to get id of λ in favo of τ) L ( dct dcτ ) = (1 m ) dct dcτ = E Mc Since the falling clock F was at infinity, whee E = Mc, so (1 m )dct F dcτ = 1 9

10 This means dcτ = dct F (1 m ) = dt F (1 m/)c (1 m/) 1 v F, v F = (1 m/) GM/ note the exta facto (1 m/) is a coection to the classical mechanics escape speed GM/. Fo the ciculating clock C, the adial motion is zeo, and its adial momentum deivative is also zeo, so its adial geodesic equation (eq 17. multiplied by L 1 on both sides) Hence 0 = dp dτ = L L = 1 [ ( dct C dcτ same as classical mechanics. ) m + ( d dcτ ) v C dφ GM = dt C ] m dφ ( ( m) dcτ ) Finally the physics of the clocks include essentially gavitational edshifts, Dopple shifts, and gavitational time delay. Same effects apply to GPS clocks. 10

11 Advanced mateial: Geneal solution of a spheical sta 1. Geneal solution of a spheical static sta has a metic ds = A()d(ct) B()d (dθ + sin θdφ ) and a stess tenso T µ ν = [ρ() + P ()/c ]u µ u ν P ()δ µ ν satisfying T µ ν = [ρ() + P ()/c ]c δ µ 0 δ0 ν P ()δ µ ν. The Einstein Equations G µ ν = κt µ ν (neglecting cosmological constant) educe to R µ ν = 8πG c 4 (T µ ν 1 δµ ν T β β ). The tt component Einstein eq. gives 8πG c 4 (ρc ) = G t t = 1 [( B ) 1], this eq. is easily solved (given the definition of an inteio mass M()), 1 B = [1 m() ], m() GM() c, M() 3. A combination of Einstein equations and tt gives 8πG c 4 (P +ρc ) = R+R t t = 1 (AB) B AB, 0 4π dρ whee we have woked out the two Ricci tensos in tem of metic deivatives. So this fixes the A() function by a simple integation. E.g., in the special case of a point mass with P = ρc = 0, we have A = 1/B = (1 m/). In the geneal case, the pessue is detemined by the consevation equation T µ ν;µ = 0, which educes to (see Hints) dp (P + ρc )d = A A GM c (1 + 4π3 P Mc )B Note coections of the hydostatic equation dp/ρd GM/. Hints: A has the meaning of gavitational potential gadient. The stess component T µ = P ()δ µ satisfies momentum consevation 0 = T µ ;µ = T µ,µ + Γ µ αµt α Γ α µt µ α = d( P ) d + Γ µ µ( P ) [Γ t t(ρc ) + i=,θ,φ Γ i i( P )] = dp d Γt t(ρc + P ), Γ α α = 1 g αα dg αα d, α 11

12 4. The above equation (called TOV Tolman-Oppenheime-Volkoff eq.) allows us to solve fo the stella stuctues of stas, including neuton stas, white dwafs, and nomal stas with elativistic coections. 5. Inside a sta of unifom density ρ = M 4π 3 /3, TOV solution: ds = A()d(ct) B()d (dθ + sin θdφ ) 1 B() = 1 m, A() = 3 [3 1 m 4 1 m 3 ] = 1 m [ P () ρ c + 1] Note that B( ) B() B(0) = 1, AB = (1.5 B()/B( ) 0.5) 1 inside the sta. B( ) becomes negative infinity fo a vey dense o massive sta, which sets uppe limit of a nomal sta; B( ) 1 = 1 m > 0, that is the sta must be bigge than its Schwazschild adius m GM c. The TOV solution finds that the pessue P () deceases fom cente outwads to zeo at the bounday. A finite cental pessue P (0) sets a limit 0 < A(0) = m size > 1.5 m (so-called Buchdahl limit). 0.5, so a stable sta has a bounday 1

13 A chaged point mass BH with cosmological constant 1. Conside a point mass M with an electic chage e. Conside also thee is a cosmological constant Λ. Then the new stess tenso is T µ ν = +Λδ ν µ + ( e ) δµ 0 + δµ 1 δµ δµ 3 δ ν µ. The tt and components of Einstein eq. satisfy 8πG c 4 0 = R + Rt t = 1 (AB) B AB, A = 1 B e (Λ + 4 ) = Gt t = 1 [( B ) 1], 1 B = 1 m + e Λ 3 3. So the metic (called Reissne-Nodstöm-deSitte metic) ds is (dcτ) = (Ldλ) = A()d(ct) 1 A() d dω, A() = 1 m + e Λ 3 Challenge: Apply cov. deiv. to show T µ ν;µ = 0 fo ν = t,, θ, φ. 4. Set the cosmological constant Λ = This metic has edshift-infinity hoizons fo a stationay souce at adius. This hoizon is set by dτ dt = A() = 0, so A() = 0 = ± m ± m e, It has been agued in the liteatue, the case of e > m is not allowed in eality, since it ceates a naked stess tenso singulaity at = 0 without a edshift hoizon suounding it. The case with m e 0 is called a chage Black Hole. No photons can escape fom the oute hoizon adius = +. In the absence of the chage and cosmological constant, and one hoizon educes to = 0, and thee is only an oute hoizon + = m at the Schwazschild adius. 6. In between the two adius < < +, the sign of the metic is evesed, and essentially the meaning of the coodinate and t exchanged. 13

14 7. Conside a neutal paticle of est enegy E 0 = E 0 µ p µp µ obiting the black hole with a conseved enegy E p t E 0, angula momentum J E 0 c 1 p φ in the equatoial plane θ = π/, then geodesics ae solutions of p t = 1 A p t = 1 A p φ = 1 p φ = 1 p = Ap = A L [ d dλ L [ dct dλ ] dct dcτ L [ dφ dλ E E 0 A dφ ] dcτ J c E 0 ] d dcτ = ± E E0 J c A V () E0 A ± c 8. Conside a adial obit (J = 0) of a paticle emitted at the coodinate and eceived at obs, combine the ct and equations, we can compute the tavel time obs 1 d dct = A 1 A E 0 E The above integal diveges when A() 0. This means that it takes infinite amount of time fo a paticle o photon inside the hoizon + to each us, i.e., it is not obsevable; hence the meaning of hoizon. 9. A cicula obit of epicyclic fequency ω must have V () = d [ d V () = 0, ω d V d = c d d ( A )J c E0 + A ]. So E E0 = (1 + J c d E0 )A, d ( A )J c E0 + A = 0. Fo photons on a cicula obit, E 0 0, it must satisfy E J c = A, d d ( A ) = 3 + 6m 4 4e 5 = 0, so a photon can have a cicula obit only at = 3m ± ( 3m ) 4e (if e < 3m/4) and this obit is unstable ω < 0. Conside e = 0 fo simplicity, then thee is only one unstable cicula obit fo photons is = 3m. Photons with impact paamete J c E > A = 3 3m will bounce to infinity, and photons with small impact paamete will infall into the black hole. 10. A Java animation of obits aound BH. 14

15 1. Synchonised Coodinates: meaning g tt = 1.. The black hole metic (in case of e = 0) is bette behaved in following coodinates [ ] ds = (dc t) d 3( ± c t) F F sdω, 3 F (, t) = 0 s whee the new coodinates ae elated to the old Schwazschild metic by ±(ct c t) = s s + s ln, = F s + s 3. In the following we take the minus sign solution of ±, which descibes things falling into a black hole. (The time-evesed solution descibes things spit out fom a white-hole) 4. Nea the Schwazschild adius = s, F = 1, so the metic is ds (dc t) d sdω, which is clealy well-behaved with no singulaity. 5. Inteestingly the spatial metic in Synchonised Coodinates is contacting in the angula diection ds = g φφ dθ = F s dθ ( c t) /3 dθ at fixed and expanding in the adial diection ds = g d = F 1/ d ( c t) 1/3 d, whee F deceases with time t. In this so-called contacting synchonized Lemaite coodinate system, the facto F has some esemblance to the scale facto in the FRW synchonized metic of the expanding univese. The stationaily co-moving obseves in both metic ae on geodesics. 6. The meaning of synchonized coodinate is that q µ = (c t,, θ, φ) = (cτ, cst, cst, cst) descibes stationay geodesic obseves (which ae adially falling geodesics in Schwazschild metic), so these co-moving fou velocity u µ = (c, 0, 0, 0). 7. The coodinate time of a stationay clock goes the same fast the as intenal clock time eveywhee in this metic, dt dτ = 1, hence synchonized, as in the FRW cosmological time. 8. Thee is no poblem fo paticles to coss the hoizon and ente the zone F 1. In fact, an obseve sitting stationaily on the geodesic = cst, θ = φ = 0 was at lage adius = when t =, eaches the Schwazschild hoizon = s at time t = cst s/3 c, and eaches the Schwazschild oigin = 0 at the time t = cst c. 9. The metic is t-dependent and non-minkowski even at lage. Not ideal coodinates fo intepetation using Newtonian intuitions. 10. Last example of nealy synchonized [ coodinate ] is: A unifom sta s metic ds = Ad(ct) d 1 k dω esembles a static closed homogenous FRW metic, except that 1 > k = m 0 is not unity, and A = (1.5 1 k k ) (1 k ) almost independent of. 15

16 1. Gavitational wave: metic with a time-changing coss tem. Slightly non-minkowski metic ds = (η ab + h ab )dx a dx b, h ab 1, η ab = (1, 1, 1, 1) with Chistoffel symbols Γ a bc h bc,a + h ab,c + h ac,b η aa 3. In vaccum so Ricci scala T µν = 0 R = κt µ µ = 0 4. Einstein Equation becomes 0 = G ab = R ab = Raeb e Γ e ab,e Γ e ae,b = h ab,e,e + h ae,b,e + h be,a,e h ae,e,b + h ae,e,b + h ee,a,b η ee Afte choosing an appopiate coodinate system to cancel some tems, we have 0 = 1 h ab η ee x e x e 5. Wave solution with an angula speed ω and wave numbe (k 1, k, k 3 ). The metic petubation is time-dependent, and can have coss-tems. h ab = H ab cos(ωt k 1 x k y k 3 z + const)), ω/c = k1 + k + k 3 6. Can simplify by futhe setting the popagation puely in the z-diection k 1 = k = 0, set h 00 = h 01 = h 0 = h 03 = h 33 = 0 and zeo the tace h 11 + h = 0. Then we have only to tansvese oscillations ds = d(ct) dz + dxdy[h 1 cos(ωt ωz/c + const)]+ [H 11 cos(ωt ωz/c) 1]dx + [ H 11 cos(ωt ωz/c) 1]dy 7. Gavitational wave involves a quaduple-like tansvese oscillation of the metic popagating at the speed of light h 11 = h, h 1 = h 1 following a tansvese (quaduple) oscillation of matte stess tenso T 11 = T, T 1 = T 1 somewhee, e.g., a sta spialing into a black hole with an angula speed ω. η ee 16